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2020-I-1
| 2,020
| 1
|
In $\triangle ABC$ with $AB=AC,$ point $D$ lies strictly between $A$ and $C$ on side $\overline{AC},$ and point $E$ lies strictly between $A$ and $B$ on side $\overline{AB}$ such that $AE=ED=DB=BC.$ The degree measure of $\angle ABC$ is $\tfrac{m}{n},$ where $m$ and $n$ are relatively prime positive integers. Find $m+n.$
| 547
|
I
|
[
"[asy] size(10cm); pair A, B, C, D, F; A = (0, tan(3 * pi / 7)); B = (1, 0); C = (-1, 0); F = rotate(90/7, A) * (A - (0, 2)); D = rotate(900/7, F) * A; draw(A -- B -- C -- cycle); draw(F -- D); draw(D -- B); label(\"$A$\", A, N); label(\"$B$\", B, E); label(\"$C$\", C, W); label(\"$D$\", D, W); label(\"$E$\", F, E); [/asy] If we set $\\angle{BAC}$ to $x$, we can find all other angles through these two properties: 1. Angles in a triangle sum to $180^{\\circ}$. 2. The base angles of an isosceles triangle are congruent. Now we angle chase. $\\angle{ADE}=\\angle{EAD}=x$, $\\angle{AED} = 180-2x$, $\\angle{BED}=\\angle{EBD}=2x$, $\\angle{EDB} = 180-4x$, $\\angle{BDC} = \\angle{BCD} = 3x$, $\\angle{CBD} = 180-6x$. Since $AB = AC$ as given by the problem, $\\angle{ABC} = \\angle{ACB}$, so $180-4x=3x$. Therefore, $x = 180/7^{\\circ}$, and our desired angle is \\[180-4\\left(\\frac{180}{7}\\right) = \\frac{540}{7}\\] for an answer of $547. See here for a video solution: https://youtu.be/4e8Hk04Ax_E",
"Let $\\angle{BAC}$ be $x$ in degrees. $\\angle{ADE}=x$. By Exterior Angle Theorem on triangle $AED$, $\\angle{BED}=2x$. By Exterior Angle Theorem on triangle $ADB$, $\\angle{BDC}=3x$. This tells us $\\angle{BCA}=\\angle{ABC}=3x$ and $3x+3x+x=180$. Thus $x=\\frac{180}{7}$ and we want $\\angle{ABC}=3x=\\frac{540}{7}$ to get an answer of $547.",
"Let $x = \\angle ABC = \\angle ACB$. Because $\\triangle BCD$ is isosceles, $\\angle CBD = 180^\\circ - 2x$. Then \\[\\angle DBE = x - \\angle CBD = x - (180^\\circ - 2x) = 3x - 180^\\circ\\!.\\]Because $\\triangle EDA$ and $\\triangle DBE$ are also isosceles, \\[\\angle BAC =\\frac12(\\angle EAD + \\angle ADE) = \\frac12(\\angle BED)= \\frac12(\\angle DBE)\\] \\[= \\frac12 (3x - 180^\\circ) = \\frac32x-90^\\circ\\!.\\] Because $\\triangle ABC$ is isosceles, $\\angle BAC$ is also $180^\\circ-2x$, so $\\frac32x - 90^\\circ = 180^\\circ - 2x$, and it follows that $\\angle ABC = x = \\left(\\frac{540}7\\right)^\\circ$. The requested sum is $540+7 = 547$. [asy] unitsize(4 cm); pair A, B, C, D, E; real a = 180/7; A = (0,0); B = dir(180 - a/2); C = dir(180 + a/2); D = extension(B, B + dir(270 + a), A, C); E = extension(D, D + dir(90 - 2*a), A, B); draw(A--B--C--cycle); draw(B--D--E); label(\"$A$\", A, dir(0)); label(\"$B$\", B, NW); label(\"$C$\", C, SW); label(\"$D$\", D, S); label(\"$E$\", E, N); [/asy] https://artofproblemsolving.com/wiki/index.php/1961_AHSME_Problems/Problem_25 (Almost Mirrored) See here for a video solution: https://youtu.be/4XkA0DwuqYk",
"Let $x = \\angle ABC = \\angle ACB$. Because $\\triangle BCD$ is isosceles, $\\angle CBD = 180^\\circ - 2x$. Then \\[\\angle DBE = x - \\angle CBD = x - (180^\\circ - 2x) = 3x - 180^\\circ\\!.\\]Because $\\triangle EDA$ and $\\triangle DBE$ are also isosceles, \\[\\angle BAC =\\frac12(\\angle EAD + \\angle ADE) = \\frac12(\\angle BED)= \\frac12(\\angle DBE)\\] \\[= \\frac12 (3x - 180^\\circ) = \\frac32x-90^\\circ\\!.\\] Because $\\triangle ABC$ is isosceles, $\\angle BAC$ is also $180^\\circ-2x$, so $\\frac32x - 90^\\circ = 180^\\circ - 2x$, and it follows that $\\angle ABC = x = \\left(\\frac{540}7\\right)^\\circ$. The requested sum is $540+7 = 547$. [asy] unitsize(4 cm); pair A, B, C, D, E; real a = 180/7; A = (0,0); B = dir(180 - a/2); C = dir(180 + a/2); D = extension(B, B + dir(270 + a), A, C); E = extension(D, D + dir(90 - 2*a), A, B); draw(A--B--C--cycle); draw(B--D--E); label(\"$A$\", A, dir(0)); label(\"$B$\", B, NW); label(\"$C$\", C, SW); label(\"$D$\", D, S); label(\"$E$\", E, N); [/asy] https://artofproblemsolving.com/wiki/index.php/1961_AHSME_Problems/Problem_25 (Almost Mirrored) See here for a video solution: https://youtu.be/4XkA0DwuqYk",
"graph soon We write equations based on the triangle sum of angles theorem. There are angles that do not need variables as the less variables the better. \\begin{align*} \\angle A &= y \\\\ \\angle B &= x+z \\\\ \\angle C &= \\frac{180-x}{2} \\\\ \\end{align*} Then, using triangle sum of angles theorem, we find that \\begin{align*} \\angle A + \\angle B + \\angle C = x+y+z+\\frac{180-x}{2}=180 \\\\ \\end{align*} Now we just need to find the variables. \\begin{align*} (180-2y)+z = 180& \\\\ (180-2z)+y+\\frac{180-x}{2} = 180& \\\\ \\end{align*} Notice how all the equations equal 180. We can use this to write \\begin{align*} (180-2y)+z = (180-2z)+y+\\frac{180-x}{2}=x+y+z+\\frac{180-x}{2} \\\\ \\end{align*} Simplifying, we get \\begin{align*} (180-2y)+z=(180-2z)+y+\\frac{180-x}{2} \\\\ 360-4y+2z=360-4z+2y+180-x \\\\ \\end{align*} \\begin{align*} 6z=6y+180-x \\\\ x=6y-6z+180 \\\\ \\end{align*} \\begin{align*} (180-2y)+z=6y-6z+180+y+z+\\frac{180-(6y-6+180)}{2} \\\\ 360-4y+2z=12y-12z+360+2y+2z+180-6y+6z-180 \\\\ \\end{align*} \\begin{align*} 6z=12y& \\\\ z=2y& \\\\ \\end{align*} Theres more. We are at a dead end right now because we forgot that the problem states that the triangle is isosceles. With this, we can write the equation \\begin{align*} \\frac{180-x}{2}=x+z \\\\ \\end{align*} Substituting $z$ with $2y$, we get \\begin{align*} \\frac{180-x}{2}=x+2y \\\\ 180-x=2x+4y \\\\ \\end{align*} \\begin{align*} 180-(6y-6z+180)=2(6y-6z+180)+4y& \\\\ 180-6y+12y-180=12y-24y+360+4y& \\\\ \\end{align*} \\begin{align*} 6y=-8y+360& \\\\ \\end{align*} With this, we get \\begin{align*} y=\\frac{180}{7} \\\\ x=\\frac{180}{7} \\\\ z=\\frac{360}{7} \\\\ \\end{align*} And a final answer of $\\frac{180}{7}+\\frac{360}{7} = \\frac{540}{7} = 547. ~orenbad",
"graph soon We write equations based on the triangle sum of angles theorem. There are angles that do not need variables as the less variables the better. \\begin{align*} \\angle A &= y \\\\ \\angle B &= x+z \\\\ \\angle C &= \\frac{180-x}{2} \\\\ \\end{align*} Then, using triangle sum of angles theorem, we find that \\begin{align*} \\angle A + \\angle B + \\angle C = x+y+z+\\frac{180-x}{2}=180 \\\\ \\end{align*} Now we just need to find the variables. \\begin{align*} (180-2y)+z = 180& \\\\ (180-2z)+y+\\frac{180-x}{2} = 180& \\\\ \\end{align*} Notice how all the equations equal 180. We can use this to write \\begin{align*} (180-2y)+z = (180-2z)+y+\\frac{180-x}{2}=x+y+z+\\frac{180-x}{2} \\\\ \\end{align*} Simplifying, we get \\begin{align*} (180-2y)+z=(180-2z)+y+\\frac{180-x}{2} \\\\ 360-4y+2z=360-4z+2y+180-x \\\\ \\end{align*} \\begin{align*} 6z=6y+180-x \\\\ x=6y-6z+180 \\\\ \\end{align*} \\begin{align*} (180-2y)+z=6y-6z+180+y+z+\\frac{180-(6y-6+180)}{2} \\\\ 360-4y+2z=12y-12z+360+2y+2z+180-6y+6z-180 \\\\ \\end{align*} \\begin{align*} 6z=12y& \\\\ z=2y& \\\\ \\end{align*} Theres more. We are at a dead end right now because we forgot that the problem states that the triangle is isosceles. With this, we can write the equation \\begin{align*} \\frac{180-x}{2}=x+z \\\\ \\end{align*} Substituting $z$ with $2y$, we get \\begin{align*} \\frac{180-x}{2}=x+2y \\\\ 180-x=2x+4y \\\\ \\end{align*} \\begin{align*} 180-(6y-6z+180)=2(6y-6z+180)+4y& \\\\ 180-6y+12y-180=12y-24y+360+4y& \\\\ \\end{align*} \\begin{align*} 6y=-8y+360& \\\\ \\end{align*} With this, we get \\begin{align*} y=\\frac{180}{7} \\\\ x=\\frac{180}{7} \\\\ z=\\frac{360}{7} \\\\ \\end{align*} And a final answer of $\\frac{180}{7}+\\frac{360}{7} = \\frac{540}{7} = 547. ~orenbad",
"vladimir.shelomovskii@gmail.com, vvsss"
] |
2020-I-2
| 2,020
| 2
|
There is a unique positive real number $x$ such that the three numbers $\log_8(2x),\log_4x,$ and $\log_2x,$ in that order, form a geometric progression with positive common ratio. The number $x$ can be written as $\tfrac{m}{n},$ where $m$ and $n$ are relatively prime positive integers. Find $m+n.$
| 17
|
I
|
[
"Since these form a geometric series, $\\frac{\\log_2{x}}{\\log_4{x}}$ is the common ratio. Rewriting this, we get $\\frac{\\log_x{4}}{\\log_x{2}} = \\log_2{4} = 2$ by base change formula. Therefore, the common ratio is 2. Now $\\frac{\\log_4{x}}{\\log_8{2x}} = 2 \\implies \\log_4{x} = 2\\log_8{2} + 2\\log_8{x} \\implies \\frac{1}{2}\\log_2{x} = \\frac{2}{3} + \\frac{2}{3}\\log_2{x}$ $\\implies -\\frac{1}{6}\\log_2{x} = \\frac{2}{3} \\implies \\log_2{x} = -4 \\implies x = \\frac{1}{16}$. Therefore, $1 + 16 = 017. ~ JHawk0224",
"If we set $x=2^y$, we can obtain three terms of a geometric sequence through logarithm properties. The three terms are \\[\\frac{y+1}{3}, \\frac{y}{2}, y.\\] In a three-term geometric sequence, the middle term squared is equal to the product of the other two terms, so we obtain the following: \\[\\frac{y^2+y}{3} = \\frac{y^2}{4},\\] which can be solved to reveal $y = -4$. Therefore, $x = 2^{-4} = \\frac{1}{16}$, so our answer is $017. -molocyxu",
"Let $r$ be the common ratio. We have \\[r = \\frac{\\log_4{(x)}}{\\log_8{(2x)}} = \\frac{\\log_2{(x)}}{\\log_4{(x)}}\\] Hence we obtain \\[(\\log_4{(x)})(\\log_4{(x)}) = (\\log_8{(2x)})(\\log_2{(x)})\\] Ideally we change everything to base $64$ and we can get: \\[(\\log_{64}{(x^3)})(\\log_{64}{(x^3)}) = (\\log_{64}{(x^6)})(\\log_{64}{(4x^2)})\\] Now divide to get: \\[\\frac{\\log_{64}{(x^3)}}{\\log_{64}{(4x^2)}} = \\frac{\\log_{64}{(x^6)}}{\\log_{64}{(x^3)}}\\] By change-of-base we obtain: \\[\\log_{(4x^2)}{(x^3)} = \\log_{(x^3)}{(x^6)} = 2\\] Hence $(4x^2)^2 = x^3 \\rightarrow 16x^4 = x^3 \\rightarrow x = \\frac{1}{16}$ and we have $1+16 = 017 as desired. ~skyscraper",
"Let $r$ be the common ratio, and let $a$ be the starting term ($a=\\log_{8}{(2x)}$). We then have: \\[\\log_{8}{(2x)}=a, \\log_{4}{(x)}=ar, \\log_{2}{(x)}=ar^2\\] Rearranging these equations gives: \\[8^a=2x, 4^{ar}=x, 2^{ar^2}=x\\] Deal with the last two equations first: Setting them equal gives: \\[4^{ar}=2^{ar^2} \\implies 2^{2ar}=2^{ar^2} \\implies 2ar=ar^2 \\implies r=2\\] Using this value of $r$, substitute into the first and second equations (or the first and third, it doesn't really matter) to get: \\[8^a=2x, 4^{2a}=x\\] Changing these to a common base gives: \\[2^{3a}=2x, 2^{4a}=x\\] Dividing the first equation by 2 on both sides yields: \\[2^{3a-1}=x\\] Setting these equations equal to each other and removing the exponent again gives: \\[3a-1=4a \\implies a=-1\\] Substituting this back into the first equation gives: \\[8^{-1}=2x \\implies 2x=\\frac{1}{8} \\implies x=\\frac{1}{16}\\] Therefore, $m+n=1+16=017 ~IAmTheHazard",
"Let $r$ be the common ratio, and let $a$ be the starting term ($a=\\log_{8}{(2x)}$). We then have: \\[\\log_{8}{(2x)}=a, \\log_{4}{(x)}=ar, \\log_{2}{(x)}=ar^2\\] Rearranging these equations gives: \\[8^a=2x, 4^{ar}=x, 2^{ar^2}=x\\] Deal with the last two equations first: Setting them equal gives: \\[4^{ar}=2^{ar^2} \\implies 2^{2ar}=2^{ar^2} \\implies 2ar=ar^2 \\implies r=2\\] Using this value of $r$, substitute into the first and second equations (or the first and third, it doesn't really matter) to get: \\[8^a=2x, 4^{2a}=x\\] Changing these to a common base gives: \\[2^{3a}=2x, 2^{4a}=x\\] Dividing the first equation by 2 on both sides yields: \\[2^{3a-1}=x\\] Setting these equations equal to each other and removing the exponent again gives: \\[3a-1=4a \\implies a=-1\\] Substituting this back into the first equation gives: \\[8^{-1}=2x \\implies 2x=\\frac{1}{8} \\implies x=\\frac{1}{16}\\] Therefore, $m+n=1+16=017 ~IAmTheHazard",
"We can relate the logarithms as follows: \\[\\frac{\\log_4{x}}{\\log_8{(2x)}}=\\frac{\\log_2{x}}{\\log_4{x}}\\] \\[\\log_8{(2x)}\\log_2{x}=\\log_4{x}\\log_4{x}\\] Now we can convert all logarithm bases to $2$ using the identity $\\log_a{b}=\\log_{a^c}{b^c}$: \\[\\log_2{\\sqrt[3]{2x}}\\log_2{x}=\\log_2{\\sqrt{x}}\\log_2{\\sqrt{x}}\\] We can solve for $x$ as follows: \\[\\frac{1}{3}\\log_2{(2x)}\\log_2{x}=\\frac{1}{4}\\log_2{x}\\log_2{x}\\] \\[\\frac{1}{3}\\log_2{(2x)}=\\frac{1}{4}\\log_2{x}\\] \\[\\frac{1}{3}\\log_2{2}+\\frac{1}{3}\\log_2{x}=\\frac{1}{4}\\log_2{x}\\] We get $x=\\frac{1}{16}$. Verifying that the common ratio is positive, we find the answer of $017. ~QIDb602",
"If the numbers are in a geometric sequence, the middle term must be the geometric mean of the surrounding terms. We can rewrite the first two logarithmic expressions as $\\frac{1+\\log_2{x}}{3}$ and $\\frac{1}{2}\\log_2{x}$, respectively. Therefore: \\[\\frac{1}{2}\\log_2{x}=\\sqrt{\\left(\\frac{1+\\log_2{x}}{3}\\right)\\left(\\log_2{x}\\right)}\\] Let $n=\\log_2{x}$. We can rewrite the expression as: \\[\\frac{n}{2}=\\sqrt{\\frac{n(n+1)}{3}}\\] \\[\\frac{n^2}{4}=\\frac{n(n+1)}{3}\\] \\[4n(n+1)=3n^2\\] \\[4n^2+4n=3n^2\\] \\[n^2+4n=0\\] \\[n(n+4)=0\\] \\[n=0 \\text{ and } -4\\] Zero does not work in this case, so we consider $n=-4$: $\\log_2{x}=-4 \\rightarrow x=\\frac{1}{16}$. Therefore, $1+16=017. ~Bowser498",
"Again, by the Change of Base Formula, obtain that the common ratio is 2. If we let $y$ be the exponent of $\\log_8 (2x)$, then we have $8^y=2x;\\:4^{2y}=x;\\:2^{4y}=x.$ Wee can then divide the first equation by two to have the right side be $x$. Also, $2^{4y}=\\left(2^{4}\\right)^y=16^y$. Setting this equal to $\\frac{8^y}{2}$, we can divide the two equations to get $2^y=\\frac12$. Therefore, $y=-1$. After that, we can raise $16$ to the $-1$th power to get $x=16^{-1}\\Rightarrow x=\\frac1{16}$. We then get our sum of $1+16=\\textbf{017}. ~SweetMango77",
"By the Change of Base Formula the common ratio of the progression is\\[\\frac{\\log_2 x}{\\log_4 x} = \\frac{\\hphantom{m}\\log_2x\\hphantom{m}}{\\frac{\\log_2x}{\\log_24}} = 2.\\]Hence $x$ must satisfy\\[2=\\frac{\\log_4 x}{\\log_8 (2x)}= \\frac{\\log_2 x}{\\log_2 4} \\div \\frac{\\log_2(2x)}{\\log_28} = \\frac 32\\cdot \\frac{\\log_2x}{1+\\log_2x}.\\]This is equivalent to $4 + 4\\log_2x = 3\\log_2x$. Hence $\\log_2x = -4$ and $x = \\frac{1}{16}$. The requested sum is $1+16 = 17$.",
"By the Change of Base Formula the common ratio of the progression is\\[\\frac{\\log_2 x}{\\log_4 x} = \\frac{\\hphantom{m}\\log_2x\\hphantom{m}}{\\frac{\\log_2x}{\\log_24}} = 2.\\]Hence $x$ must satisfy\\[2=\\frac{\\log_4 x}{\\log_8 (2x)}= \\frac{\\log_2 x}{\\log_2 4} \\div \\frac{\\log_2(2x)}{\\log_28} = \\frac 32\\cdot \\frac{\\log_2x}{1+\\log_2x}.\\]This is equivalent to $4 + 4\\log_2x = 3\\log_2x$. Hence $\\log_2x = -4$ and $x = \\frac{1}{16}$. The requested sum is $1+16 = 17$."
] |
2020-I-3
| 2,020
| 3
|
A positive integer $N$ has base-eleven representation $\underline{a}\kern 0.1em\underline{b}\kern 0.1em\underline{c}$ and base-eight representation $\underline1\kern 0.1em\underline{b}\kern 0.1em\underline{c}\kern 0.1em\underline{a},$ where $a,b,$ and $c$ represent (not necessarily distinct) digits. Find the least such $N$ expressed in base ten.
| 621
|
I
|
[
"From the given information, $121a+11b+c=512+64b+8c+a \\implies 120a=512+53b+7c$. Since $a$, $b$, and $c$ have to be positive, $a \\geq 5$. Since we need to minimize the value of $n$, we want to minimize $a$, so we have $a = 5$. Then we know $88=53b+7c$, and we can see the only solution is $b=1$, $c=5$. Finally, $515_{11} = 621_{10}$, so our answer is $621. ~ JHawk0224",
"The conditions of the problem imply that $121a + 11b + c = 512 + 64b + 8 c + a$, so $120 a = 512+ 53b+7c$. The maximum digit in base eight is $7,$ and because $120a \\ge 512$, it must be that $a$ is $5, 6,$ or $7.$ When $a = 5$, it follows that $600=512 + 53b+7c$, which implies that $88 = 53b+7c$. Then $b$ must be $0$ or $1.$ If $b = 0$, then $c$ is not an integer, and if $b = 1$, then $7c = 35$, so $c = 5$. Thus $N = 515_{11}$, and $N=5\\cdot 121 + 1\\cdot 11 + 5 = 621$. The number $637_{11} =1376_{8} = 766$ also satisfies the conditions of the problem, but $621$ is the least such number.",
"The conditions of the problem imply that $121a + 11b + c = 512 + 64b + 8 c + a$, so $120 a = 512+ 53b+7c$. The maximum digit in base eight is $7,$ and because $120a \\ge 512$, it must be that $a$ is $5, 6,$ or $7.$ When $a = 5$, it follows that $600=512 + 53b+7c$, which implies that $88 = 53b+7c$. Then $b$ must be $0$ or $1.$ If $b = 0$, then $c$ is not an integer, and if $b = 1$, then $7c = 35$, so $c = 5$. Thus $N = 515_{11}$, and $N=5\\cdot 121 + 1\\cdot 11 + 5 = 621$. The number $637_{11} =1376_{8} = 766$ also satisfies the conditions of the problem, but $621$ is the least such number."
] |
2020-I-4
| 2,020
| 4
|
Let $S$ be the set of positive integers $N$ with the property that the last four digits of $N$ are $2020,$ and when the last four digits are removed, the result is a divisor of $N.$ For example, $42{,}020$ is in $S$ because $4$ is a divisor of $42{,}020.$ Find the sum of all the digits of all the numbers in $S.$ For example, the number $42{,}020$ contributes $4+2+0+2+0=8$ to this total.
| 93
|
I
|
[
"We note that any number in $S$ can be expressed as $a(10,000) + 2,020$ for some integer $a$. The problem requires that $a$ divides this number, and since we know $a$ divides $a(10,000)$, we need that $a$ divides 2020. Each number contributes the sum of the digits of $a$, as well as $2 + 0 + 2 +0 = 4$. Since $2020$ can be prime factorized as $2^2 \\cdot 5 \\cdot 101$, it has $(2+1)(1+1)(1+1) = 12$ factors. So if we sum all the digits of all possible $a$ values, and add $4 \\cdot 12 = 48$, we obtain the answer. Now we list out all factors of $2,020$, or all possible values of $a$. $1,2,4,5,10,20,101,202,404,505,1010,2020$. If we add up these digits, we get $45$, for a final answer of $45+48=093. -molocyxu",
"Suppose that $N$ has the required property. Then there are positive integers $k$ and $m$ such that $N = 10^4m + 2020 = k\\cdot m$. Thus $(k - 10^4)m = 2020$, which holds exactly when $m$ is a positive divisor of $2020.$ The number $2020 = 2^2\\cdot 5\\cdot 101$ has $12$ divisors: $1, 2, 4, 5, 10, 20, 101, 202, 404, 505, 1010$, and $2020.$ The requested sum is therefore the sum of the digits in these divisors plus $12$ times the sum of the digits in $2020,$ which is \\[(1+2+4+5+1+2+2+4+8+10+2+4)+12\\cdot4 = 93.\\]",
"Suppose that $N$ has the required property. Then there are positive integers $k$ and $m$ such that $N = 10^4m + 2020 = k\\cdot m$. Thus $(k - 10^4)m = 2020$, which holds exactly when $m$ is a positive divisor of $2020.$ The number $2020 = 2^2\\cdot 5\\cdot 101$ has $12$ divisors: $1, 2, 4, 5, 10, 20, 101, 202, 404, 505, 1010$, and $2020.$ The requested sum is therefore the sum of the digits in these divisors plus $12$ times the sum of the digits in $2020,$ which is \\[(1+2+4+5+1+2+2+4+8+10+2+4)+12\\cdot4 = 93.\\]",
"Note that for all $N \\in S$, $N$ can be written as $N=10000x+2020=20(500x+101)$ for some positive integer $x$. Because $N$ must be divisible by $x$, $\\frac{20(500x+101)}{x}$ is an integer. We now let $x=ab$, where $a$ is a divisor of $20$. Then $\\frac{20(500x+101)}{x}=(\\frac{20}{a})( \\frac{500x}{b}+\\frac{101}{b})$. We know $\\frac{20}{a}$ and $\\frac{500x}{b}$ are integers, so for $N$ to be an integer, $\\frac{101}{b}$ must be an integer. For this to happen, $b$ must be a divisor of $101$. $101$ is prime, so $b\\in \\left \\{ 1, 101 \\right \\}$. Because $a$ is a divisor of $20$, $a \\in \\left \\{ 1,2,4,5,10,20\\right\\}$. So $x \\in \\left\\{1,2,4,5,10,20,101,202,404,505,1010,2020\\right\\}$. Be know that all $N$ end in $2020$, so the sum of the digits of each $N$ is the sum of the digits of each $x$ plus $2+0+2+0=4$. Hence the sum of all of the digits of the numbers in $S$ is $12 \\cdot 4 +45=093."
] |
2020-I-5
| 2,020
| 5
|
Six cards numbered $1$ through $6$ are to be lined up in a row. Find the number of arrangements of these six cards where one of the cards can be removed leaving the remaining five cards in either ascending or descending order.
| 52
|
I
|
[
"Realize that any sequence that works (ascending) can be reversed for descending, so we can just take the amount of sequences that satisfy the ascending condition and multiply by two. If we choose any of the numbers $1$ through $6$, there are five other spots to put them, so we get $6 \\cdot 5 = 30$. However, we overcount some cases. Take the example of $132456$. We overcount this case because we can remove the $3$ or the $2$. Therefore, any cases with two adjacent numbers swapped is overcounted, so we subtract $5$ cases (namely, $213456, 132456, 124356, 123546, 123465$,) to get $30-5=25$, but we have to add back one more for the original case, $123456$. Therefore, there are $26$ cases. Multiplying by $2$ gives the desired answer, $052. ~molocyxu",
"Similar to above, a $1-1$ correspondence between ascending and descending is established by subtracting each number from $7$. We note that the given condition is equivalent to \"cycling\" $123456$ for a contiguous subset of it. For example, $12(345)6 \\rightarrow 125346, 124536$ It's not hard to see that no overcount is possible, and that the cycle is either $1$ \"right\" or $1$ \"left.\" Therefore, we consider how many elements we flip by. If we flip $1$ or $2$ such elements, then there is one way to cycle them. Otherwise, we have $2$ ways. Therefore, the total number of ascending is $1 + 5 + 2(4 + 3 + 2 + 1) = 26$, and multiplying by two gives $052 ~awang11",
"Similar to above, a $1-1$ correspondence between ascending and descending is established by subtracting each number from $7$. We note that the given condition is equivalent to \"cycling\" $123456$ for a contiguous subset of it. For example, $12(345)6 \\rightarrow 125346, 124536$ It's not hard to see that no overcount is possible, and that the cycle is either $1$ \"right\" or $1$ \"left.\" Therefore, we consider how many elements we flip by. If we flip $1$ or $2$ such elements, then there is one way to cycle them. Otherwise, we have $2$ ways. Therefore, the total number of ascending is $1 + 5 + 2(4 + 3 + 2 + 1) = 26$, and multiplying by two gives $052 ~awang11",
"Similarly to above, we find the number of ascending arrangements and multiply by 2. We can choose $5$ cards to be the ascending cards, therefore leaving $6$ places to place the remaining card. There are $\\binom{6}{5}\\cdot 6=36$ to do this. However, since the problem is asking for the number of arrangements, we overcount cases such as $123456$. Notice that the only arrangements that overcount are $123456$ (case 1) or if two adjacent numbers of $123456$ are switched (case 2). $\\text{Case 1: }$ This arrangement is counted $6$ times. Each time it is counted for any of the $5$ numbers selected. Therefore we need to subtract $5$ cases of overcounting. $\\text{Case 2: }$ Each time $2$ adjacent numbers of switched, there is one overcount. For example, if we have $213456$, both $1$ or $2$ could be removed. Since there are $5$ possible switches, we need to subtract $5$ cases of overcounting. Therefore, we have $36-5-5=26$ total arrangements of ascending numbers. We multiply by two (for descending) to get the answer of $052 ~PCChess",
"Like in previous solutions, we will count the number of ascending arrangements and multiply by 2. First, consider the arrangement 1-2-3-4-5-6. That gives us 1 arrangement which works. Next, we can switch two adjacent cards. There are 5 ways to pick two adjacent cards, so this gives us 5 arrangements. Now, we can \"cycle\" 3 adjacent cards. For example, 1-2-3 becomes 2-3-1 which becomes 3-1-2. There are 4 ways to pick a set of 3 adjacent cards, so this gives us 4x2=8 arrangements. Cycling 4 adjacent cards, we get the new arrangements 2-3-4-1 (which works), 3-4-1-2 (which doesn't work), and 4-1-2-3 (which does work). We get 6 arrangements. Similarly, when cycling 5 cards, we find 2x2=4 arrangements, and when cycling 6 cards, we find 2x1=2 arrangements. Adding, we figure out that there are 1+5+8+6+4+2=26 ascending arrangements. Multiplying by 2, we get the answer $052 ~i8Pie",
"Like in previous solutions, we will count the number of ascending arrangements and multiply by 2. First, consider the arrangement 1-2-3-4-5-6. That gives us 1 arrangement which works. Next, we can switch two adjacent cards. There are 5 ways to pick two adjacent cards, so this gives us 5 arrangements. Now, we can \"cycle\" 3 adjacent cards. For example, 1-2-3 becomes 2-3-1 which becomes 3-1-2. There are 4 ways to pick a set of 3 adjacent cards, so this gives us 4x2=8 arrangements. Cycling 4 adjacent cards, we get the new arrangements 2-3-4-1 (which works), 3-4-1-2 (which doesn't work), and 4-1-2-3 (which does work). We get 6 arrangements. Similarly, when cycling 5 cards, we find 2x2=4 arrangements, and when cycling 6 cards, we find 2x1=2 arrangements. Adding, we figure out that there are 1+5+8+6+4+2=26 ascending arrangements. Multiplying by 2, we get the answer $052 ~i8Pie",
"First count the number of permutations of the cards such that if one card is removed, the remaining cards will be in ascending order. There is $1$ such permutation where all the cards appear in order: $123456.$ There are $5$ such permutations where two adjacent cards are interchanged, as in $124356.$ The other such permutations arise from removing one card from $123456$ and placing it in a position at least two away from its starting location. There are $4$ such positions to place each of the cards numbered $1$ and $6,$ and $3$ such positions for each of the cards numbered $2, 3, 4,$ and $5.$ This accounts for $2\\cdot4 + 4\\cdot3 =20$ permutations. Thus there are $1 + 5 + 20 = 26$ permutations where one card can be removed so that the remaining cards are in ascending order. There is an equal number of permutations that result in the cards' being in descending order. This gives the total $26 + 26 = 52.",
"First count the number of permutations of the cards such that if one card is removed, the remaining cards will be in ascending order. There is $1$ such permutation where all the cards appear in order: $123456.$ There are $5$ such permutations where two adjacent cards are interchanged, as in $124356.$ The other such permutations arise from removing one card from $123456$ and placing it in a position at least two away from its starting location. There are $4$ such positions to place each of the cards numbered $1$ and $6,$ and $3$ such positions for each of the cards numbered $2, 3, 4,$ and $5.$ This accounts for $2\\cdot4 + 4\\cdot3 =20$ permutations. Thus there are $1 + 5 + 20 = 26$ permutations where one card can be removed so that the remaining cards are in ascending order. There is an equal number of permutations that result in the cards' being in descending order. This gives the total $26 + 26 = 52.",
"More generally, suppose there are $n \\geq 4$ cards numbered $1, 2, 3, \\dots, n$ arranged in ascending order. If any one of the $n$ cards is removed and placed in one of the $n$ positions in the arrangement, the resulting permutation will have the property that one card can be removed so that the remaining cards are in ascending order. This accounts for $n\\cdot n = n^2$ permutations. However, the original ascending order has been counted $n$ times, and each order that arises by switching two neighboring cards has been counted twice. Hence the number of arrangements where one card can be removed resulting in the remaining cards' being in ascending order is $n^2-(n-1)-(n-1)=(n-1)^2+1.$ When $n = 6$, this is $(6-1)^2+1 = 26$, and the final answer is $2\\cdot26 = 52.",
"More generally, suppose there are $n \\geq 4$ cards numbered $1, 2, 3, \\dots, n$ arranged in ascending order. If any one of the $n$ cards is removed and placed in one of the $n$ positions in the arrangement, the resulting permutation will have the property that one card can be removed so that the remaining cards are in ascending order. This accounts for $n\\cdot n = n^2$ permutations. However, the original ascending order has been counted $n$ times, and each order that arises by switching two neighboring cards has been counted twice. Hence the number of arrangements where one card can be removed resulting in the remaining cards' being in ascending order is $n^2-(n-1)-(n-1)=(n-1)^2+1.$ When $n = 6$, this is $(6-1)^2+1 = 26$, and the final answer is $2\\cdot26 = 52.",
"For ascending, if the $1$ goes in anything but the first two slots, the rest of the numbers have to go in ascending from $2$, which are $4$ cases if there are $6$ cards. If $1$ goes in the second spot, then you can put any of the rest in the first slot but then the rest are determined, so in the case of $6$ cards, that gives $5$ more. If $1$ goes in the first slot, that means that you are doing the same problem with $n-1$ cards. So the recursion is $a_n=(n-2)+(n-1)+a_{n-1}$. There's $a_1=1$ and $a_2=2$, so you get $a_3=2+3=5$, $a_4=5+5=10$, $a_5=7+10=17$, and $a_6=9+17=26$. Or you can see that $a_n=(n-1)^2+1$. We double to account for descending and get $052. ~ahclark11",
"For ascending, if the $1$ goes in anything but the first two slots, the rest of the numbers have to go in ascending from $2$, which are $4$ cases if there are $6$ cards. If $1$ goes in the second spot, then you can put any of the rest in the first slot but then the rest are determined, so in the case of $6$ cards, that gives $5$ more. If $1$ goes in the first slot, that means that you are doing the same problem with $n-1$ cards. So the recursion is $a_n=(n-2)+(n-1)+a_{n-1}$. There's $a_1=1$ and $a_2=2$, so you get $a_3=2+3=5$, $a_4=5+5=10$, $a_5=7+10=17$, and $a_6=9+17=26$. Or you can see that $a_n=(n-1)^2+1$. We double to account for descending and get $052. ~ahclark11",
"First, we know that ascending order and descending order are symmetrical to each other (namely, if we get 132456 where after we take out 3, it will be one scenario; and if we flip it and write 654231, it will be another scenario) Thus, we only need to consider either descending or ascending and then times 2. WLOG let us consider ascending order Case 1: after we take out 1, the rest will be in ascending order: 2 3 4 5 6 Notice that 1 can be tucked in any one of the 6 spaces, thus there are 6 scenarios. Case 2: after we take out 2, the rest will be in ascending order: 1 3 4 5 6 Notice that if we put 2 next to 1 (to the right or to the left of 1), it will be an overcount, so there are only 4 cases for 2. It is easy to see that this is the same for 3, 4, 5, and 6. Thus, in total, we have \\[(6+4\\times5)\\times2=052\\] ~Adali",
"First, we know that ascending order and descending order are symmetrical to each other (namely, if we get 132456 where after we take out 3, it will be one scenario; and if we flip it and write 654231, it will be another scenario) Thus, we only need to consider either descending or ascending and then times 2. WLOG let us consider ascending order Case 1: after we take out 1, the rest will be in ascending order: 2 3 4 5 6 Notice that 1 can be tucked in any one of the 6 spaces, thus there are 6 scenarios. Case 2: after we take out 2, the rest will be in ascending order: 1 3 4 5 6 Notice that if we put 2 next to 1 (to the right or to the left of 1), it will be an overcount, so there are only 4 cases for 2. It is easy to see that this is the same for 3, 4, 5, and 6. Thus, in total, we have \\[(6+4\\times5)\\times2=052\\] ~Adali",
"We start with five cards in ascending order, then insert the sixth card to obtain a valid arrangement. Based on the card to be inserted, we have six cases. As shown below, the red squares indicate the possible positions to insert the sixth card. [asy] /* Made by MRENTHUSIASM */ unitsize(7mm); void drawSquare(real x, real y) { draw((x+0.5,y+0.5)--(x-0.5,y+0.5)--(x-0.5,y-0.5)--(x+0.5,y-0.5)--cycle,red); } label(\"$2$\",(0.5,8)); label(\"$3$\",(2.5,8)); label(\"$4$\",(4.5,8)); label(\"$5$\",(6.5,8)); label(\"$6$\",(8.5,8)); label(\"Insert $1.$\",(13.5,8),red); drawSquare(-0.5,8); drawSquare(1.5,8); drawSquare(3.5,8); drawSquare(5.5,8); drawSquare(7.5,8); drawSquare(9.5,8); label(\"$1$\",(0.5,6.5)); label(\"$3$\",(2.5,6.5)); label(\"$4$\",(4.5,6.5)); label(\"$5$\",(6.5,6.5)); label(\"$6$\",(8.5,6.5)); label(\"Insert $2.$\",(13.5,6.5),red); drawSquare(3.5,6.5); drawSquare(5.5,6.5); drawSquare(7.5,6.5); drawSquare(9.5,6.5); label(\"$1$\",(0.5,5)); label(\"$2$\",(2.5,5)); label(\"$4$\",(4.5,5)); label(\"$5$\",(6.5,5)); label(\"$6$\",(8.5,5)); label(\"Insert $3.$\",(13.5,5),red); drawSquare(-0.5,5); drawSquare(5.5,5); drawSquare(7.5,5); drawSquare(9.5,5); label(\"$1$\",(0.5,3.5)); label(\"$2$\",(2.5,3.5)); label(\"$3$\",(4.5,3.5)); label(\"$5$\",(6.5,3.5)); label(\"$6$\",(8.5,3.5)); label(\"Insert $4.$\",(13.5,3.5),red); drawSquare(-0.5,3.5); drawSquare(1.5,3.5); drawSquare(7.5,3.5); drawSquare(9.5,3.5); label(\"$1$\",(0.5,2)); label(\"$2$\",(2.5,2)); label(\"$3$\",(4.5,2)); label(\"$4$\",(6.5,2)); label(\"$6$\",(8.5,2)); label(\"Insert $5.$\",(13.5,2),red); drawSquare(-0.5,2); drawSquare(1.5,2); drawSquare(3.5,2); drawSquare(9.5,2); label(\"$1$\",(0.5,0.5)); label(\"$2$\",(2.5,0.5)); label(\"$3$\",(4.5,0.5)); label(\"$4$\",(6.5,0.5)); label(\"$5$\",(8.5,0.5)); label(\"Insert $6.$\",(13.5,0.5),red); drawSquare(-0.5,0.5); drawSquare(1.5,0.5); drawSquare(3.5,0.5); drawSquare(5.5,0.5); [/asy] Note that all arrangements of the six cards are distinct. There are $26$ arrangements in which removing one card will leave the remaining five cards in ascending order. By symmetry, there are $26$ arrangements in which removing one card will leave the remaining five cards in descending order. So, the answer is $26\\cdot2=052 ~MRENTHUSIASM",
"We start with five cards in ascending order, then insert the sixth card to obtain a valid arrangement. Based on the card to be inserted, we have six cases. As shown below, the red squares indicate the possible positions to insert the sixth card. [asy] /* Made by MRENTHUSIASM */ unitsize(7mm); void drawSquare(real x, real y) { draw((x+0.5,y+0.5)--(x-0.5,y+0.5)--(x-0.5,y-0.5)--(x+0.5,y-0.5)--cycle,red); } label(\"$2$\",(0.5,8)); label(\"$3$\",(2.5,8)); label(\"$4$\",(4.5,8)); label(\"$5$\",(6.5,8)); label(\"$6$\",(8.5,8)); label(\"Insert $1.$\",(13.5,8),red); drawSquare(-0.5,8); drawSquare(1.5,8); drawSquare(3.5,8); drawSquare(5.5,8); drawSquare(7.5,8); drawSquare(9.5,8); label(\"$1$\",(0.5,6.5)); label(\"$3$\",(2.5,6.5)); label(\"$4$\",(4.5,6.5)); label(\"$5$\",(6.5,6.5)); label(\"$6$\",(8.5,6.5)); label(\"Insert $2.$\",(13.5,6.5),red); drawSquare(3.5,6.5); drawSquare(5.5,6.5); drawSquare(7.5,6.5); drawSquare(9.5,6.5); label(\"$1$\",(0.5,5)); label(\"$2$\",(2.5,5)); label(\"$4$\",(4.5,5)); label(\"$5$\",(6.5,5)); label(\"$6$\",(8.5,5)); label(\"Insert $3.$\",(13.5,5),red); drawSquare(-0.5,5); drawSquare(5.5,5); drawSquare(7.5,5); drawSquare(9.5,5); label(\"$1$\",(0.5,3.5)); label(\"$2$\",(2.5,3.5)); label(\"$3$\",(4.5,3.5)); label(\"$5$\",(6.5,3.5)); label(\"$6$\",(8.5,3.5)); label(\"Insert $4.$\",(13.5,3.5),red); drawSquare(-0.5,3.5); drawSquare(1.5,3.5); drawSquare(7.5,3.5); drawSquare(9.5,3.5); label(\"$1$\",(0.5,2)); label(\"$2$\",(2.5,2)); label(\"$3$\",(4.5,2)); label(\"$4$\",(6.5,2)); label(\"$6$\",(8.5,2)); label(\"Insert $5.$\",(13.5,2),red); drawSquare(-0.5,2); drawSquare(1.5,2); drawSquare(3.5,2); drawSquare(9.5,2); label(\"$1$\",(0.5,0.5)); label(\"$2$\",(2.5,0.5)); label(\"$3$\",(4.5,0.5)); label(\"$4$\",(6.5,0.5)); label(\"$5$\",(8.5,0.5)); label(\"Insert $6.$\",(13.5,0.5),red); drawSquare(-0.5,0.5); drawSquare(1.5,0.5); drawSquare(3.5,0.5); drawSquare(5.5,0.5); [/asy] Note that all arrangements of the six cards are distinct. There are $26$ arrangements in which removing one card will leave the remaining five cards in ascending order. By symmetry, there are $26$ arrangements in which removing one card will leave the remaining five cards in descending order. So, the answer is $26\\cdot2=052 ~MRENTHUSIASM",
"We first realize that as long as we have an ascending sequence of $5$ numbers, we can just plug in a $6$th to make a sequence that works. For example, if we have $12345$, we can plug in a $6$ in any of $6$ spaces, before $1$, between $1$ and $2$, and so on to after $5$. We can also realize that this is completely symmetrical if the sequence is in descending order. For example, we could have $54321$, and we could plug in a 6 in 6 of the spaces. For the total number of combinations, we have 6 ascending cases multiplied by the $6$ places that can hold whatever number is missing, and multiply by 2 because there is descending and ascending number cases. But wait, what if we have $12345$ and plug in a $6$ at the end and $12346$ and plug in a $5$ at the $5$th spot? We have overcounted, so we need to subtract off the identical pairs. Assume that one of them is the \"right\" combination. That means that there are $1$ \"right\" combination because all the rest will have $2$ of their combinations taken by the previous one. For example, if I have $12356$ and $12346$, I would count those $123456$ as $12356$'s instead of $12346$'s. Therefore, we have $6+4+4+4+4+4 = 26$ combinations of ascending order, and since we need to count those of descending order as well, we have $26\\cdot2=052 Side note: We could have just realized that the first part was unnecessary, but it does help you understand the problem better. ~Arcticturn",
"Like the previous solutions, we calculate ascending possibilities then multiply by two. Then, the configuration can look like +++++ or +-+++ (the minus can go anywhere), where \"+\" indicates an increase, and \"-\" indicates a decrease. Obviously, for the first case, the only possibility is 123456. For the second case, we can start with 123456, and then take either a card after the \"-\" and put it before the \"-\", or we could take a card before the \"-\" and put it after the \"-\". This means that every card other than the one at the position of the \"-\" can be used in one way. There are five ways to have 4 plus's, and one minus, and for each way, there are 5 ways to rearrange 123456 to achieve that. So, our answer is $2(5\\cdot5+1)=052",
"First, we select $5$ of the $6$ cards to put in ascending or descending order. Then, we must add the other card to the group to arrange the six cards such that one can be removed so that the remaining cards are in ascending or descending order. Then, it would seem that the answer is $(6C5)*2$(ascending or descending)*$6$(the number of ways to place the last card). However, this overcounts some cases. For example, 123456 and 654321 can be made with 12345 and 6, 12346 and 5, etc. Furthermore, 654312 can be made with 65432 and 1 or 65431 and 2. Therefore, we first find that $(6C5)*2*4$=$48$. Then, we add the special cases ($123456$, $654321$, $654312$, $123465$) to obtain $052. ~mathMagicOPS",
"First, we select $5$ of the $6$ cards to put in ascending or descending order. Then, we must add the other card to the group to arrange the six cards such that one can be removed so that the remaining cards are in ascending or descending order. Then, it would seem that the answer is $(6C5)*2$(ascending or descending)*$6$(the number of ways to place the last card). However, this overcounts some cases. For example, 123456 and 654321 can be made with 12345 and 6, 12346 and 5, etc. Furthermore, 654312 can be made with 65432 and 1 or 65431 and 2. Therefore, we first find that $(6C5)*2*4$=$48$. Then, we add the special cases ($123456$, $654321$, $654312$, $123465$) to obtain $052. ~mathMagicOPS"
] |
2020-I-6
| 2,020
| 6
|
A flat board has a circular hole with radius $1$ and a circular hole with radius $2$ such that the distance between the centers of the two holes is $7$ . Two spheres with equal radii sit in the two holes such that the spheres are tangent to each other. The square of the radius of the spheres is $\tfrac{m}{n}$ , where $m$ and $n$ are relatively prime positive integers. Find $m+n$ .
| 173
|
I
|
[
"[asy] size(10cm); pair A, B, C, D, O, P, H, L, X, Y; A = (-1, 0); B = (1, 0); H = (0, 0); C = (5, 0); D = (9, 0); L = (7, 0); O = (0, sqrt(160/13 - 1)); P = (7, sqrt(160/13 - 4)); X = (0, sqrt(160/13 - 4)); Y = (O + P) / 2; draw(A -- O -- B -- cycle); draw(C -- P -- D -- cycle); draw(B -- C); draw(O -- P); draw(P -- X, dashed); draw(O -- H, dashed); draw(P -- L, dashed); draw(circle(O, sqrt(160/13))); draw(circle(P, sqrt(160/13))); path b = brace(L-(0,1), H-(0,1),0.5); draw(b); label(\"$r$\", O -- Y, N); label(\"$r$\", Y -- P, N); label(\"$r$\", O -- A, NW); label(\"$r$\", P -- D, NE); label(\"$1$\", A -- H, N); label(\"$2$\", L -- D, N); label(\"$7$\", b, S); dot(A^^B^^C^^D^^O^^P^^H^^L^^X^^Y,linewidth(4)); [/asy] Set the common radius to $r$. First, take the cross section of the sphere sitting in the hole of radius $1$. If we draw the perpendicular bisector of the chord (the hole) through the circle, this line goes through the center. Connect the center also to where the chord hits the circle, for a right triangle with hypotenuse $r$ and base $1$. Therefore, the height of this circle outside of the hole is $\\sqrt{r^2-1}$. The other circle follows similarly for a height (outside the hole) of $\\sqrt{r^2-4}$. Now, if we take the cross section of the entire board, essentially making it two-dimensional, we can connect the centers of the two spheres, then form another right triangle with base $7$, as given by the problem. The height of this triangle is the difference between the heights of the parts of the two spheres outside the holes, which is $\\sqrt{r^2-1} - \\sqrt{r^2-4}$. Now we can set up an equation in terms of $r$ with the Pythagorean theorem: \\[\\left(\\sqrt{r^2-1} - \\sqrt{r^2-4}\\right)^2 + 7^2 = (2r)^2.\\] Simplifying a few times, \\begin{align*} r^2 - 1 - 2\\left(\\sqrt{(r^2-1)(r^2-4)}\\right) + r^2 - 4 + 49 &= 4r^2 \\\\ 2r^2-44 &= -2\\left(\\sqrt{(r^2-1)(r^2-4)}\\right) \\\\ 22-r^2 &= \\left(\\sqrt{r^4 - 5r^2 + 4}\\right) \\\\ r^4 -44r^2 + 484 &= r^4 - 5r^2 + 4 \\\\ 39r^2&=480 \\\\ r^2&=\\frac{480}{39} = \\frac{160}{13}. \\end{align*} Therefore, our answer is $173. ~molocyxu",
"[asy] size(10cm); pair A, B, C, D, O, P, H, L, X, Y; A = (-1, 0); B = (1, 0); H = (0, 0); C = (5, 0); D = (9, 0); L = (7, 0); O = (0, sqrt(160/13 - 1)); P = (7, sqrt(160/13 - 4)); X = (0, sqrt(160/13 - 4)); Y = (O + P) / 2; draw(A -- O -- B -- cycle); draw(C -- P -- D -- cycle); draw(B -- C); draw(O -- P); draw(P -- X, dashed); draw(O -- H, dashed); draw(P -- L, dashed); draw(circle(O, sqrt(160/13))); draw(circle(P, sqrt(160/13))); path b = brace(L-(0,1), H-(0,1),0.5); draw(b); label(\"$r$\", O -- Y, N); label(\"$r$\", Y -- P, N); label(\"$r$\", O -- A, NW); label(\"$r$\", P -- D, NE); label(\"$1$\", A -- H, N); label(\"$2$\", L -- D, N); label(\"$7$\", b, S); dot(A^^B^^C^^D^^O^^P^^H^^L^^X^^Y,linewidth(4)); [/asy] Set the common radius to $r$. First, take the cross section of the sphere sitting in the hole of radius $1$. If we draw the perpendicular bisector of the chord (the hole) through the circle, this line goes through the center. Connect the center also to where the chord hits the circle, for a right triangle with hypotenuse $r$ and base $1$. Therefore, the height of this circle outside of the hole is $\\sqrt{r^2-1}$. The other circle follows similarly for a height (outside the hole) of $\\sqrt{r^2-4}$. Now, if we take the cross section of the entire board, essentially making it two-dimensional, we can connect the centers of the two spheres, then form another right triangle with base $7$, as given by the problem. The height of this triangle is the difference between the heights of the parts of the two spheres outside the holes, which is $\\sqrt{r^2-1} - \\sqrt{r^2-4}$. Now we can set up an equation in terms of $r$ with the Pythagorean theorem: \\[\\left(\\sqrt{r^2-1} - \\sqrt{r^2-4}\\right)^2 + 7^2 = (2r)^2.\\] Simplifying a few times, \\begin{align*} r^2 - 1 - 2\\left(\\sqrt{(r^2-1)(r^2-4)}\\right) + r^2 - 4 + 49 &= 4r^2 \\\\ 2r^2-44 &= -2\\left(\\sqrt{(r^2-1)(r^2-4)}\\right) \\\\ 22-r^2 &= \\left(\\sqrt{r^4 - 5r^2 + 4}\\right) \\\\ r^4 -44r^2 + 484 &= r^4 - 5r^2 + 4 \\\\ 39r^2&=480 \\\\ r^2&=\\frac{480}{39} = \\frac{160}{13}. \\end{align*} Therefore, our answer is $173. ~molocyxu",
"Let $A$ and $B$ be the centers of the holes, let $C$ be the point of crossing $AB$ and radical axes of the circles. So $C$ has equal tangential distance to any point of both spheres. In particular to the circles (https://en.wikipedia.org/wiki/Radical_axis.) \\[CA = \\frac {AB} {2} – \\frac {r_A^2 – r_B^2}{2 AB} = \\frac{23}{7}, CB = AB - AC =\\frac{26}{7},\\] \\[CA' = CB'= \\sqrt{BC^2 – r_B^2} = \\frac {4}{7} \\sqrt{30}.\\] Let $D$ be the point of tangency of the spheres common radius $R$ centered at $O$ and $O'.$ Let $\\alpha$ be the angle between $OO'$ and flat board. In the plane, perpendicular to board \\[DC \\perp OO', DC = \\frac {4}{7} \\sqrt{30}.\\] Distance between $C$ and midpoint $M$ of $AB$ is \\[d = \\frac {26}{7} - \\frac {7}{2} = \\frac{3}{14} \\implies \\sin \\alpha = \\frac {d}{DC} = \\frac {3}{8\\sqrt {30}}.\\] \\[\\cos \\alpha = \\sqrt {1 – \\frac {9}{64 \\cdot 30}} = \\sqrt{ \\frac {637}{640}} = \\frac {7}{2} \\sqrt {\\frac{13}{160}}.\\] \\[2R \\cos \\alpha = AB = 7 \\implies R = \\frac {\\frac{7}{2} } {\\frac{7}{2}\\sqrt \\frac{13}{160}} = \\sqrt {\\frac{160}{13}}.\\] Hence the answer is \\[160+13 = 173.\\] vladimir.shelomovskii@gmail.com, vvsss",
"Let $A$ and $B$ be the centers of the holes, let $C$ be the point of crossing $AB$ and radical axes of the circles. So $C$ has equal tangential distance to any point of both spheres. In particular to the circles (https://en.wikipedia.org/wiki/Radical_axis.) \\[CA = \\frac {AB} {2} – \\frac {r_A^2 – r_B^2}{2 AB} = \\frac{23}{7}, CB = AB - AC =\\frac{26}{7},\\] \\[CA' = CB'= \\sqrt{BC^2 – r_B^2} = \\frac {4}{7} \\sqrt{30}.\\] Let $D$ be the point of tangency of the spheres common radius $R$ centered at $O$ and $O'.$ Let $\\alpha$ be the angle between $OO'$ and flat board. In the plane, perpendicular to board \\[DC \\perp OO', DC = \\frac {4}{7} \\sqrt{30}.\\] Distance between $C$ and midpoint $M$ of $AB$ is \\[d = \\frac {26}{7} - \\frac {7}{2} = \\frac{3}{14} \\implies \\sin \\alpha = \\frac {d}{DC} = \\frac {3}{8\\sqrt {30}}.\\] \\[\\cos \\alpha = \\sqrt {1 – \\frac {9}{64 \\cdot 30}} = \\sqrt{ \\frac {637}{640}} = \\frac {7}{2} \\sqrt {\\frac{13}{160}}.\\] \\[2R \\cos \\alpha = AB = 7 \\implies R = \\frac {\\frac{7}{2} } {\\frac{7}{2}\\sqrt \\frac{13}{160}} = \\sqrt {\\frac{160}{13}}.\\] Hence the answer is \\[160+13 = 173.\\] vladimir.shelomovskii@gmail.com, vvsss"
] |
2020-I-7
| 2,020
| 7
|
A club consisting of $11$ men and $12$ women needs to choose a committee from among its members so that the number of women on the committee is one more than the number of men on the committee. The committee could have as few as $1$ member or as many as $23$ members. Let $N$ be the number of such committees that can be formed. Find the sum of the prime numbers that divide $N.$
| 81
|
I
|
[
"Let $k$ be the number of women selected. Then, the number of men not selected is $11-(k-1)=12-k$. Note that the sum of the number of women selected and the number of men not selected is constant at $12$. Each combination of women selected and men not selected corresponds to a committee selection. Since choosing 12 individuals from the total of 23 would give $k$ women and $12-k$ men, the number of committee selections is $\\binom{23}{12}$. The answer is $081. ~awang11's sol",
"We casework on the amount of men on the committee. If there are no men in the committee, there are $\\dbinom{12}{1}$ ways to pick the women on the committee, for a total of $\\dbinom{11}{0} \\cdot \\dbinom{12}{1}$. Notice that $\\dbinom{11}{0}$ is equal to $\\dbinom{11}{11}$, so the case where no men are picked can be grouped with the case where all men are picked. When all men are picked, all women must also be picked, for a total of $\\dbinom{12}{12}$. Therefore, these cases can be combined to \\[\\dbinom{11}{0} \\cdot \\left(\\dbinom{12}{1} + \\dbinom{12}{12}\\right)\\] Since $\\dbinom{12}{12} = \\dbinom{12}{0}$, and $\\dbinom{12}{0} + \\dbinom{12}{1} = \\dbinom{13}{1}$, we can further simplify this to \\[\\dbinom{11}{0} \\cdot \\dbinom{13}{1}\\] All other cases proceed similarly. For example, the case with one men or ten men is equal to $\\dbinom{11}{1} \\cdot \\dbinom{13}{2}$. Now, if we factor out a $13$, then all cases except the first two have a factor of $121$, so we can factor this out too to make our computation slightly easier. The first two cases (with $13$ factored out) give $1+66=67$, and the rest gives $121(10+75+270+504) = 103,939$. Adding the $67$ gives $104,006$. Now, we can test for prime factors. We know there is a factor of $2$, and the rest is $52,003$. We can also factor out a $7$, for $7,429$, and the rest is $17 \\cdot 19 \\cdot 23$. Adding up all the prime factors gives $2+7+13+17+19+23 = 081. Video Solution: https://youtu.be/MVxsY8DwHVk ~ avn",
"We casework on the amount of men on the committee. If there are no men in the committee, there are $\\dbinom{12}{1}$ ways to pick the women on the committee, for a total of $\\dbinom{11}{0} \\cdot \\dbinom{12}{1}$. Notice that $\\dbinom{11}{0}$ is equal to $\\dbinom{11}{11}$, so the case where no men are picked can be grouped with the case where all men are picked. When all men are picked, all women must also be picked, for a total of $\\dbinom{12}{12}$. Therefore, these cases can be combined to \\[\\dbinom{11}{0} \\cdot \\left(\\dbinom{12}{1} + \\dbinom{12}{12}\\right)\\] Since $\\dbinom{12}{12} = \\dbinom{12}{0}$, and $\\dbinom{12}{0} + \\dbinom{12}{1} = \\dbinom{13}{1}$, we can further simplify this to \\[\\dbinom{11}{0} \\cdot \\dbinom{13}{1}\\] All other cases proceed similarly. For example, the case with one men or ten men is equal to $\\dbinom{11}{1} \\cdot \\dbinom{13}{2}$. Now, if we factor out a $13$, then all cases except the first two have a factor of $121$, so we can factor this out too to make our computation slightly easier. The first two cases (with $13$ factored out) give $1+66=67$, and the rest gives $121(10+75+270+504) = 103,939$. Adding the $67$ gives $104,006$. Now, we can test for prime factors. We know there is a factor of $2$, and the rest is $52,003$. We can also factor out a $7$, for $7,429$, and the rest is $17 \\cdot 19 \\cdot 23$. Adding up all the prime factors gives $2+7+13+17+19+23 = 081. Video Solution: https://youtu.be/MVxsY8DwHVk ~ avn",
"Applying [1] by setting $m=12$, $n=11$, and $r=11$, we obtain $\\binom{23}{11}\\implies$ $081. ~programjames1",
"Applying [1] by setting $m=12$, $n=11$, and $r=11$, we obtain $\\binom{23}{11}\\implies$ $081. ~programjames1",
"Notice that the committee can consist of $k$ boys and $k+1$ girls. Summing over all possible $k$ gives \\[\\sum_{k=0}^{11}\\binom{11}{k}\\binom{12}{k+1}=\\binom{11}{0}\\binom{12}{1}+\\binom{11}{1}\\binom{12}{2}+\\cdots + \\binom{11}{11}\\binom{12}{12}\\] Using the identity $\\binom{n}{k}=\\binom{n}{n-k}$, and Pascal's Identity $\\binom{n}{k}+\\binom{n}{k+1}=\\binom{n+1}{k+1}$, we get \\[\\sum_{k=0}^{11}\\binom{11}{k}\\binom{12}{k+1}=\\binom{12}{12}+\\binom{12}{1}\\left(\\binom{11}{0}+\\binom{11}{1}\\right)+\\cdots\\] \\[=\\binom{12}{0}^2+\\binom{12}{1}^2+\\binom{12}{2}^2+\\binom{12}{3}^2+\\binom{12}{4}^2+\\binom{12}{5}^2+\\frac{\\binom{12}{6}^2}{2}\\] \\[=\\frac{1}{2}\\sum_{k=0}^{12}\\binom{12}{k}^2\\] Using the identity $\\sum_{k=0}^n\\binom{n}{k}^2=\\binom{2n}{n}$, this simplifies to \\[\\frac{1}{2}\\cdot \\binom{24}{12}=\\frac{24\\cdot 23\\cdot 22\\cdot 21\\cdot 20\\cdot 19\\cdot 18\\cdot 17\\cdot 16\\cdot 15\\cdot 14\\cdot 13}{2\\cdot 12\\cdot 11\\cdot 10\\cdot 9\\cdot 8\\cdot 7\\cdot 6\\cdot 5\\cdot 4\\cdot 3\\cdot 2}=2\\cdot 7\\cdot 13\\cdot 17\\cdot 19\\cdot 23\\] so the desired answer is $2+7+13+17+19+23=081 ~ktong",
"Select any $11$ club members. That group will have $i$ men and $11-i$ women, so the number of women in the club not selected in that group is $12 - (11-i) = i+1$. Thus, if the committee includes the men who were selected and the women who were not selected, the committee would have the correct number of men and women. Conversely, for every committee that could be formed with $i$ men and $i+1$ women, the men on this committee together with the women not on the committee comprise a subset of $i + (12 - (i+1)) = 11$ club members. Thus\\[N = \\binom{23}{11}= \\frac{23\\cdot22\\cdot21\\cdot20\\cdot19\\cdot18\\cdot17\\cdot16\\cdot15\\cdot14\\cdot13}{11\\cdot10\\cdot9\\cdot8\\cdot7\\cdot6\\cdot5\\cdot4\\cdot3\\cdot2\\cdot1}=23\\cdot19\\cdot17\\cdot13\\cdot7\\cdot2.\\]The requested sum is $23+19+17+13+7+2=081.",
"Select any $11$ club members. That group will have $i$ men and $11-i$ women, so the number of women in the club not selected in that group is $12 - (11-i) = i+1$. Thus, if the committee includes the men who were selected and the women who were not selected, the committee would have the correct number of men and women. Conversely, for every committee that could be formed with $i$ men and $i+1$ women, the men on this committee together with the women not on the committee comprise a subset of $i + (12 - (i+1)) = 11$ club members. Thus\\[N = \\binom{23}{11}= \\frac{23\\cdot22\\cdot21\\cdot20\\cdot19\\cdot18\\cdot17\\cdot16\\cdot15\\cdot14\\cdot13}{11\\cdot10\\cdot9\\cdot8\\cdot7\\cdot6\\cdot5\\cdot4\\cdot3\\cdot2\\cdot1}=23\\cdot19\\cdot17\\cdot13\\cdot7\\cdot2.\\]The requested sum is $23+19+17+13+7+2=081.",
"Notice that if $k$ men are picked, then $k+1$ women must be picked. Furthermore, $k$ can range from $0$ to $11$. Then, \\[N=\\sum_{k=0}^{11} \\binom{11}{k}\\binom{12}{k+1}=\\binom{11}{0}\\binom{12}{1}+\\binom{11}{1}\\binom{12}{2}+\\dots+\\binom{11}{11}\\binom{12}{12}\\] Since $\\binom{n}{k}=\\binom{n}{n-k}$, this equals \\[N=\\binom{11}{0}\\binom{12}{11}+\\binom{11}{1}\\binom{12}{10}+\\dots+\\binom{11}{11}\\binom{12}{0}\\] According to Vandermonde's Identity, \\[N=\\binom{11+12}{11}=\\binom{23}{11}\\] \\[N=\\frac{23!}{11!\\cdot 12!}=\\frac{2^{19}\\cdot 3^9\\cdot 5^4\\cdot 7^3\\cdot 11^2\\cdot 13\\cdot 17\\cdot 19\\cdot 23}{2^{10}\\cdot 3^5\\cdot 5^2\\cdot 7\\cdot 11\\times 2^8\\cdot 3^4\\cdot 5^2\\cdot 7\\cdot 11}=2\\cdot 7\\cdot 13\\cdot 17\\cdot 19\\cdot 23\\rightarrow 081\\] ~sid2012",
"Test the cases where there are $0$ men $1$ woman, $1$ man $2$ women, $2$ men $3$ women ... you will get the sequence $1$, $3$, $10$, $35$. Multiply all these numbers by $2$ to get $2$, $6$, $20$, $70$, which is also $\\binom{2}{1}$, $\\binom{4}{2}$, $\\binom{6}{3}$, $\\binom{8}{4}$. Thus, continuing this pattern, the case with $11$ men and $12$ women should have $\\frac{\\binom{24}{12}}{2}$ ways to select the committee. (someone prove this rigorously im too lazy ) -Kevin2010",
"Test the cases where there are $0$ men $1$ woman, $1$ man $2$ women, $2$ men $3$ women ... you will get the sequence $1$, $3$, $10$, $35$. Multiply all these numbers by $2$ to get $2$, $6$, $20$, $70$, which is also $\\binom{2}{1}$, $\\binom{4}{2}$, $\\binom{6}{3}$, $\\binom{8}{4}$. Thus, continuing this pattern, the case with $11$ men and $12$ women should have $\\frac{\\binom{24}{12}}{2}$ ways to select the committee. (someone prove this rigorously im too lazy ) -Kevin2010"
] |
2020-I-8
| 2,020
| 8
|
A bug walks all day and sleeps all night. On the first day, it starts at point $O,$ faces east, and walks a distance of $5$ units due east. Each night the bug rotates $60^\circ$ counterclockwise. Each day it walks in this new direction half as far as it walked the previous day. The bug gets arbitrarily close to the point $P.$ Then $OP^2=\tfrac{m}{n},$ where $m$ and $n$ are relatively prime positive integers. Find $m+n.$
| 103
|
I
|
[
"[asy] size(8cm); pair O, A, B, C, D, F, G, H, I, P, X; O = (0, 0); A = (5, 0); X = (8, 0); P = (5, 5 / sqrt(3)); B = rotate(-120, A) * ((O + A) / 2); C = rotate(-120, B) * ((A + B) / 2); D = rotate(-120, C) * ((B + C) / 2); F = rotate(-120, D) * ((C + D) / 2); G = rotate(-120, F) * ((D + F) / 2); H = rotate(-120, G) * ((F + G) / 2); I = rotate(-120, H) * ((G + H) / 2); draw(O -- A -- B -- C -- D -- F -- G -- H -- I); draw(A -- X, dashed); markscalefactor = 0.05; path angle = anglemark(X, A, B); draw(angle); dot(P); dot(O); dot(A); dot(B); dot(C); dot(D); label(\"$O$\", O, W); label(\"$P$\", P, E); label(\"$A$\", A, S); label(\"$B$\", B, E); label(\"$C$\", C, E); label(\"$D$\", D, W); label(\"$60^\\circ$\", angle, ENE*3); [/asy] We notice that the moves cycle every 6 moves, so we plot out the first 6 moves on the coordinate grid with point $O$ as the origin. We will keep a tally of the x-coordinate and y-coordinate separately. Then, we will combine them and account for the cycling using the formula for an infinite geometric series. First move: The ant moves right $5$. Second move: We use properties of a $30-60-90$ triangle to get $\\frac{5}{4}$ right, $\\frac{5\\sqrt{3}}{4}$ up. Third move: $\\frac{5}{8}$ left, $\\frac{5\\sqrt{3}}{8}$ up. Fourth move: $\\frac{5}{8}$ left. Fifth move: $\\frac{5}{32}$ left, $\\frac{5\\sqrt{3}}{32}$ down. Sixth move: $\\frac{5}{64}$ right, $\\frac{5\\sqrt{3}}{64}$ down. Total of x-coordinate: $5 + \\frac{5}{4} - \\frac{5}{8} - \\frac{5}{8} - \\frac{5}{32} + \\frac{5}{64} = \\frac{315}{64}$. Total of y-coordinate: $0 + \\frac{5\\sqrt{3}}{4} + \\frac{5\\sqrt{3}}{8} + 0 - \\frac{5\\sqrt{3}}{32} - \\frac{5\\sqrt{3}}{64} = \\frac{105\\sqrt{3}}{64}$. After this cycle of six moves, all moves repeat with a factor of $(\\frac{1}{2})^6 = \\frac{1}{64}$. Using the formula for a geometric series, multiplying each sequence by $\\frac{1}{1-\\frac{1}{64}} = \\frac{64}{63}$ will give us the point $P$. Now, knowing the initial $x$ and $y,$ we plug this into the geometric series formula ($\\frac{a}{1-r}$), and we get $\\frac{315}{64} \\cdot \\frac{64}{63} = 5$, $\\frac{105\\sqrt{3}}{64} \\cdot \\frac{64}{63} = \\frac{5\\sqrt{3}}{3}$. Therefore, the coordinates of point $P$ are $(5,\\frac{5\\sqrt{3}}{3})$, so using the Pythagorean Theorem, $OP^2 = \\frac{100}{3}$, for an answer of $103. -molocyxu",
"[asy] size(8cm); pair O, A, B, C, D, F, G, H, I, P, X; O = (0, 0); A = (5, 0); X = (8, 0); P = (5, 5 / sqrt(3)); B = rotate(-120, A) * ((O + A) / 2); C = rotate(-120, B) * ((A + B) / 2); D = rotate(-120, C) * ((B + C) / 2); F = rotate(-120, D) * ((C + D) / 2); G = rotate(-120, F) * ((D + F) / 2); H = rotate(-120, G) * ((F + G) / 2); I = rotate(-120, H) * ((G + H) / 2); draw(O -- A -- B -- C -- D -- F -- G -- H -- I); draw(A -- X, dashed); markscalefactor = 0.05; path angle = anglemark(X, A, B); draw(angle); dot(P); dot(O); dot(A); dot(B); dot(C); dot(D); label(\"$O$\", O, W); label(\"$P$\", P, E); label(\"$A$\", A, S); label(\"$B$\", B, E); label(\"$C$\", C, E); label(\"$D$\", D, W); label(\"$60^\\circ$\", angle, ENE*3); [/asy] We notice that the moves cycle every 6 moves, so we plot out the first 6 moves on the coordinate grid with point $O$ as the origin. We will keep a tally of the x-coordinate and y-coordinate separately. Then, we will combine them and account for the cycling using the formula for an infinite geometric series. First move: The ant moves right $5$. Second move: We use properties of a $30-60-90$ triangle to get $\\frac{5}{4}$ right, $\\frac{5\\sqrt{3}}{4}$ up. Third move: $\\frac{5}{8}$ left, $\\frac{5\\sqrt{3}}{8}$ up. Fourth move: $\\frac{5}{8}$ left. Fifth move: $\\frac{5}{32}$ left, $\\frac{5\\sqrt{3}}{32}$ down. Sixth move: $\\frac{5}{64}$ right, $\\frac{5\\sqrt{3}}{64}$ down. Total of x-coordinate: $5 + \\frac{5}{4} - \\frac{5}{8} - \\frac{5}{8} - \\frac{5}{32} + \\frac{5}{64} = \\frac{315}{64}$. Total of y-coordinate: $0 + \\frac{5\\sqrt{3}}{4} + \\frac{5\\sqrt{3}}{8} + 0 - \\frac{5\\sqrt{3}}{32} - \\frac{5\\sqrt{3}}{64} = \\frac{105\\sqrt{3}}{64}$. After this cycle of six moves, all moves repeat with a factor of $(\\frac{1}{2})^6 = \\frac{1}{64}$. Using the formula for a geometric series, multiplying each sequence by $\\frac{1}{1-\\frac{1}{64}} = \\frac{64}{63}$ will give us the point $P$. Now, knowing the initial $x$ and $y,$ we plug this into the geometric series formula ($\\frac{a}{1-r}$), and we get $\\frac{315}{64} \\cdot \\frac{64}{63} = 5$, $\\frac{105\\sqrt{3}}{64} \\cdot \\frac{64}{63} = \\frac{5\\sqrt{3}}{3}$. Therefore, the coordinates of point $P$ are $(5,\\frac{5\\sqrt{3}}{3})$, so using the Pythagorean Theorem, $OP^2 = \\frac{100}{3}$, for an answer of $103. -molocyxu",
"We place the ant at the origin of the complex plane with its first move being in the positive real direction. Then the ant's journey can be represented as the infinite series \\[5\\left(1 + \\frac{e^{\\frac{i\\pi}{3}}}{2} + \\left(\\frac{e^{\\frac{i\\pi}{3}}}{2}\\right)^2 + \\cdots\\right)\\] Using the formula for an infinite geometric series, this is equal to \\[\\frac{5}{1 - \\frac12e^{\\frac{i\\pi}{3}}} = \\frac{5}{1 - \\frac{1 + i\\sqrt{3}}{4}} = \\frac{20}{3 - i\\sqrt{3}} = 5 + \\frac{5i\\sqrt{3}}{3}\\] We are looking for the square of the modulus of this value: \\[\\left|\\frac{5 + 5i\\sqrt{3}}{3}\\right|^2 = 25 + \\frac{25}{3} = \\frac{100}{3}\\] so the answer is $100 + 3 = 103.",
"We place the ant at the origin of the complex plane with its first move being in the positive real direction. Then the ant's journey can be represented as the infinite series \\[5\\left(1 + \\frac{e^{\\frac{i\\pi}{3}}}{2} + \\left(\\frac{e^{\\frac{i\\pi}{3}}}{2}\\right)^2 + \\cdots\\right)\\] Using the formula for an infinite geometric series, this is equal to \\[\\frac{5}{1 - \\frac12e^{\\frac{i\\pi}{3}}} = \\frac{5}{1 - \\frac{1 + i\\sqrt{3}}{4}} = \\frac{20}{3 - i\\sqrt{3}} = 5 + \\frac{5i\\sqrt{3}}{3}\\] We are looking for the square of the modulus of this value: \\[\\left|\\frac{5 + 5i\\sqrt{3}}{3}\\right|^2 = 25 + \\frac{25}{3} = \\frac{100}{3}\\] so the answer is $100 + 3 = 103.",
"The ant goes in the opposite direction every $3$ moves, going $(1/2)^3=1/8$ the distance backwards. Using geometric series, he travels $1-1/8+1/64-1/512...=(7/8)(1+1/64+1/4096...)=(7/8)(64/63)=8/9$ the distance of the first three moves over infinity moves. Now, we use coordinates meaning $(5+5/4-5/8, 0+5\\sqrt3/4+5\\sqrt3/8)$ or $(45/8, 15\\sqrt3/8)$. Multiplying these by $8/9$, we get $(5, 5\\sqrt3/3)$ $\\implies$ $103 . ~Lcz",
"The ant goes in the opposite direction every $3$ moves, going $(1/2)^3=1/8$ the distance backwards. Using geometric series, he travels $1-1/8+1/64-1/512...=(7/8)(1+1/64+1/4096...)=(7/8)(64/63)=8/9$ the distance of the first three moves over infinity moves. Now, we use coordinates meaning $(5+5/4-5/8, 0+5\\sqrt3/4+5\\sqrt3/8)$ or $(45/8, 15\\sqrt3/8)$. Multiplying these by $8/9$, we get $(5, 5\\sqrt3/3)$ $\\implies$ $103 . ~Lcz",
"Suppose that the bug starts at the origin $(0,0)$ and travels a distance of $a$ units due east on the first day, and that there is a real number $r$ with $0<r < 1$ such that each day after the first, the bug walks $r$ times as far as the previous day. On day $n$, the bug travels along the vector $\\pmb v_{n}$ that has magnitude $ar^{n-1}$ and direction $\\langle\\cos(n\\cdot 60^\\circ),\\sin(n\\cdot 60^\\circ)\\rangle$. Then $P$ is the terminal point of the infinite sum of the vectors $\\pmb v_{1}+\\pmb v_{2}+\\pmb v_3+\\cdots$. The $x$-coordinate of this sum is \\[a\\big(\\!\\cos0^\\circ+r\\cos60^\\circ + r^{2}\\cos120^\\circ+r^{3}\\cos180^\\circ+r^{4}\\cos240^\\circ\\] \\[+r^{5}\\cos300^\\circ+r^{6}\\cos360^\\circ+\\cdots\\big).\\] Because the angles repeat after 6 terms, this sum is equal to \\[aS(1+r^{6}+r^{12}+r^{18}+\\cdots)=\\frac{aS}{1-r^{6}},\\] where \\[S=\\cos0^\\circ+r\\cos60^\\circ + r^{2}\\cos120^\\circ+r^{3}\\cos180^\\circ+r^{4}\\cos240^\\circ+ r^{5}\\cos300^\\circ.\\] Similarly, the $y$-coordinate of $P$ will be ${\\frac{aT}{1-r^{6}}}$, where \\[T=\\sin0^\\circ+r\\sin60^\\circ + r^{2}\\sin120^\\circ+r^{3}\\sin180^\\circ+r^{4}\\sin240^\\circ+ r^{5}\\sin300^\\circ.\\]In this case $r=\\frac12$ and $a = 5$, so \\[S=1+\\frac14-\\frac18-\\frac18-\\frac1{32}+\\frac1{64}=\\frac{63}{64},\\] \\[T=0+\\frac{\\sqrt3}4+\\frac{\\sqrt3}8+0-\\frac{\\sqrt3}{32}-\\frac{\\sqrt3}{64}=\\frac{21\\sqrt3}{64},\\]and the coordinates of $P$ are \\[\\left(\\frac{5S}{1-\\frac1{64}}, \\frac{5T}{1-\\frac1{64}}\\right)=\\left(5,\\frac{5\\sqrt3}{3}\\right).\\]Thus the square of the distance from the origin to $P$ is $25+\\frac{25}3=\\frac{100}3$. The requested sum is $100+3=103.",
"Suppose that the bug starts at the origin $(0,0)$ and travels a distance of $a$ units due east on the first day, and that there is a real number $r$ with $0<r < 1$ such that each day after the first, the bug walks $r$ times as far as the previous day. On day $n$, the bug travels along the vector $\\pmb v_{n}$ that has magnitude $ar^{n-1}$ and direction $\\langle\\cos(n\\cdot 60^\\circ),\\sin(n\\cdot 60^\\circ)\\rangle$. Then $P$ is the terminal point of the infinite sum of the vectors $\\pmb v_{1}+\\pmb v_{2}+\\pmb v_3+\\cdots$. The $x$-coordinate of this sum is \\[a\\big(\\!\\cos0^\\circ+r\\cos60^\\circ + r^{2}\\cos120^\\circ+r^{3}\\cos180^\\circ+r^{4}\\cos240^\\circ\\] \\[+r^{5}\\cos300^\\circ+r^{6}\\cos360^\\circ+\\cdots\\big).\\] Because the angles repeat after 6 terms, this sum is equal to \\[aS(1+r^{6}+r^{12}+r^{18}+\\cdots)=\\frac{aS}{1-r^{6}},\\] where \\[S=\\cos0^\\circ+r\\cos60^\\circ + r^{2}\\cos120^\\circ+r^{3}\\cos180^\\circ+r^{4}\\cos240^\\circ+ r^{5}\\cos300^\\circ.\\] Similarly, the $y$-coordinate of $P$ will be ${\\frac{aT}{1-r^{6}}}$, where \\[T=\\sin0^\\circ+r\\sin60^\\circ + r^{2}\\sin120^\\circ+r^{3}\\sin180^\\circ+r^{4}\\sin240^\\circ+ r^{5}\\sin300^\\circ.\\]In this case $r=\\frac12$ and $a = 5$, so \\[S=1+\\frac14-\\frac18-\\frac18-\\frac1{32}+\\frac1{64}=\\frac{63}{64},\\] \\[T=0+\\frac{\\sqrt3}4+\\frac{\\sqrt3}8+0-\\frac{\\sqrt3}{32}-\\frac{\\sqrt3}{64}=\\frac{21\\sqrt3}{64},\\]and the coordinates of $P$ are \\[\\left(\\frac{5S}{1-\\frac1{64}}, \\frac{5T}{1-\\frac1{64}}\\right)=\\left(5,\\frac{5\\sqrt3}{3}\\right).\\]Thus the square of the distance from the origin to $P$ is $25+\\frac{25}3=\\frac{100}3$. The requested sum is $100+3=103.",
"Let point $O$ be the origin in the complex plane. Point $P$ is the complex sum $5(1+z+z^2+\\cdots) = \\frac{5}{1-z}$, where $z=\\frac{1+i\\sqrt3}4$. The distance squared is\\[{OP}^2=\\left|\\frac5{1-\\frac{1+i\\sqrt3}4}\\right|^{2}= \\frac{(4\\cdot5)^2}{\\left|4-(1+i\\sqrt3)\\right|^2}=\\frac{400}{9+3}=\\frac{100}3.\\] Hence the answer is $100 + 3 = 103.",
"Let point $O$ be the origin in the complex plane. Point $P$ is the complex sum $5(1+z+z^2+\\cdots) = \\frac{5}{1-z}$, where $z=\\frac{1+i\\sqrt3}4$. The distance squared is\\[{OP}^2=\\left|\\frac5{1-\\frac{1+i\\sqrt3}4}\\right|^{2}= \\frac{(4\\cdot5)^2}{\\left|4-(1+i\\sqrt3)\\right|^2}=\\frac{400}{9+3}=\\frac{100}3.\\] Hence the answer is $100 + 3 = 103.",
"The bug goes forward and backward in three directions: straight east, northeast, and northwest. It travels $5-\\frac{5}{8}+\\frac{5}{64}-\\cdots=\\frac{40}{9}$ units east. Thus, it goes northeast $\\frac{\\frac{40}{9}}{2}=\\frac{20}{9}$ units northeast and $\\frac{10}{9}$ units northwest. Now, the bug travels a total of $\\frac{40}{9}+\\frac{10}{9}-\\frac{5}{9}=\\frac{45}{9}=5$ units east and a total of $\\frac{10\\sqrt{3}}{9}+\\frac{5\\sqrt{3}}{9}=\\frac{15\\sqrt{3}}{9}=\\frac{5\\sqrt{3}}{3}$ units north because of the 30-60-90 right triangles formed. Now, $OP^2=5^2+\\frac{5^2}{3}=\\frac{100}{3}$ by the Pythagorean Theorem, and the answer is $103. -integralarefun",
"The bug goes forward and backward in three directions: straight east, northeast, and northwest. It travels $5-\\frac{5}{8}+\\frac{5}{64}-\\cdots=\\frac{40}{9}$ units east. Thus, it goes northeast $\\frac{\\frac{40}{9}}{2}=\\frac{20}{9}$ units northeast and $\\frac{10}{9}$ units northwest. Now, the bug travels a total of $\\frac{40}{9}+\\frac{10}{9}-\\frac{5}{9}=\\frac{45}{9}=5$ units east and a total of $\\frac{10\\sqrt{3}}{9}+\\frac{5\\sqrt{3}}{9}=\\frac{15\\sqrt{3}}{9}=\\frac{5\\sqrt{3}}{3}$ units north because of the 30-60-90 right triangles formed. Now, $OP^2=5^2+\\frac{5^2}{3}=\\frac{100}{3}$ by the Pythagorean Theorem, and the answer is $103. -integralarefun",
"The bug's bearings on each traversal are $0^\\circ, 60^\\circ, 120^\\circ,$ and so on; in general, the $n-$th traversal has length $5\\cdot (1/2)^{n-1}$ and bearing $60(n-1).$ This means that the $x$ and $y$ displacements for the $n-$th traversal are \\[(\\Delta x_n, \\Delta y_n)=(5\\cdot (1/2)^{n-1}\\cos (60(n-1))^\\circ,5\\cdot (1/2)^{n-1}\\sin (60(n-1))^\\circ).\\] Summing this over all the displacements, we get \\[x_P=\\sum_{n=1}^{\\infty} 5\\cdot (1/2)^{n-1}\\cos (60(n-1))^\\circ, y_P=\\sum_{n=1}^{\\infty} 5\\cdot (1/2)^{n-1}\\sin (60(n-1))^\\circ.\\] We then have \\begin{align} OP^2 &= x_P^2+y_P^2 \\\\ &=\\sum_{n=1}^{\\infty} (5\\cdot (1/2)^{n-1}\\cos ^2(60(n-1))^\\circ) + (5\\cdot (1/2)^{n-1}\\sin ^2(60(n-1))^\\circ) \\\\ &= \\sum_{n=1}^{\\infty} (5\\cdot (1/2)^{n-1})^2(\\cos ^2 (60(n-1))^\\circ+\\sin ^2 (60(n-1))^\\circ) \\\\ &= \\sum_{n=1}^{\\infty} (25\\cdot (1/4)^{n-1}) \\\\ &= \\dfrac{25}{1-1/4} \\\\ &= 100/3. \\end{align} Thus, the answer is $100+3=103 --MenuThreeOne"
] |
2020-I-9
| 2,020
| 9
|
Let $S$ be the set of positive integer divisors of $20^9.$ Three numbers are chosen independently and at random with replacement from the set $S$ and labeled $a_1,a_2,$ and $a_3$ in the order they are chosen. The probability that both $a_1$ divides $a_2$ and $a_2$ divides $a_3$ is $\tfrac{m}{n},$ where $m$ and $n$ are relatively prime positive integers. Find $m.$
| 77
|
I
|
[
"[asy] size(12cm); for (int x = 1; x < 18; ++x) { draw((x, 0) -- (x, 9), dotted); } for (int y = 1; y < 9; ++y) { draw((0, y) -- (18, y), dotted); } draw((0, 0) -- (18, 0) -- (18, 9) -- (0, 9) -- cycle); pair b1, b2, b3; pair c1, c2, c3; pair a1, a2, a3; b1 = (3, 0); b2 = (12, 0); b3 = (16, 0); c1 = (0, 2); c2 = (0, 4); c3 = (0, 8); a1 = b1 + c1; a2 = b2 + c2; a3 = b3 + c3; draw(b1 -- a1 -- c1); draw(b2 -- a2 -- c2); draw(b3 -- a3 -- c3); dot(a1); dot(a2); dot(a3); label(\"$a_1$\", a1, NE); label(\"$a_2$\", a2, NE); label(\"$a_3$\", a3, NE); label(\"$b_1$\", b1, S); label(\"$b_2$\", b2, S); label(\"$b_3$\", b3, S); label(\"$c_1$\", c1, W); label(\"$c_2$\", c2, W); label(\"$c_3$\", c3, W); [/asy] First, prime factorize $20^9$ as $2^{18} \\cdot 5^9$. Denote $a_1$ as $2^{b_1} \\cdot 5^{c_1}$, $a_2$ as $2^{b_2} \\cdot 5^{c_2}$, and $a_3$ as $2^{b_3} \\cdot 5^{c_3}$. In order for $a_1$ to divide $a_2$, and for $a_2$ to divide $a_3$, $b_1\\le b_2\\le b_3$, and $c_1\\le c_2\\le c_3$. We will consider each case separately. Note that the total amount of possibilities is $190^3$, as there are $(18+1)(9+1)=190$ choices for each factor. We notice that if we add $1$ to $b_2$ and $2$ to $b_3$, then we can reach the stronger inequality $0\\le b_1<b_2+1<b_3+2\\le 20$. Therefore, if we pick $3$ integers from $0$ to $20$, they will correspond to a unique solution, forming a 1-1 correspondence between the numbers $b_1$, $b_2+1$, and $b_3+2$. This is also equivalent to applying stars and bars on distributing the powers of 2 and 5 through differences. The amount of solutions to this inequality is $\\dbinom{21}{3}$. The case for $c_1$,$c_2$, and $c_3$ proceeds similarly for a result of $\\dbinom{12}{3}$. Therefore, the probability of choosing three such factors is \\[\\frac{\\dbinom{21}{3} \\cdot \\dbinom{12}{3}}{190^3}.\\] Simplification gives $\\frac{77}{1805}$, and therefore the answer is $077. -molocyxu",
"Same as before, say the factors have powers of $b$ and $c$. $b_1, b_2, b_3$ can either be all distinct, all equal, or two of the three are equal. As well, we must have $b_1 \\leq b_2 \\leq b_3$. If they are all distinct, the number of cases is simply ${19 \\choose 3}$. If they are all equal, there are only $19$ cases for the general value. If we have a pair equal, then we have $2 \\cdot {19\\choose 2}$. We need to multiply by $2$ because if we have two values $b_i < b_j$, we can have either $(b_i, b_i, b_j)$ or $(b_i, b_j, b_j)$. \\[{19 \\choose 3} + 2 \\cdot {19 \\choose 2} + 19 = 1330\\] Likewise for $c$, we get \\[{10 \\choose 3} + 2 \\cdot {10 \\choose 2} + 10 = 220\\] The final probability is simply $\\frac{1330 \\cdot 220}{190^3}$. Simplification gives $\\frac{77}{1805}$, and therefore the answer is $077.",
"Similar to before, we calculate that there are $190^3$ ways to choose $3$ factors with replacement. Then, we figure out the number of triplets ${a,b,c}$ and ${d,f,g}$, where $a$, $b$, and $c$ represent powers of $2$ and $d$, $f$, and $g$ represent powers of $5$, such that the triplets are in non-descending order. The maximum power of $2$ is $18$, and the maximum power of $5$ is $9$. Using the Hockey Stick identity, we figure out that there are $\\dbinom{12}{3}$ ways to choose $d$, $f$ and $g$, and $\\dbinom{21}{3}$ ways to choose $a$, $b$, and $c$. Therefore, the probability of choosing $3$ factors which satisfy the conditions is \\[\\frac{\\dbinom{21}{3} \\cdot \\dbinom{12}{3}}{190^3}.\\] This simplifies to $\\frac{77}{1805}$, therefore $m =$ $077.",
"Note that the prime factorization of $20^9$ is $2^{18}\\cdot5^{9}.$ The problem reduces to selecting independently the powers of $2$ and the powers of $5$ in the numbers $a_1$, $a_2$, and $a_3$. This is equivalent to selecting $3$ exponents for the powers of $2$ and $3$ exponents for the powers of $5$ and determining in each case the probability that the 3 exponents are chosen in nondecreasing order. Given a positive integer $k$, the probability that three integers $a$, $b$, and $c$ are chosen such that $0\\le a \\le b\\le c \\le k$ is the probability that $a$, $b+1$, and $c+2$ are chosen such that $0 \\le a < b+1 < c+2 \\le k+2$. There are $\\binom {k+3}3$ ways to choose $a$, $b+1$, and $c+2$, so the probability that the integers are chosen in order is \\[\\frac{\\binom{k+3}3}{(k+1)^3}.\\]Thus the probability that three chosen divisors of $20^9$ satisfy the divisibility requirement is \\[\\frac{\\binom{12}3}{10^3}\\cdot\\frac{\\binom{21}3}{19^3}=\\frac{12\\cdot11\\cdot10}{6\\cdot10\\cdot10\\cdot10}\\cdot\\frac{21\\cdot20\\cdot19}{6\\cdot19\\cdot19\\cdot19}=\\frac{77}{1805}.\\]The requested numerator is $77.$",
"Note that the prime factorization of $20^9$ is $2^{18}\\cdot5^{9}.$ The problem reduces to selecting independently the powers of $2$ and the powers of $5$ in the numbers $a_1$, $a_2$, and $a_3$. This is equivalent to selecting $3$ exponents for the powers of $2$ and $3$ exponents for the powers of $5$ and determining in each case the probability that the 3 exponents are chosen in nondecreasing order. Given a positive integer $k$, the probability that three integers $a$, $b$, and $c$ are chosen such that $0\\le a \\le b\\le c \\le k$ is the probability that $a$, $b+1$, and $c+2$ are chosen such that $0 \\le a < b+1 < c+2 \\le k+2$. There are $\\binom {k+3}3$ ways to choose $a$, $b+1$, and $c+2$, so the probability that the integers are chosen in order is \\[\\frac{\\binom{k+3}3}{(k+1)^3}.\\]Thus the probability that three chosen divisors of $20^9$ satisfy the divisibility requirement is \\[\\frac{\\binom{12}3}{10^3}\\cdot\\frac{\\binom{21}3}{19^3}=\\frac{12\\cdot11\\cdot10}{6\\cdot10\\cdot10\\cdot10}\\cdot\\frac{21\\cdot20\\cdot19}{6\\cdot19\\cdot19\\cdot19}=\\frac{77}{1805}.\\]The requested numerator is $77.$"
] |
2020-I-10
| 2,020
| 10
|
Let $m$ and $n$ be positive integers satisfying the conditions $\quad\bullet\ \gcd(m+n,210)=1,$ $\quad\bullet\ m^m$ is a multiple of $n^n,$ and $\quad\bullet\ m$ is not a multiple of $n.$ Find the least possible value of $m+n.$
| 407
|
I
|
[
"Taking inspiration from $4^4 \\mid 10^{10}$ we are inspired to take $n$ to be $p^2$, the lowest prime not dividing $210$, or $11 \\implies n = 121$. Now, there are $242$ factors of $11$, so $11^{242} \\mid m^m$, and then $m = 11k$ for $k \\geq 22$. Now, $\\gcd(m+n, 210) = \\gcd(11+k,210) = 1$. Noting $k = 26$ is the minimal that satisfies this, we get $(n,m) = (121,286)$. Thus, it is easy to verify this is minimal and we get $407. ~awang11",
"Assume for the sake of contradiction that $n$ is a multiple of a single digit prime number, then $m$ must also be a multiple of that single digit prime number to accommodate for $n^n | m^m$. However that means that $m+n$ is divisible by that single digit prime number, which violates $\\gcd(m+n,210) = 1$, so contradiction. $n$ is also not 1 because then $m$ would be a multiple of it. Thus, $n$ is a multiple of 11 and/or 13 and/or 17 and/or... Assume for the sake of contradiction that $n$ has at most 1 power of 11, at most 1 power of 13...and so on... Then, for $n^n | m^m$ to be satisfied, $m$ must contain at least the same prime factors that $n$ has. This tells us that for the primes where $n$ has one power of, $m$ also has at least one power, and since this holds true for all the primes of $n$, $n|m$. Contradiction. Thus $n$ needs more than one power of some prime. The obvious smallest possible value of $n$ now is $11^2 =121$. Since $121^{121}=11^{242}$, we need $m$ to be a multiple of 11 at least $242$ that is not divisible by $121$ and most importantly, $\\gcd(m+n,210) = 1$. $242$ is divisible by $121$, out. $253+121$ is divisible by 2, out. $264+121$ is divisible by 5, out. $275+121$ is divisible by 2, out. $286+121=37\\cdot 11$ and satisfies all the conditions in the given problem, and the next case $n=169$ will give us at least $169\\cdot 3$, so we get $407.",
"Let $m$ and $n$ be positive integers where $m^m$ is a multiple of $n^n$ and $m$ is not a multiple of $n$. If a prime $p$ divides $n$, then $p$ divides $n^n$, so it also divides $m^m$, and thus $p$ divides $m$. Therefore any prime $p$ dividing $n$ also divides both $m$ and $k = m + n$. Because $k$ is relatively prime to $210=2\\cdot3\\cdot5\\cdot7$, the primes $2$, $3$, $5$, and $7$ cannot divide $n$. Furthermore, because $m$ is divisible by every prime factor of $n$, but $m$ is not a multiple of $n$, the integer $n$ must be divisible by the square of some prime, and that prime must be at least $11$. Thus $n$ must be at least $11^2 = 121$. If $n=11^2$, then $m$ must be a multiple of $11$ but not a multiple of $121$, and $m^m$ must be divisible by $n^n = 121^{121} = 11^{242}$. Therefore $m$ must be a multiple of $11$ that is greater than $242$. Let $m = 11m_0$, with $m_0 > 22$. Then $k = m + n = 11(m_0 + 11)$. The least $m_0 > 22$ for which $m_0 + 11$ is not divisible by any of the primes $2$, $3$, $5$, or $7$ is $m_0 = 26$, giving the prime $m_0 + 11 = 37$. Hence the least possible $k$ when $n = 121$ is $k = 11 \\cdot 37 = 407$. It remains to consider other possible values for $n$. If $n = 13^2 = 169$, then $m$ must be divisible by $13$ but not $169$, and $m^m$ must be a multiple of $n^n = 169^{169} = 13^{338}$, so $m > 338$. Then $k = m + n > 169 + 338 = 507$. All other possible values for $n$ have $n \\ge 242$, and in this case $m > n \\ge 242$, so $k \\ge 2 \\cdot 242 = 484$. Hence no greater values of $n$ can produce lesser values for $k$, and the least possible $k$ is indeed $407$.",
"Let $m$ and $n$ be positive integers where $m^m$ is a multiple of $n^n$ and $m$ is not a multiple of $n$. If a prime $p$ divides $n$, then $p$ divides $n^n$, so it also divides $m^m$, and thus $p$ divides $m$. Therefore any prime $p$ dividing $n$ also divides both $m$ and $k = m + n$. Because $k$ is relatively prime to $210=2\\cdot3\\cdot5\\cdot7$, the primes $2$, $3$, $5$, and $7$ cannot divide $n$. Furthermore, because $m$ is divisible by every prime factor of $n$, but $m$ is not a multiple of $n$, the integer $n$ must be divisible by the square of some prime, and that prime must be at least $11$. Thus $n$ must be at least $11^2 = 121$. If $n=11^2$, then $m$ must be a multiple of $11$ but not a multiple of $121$, and $m^m$ must be divisible by $n^n = 121^{121} = 11^{242}$. Therefore $m$ must be a multiple of $11$ that is greater than $242$. Let $m = 11m_0$, with $m_0 > 22$. Then $k = m + n = 11(m_0 + 11)$. The least $m_0 > 22$ for which $m_0 + 11$ is not divisible by any of the primes $2$, $3$, $5$, or $7$ is $m_0 = 26$, giving the prime $m_0 + 11 = 37$. Hence the least possible $k$ when $n = 121$ is $k = 11 \\cdot 37 = 407$. It remains to consider other possible values for $n$. If $n = 13^2 = 169$, then $m$ must be divisible by $13$ but not $169$, and $m^m$ must be a multiple of $n^n = 169^{169} = 13^{338}$, so $m > 338$. Then $k = m + n > 169 + 338 = 507$. All other possible values for $n$ have $n \\ge 242$, and in this case $m > n \\ge 242$, so $k \\ge 2 \\cdot 242 = 484$. Hence no greater values of $n$ can produce lesser values for $k$, and the least possible $k$ is indeed $407$."
] |
2020-I-11
| 2,020
| 11
|
For integers $a,b,c$ and $d,$ let $f(x)=x^2+ax+b$ and $g(x)=x^2+cx+d.$ Find the number of ordered triples $(a,b,c)$ of integers with absolute values not exceeding $10$ for which there is an integer $d$ such that $g(f(2))=g(f(4))=0.$
| 510
|
I
|
[
"There can be two different cases for this problem, either $f(2)=f(4)$ or not. If it is, note that by Vieta's formulas $a = -6$. Then, $b$ can be anything. However, $c$ can also be anything, as we can set the root of $g$ (not equal to $f(2) = f(4)$) to any integer, producing a possible integer value of $d$. Therefore there are $21^2 = 441$ in this case*. If it isn't, then $f(2),f(4)$ are the roots of $g$. This means by Vieta's, that: \\[f(2)+f(4) = -c \\in [-10,10]\\] \\[20 + 6a + 2b \\in [-10,10]\\] \\[3a + b \\in [-15,-5].\\] Solving these inequalities while considering that $a \\neq -6$ to prevent $f(2) = f(4)$, we obtain $69$ possible tuples and adding gives $441+69=510. ~awang11",
"There can be two different cases for this problem, either $f(2)=f(4)$ or not. If it is, note that by Vieta's formulas $a = -6$. Then, $b$ can be anything. However, $c$ can also be anything, as we can set the root of $g$ (not equal to $f(2) = f(4)$) to any integer, producing a possible integer value of $d$. Therefore there are $21^2 = 441$ in this case*. If it isn't, then $f(2),f(4)$ are the roots of $g$. This means by Vieta's, that: \\[f(2)+f(4) = -c \\in [-10,10]\\] \\[20 + 6a + 2b \\in [-10,10]\\] \\[3a + b \\in [-15,-5].\\] Solving these inequalities while considering that $a \\neq -6$ to prevent $f(2) = f(4)$, we obtain $69$ possible tuples and adding gives $441+69=510. ~awang11",
"Define $h(x)=x^2+cx$. Since $g(f(2))=g(f(4))=0$, we know $h(f(2))=h(f(4))=-d$. Plugging in $f(x)$ into $h(x)$, we get $h(f(x))=x^4+2ax^3+(2b+a^2+c)x^2+(2ab+ac)x+(b^2+bc)$. Setting $h(f(2))=h(f(4))$, \\[16+16a+8b+4a^2+4ab+b^2+4c+2ac+bc=256+128a+32b+16a^2+8ab+b^2+16c+4ac+bc\\]. Simplifying and cancelling terms, \\[240+112a+24b+12a^2+4ab+12c+2ac=0\\] \\[120+56a+12b+6a^2+2ab+6c+ac=0\\] \\[6a^2+2ab+ac+56a+12b+6c+120=0\\] \\[6a^2+2ab+ac+20a+36a+12b+6c+120=0\\] \\[a(6a+2b+c+20)+6(6a+2b+c+20)=0\\] \\[(a+6)(6a+2b+c+20)=0\\] Therefore, either $a+6=0$ or $6a+2b+c=-20$. The first case is easy: $a=-6$ and there are $441$ tuples in that case. In the second case, we simply perform casework on even values of $c$, to get $77$ tuples, subtracting the $8$ tuples in both cases we get $441+77-8=510. -EZmath2006",
"Define $h(x)=x^2+cx$. Since $g(f(2))=g(f(4))=0$, we know $h(f(2))=h(f(4))=-d$. Plugging in $f(x)$ into $h(x)$, we get $h(f(x))=x^4+2ax^3+(2b+a^2+c)x^2+(2ab+ac)x+(b^2+bc)$. Setting $h(f(2))=h(f(4))$, \\[16+16a+8b+4a^2+4ab+b^2+4c+2ac+bc=256+128a+32b+16a^2+8ab+b^2+16c+4ac+bc\\]. Simplifying and cancelling terms, \\[240+112a+24b+12a^2+4ab+12c+2ac=0\\] \\[120+56a+12b+6a^2+2ab+6c+ac=0\\] \\[6a^2+2ab+ac+56a+12b+6c+120=0\\] \\[6a^2+2ab+ac+20a+36a+12b+6c+120=0\\] \\[a(6a+2b+c+20)+6(6a+2b+c+20)=0\\] \\[(a+6)(6a+2b+c+20)=0\\] Therefore, either $a+6=0$ or $6a+2b+c=-20$. The first case is easy: $a=-6$ and there are $441$ tuples in that case. In the second case, we simply perform casework on even values of $c$, to get $77$ tuples, subtracting the $8$ tuples in both cases we get $441+77-8=510. -EZmath2006",
"For a given ordered triple $(a, b, c)$, the value of $g(f(4)) - g(f(2))$ is uniquely determined, and a value of $d$ can be found to give $g(f(2))$ any prescribed integer value. Hence the required condition can be satisfied provided that $a$, $b$, and $c$ are chosen so that \\begin{align*} 0 &= g(f(4)) - g(f(2)) = \\big(f(4)\\big)^{\\!2} - \\big(f(2)\\big)^{\\!2} + c\\big(f(4) - f(2)\\big)\\\\ &= \\big(f(4) - f(2)\\big)\\big(f(4) + f(2) + c\\big)\\\\ &= (12 + 2a)(20 + 6a + 2b + c). \\end{align*} First suppose that $12 + 2a = 0$, so $a = -6$. In this case there are $21$ choices for each of $b$ and $c$ with $-10 \\le b \\le 10$ and $-10 \\le c \\le 10$, so this case accounts for $21^2 = 441$ ordered triples. Next suppose that $a \\ne -6$ and $20 + 6a + 2b + c = 0$, so $c = -6a - 2b - 20$. Because $-10 \\le c \\le 10$, it follows that $-30 \\le 6a + 2b \\le -10$, and because $-10\\le b \\le10$, it follows that $-8 \\le a \\le 1$. Then $-15 - 3a \\le b \\le -5 - 3a$. The number of ordered triples for various values of $a$ are presented in the following table. \\[\\begin{tabular}{|c|c|c|r|} \\hline a & b & c & triples \\\\ \\hline -8 & \\{9,10\\} & -6a - 2b - 20 & 2 \\\\ -7 & \\{6, 7, 8, 9, 10\\} & -6a - 2b - 20 & 5 \\\\ -6 & \\{-10, \\ldots, 10\\} & \\{-10, \\ldots, 10\\} & 441 \\\\ -5 & \\{0, \\ldots, 10\\} & -6a - 2b - 20 & 11 \\\\ -4 & \\{-3, \\ldots, 7\\} & -6a - 2b - 20 & 11 \\\\ -3 & \\{-6, \\ldots, 4\\} & -6a - 2b - 20 & 11 \\\\ -2 & \\{-9, \\ldots, 1\\} & -6a - 2b - 20 & 11 \\\\ -1 & \\{-10, \\ldots, -2\\} & -6a - 2b - 20 & 9 \\\\ 0 & \\{-10, \\ldots, -5\\} & -6a - 2b - 20 & 6 \\\\ 1 & \\{-10, -9, -8\\} & -6a - 2b - 20 & 3 \\\\ \\hline Total & & & 510\\\\ \\hline \\end{tabular}\\] The total number of ordered triples that satisfy the required condition is $510.",
"For a given ordered triple $(a, b, c)$, the value of $g(f(4)) - g(f(2))$ is uniquely determined, and a value of $d$ can be found to give $g(f(2))$ any prescribed integer value. Hence the required condition can be satisfied provided that $a$, $b$, and $c$ are chosen so that \\begin{align*} 0 &= g(f(4)) - g(f(2)) = \\big(f(4)\\big)^{\\!2} - \\big(f(2)\\big)^{\\!2} + c\\big(f(4) - f(2)\\big)\\\\ &= \\big(f(4) - f(2)\\big)\\big(f(4) + f(2) + c\\big)\\\\ &= (12 + 2a)(20 + 6a + 2b + c). \\end{align*} First suppose that $12 + 2a = 0$, so $a = -6$. In this case there are $21$ choices for each of $b$ and $c$ with $-10 \\le b \\le 10$ and $-10 \\le c \\le 10$, so this case accounts for $21^2 = 441$ ordered triples. Next suppose that $a \\ne -6$ and $20 + 6a + 2b + c = 0$, so $c = -6a - 2b - 20$. Because $-10 \\le c \\le 10$, it follows that $-30 \\le 6a + 2b \\le -10$, and because $-10\\le b \\le10$, it follows that $-8 \\le a \\le 1$. Then $-15 - 3a \\le b \\le -5 - 3a$. The number of ordered triples for various values of $a$ are presented in the following table. \\[\\begin{tabular}{|c|c|c|r|} \\hline a & b & c & triples \\\\ \\hline -8 & \\{9,10\\} & -6a - 2b - 20 & 2 \\\\ -7 & \\{6, 7, 8, 9, 10\\} & -6a - 2b - 20 & 5 \\\\ -6 & \\{-10, \\ldots, 10\\} & \\{-10, \\ldots, 10\\} & 441 \\\\ -5 & \\{0, \\ldots, 10\\} & -6a - 2b - 20 & 11 \\\\ -4 & \\{-3, \\ldots, 7\\} & -6a - 2b - 20 & 11 \\\\ -3 & \\{-6, \\ldots, 4\\} & -6a - 2b - 20 & 11 \\\\ -2 & \\{-9, \\ldots, 1\\} & -6a - 2b - 20 & 11 \\\\ -1 & \\{-10, \\ldots, -2\\} & -6a - 2b - 20 & 9 \\\\ 0 & \\{-10, \\ldots, -5\\} & -6a - 2b - 20 & 6 \\\\ 1 & \\{-10, -9, -8\\} & -6a - 2b - 20 & 3 \\\\ \\hline Total & & & 510\\\\ \\hline \\end{tabular}\\] The total number of ordered triples that satisfy the required condition is $510."
] |
2020-I-12
| 2,020
| 12
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Let $n$ be the least positive integer for which $149^n-2^n$ is divisible by $3^3\cdot5^5\cdot7^7.$ Find the number of positive integer divisors of $n.$
| 270
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I
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[
"As usual, denote $v_p(n)$ the highest power of prime $p$ that divides $n$. Lifting the Exponent shows that \\[3=v_3(149^n-2^n) = v_3(n) + v_3(147) = v_3(n)+1\\] so thus, $3^2$ divides $n$. It also shows that \\[7=v_7(149^n-2^n) = v_7(n) + v_7(147) = v_7(n)+2\\] so thus, $7^5$ divides $n$. Now, setting $n = 4c$ (necessitated by $149^n \\equiv 2^n \\pmod 5$ in order to set up LTE), we see \\[v_5(149^{4c}-2^{4c}) = v_5(149^{4c}-16^{c})\\] and since $149^{4} \\equiv 1 \\pmod{25}$ and $16^1 \\equiv 16 \\pmod{25}$ then $v_5(149^{4c}-2^{4c})=v_5(149^4-16)+v_5(c)=1+v_5(c)$ meaning that we have that by LTE, $5^4 | c$ and $4 \\cdot 5^4$ divides $n$. Since $3^2$, $7^5$ and $4\\cdot 5^4$ all divide $n$, the smallest value of $n$ working is their LCM, also $3^2 \\cdot 7^5 \\cdot 4 \\cdot 5^4 = 2^2 \\cdot 3^2 \\cdot 5^4 \\cdot 7^5$. Thus the number of divisors is $(2+1)(2+1)(4+1)(5+1) = 270. ~kevinmathz clarified by another user notation note from another user Note We were able to use LTE with 3 and 7 but not 5 because in order to use LTE, we need \\( p \\mid x-y \\). Obviously, \\( 149^n \\equiv 2^n \\pmod{3} \\) implies \\( 149^n - 2^n \\equiv 0 \\pmod{3} \\), so LTE works here. Furthermore, \\( 149^n \\equiv 2^n \\pmod{7} \\) implies \\( 149^n - 2^n \\equiv 0 \\pmod{7} \\), so LTE works here. However, when we get to the case of 5, we see that \\( 149^n \\equiv 2^n \\pmod{5} \\) doesn't always hold; specifically, this is only valid when \\( n \\) is a multiple of \\( 4 \\), which is why we let \\( n = 4c \\) in the solution. mathboy282",
"Note that for all $n$, $149^n - 2^n$ is divisible by $149-2 = 147$ by difference of $n$th powers. That is $3\\cdot7^2$, so now we can clearly see that the smallest $n$ to make the expression divisible by $3^3$ is just $3^2$. Similarly, we can reason that the smallest $n$ to make the expression divisible by $7^7$ is just $7^5$. Finally, for $5^5$, take $\\pmod {5}$ and $\\pmod {25}$ of each quantity (They happen to both be $-1$ and $2$ respectively, so you only need to compute once). One knows from Fermat's theorem that the maximum possible minimum $n$ for divisibility by $5$ is $4$, and other values are factors of $4$. Testing all of them(just $1$,$2$,$4$ using mods-not too bad), $4$ is indeed the smallest value to make the expression divisible by $5$, and this clearly is NOT divisible by $25$. Therefore, the smallest $n$ to make this expression divisible by $5^5$ is $2^2 \\cdot 5^4$. Calculating the LCM of all these, one gets $2^2 \\cdot 3^2 \\cdot 5^4 \\cdot 7^5$. Using the factor counting formula, the answer is $3\\cdot3\\cdot5\\cdot6$ = $270. ~Solution by thanosaops ~formatted by MY-2 and pandyhu2001",
"Note that for all $n$, $149^n - 2^n$ is divisible by $149-2 = 147$ by difference of $n$th powers. That is $3\\cdot7^2$, so now we can clearly see that the smallest $n$ to make the expression divisible by $3^3$ is just $3^2$. Similarly, we can reason that the smallest $n$ to make the expression divisible by $7^7$ is just $7^5$. Finally, for $5^5$, take $\\pmod {5}$ and $\\pmod {25}$ of each quantity (They happen to both be $-1$ and $2$ respectively, so you only need to compute once). One knows from Fermat's theorem that the maximum possible minimum $n$ for divisibility by $5$ is $4$, and other values are factors of $4$. Testing all of them(just $1$,$2$,$4$ using mods-not too bad), $4$ is indeed the smallest value to make the expression divisible by $5$, and this clearly is NOT divisible by $25$. Therefore, the smallest $n$ to make this expression divisible by $5^5$ is $2^2 \\cdot 5^4$. Calculating the LCM of all these, one gets $2^2 \\cdot 3^2 \\cdot 5^4 \\cdot 7^5$. Using the factor counting formula, the answer is $3\\cdot3\\cdot5\\cdot6$ = $270. ~Solution by thanosaops ~formatted by MY-2 and pandyhu2001",
"As usual, denote $v_p(n)$ the highest power of prime $p$ that divides $n$. For divisibility by $3^3$, notice that $v_3(149^3 - 2^3) = 2$ as $149^3 - 2^3 =$ $(147)(149^2 + 2\\cdot149 + 2^2)$, and upon checking mods, $149^2 + 2\\cdot149 + 2^2$ is divisible by $3$ but not $9$. In addition, $149^9 - 2^9$ is divisible by $3^3$ because $149^9 - 2^9 = (149^3 - 2^3)(149^6 + 149^3\\cdot2^3 + 2^6)$, and the rightmost factor equates to $1 + 1 + 1 \\pmod{3} \\equiv 0 \\pmod{3}$. In fact, $n = 9 = 3^2$ is the least possible choice to ensure divisibility by $3^3$ because if $n = a \\cdot 3^b$, with $3 \\nmid a$ and $b < 2$, we write \\[149^{a \\cdot 3^b} - 2^{a \\cdot 3^b} = (149^{3^b} - 2^{3^b})(149^{3^b(a - 1)} + 149^{3^b(a - 2)}\\cdot2^{3^b}+\\cdots2^{3^b(a - 1)}).\\] Then, the rightmost factor is equivalent to $\\pm a \\pmod{3} \\not\\equiv 0 \\pmod{3}$, and $v_3(149^{3^b} - 2^{3^b}) = b + 1 < 3$. For divisibility by $7^7$, we'll induct, claiming that $v_7(149^{7^k} - 2^{7^k}) = k + 2$ for whole numbers $k$. The base case is clear. Then, \\[v_7(149^{7^{k+1}} - 2^{7^{k+1}}) = v_7(149^{7^k} - 2^{7^k}) + v_7(149^{6\\cdot7^k} + 2^{7^k}\\cdot149^{5\\cdot7^k} + \\cdots + 2^{5\\cdot7^k}\\cdot149^{7^k} + 2^{6\\cdot7^k}).\\] By the induction hypothesis, $v_7(149^{7^k} - 2^{7^k}) = k + 2$. Then, notice that \\[S(k) = 149^{6\\cdot7^k} + 2^{7^k}\\cdot149^{5\\cdot7^k} + \\cdots + 2^{5\\cdot7^k}\\cdot149^{7^k} + 2^{6\\cdot7^k} \\equiv 7 \\cdot 2^{6\\cdot7^k}\\pmod{7} \\equiv 7 \\cdot 2^{6\\cdot7^k}\\pmod{49}.\\] This tells us that $S(k)$ is divisible by $7$, but not $49$ so that $v_7\\left(S(k)\\right) = 1$, completing our induction. We can verify that $7^5$ is the least choice of $n$ to ensure divisibility by $7^7$ by arguing similarly to the $3^3$ case. Finally, for $5^5$, we take the powers of $149$ and $2$ in mod $5$ and mod $25$. Writing out these mods, we have that $149^n \\equiv 2^n \\pmod{5}$ if and only if $4 | n$, in which $149^n \\equiv 2^n \\equiv 1 \\pmod{5}$. So here we claim that $v_5(149^{4\\cdot5^k} - 2^{4\\cdot5^k}) = k + 1$ and perform yet another induction. The base case is true: $5 | 149^4 - 2^4$, but $149^4 - 2^4 \\equiv 1 - 16 \\pmod{25}$. Now then, assuming the induction statement to hold for some $k$, \\[v_5(149^{4\\cdot5^{k+1}} - 2^{4\\cdot5^{k+1}}) = (k+1) + v_5(149^{4\\cdot4\\cdot5^k}+2^{4\\cdot5^k}\\cdot149^{3\\cdot4\\cdot5^k}+\\cdots+2^{3\\cdot4\\cdot5^k}\\cdot149^{4\\cdot5^k}+2^{4\\cdot4\\cdot5^k}).\\] Note that $S'(k) = 149^{4\\cdot4\\cdot5^k}+2^{4\\cdot5^k}\\cdot149^{3\\cdot4\\cdot5^k}+\\cdots+2^{3\\cdot4\\cdot5^k}\\cdot149^{4\\cdot5^k}+2^{4\\cdot4\\cdot5^k}$ equates to $S''(k) = 1 + 2^{4\\cdot5^k} + \\cdots + 2^{16\\cdot5^k}$ in both mod $5$ and mod $25$. We notice that $S''(k) \\equiv 0 \\pmod{5}$. Writing out the powers of $2$ mod $25$, we have $S''(0) \\equiv 5 \\pmod{25}$. Also $2^n \\equiv 1 \\pmod{25}$ when $n$ is a multiple of $20$. Hence for $k > 0$, $S''(k) \\equiv 5 \\mod{25}$. Thus, $v_5\\left(S'(k)\\right) = 1$, completing our induction. Applying the same argument from the previous two cases, $4\\cdot5^4$ is the least choice to ensure divisibility by $5^5$. Our answer is the number of divisors of $\\text{lcm}(3^2, 7^5, 2^2\\cdot5^4) = 2^2 \\cdot 3^2 \\cdot 5^4 \\cdot 7^5$. It is $(2 + 1)(2 + 1)(4 + 1)(5 + 1) = 270. ~hnkevin42",
"As usual, denote $v_p(n)$ the highest power of prime $p$ that divides $n$. For divisibility by $3^3$, notice that $v_3(149^3 - 2^3) = 2$ as $149^3 - 2^3 =$ $(147)(149^2 + 2\\cdot149 + 2^2)$, and upon checking mods, $149^2 + 2\\cdot149 + 2^2$ is divisible by $3$ but not $9$. In addition, $149^9 - 2^9$ is divisible by $3^3$ because $149^9 - 2^9 = (149^3 - 2^3)(149^6 + 149^3\\cdot2^3 + 2^6)$, and the rightmost factor equates to $1 + 1 + 1 \\pmod{3} \\equiv 0 \\pmod{3}$. In fact, $n = 9 = 3^2$ is the least possible choice to ensure divisibility by $3^3$ because if $n = a \\cdot 3^b$, with $3 \\nmid a$ and $b < 2$, we write \\[149^{a \\cdot 3^b} - 2^{a \\cdot 3^b} = (149^{3^b} - 2^{3^b})(149^{3^b(a - 1)} + 149^{3^b(a - 2)}\\cdot2^{3^b}+\\cdots2^{3^b(a - 1)}).\\] Then, the rightmost factor is equivalent to $\\pm a \\pmod{3} \\not\\equiv 0 \\pmod{3}$, and $v_3(149^{3^b} - 2^{3^b}) = b + 1 < 3$. For divisibility by $7^7$, we'll induct, claiming that $v_7(149^{7^k} - 2^{7^k}) = k + 2$ for whole numbers $k$. The base case is clear. Then, \\[v_7(149^{7^{k+1}} - 2^{7^{k+1}}) = v_7(149^{7^k} - 2^{7^k}) + v_7(149^{6\\cdot7^k} + 2^{7^k}\\cdot149^{5\\cdot7^k} + \\cdots + 2^{5\\cdot7^k}\\cdot149^{7^k} + 2^{6\\cdot7^k}).\\] By the induction hypothesis, $v_7(149^{7^k} - 2^{7^k}) = k + 2$. Then, notice that \\[S(k) = 149^{6\\cdot7^k} + 2^{7^k}\\cdot149^{5\\cdot7^k} + \\cdots + 2^{5\\cdot7^k}\\cdot149^{7^k} + 2^{6\\cdot7^k} \\equiv 7 \\cdot 2^{6\\cdot7^k}\\pmod{7} \\equiv 7 \\cdot 2^{6\\cdot7^k}\\pmod{49}.\\] This tells us that $S(k)$ is divisible by $7$, but not $49$ so that $v_7\\left(S(k)\\right) = 1$, completing our induction. We can verify that $7^5$ is the least choice of $n$ to ensure divisibility by $7^7$ by arguing similarly to the $3^3$ case. Finally, for $5^5$, we take the powers of $149$ and $2$ in mod $5$ and mod $25$. Writing out these mods, we have that $149^n \\equiv 2^n \\pmod{5}$ if and only if $4 | n$, in which $149^n \\equiv 2^n \\equiv 1 \\pmod{5}$. So here we claim that $v_5(149^{4\\cdot5^k} - 2^{4\\cdot5^k}) = k + 1$ and perform yet another induction. The base case is true: $5 | 149^4 - 2^4$, but $149^4 - 2^4 \\equiv 1 - 16 \\pmod{25}$. Now then, assuming the induction statement to hold for some $k$, \\[v_5(149^{4\\cdot5^{k+1}} - 2^{4\\cdot5^{k+1}}) = (k+1) + v_5(149^{4\\cdot4\\cdot5^k}+2^{4\\cdot5^k}\\cdot149^{3\\cdot4\\cdot5^k}+\\cdots+2^{3\\cdot4\\cdot5^k}\\cdot149^{4\\cdot5^k}+2^{4\\cdot4\\cdot5^k}).\\] Note that $S'(k) = 149^{4\\cdot4\\cdot5^k}+2^{4\\cdot5^k}\\cdot149^{3\\cdot4\\cdot5^k}+\\cdots+2^{3\\cdot4\\cdot5^k}\\cdot149^{4\\cdot5^k}+2^{4\\cdot4\\cdot5^k}$ equates to $S''(k) = 1 + 2^{4\\cdot5^k} + \\cdots + 2^{16\\cdot5^k}$ in both mod $5$ and mod $25$. We notice that $S''(k) \\equiv 0 \\pmod{5}$. Writing out the powers of $2$ mod $25$, we have $S''(0) \\equiv 5 \\pmod{25}$. Also $2^n \\equiv 1 \\pmod{25}$ when $n$ is a multiple of $20$. Hence for $k > 0$, $S''(k) \\equiv 5 \\mod{25}$. Thus, $v_5\\left(S'(k)\\right) = 1$, completing our induction. Applying the same argument from the previous two cases, $4\\cdot5^4$ is the least choice to ensure divisibility by $5^5$. Our answer is the number of divisors of $\\text{lcm}(3^2, 7^5, 2^2\\cdot5^4) = 2^2 \\cdot 3^2 \\cdot 5^4 \\cdot 7^5$. It is $(2 + 1)(2 + 1)(4 + 1)(5 + 1) = 270. ~hnkevin42",
"Analyze each prime power separately. Start with the case of $3^3$. By the Binomial Theorem, \\begin{align*} 149^n - 2^n &= (147+2)^n - 2^n \\\\ &= \\binom n1 \\cdot 147 \\cdot 2^{n-1} + \\binom n2 \\cdot 147^2 \\cdot 2^{n-2}\\\\ &\\qquad+ \\binom n3 \\cdot 147^3 \\cdot 2^{n-3} + \\cdots. \\end{align*}Because $147$ is divisible by $3$, all terms after the first two are divisible by $3^3$, and the exponent of $3$ in the first term is less than that in the second term. Hence it is necessary and sufficient that $3^3 \\mid 147n$, that is, $3^2 \\mid n$. For the $7^7$ case, consider the same expansion as in the previous case. Because $147$ is divisible by $49 = 7^2$, all terms after the first three are divisible by $7^7$, and the exponent of $7$ in the first term is less than that in the second and third term. Hence it is necessary and sufficient that $7^7 \\mid 147n$, that is, $7^5 \\mid n$. For the $5^5$ case, working modulo $5$ gives $149^n - 2^n \\equiv 4^n - 2^n = 2^n(2^n-1) \\pmod 5$, so it must be that $4 \\mid n$. Let $n = 4m$, and let $c = 149^4 - 2^4 = (149^2-2^2)(149^2+2^2) = 147 \\cdot 151 \\cdot 22205$. Note that $\\frac c5$ is an integer not divisible by $5$. Expand by the Binomial Theorem again to get \\begin{align*} (149^4)^m - (2^4)^m &= (c+16)^m - (16)^m \\\\ &= \\binom m1 \\cdot c \\cdot 16^{m-1} + \\binom m2 \\cdot c^2 \\cdot 16^{m-2} \\\\ &\\qquad+ \\binom m3 \\cdot c^3 \\cdot 16^{m-3} + \\binom m4 \\cdot c^4 \\cdot 16^{m-4} + \\cdots. \\end{align*}All terms after the first four are divisible by $5^5$, and the exponent of $5$ in the first term is less than that in the second, third, or fourth term. Hence it is necessary and sufficient that $5^5 \\mid cm$. Thus $5^4 \\mid m$, and it follows that $4 \\cdot 5^4 \\mid n$. Therefore the least $n$ is $3^2 \\cdot (2^2 \\cdot 5^4) \\cdot 7^5$. The requested number of divisors is $(1+2)(1+2)(1+4)(1+5) = 270$. The results of the above cases can be generalized using the following lemma. Lifting the Exponent Lemma: Let $p$ be an odd prime, and let $a$ and $b$ be integers relatively prime to $p$ such that $p \\mid (a-b)$. Let $n$ be a positive integer. Then the number of factors of $p$ that divide $a^n - b^n$ is equal to the number of factors of $p$ that divide $a-b$ plus the number of factors of $p$ that divide $n$.",
"Analyze each prime power separately. Start with the case of $3^3$. By the Binomial Theorem, \\begin{align*} 149^n - 2^n &= (147+2)^n - 2^n \\\\ &= \\binom n1 \\cdot 147 \\cdot 2^{n-1} + \\binom n2 \\cdot 147^2 \\cdot 2^{n-2}\\\\ &\\qquad+ \\binom n3 \\cdot 147^3 \\cdot 2^{n-3} + \\cdots. \\end{align*}Because $147$ is divisible by $3$, all terms after the first two are divisible by $3^3$, and the exponent of $3$ in the first term is less than that in the second term. Hence it is necessary and sufficient that $3^3 \\mid 147n$, that is, $3^2 \\mid n$. For the $7^7$ case, consider the same expansion as in the previous case. Because $147$ is divisible by $49 = 7^2$, all terms after the first three are divisible by $7^7$, and the exponent of $7$ in the first term is less than that in the second and third term. Hence it is necessary and sufficient that $7^7 \\mid 147n$, that is, $7^5 \\mid n$. For the $5^5$ case, working modulo $5$ gives $149^n - 2^n \\equiv 4^n - 2^n = 2^n(2^n-1) \\pmod 5$, so it must be that $4 \\mid n$. Let $n = 4m$, and let $c = 149^4 - 2^4 = (149^2-2^2)(149^2+2^2) = 147 \\cdot 151 \\cdot 22205$. Note that $\\frac c5$ is an integer not divisible by $5$. Expand by the Binomial Theorem again to get \\begin{align*} (149^4)^m - (2^4)^m &= (c+16)^m - (16)^m \\\\ &= \\binom m1 \\cdot c \\cdot 16^{m-1} + \\binom m2 \\cdot c^2 \\cdot 16^{m-2} \\\\ &\\qquad+ \\binom m3 \\cdot c^3 \\cdot 16^{m-3} + \\binom m4 \\cdot c^4 \\cdot 16^{m-4} + \\cdots. \\end{align*}All terms after the first four are divisible by $5^5$, and the exponent of $5$ in the first term is less than that in the second, third, or fourth term. Hence it is necessary and sufficient that $5^5 \\mid cm$. Thus $5^4 \\mid m$, and it follows that $4 \\cdot 5^4 \\mid n$. Therefore the least $n$ is $3^2 \\cdot (2^2 \\cdot 5^4) \\cdot 7^5$. The requested number of divisors is $(1+2)(1+2)(1+4)(1+5) = 270$. The results of the above cases can be generalized using the following lemma. Lifting the Exponent Lemma: Let $p$ be an odd prime, and let $a$ and $b$ be integers relatively prime to $p$ such that $p \\mid (a-b)$. Let $n$ be a positive integer. Then the number of factors of $p$ that divide $a^n - b^n$ is equal to the number of factors of $p$ that divide $a-b$ plus the number of factors of $p$ that divide $n$."
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2020-I-13
| 2,020
| 13
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Point $D$ lies on side $\overline{BC}$ of $\triangle ABC$ so that $\overline{AD}$ bisects $\angle BAC.$ The perpendicular bisector of $\overline{AD}$ intersects the bisectors of $\angle ABC$ and $\angle ACB$ in points $E$ and $F,$ respectively. Given that $AB=4,BC=5,$ and $CA=6,$ the area of $\triangle AEF$ can be written as $\tfrac{m\sqrt{n}}p,$ where $m$ and $p$ are relatively prime positive integers, and $n$ is a positive integer not divisible by the square of any prime. Find $m+n+p$ .
| 36
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I
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[
"Points are defined as shown. It is pretty easy to show that $\\triangle AFE \\sim \\triangle AGH$ by spiral similarity at $A$ by some short angle chasing. Now, note that $AD$ is the altitude of $\\triangle AFE$, as the altitude of $AGH$. We need to compare these altitudes in order to compare their areas. Note that Stewart's theorem implies that $AD/2 = \\frac{\\sqrt{18}}{2}$, the altitude of $\\triangle AFE$. Similarly, the altitude of $\\triangle AGH$ is the altitude of $\\triangle ABC$, or $\\frac{3\\sqrt{7}}{2}$. However, it's not too hard to see that $GB = HC = 1$, and therefore $[AGH] = [ABC]$. From here, we get that the area of $\\triangle ABC$ is $\\frac{15\\sqrt{7}}{14} \\implies 036, by similarity. ~awang11",
"Let $M_A$, $M_B$, $M_C$ be the midpoints of arcs $BC$, $CA$, $AB$. By Fact 5, we know that $M_AB = M_AC = M_AI$, and so by Ptolmey's theorem, we deduce that \\[AB\\cdot M_AC + AC\\cdot M_AB = BC\\cdot M_AA \\implies M_AA = 2M_AI.\\] In particular, we have $AI = IM_A$. [asy] defaultpen(fontsize(10pt)); size(200); pair A, B, C, D, E, F, I, P, MA, MB, MC; B = (0,0); C = (5,0); A = IP(Circle(B, 4), Circle(C, 6), 0); I = incenter(A, B, C); D = extension(A, I, B, C); P = midpoint(A--D); E = extension(P, rotate(90, P)*A, B, I); F = extension(P, rotate(90, P)*A, C, I); MA = IP(Line(A, I, 20), circumcircle(A, B, C), 1); MB = IP(Line(B, I, 20), circumcircle(A, B, C), 1); MC = IP(Line(C, I, 20), circumcircle(A, B, C), 1); draw(A--B--C--cycle, orange); draw(circumcircle(A, B, C), red); draw(A--E--F--cycle, lightblue); draw(E--D--F, lightblue); draw(A--MA^^B--MB^^C--MC, heavygreen); draw(MA--MB--MC--cycle, magenta); dot(\"$A$\", A, dir(120)); dot(\"$B$\", B, dir(220)); dot(\"$C$\", C, dir(320)); dot(\"$D$\", D, dir(230)); dot(\"$E$\", E, dir(330)); dot(\"$F$\", F, dir(250)); dot(\"$I$\", I, dir(80)); dot(\"$M_A$\", MA, dir(270)); dot(\"$M_B$\", MB, dir(60)); dot(\"$M_C$\", MC, dir(150)); [/asy] Now the key claim is that: Claim: $\\triangle DEF$ and $\\triangle M_AM_BM_C$ are homothetic at $I$ with ratio $2$. Proof. First, we show that $D$ is the midpoint of $M_AI$. Indeed, we have \\[\\frac{ID}{DM_A} = \\frac{BI}{BM_A}\\cdot \\frac{\\sin\\angle IBC}{\\sin \\angle CBM_A} = \\frac{BI}{AI}\\cdot\\frac{\\sin \\angle B/2}{\\sin \\angle A/2} = 1\\] by Ratio lemma and Law of Sines. Now observe that: $\\overline{M_BM_C}$ is the perpendicular bisector of $\\overline{AI}$, $\\overline{EF}$ is the perpendicular bisector of $\\overline{AD}$, and $ID = AI/2$. Combining these facts gives that $\\overline{EF}$ is a midline in $\\triangle IM_BM_C$, which proves the claim. $\\blacksquare$ To finish, we compute $[M_AM_BM_C]$, noting that $[AEF] = [DEF] = \\tfrac{1}{4}[M_AM_BM_C]$. By Heron's, we can calculate the circumradius $R = 8/\\sqrt{7}$, and by Law of Cosines, we get \\begin{align*}\\cos A &= \\frac{9}{16}\\implies \\cos A/2 = \\frac{5}{\\sqrt{32}} \\\\ \\cos B &= \\frac{1}{8} \\implies \\cos B/2 = \\frac{3}{4} \\\\ \\cos C &= \\frac{3}{4} \\implies \\cos C/2 = \\sqrt{\\frac{7}{8}}.\\end{align*} Then using $[XYZ] = 2R^2\\sin X\\sin Y\\sin Z$, we can compute \\[[M_AM_BM_C] = 2\\cdot \\frac{64}{7}\\cdot \\frac{5}{\\sqrt{32}}\\cdot \\frac{3}{4}\\cdot \\frac{\\sqrt{7}}{\\sqrt{8}} = \\frac{30\\sqrt{7}}{7}.\\] Thus $[AEF] = 15\\sqrt{7}/14$, which gives a final answer of $036. ~pinetree1",
"Let $\\overline{BC}$ lie on the x-axis and $B$ be the origin. $C$ is $(5,0)$. Use Heron's formula to compute the area of triangle $ABC$. We have $s=\\frac{15}{2}$. and $[ABC]=\\sqrt{\\frac{15 \\cdot 7 \\cdot 5 \\cdot 3}{2^4}}=\\frac{15\\sqrt{7}}{4}$. We now find the altitude, which is $\\frac{\\frac{15\\sqrt{7}}{2}}{5}=\\frac{3\\sqrt{7}}{2}$, which is the y-coordinate of $A$. We now find the x-coordinate of $A$, which satisfies $x^2 + (\\frac{3\\sqrt{7}}{2})^{2}=16$, which gives $x=\\frac{1}{2}$ since the triangle is acute. Now using the Angle Bisector Theorem, we have $\\frac{4}{6}=\\frac{BD}{CD}$ and $BD+CD=5$ to get $BD=2$. The coordinates of D are $(2,0)$. Since we want the area of triangle $AEF$, we will find equations for perpendicular bisector of AD, and the other two angle bisectors. The perpendicular bisector is not too challenging: the midpoint of AD is $(\\frac{5}{4}, \\frac{3\\sqrt{7}}{4})$ and the slope of AD is $-\\sqrt{7}$. The slope of the perpendicular bisector is $\\frac{1}{\\sqrt{7}}$. The equation is(in point slope form) $y-\\frac{3\\sqrt{7}}{4}=\\frac{1}{\\sqrt{7}}(x-\\frac{5}{4})$. The slope of AB, or in trig words, the tangent of $\\angle ABC$ is $3\\sqrt{7}$. Finding $\\sin{\\angle ABC}=\\frac{\\frac{3\\sqrt{7}}{2}}{4}=\\frac{3\\sqrt{7}}{8}$ and $\\cos{\\angle ABC}=\\frac{\\frac{1}{2}}{4}=\\frac{1}{8}$. Plugging this in to half angle tangent, it gives $\\frac{\\frac{3\\sqrt{7}}{8}}{1+\\frac{1}{8}}=\\frac{\\sqrt{7}}{3}$ as the slope of the angle bisector, since it passes through $B$, the equation is $y=\\frac{\\sqrt{7}}{3}x$. Similarly, the equation for the angle bisector of $C$ will be $y=-\\frac{1}{\\sqrt{7}}(x-5)$. For $E$ use the B-angle bisector and the perpendicular bisector of AD equations to intersect at $(3,\\sqrt{7})$. For $F$ use the C-angle bisector and the perpendicular bisector of AD equations to intersect at $(\\frac{1}{2}, \\frac{9}{2\\sqrt{7}})$. The area of AEF is equal to $\\frac{EF \\cdot \\frac{AD}{2}}{2}$ since AD is the altitude of that triangle with EF as the base, with $\\frac{AD}{2}$ being the height. $EF=\\frac{5\\sqrt{2}}{\\sqrt{7}}$ and $AD=3\\sqrt{2}$, so $[AEF]=\\frac{15}{2\\sqrt{7}}=\\frac{15\\sqrt{7}}{14}$ which gives $036. NEVER overlook coordinate bash in combination with beginner synthetic techniques.~vvluo",
"Let $\\overline{BC}$ lie on the x-axis and $B$ be the origin. $C$ is $(5,0)$. Use Heron's formula to compute the area of triangle $ABC$. We have $s=\\frac{15}{2}$. and $[ABC]=\\sqrt{\\frac{15 \\cdot 7 \\cdot 5 \\cdot 3}{2^4}}=\\frac{15\\sqrt{7}}{4}$. We now find the altitude, which is $\\frac{\\frac{15\\sqrt{7}}{2}}{5}=\\frac{3\\sqrt{7}}{2}$, which is the y-coordinate of $A$. We now find the x-coordinate of $A$, which satisfies $x^2 + (\\frac{3\\sqrt{7}}{2})^{2}=16$, which gives $x=\\frac{1}{2}$ since the triangle is acute. Now using the Angle Bisector Theorem, we have $\\frac{4}{6}=\\frac{BD}{CD}$ and $BD+CD=5$ to get $BD=2$. The coordinates of D are $(2,0)$. Since we want the area of triangle $AEF$, we will find equations for perpendicular bisector of AD, and the other two angle bisectors. The perpendicular bisector is not too challenging: the midpoint of AD is $(\\frac{5}{4}, \\frac{3\\sqrt{7}}{4})$ and the slope of AD is $-\\sqrt{7}$. The slope of the perpendicular bisector is $\\frac{1}{\\sqrt{7}}$. The equation is(in point slope form) $y-\\frac{3\\sqrt{7}}{4}=\\frac{1}{\\sqrt{7}}(x-\\frac{5}{4})$. The slope of AB, or in trig words, the tangent of $\\angle ABC$ is $3\\sqrt{7}$. Finding $\\sin{\\angle ABC}=\\frac{\\frac{3\\sqrt{7}}{2}}{4}=\\frac{3\\sqrt{7}}{8}$ and $\\cos{\\angle ABC}=\\frac{\\frac{1}{2}}{4}=\\frac{1}{8}$. Plugging this in to half angle tangent, it gives $\\frac{\\frac{3\\sqrt{7}}{8}}{1+\\frac{1}{8}}=\\frac{\\sqrt{7}}{3}$ as the slope of the angle bisector, since it passes through $B$, the equation is $y=\\frac{\\sqrt{7}}{3}x$. Similarly, the equation for the angle bisector of $C$ will be $y=-\\frac{1}{\\sqrt{7}}(x-5)$. For $E$ use the B-angle bisector and the perpendicular bisector of AD equations to intersect at $(3,\\sqrt{7})$. For $F$ use the C-angle bisector and the perpendicular bisector of AD equations to intersect at $(\\frac{1}{2}, \\frac{9}{2\\sqrt{7}})$. The area of AEF is equal to $\\frac{EF \\cdot \\frac{AD}{2}}{2}$ since AD is the altitude of that triangle with EF as the base, with $\\frac{AD}{2}$ being the height. $EF=\\frac{5\\sqrt{2}}{\\sqrt{7}}$ and $AD=3\\sqrt{2}$, so $[AEF]=\\frac{15}{2\\sqrt{7}}=\\frac{15\\sqrt{7}}{14}$ which gives $036. NEVER overlook coordinate bash in combination with beginner synthetic techniques.~vvluo",
"[asy] size(8cm); defaultpen(fontsize(10pt)); pair A,B,C,I,D,M,T,Y,Z,EE,F; A=(0,3sqrt(7)); B=(-1,0); C=(9,0); I=incenter(A,B,C); D=extension(A,I,B,C); M=(A+D)/2; draw(B--EE,gray+dashed); draw(C--F,gray+dashed); draw(A--B--C--A); draw(A--D); draw(B--(5,sqrt(28))); draw(M--(5,sqrt(28))); draw(C--(0,9sqrt(7)/7)); draw(M--(0,9sqrt(7)/7)); dot(\"$A$\",A,NW); dot(\"$B$\",B,SW); dot(\"$C$\",C,SE); dot(\"$D$\",D,S); dot(\"$E$\",(5,sqrt(28)),N); dot(\"$M$\",M,dir(70)); dot(\"$F$\",(0,9sqrt(7)/7),N); label(\"$2$\",B--D,S); label(\"$3$\",D--C,S); label(\"$6$\",A--C,N); label(\"$4$\",A--B,W); [/asy] Let $B=(0,0)$ and $BC$ be the line $y=0$. We compute that $\\cos{\\angle{ABC}}=\\frac{1}{8}$, so $\\tan{\\angle{ABC}}=3\\sqrt{7}$. Thus, $A$ lies on the line $y=3x\\sqrt{7}$. The length of $AB$ at a point $x$ is $8x$, so $x=\\frac{1}{2}$. We now have the coordinates $A=\\left(\\frac{1}{2},\\frac{3\\sqrt{7}}{2}\\right)$, $B=(0,0)$ and $C=(5,0)$. We also have $D=(2,0)$ by the angle-bisector theorem and $M=\\left(\\frac{5}{4},\\frac{3\\sqrt{7}}{4}\\right)$ by taking the midpoint. We have that because $\\cos{\\angle{ABC}}=\\frac{1}{8}$, $\\cos{\\frac{\\angle{ABC}}{2}}=\\frac{3}{4}$ by half angle formula. We also compute $\\cos{\\angle{ACB}}=\\frac{3}{4}$, so $\\cos{\\frac{\\angle{ACB}}{2}}=\\frac{\\sqrt{14}}{4}$. Now, $AD$ has slope $-\\frac{\\frac{3\\sqrt{7}}{2}}{2-\\frac{1}{2}}=-\\sqrt{7}$, so it's perpendicular bisector has slope $\\frac{\\sqrt{7}}{7}$ and goes through $\\left(\\frac{5}{4},\\frac{3\\sqrt{7}}{4}\\right)$. We find that this line has equation $y=\\frac{\\sqrt{7}}{7}x+\\frac{4\\sqrt{7}}{7}$. As $\\cos{\\angle{CBI}}=\\frac{3}{4}$, we have that line $BI$ has form $y=\\frac{\\sqrt{7}}{3}x$. Solving for the intersection point of these two lines, we get $x=3$ and thus $E=\\left(3, \\sqrt{7}\\right)$ We also have that because $\\cos{\\angle{ICB}}=\\frac{\\sqrt{14}}{4}$, $CI$ has form $y=-\\frac{x\\sqrt{7}}{7}+\\frac{5\\sqrt{7}}{7}$. Intersecting the line $CI$ and the perpendicular bisector of $AD$ yields $-\\frac{x\\sqrt{7}}{7}+\\frac{5\\sqrt{7}}{7}=\\frac{x\\sqrt{7}}{7}+\\frac{4\\sqrt{7}}{7}$. Solving this, we get $x=\\frac{1}{2}$ and so $F=\\left(\\frac{1}{2},\\frac{9\\sqrt{7}}{14}\\right)$. We now compute $EF=\\sqrt{\\left(\\frac{5}{2}\\right)^2+\\left(\\frac{5\\sqrt{7}}{14}\\right)^2}=\\frac{5\\sqrt{14}}{7}$. We also have $MA=\\sqrt{\\left(\\frac{3}{4}\\right)^2+\\left(\\frac{3\\sqrt{7}}{4}\\right)^2}=\\frac{3\\sqrt{2}}{2}$. As ${MA}\\perp{EF}$, we have $[\\triangle{AEF}]=\\frac{1}{2}\\left(\\frac{3\\sqrt{2}}{2}\\times\\frac{5\\sqrt{14}}{7}\\right)=\\frac{15\\sqrt{7}}{14}$. The desired answer is $15+7+14=036 ~Imayormaynotknowcalculus",
"[asy] size(8cm); defaultpen(fontsize(10pt)); pair A,B,C,I,D,M,T,Y,Z,EE,F; A=(0,3sqrt(7)); B=(-1,0); C=(9,0); I=incenter(A,B,C); D=extension(A,I,B,C); M=(A+D)/2; draw(B--EE,gray+dashed); draw(C--F,gray+dashed); draw(A--B--C--A); draw(A--D); draw(B--(5,sqrt(28))); draw(M--(5,sqrt(28))); draw(C--(0,9sqrt(7)/7)); draw(M--(0,9sqrt(7)/7)); dot(\"$A$\",A,NW); dot(\"$B$\",B,SW); dot(\"$C$\",C,SE); dot(\"$D$\",D,S); dot(\"$E$\",(5,sqrt(28)),N); dot(\"$M$\",M,dir(70)); dot(\"$F$\",(0,9sqrt(7)/7),N); label(\"$2$\",B--D,S); label(\"$3$\",D--C,S); label(\"$6$\",A--C,N); label(\"$4$\",A--B,W); [/asy] Let $B=(0,0)$ and $BC$ be the line $y=0$. We compute that $\\cos{\\angle{ABC}}=\\frac{1}{8}$, so $\\tan{\\angle{ABC}}=3\\sqrt{7}$. Thus, $A$ lies on the line $y=3x\\sqrt{7}$. The length of $AB$ at a point $x$ is $8x$, so $x=\\frac{1}{2}$. We now have the coordinates $A=\\left(\\frac{1}{2},\\frac{3\\sqrt{7}}{2}\\right)$, $B=(0,0)$ and $C=(5,0)$. We also have $D=(2,0)$ by the angle-bisector theorem and $M=\\left(\\frac{5}{4},\\frac{3\\sqrt{7}}{4}\\right)$ by taking the midpoint. We have that because $\\cos{\\angle{ABC}}=\\frac{1}{8}$, $\\cos{\\frac{\\angle{ABC}}{2}}=\\frac{3}{4}$ by half angle formula. We also compute $\\cos{\\angle{ACB}}=\\frac{3}{4}$, so $\\cos{\\frac{\\angle{ACB}}{2}}=\\frac{\\sqrt{14}}{4}$. Now, $AD$ has slope $-\\frac{\\frac{3\\sqrt{7}}{2}}{2-\\frac{1}{2}}=-\\sqrt{7}$, so it's perpendicular bisector has slope $\\frac{\\sqrt{7}}{7}$ and goes through $\\left(\\frac{5}{4},\\frac{3\\sqrt{7}}{4}\\right)$. We find that this line has equation $y=\\frac{\\sqrt{7}}{7}x+\\frac{4\\sqrt{7}}{7}$. As $\\cos{\\angle{CBI}}=\\frac{3}{4}$, we have that line $BI$ has form $y=\\frac{\\sqrt{7}}{3}x$. Solving for the intersection point of these two lines, we get $x=3$ and thus $E=\\left(3, \\sqrt{7}\\right)$ We also have that because $\\cos{\\angle{ICB}}=\\frac{\\sqrt{14}}{4}$, $CI$ has form $y=-\\frac{x\\sqrt{7}}{7}+\\frac{5\\sqrt{7}}{7}$. Intersecting the line $CI$ and the perpendicular bisector of $AD$ yields $-\\frac{x\\sqrt{7}}{7}+\\frac{5\\sqrt{7}}{7}=\\frac{x\\sqrt{7}}{7}+\\frac{4\\sqrt{7}}{7}$. Solving this, we get $x=\\frac{1}{2}$ and so $F=\\left(\\frac{1}{2},\\frac{9\\sqrt{7}}{14}\\right)$. We now compute $EF=\\sqrt{\\left(\\frac{5}{2}\\right)^2+\\left(\\frac{5\\sqrt{7}}{14}\\right)^2}=\\frac{5\\sqrt{14}}{7}$. We also have $MA=\\sqrt{\\left(\\frac{3}{4}\\right)^2+\\left(\\frac{3\\sqrt{7}}{4}\\right)^2}=\\frac{3\\sqrt{2}}{2}$. As ${MA}\\perp{EF}$, we have $[\\triangle{AEF}]=\\frac{1}{2}\\left(\\frac{3\\sqrt{2}}{2}\\times\\frac{5\\sqrt{14}}{7}\\right)=\\frac{15\\sqrt{7}}{14}$. The desired answer is $15+7+14=036 ~Imayormaynotknowcalculus",
"[asy] size(8cm); defaultpen(fontsize(10pt)); pair A,B,C,I,D,M,T,Y,Z,EE,F; A=(0,3sqrt(7)); B=(-1,0); C=(9,0); I=incenter(A,B,C); D=extension(A,I,B,C); M=(A+D)/2; draw(B--EE,gray+dashed); draw(C--F,gray+dashed); draw(A--B--C--A); draw(A--D); draw(B--(5,sqrt(28))); draw(M--(5,sqrt(28))); draw(C--(0,9sqrt(7)/7)); draw(M--(0,9sqrt(7)/7)); dot(\"$A$\",A,NW); dot(\"$B$\",B,SW); dot(\"$C$\",C,SE); dot(\"$D$\",D,S); dot(\"$E$\",(5,sqrt(28)),N); dot(\"$M$\",M,dir(70)); dot(\"$F$\",(0,9sqrt(7)/7),N); label(\"$2$\",B--D,S); label(\"$3$\",D--C,S); label(\"$6$\",A--C,N); label(\"$4$\",A--B,W); [/asy] As usual, we will use homogenized barycentric coordinates. We have that $AD$ will have form $3z=2y$. Similarly, $CF$ has form $5y=6x$ and $BE$ has form $5z=4x$. Since $A=(1,0,0)$ and $D=\\left(0,\\frac{3}{5},\\frac{2}{5}\\right)$, we also have $M=\\left(\\frac{1}{2},\\frac{3}{10},\\frac{1}{5}\\right)$. It remains to determine the equation of the line formed by the perpendicular bisector of $AD$. This can be found using EFFT. Let a point $T$ on $EF$ have coordinates $(x, y, z)$. We then have that the displacement vector $\\overrightarrow{AD}=\\left(-1, \\frac{3}{5}, \\frac{2}{5}\\right)$ and that the displacement vector $\\overrightarrow{TM}$ has form $\\left(x-\\frac{1}{2},y-\\frac{3}{10},z-\\frac{1}{5}\\right)$. Now, by EFFT, we have $5^2\\left(\\frac{3}{5}\\times\\left(z-\\frac{1}{5}\\right)+\\frac{2}{5}\\times\\left(y-\\frac{3}{10}\\right)\\right)+6^2\\left(-1\\times\\left(z-\\frac{1}{5}\\right)+\\frac{2}{5}\\times\\left(x-\\frac{1}{2}\\right)\\right)+4^2\\left(-1\\times\\left(y-\\frac{3}{10}\\right)+\\frac{3}{5}\\times\\left(x-\\frac{1}{2}\\right)\\right)=0$. This equates to $8x-2y-7z=2$. Now, intersecting this with $BE$, we have $5z=4x$, $8x-2y-7z=2$, and $x+y+z=1$. This yields $x=\\frac{2}{3}$, $y=-\\frac{1}{5}$, and $z=\\frac{8}{15}$, or $E=\\left(\\frac{2}{3},-\\frac{1}{5},\\frac{8}{15}\\right)$. Similarly, intersecting this with $CF$, we have $5y=6x$, $8x-2y-7z=2$, and $x+y+z=1$. Solving this, we obtain $x=\\frac{3}{7}$, $y=\\frac{18}{35}$, and $z=\\frac{2}{35}$, or $F=\\left(\\frac{3}{7},\\frac{18}{35},\\frac{2}{35}\\right)$. We finish by invoking the Barycentric Distance Formula twice; our first displacement vector being $\\overrightarrow{FE}=\\left(\\frac{5}{21},-\\frac{5}{7},\\frac{10}{21}\\right)$. We then have $FE^2=-25\\left(-\\frac{5}{7}\\cdot\\frac{10}{21}\\right)-36\\left(\\frac{5}{21}\\cdot\\frac{10}{21}\\right)-16\\left(\\frac{5}{21}\\cdot-\\frac{5}{7}\\right)=\\frac{50}{7}$, thus $FE=\\frac{5\\sqrt{14}}{7}$. Our second displacement vector is $\\overrightarrow{AM}=\\left(-\\frac{1}{2},\\frac{3}{10},\\frac{1}{5}\\right)$. As a result, $AM^2=-25\\left(\\frac{3}{10}\\cdot\\frac{1}{5}\\right)-36\\left(-\\frac{1}{2}\\cdot\\frac{1}{5}\\right)-16\\left(-\\frac{1}{2}\\cdot\\frac{3}{10}\\right)=\\frac{9}{2}$, so $AM=\\frac{3\\sqrt{2}}{2}$. As ${AM}\\perp{EF}$, the desired area is $\\frac{\\frac{5\\sqrt{14}}{7}\\times\\frac{3\\sqrt{2}}{2}}{2}={\\frac{15\\sqrt{7}}{14}}\\implies{m+n+p=036 can also be computed using the Barycentric Area Formula, although it may increase the risk of computational errors; there are also many different ways to proceed once the coordinates are determined.",
"[asy] size(8cm); defaultpen(fontsize(10pt)); pair A,B,C,I,D,M,T,Y,Z,EE,F; A=(0,3sqrt(7)); B=(-1,0); C=(9,0); I=incenter(A,B,C); D=extension(A,I,B,C); M=(A+D)/2; draw(B--EE,gray+dashed); draw(C--F,gray+dashed); draw(A--B--C--A); draw(A--D); draw(B--(5,sqrt(28))); draw(M--(5,sqrt(28))); draw(C--(0,9sqrt(7)/7)); draw(M--(0,9sqrt(7)/7)); dot(\"$A$\",A,NW); dot(\"$B$\",B,SW); dot(\"$C$\",C,SE); dot(\"$D$\",D,S); dot(\"$E$\",(5,sqrt(28)),N); dot(\"$M$\",M,dir(70)); dot(\"$F$\",(0,9sqrt(7)/7),N); label(\"$2$\",B--D,S); label(\"$3$\",D--C,S); label(\"$6$\",A--C,N); label(\"$4$\",A--B,W); [/asy] As usual, we will use homogenized barycentric coordinates. We have that $AD$ will have form $3z=2y$. Similarly, $CF$ has form $5y=6x$ and $BE$ has form $5z=4x$. Since $A=(1,0,0)$ and $D=\\left(0,\\frac{3}{5},\\frac{2}{5}\\right)$, we also have $M=\\left(\\frac{1}{2},\\frac{3}{10},\\frac{1}{5}\\right)$. It remains to determine the equation of the line formed by the perpendicular bisector of $AD$. This can be found using EFFT. Let a point $T$ on $EF$ have coordinates $(x, y, z)$. We then have that the displacement vector $\\overrightarrow{AD}=\\left(-1, \\frac{3}{5}, \\frac{2}{5}\\right)$ and that the displacement vector $\\overrightarrow{TM}$ has form $\\left(x-\\frac{1}{2},y-\\frac{3}{10},z-\\frac{1}{5}\\right)$. Now, by EFFT, we have $5^2\\left(\\frac{3}{5}\\times\\left(z-\\frac{1}{5}\\right)+\\frac{2}{5}\\times\\left(y-\\frac{3}{10}\\right)\\right)+6^2\\left(-1\\times\\left(z-\\frac{1}{5}\\right)+\\frac{2}{5}\\times\\left(x-\\frac{1}{2}\\right)\\right)+4^2\\left(-1\\times\\left(y-\\frac{3}{10}\\right)+\\frac{3}{5}\\times\\left(x-\\frac{1}{2}\\right)\\right)=0$. This equates to $8x-2y-7z=2$. Now, intersecting this with $BE$, we have $5z=4x$, $8x-2y-7z=2$, and $x+y+z=1$. This yields $x=\\frac{2}{3}$, $y=-\\frac{1}{5}$, and $z=\\frac{8}{15}$, or $E=\\left(\\frac{2}{3},-\\frac{1}{5},\\frac{8}{15}\\right)$. Similarly, intersecting this with $CF$, we have $5y=6x$, $8x-2y-7z=2$, and $x+y+z=1$. Solving this, we obtain $x=\\frac{3}{7}$, $y=\\frac{18}{35}$, and $z=\\frac{2}{35}$, or $F=\\left(\\frac{3}{7},\\frac{18}{35},\\frac{2}{35}\\right)$. We finish by invoking the Barycentric Distance Formula twice; our first displacement vector being $\\overrightarrow{FE}=\\left(\\frac{5}{21},-\\frac{5}{7},\\frac{10}{21}\\right)$. We then have $FE^2=-25\\left(-\\frac{5}{7}\\cdot\\frac{10}{21}\\right)-36\\left(\\frac{5}{21}\\cdot\\frac{10}{21}\\right)-16\\left(\\frac{5}{21}\\cdot-\\frac{5}{7}\\right)=\\frac{50}{7}$, thus $FE=\\frac{5\\sqrt{14}}{7}$. Our second displacement vector is $\\overrightarrow{AM}=\\left(-\\frac{1}{2},\\frac{3}{10},\\frac{1}{5}\\right)$. As a result, $AM^2=-25\\left(\\frac{3}{10}\\cdot\\frac{1}{5}\\right)-36\\left(-\\frac{1}{2}\\cdot\\frac{1}{5}\\right)-16\\left(-\\frac{1}{2}\\cdot\\frac{3}{10}\\right)=\\frac{9}{2}$, so $AM=\\frac{3\\sqrt{2}}{2}$. As ${AM}\\perp{EF}$, the desired area is $\\frac{\\frac{5\\sqrt{14}}{7}\\times\\frac{3\\sqrt{2}}{2}}{2}={\\frac{15\\sqrt{7}}{14}}\\implies{m+n+p=036 can also be computed using the Barycentric Area Formula, although it may increase the risk of computational errors; there are also many different ways to proceed once the coordinates are determined.",
"[asy] size(8cm); defaultpen(fontsize(10pt)); pair A,B,C,I,D,M,T,Y,Z,EE,F; A=(0,3sqrt(7)); B=(-1,0); C=(9,0); I=incenter(A,B,C); D=extension(A,I,B,C); M=(A+D)/2; draw(B--EE,gray+dashed); draw(C--F,gray+dashed); draw(A--B--C--A); draw(A--D); draw(A--(5,sqrt(28))); draw(A--(0,9sqrt(7)/7)); draw(D--(0,9sqrt(7)/7)); draw(D--(5,sqrt(28))); draw(B--(5,sqrt(28))); draw(M--(5,sqrt(28))); draw(C--(0,9sqrt(7)/7)); draw(M--(0,9sqrt(7)/7)); dot(\"$A$\",A,NW); dot(\"$B$\",B,SW); dot(\"$C$\",C,SE); dot(\"$D$\",D,S); dot(\"$E$\",(5,sqrt(28)),N); dot(\"$M$\",M,dir(70)); dot(\"$F$\",(0,9sqrt(7)/7),N); label(\"$2$\",B--D,S); label(\"$3$\",D--C,S); label(\"$6$\",A--C,N); label(\"$4$\",A--B,W); [/asy] To get the area of $\\triangle AEF$, we try to find $AM$ and $\\angle EAF$. Since $AD$ is the angle bisector, we can get that $BD=2$ and $CD=3$. By applying Stewart's Theorem, we can get that $AD=3\\sqrt{2}$. Therefore $AM=\\frac{3\\sqrt{2}}{2}$. Since $EF$ is the perpendicular bisector of $AD$, we know that $AE = DE$. Since $BE$ is the angle bisector of $\\angle BAC$, we know that $\\angle ABE = \\angle DBE$. By applying the Law of Sines to $\\triangle ABE$ and $\\triangle DBE$, we know that $\\sin \\angle BAE = \\sin \\angle BDE$. Since $BD$ is not equal to $AB$ and therefore these two triangles are not congruent, we know that $\\angle BAE$ and $\\angle BDE$ are supplementary. Then we know that $\\angle ABD$ and $\\angle AED$ are also supplementary. Given that $AE=DE$, we can get that $\\angle DAE$ is half of $\\angle ABC$. Similarly, we have $\\angle DAF$ is half of $\\angle ACB$. By applying the Law of Cosines, we get $\\cos \\angle ABC = \\frac{1}{8}$, and then $\\sin \\angle ABC = \\frac{3\\sqrt{7}}{8}$. Similarly, we can get $\\cos \\angle ACB = \\frac{3}{4}$ and $\\sin \\angle ACB = \\frac{\\sqrt{7}}{4}$. Based on some trig identities, we can compute that $\\tan \\angle DAE = \\frac{\\sin \\angle ABC}{1 + \\cos \\angle ABC} = \\frac{\\sqrt{7}}{3}$, and $\\tan \\angle DAF = \\frac{\\sqrt{7}}{7}$. Finally, the area of $\\triangle AEF$ equals $\\frac{1}{2}AM^2(\\tan \\angle DAE + \\tan \\angle DAF)=\\frac{15\\sqrt{7}}{14}$. Therefore, the final answer is $15+7+14=036. Can someone please help add these segments? (Added :) ~Math_Genius_164)",
"[asy] size(8cm); defaultpen(fontsize(10pt)); pair A,B,C,I,D,M,T,Y,Z,EE,F; A=(0,3sqrt(7)); B=(-1,0); C=(9,0); I=incenter(A,B,C); D=extension(A,I,B,C); M=(A+D)/2; draw(B--EE,gray+dashed); draw(C--F,gray+dashed); draw(A--B--C--A); draw(A--D); draw(A--(5,sqrt(28))); draw(A--(0,9sqrt(7)/7)); draw(D--(0,9sqrt(7)/7)); draw(D--(5,sqrt(28))); draw(B--(5,sqrt(28))); draw(M--(5,sqrt(28))); draw(C--(0,9sqrt(7)/7)); draw(M--(0,9sqrt(7)/7)); dot(\"$A$\",A,NW); dot(\"$B$\",B,SW); dot(\"$C$\",C,SE); dot(\"$D$\",D,S); dot(\"$E$\",(5,sqrt(28)),N); dot(\"$M$\",M,dir(70)); dot(\"$F$\",(0,9sqrt(7)/7),N); label(\"$2$\",B--D,S); label(\"$3$\",D--C,S); label(\"$6$\",A--C,N); label(\"$4$\",A--B,W); [/asy] To get the area of $\\triangle AEF$, we try to find $AM$ and $\\angle EAF$. Since $AD$ is the angle bisector, we can get that $BD=2$ and $CD=3$. By applying Stewart's Theorem, we can get that $AD=3\\sqrt{2}$. Therefore $AM=\\frac{3\\sqrt{2}}{2}$. Since $EF$ is the perpendicular bisector of $AD$, we know that $AE = DE$. Since $BE$ is the angle bisector of $\\angle BAC$, we know that $\\angle ABE = \\angle DBE$. By applying the Law of Sines to $\\triangle ABE$ and $\\triangle DBE$, we know that $\\sin \\angle BAE = \\sin \\angle BDE$. Since $BD$ is not equal to $AB$ and therefore these two triangles are not congruent, we know that $\\angle BAE$ and $\\angle BDE$ are supplementary. Then we know that $\\angle ABD$ and $\\angle AED$ are also supplementary. Given that $AE=DE$, we can get that $\\angle DAE$ is half of $\\angle ABC$. Similarly, we have $\\angle DAF$ is half of $\\angle ACB$. By applying the Law of Cosines, we get $\\cos \\angle ABC = \\frac{1}{8}$, and then $\\sin \\angle ABC = \\frac{3\\sqrt{7}}{8}$. Similarly, we can get $\\cos \\angle ACB = \\frac{3}{4}$ and $\\sin \\angle ACB = \\frac{\\sqrt{7}}{4}$. Based on some trig identities, we can compute that $\\tan \\angle DAE = \\frac{\\sin \\angle ABC}{1 + \\cos \\angle ABC} = \\frac{\\sqrt{7}}{3}$, and $\\tan \\angle DAF = \\frac{\\sqrt{7}}{7}$. Finally, the area of $\\triangle AEF$ equals $\\frac{1}{2}AM^2(\\tan \\angle DAE + \\tan \\angle DAF)=\\frac{15\\sqrt{7}}{14}$. Therefore, the final answer is $15+7+14=036. Can someone please help add these segments? (Added :) ~Math_Genius_164)",
"First and foremost $\\big[\\triangle{AEF}\\big]=\\big[\\triangle{DEF}\\big]$ as $EF$ is the perpendicular bisector of $AD$. Now note that quadrilateral $ABDF$ is cyclic, because $\\angle{ABF}=\\angle{FBD}$ and $FA=FD$. Similarly quadrilateral $AEDC$ is cyclic, \\[\\implies \\angle{EDA}=\\dfrac{C}{2}, \\quad \\angle{FDA}=\\dfrac{B}{2}\\] Let $A'$,$B'$, $C'$ be the $A$,$B$, and $C$ excenters of $\\triangle{ABC}$ respectively. Then it follows that $\\triangle{DEF} \\sim \\triangle{A'C'B'}$. By angle bisector theorem we have $BD=2 \\implies \\dfrac{ID}{IA}=\\dfrac{BD}{BA}=\\dfrac{1}{2}$. Now let the feet of the perpendiculars from $I$ and $A'$ to $BC$ be $X$ and $Y$ resptively. Then by tangents we have \\[BX=s-AC=\\dfrac{3}{2} \\implies XD=2-\\dfrac{3}{2}=\\dfrac{1}{2}\\] \\[CY=s-AC \\implies YD=3-\\dfrac{3}{2}=\\dfrac{3}{2} \\implies \\dfrac{ID}{DA'}=\\dfrac{XD}{YD}=\\dfrac{1}{3} \\implies \\big[\\triangle{DEF}\\big]=\\dfrac{1}{16}\\big[\\triangle{A'C'B'}\\big]\\] From the previous ratios, $AI:ID:DA'=2:1:3 \\implies AD=DA' \\implies \\big[\\triangle{ABC}\\big]=\\big[\\triangle{A'BC}\\big]$ Similarly we can find that $\\big[\\triangle{B'AC}\\big]=2\\big[\\triangle{ABC}\\big]$ and $\\big[\\triangle{C'AB}\\big]=\\dfrac{4}{7}\\big[\\triangle{ABC}\\big]$ and thus \\[\\big[\\triangle{A'B'C'}\\big]=\\bigg(1+1+2+\\dfrac{4}{7}\\bigg)\\big[\\triangle{ABC}\\big]=\\dfrac{32}{7}\\big[\\triangle{ABC}\\big] \\implies \\big[\\triangle{DEF}\\big]=\\dfrac{2}{7}\\big[\\triangle{ABC}\\big]=\\dfrac{15\\sqrt{7}}{14} \\implies m+n+p = 036\\] -tkhalid",
"Trig values we use here: $\\cos A = \\frac{9}{16}$ $\\cos \\frac{A}{2} = \\frac{5}{4\\sqrt2}$ $\\sin \\frac{A}{2} = \\frac{\\sqrt7}{4\\sqrt2}$ $\\cos \\frac{B}{2} = \\frac{3}{4}$ $\\cos \\frac{C}{2} = \\frac{\\sqrt7}{2\\sqrt2}$ First let the incenter be $I$. Let $M$ be the midpoint of minor arc $BC$ on $(ABC)$ and let $K$ be the foot of $M$ to $BC$. We can find $AD$ using Stewart's Theorem: from Angle Bisector Theorem $BD = 2$ and $CD = 3$. Then it is easy to find that $AD = 3\\sqrt3$. Now we trig bash for $DI = MI - MD$. Notice that $MI = MB$ from the Incenter Excenter Lemma. We obtain that $MB = \\frac{BK}{\\cos \\frac{A}{2}} = \\frac{\\frac{5}{2}}{\\frac{5}{4\\sqrt2}}=2\\sqrt2$. To get $MD$ we angle chase to get $\\angle KDM = \\frac{A}{2}+C$. Then \\[\\cos(\\frac{A}{2}+C) = \\cos\\frac{A}{2}\\cos C - \\sin\\frac{A}{2}\\sin C = \\frac{1}{2\\sqrt2} = \\frac{\\frac{1}{2}}{MD}\\] gives $MD = \\sqrt{2}$. This means $DI = \\sqrt2$. Now let $AI \\cap EF = G$. It is easy to angle chase $\\angle GIE = 90- \\frac{B}{2}$ and $\\angle GIF = 90- \\frac{C}{2}$. Since $GI = GD - ID = \\frac{3\\sqrt2}{2}-\\sqrt2=\\frac{\\sqrt2}{2}$, we compute that \\[EF = EG + FG = \\frac{\\sqrt2}{2}(\\cot B/2 + \\cot C/2) = \\frac{\\sqrt2}{2}(\\frac{3}{\\sqrt7}+\\sqrt7) = \\frac{5\\sqrt{14}}{7}\\] which implies \\[[AEF] = AG*EF/2 = \\frac{3\\sqrt2}{2} * \\frac{5\\sqrt{14}}{7} / 2 = \\frac{15\\sqrt7}{14}\\] which gives an answer of $36. ~Leonard_my_dude~",
"Trig values we use here: $\\cos A = \\frac{9}{16}$ $\\cos \\frac{A}{2} = \\frac{5}{4\\sqrt2}$ $\\sin \\frac{A}{2} = \\frac{\\sqrt7}{4\\sqrt2}$ $\\cos \\frac{B}{2} = \\frac{3}{4}$ $\\cos \\frac{C}{2} = \\frac{\\sqrt7}{2\\sqrt2}$ First let the incenter be $I$. Let $M$ be the midpoint of minor arc $BC$ on $(ABC)$ and let $K$ be the foot of $M$ to $BC$. We can find $AD$ using Stewart's Theorem: from Angle Bisector Theorem $BD = 2$ and $CD = 3$. Then it is easy to find that $AD = 3\\sqrt3$. Now we trig bash for $DI = MI - MD$. Notice that $MI = MB$ from the Incenter Excenter Lemma. We obtain that $MB = \\frac{BK}{\\cos \\frac{A}{2}} = \\frac{\\frac{5}{2}}{\\frac{5}{4\\sqrt2}}=2\\sqrt2$. To get $MD$ we angle chase to get $\\angle KDM = \\frac{A}{2}+C$. Then \\[\\cos(\\frac{A}{2}+C) = \\cos\\frac{A}{2}\\cos C - \\sin\\frac{A}{2}\\sin C = \\frac{1}{2\\sqrt2} = \\frac{\\frac{1}{2}}{MD}\\] gives $MD = \\sqrt{2}$. This means $DI = \\sqrt2$. Now let $AI \\cap EF = G$. It is easy to angle chase $\\angle GIE = 90- \\frac{B}{2}$ and $\\angle GIF = 90- \\frac{C}{2}$. Since $GI = GD - ID = \\frac{3\\sqrt2}{2}-\\sqrt2=\\frac{\\sqrt2}{2}$, we compute that \\[EF = EG + FG = \\frac{\\sqrt2}{2}(\\cot B/2 + \\cot C/2) = \\frac{\\sqrt2}{2}(\\frac{3}{\\sqrt7}+\\sqrt7) = \\frac{5\\sqrt{14}}{7}\\] which implies \\[[AEF] = AG*EF/2 = \\frac{3\\sqrt2}{2} * \\frac{5\\sqrt{14}}{7} / 2 = \\frac{15\\sqrt7}{14}\\] which gives an answer of $36. ~Leonard_my_dude~",
"Let $x = \\angle BAD = \\angle CAD$, $y = \\angle CBE = \\angle ABE$, and $z = \\angle BCF = \\angle ACF$. Notice that $x+y+z = 90^\\circ$. In $\\triangle ABD$, segment $\\overline{BE}$ is the bisector of $\\angle ABD$, and $E$ lies on the perpendicular bisector of side $\\overline{AD}$. Therefore $E$ is the midpoint of arc $\\stackrel{\\textstyle\\frown}{AD}$ on the circumcircle of $\\triangle ABD$. It follows that $\\angle BED = \\angle BAD = x$ and $\\angle EDA = \\angle EBA = y$. Likewise, $ACDF$ is cyclic, $\\angle CFD = \\angle CAD = x$, and $\\angle FDA = \\angle FCA = z$. Because $\\overline{EF}$ is the perpendicular bisector of $\\overline{AD}$, triangles $AEF$ and $DEF$ are congruent, implying that \\begin{align*} [\\triangle AEF] = [\\triangle DEF] &= \\frac{DE\\cdot DF\\cdot\\sin(\\angle EDF)}{2}\\\\ &= \\frac{DE\\cdot DF\\cdot\\sin(y+z)}{2} = \\frac{DE\\cdot DF\\cdot\\cos x}{2}. \\end{align*} [asy] unitsize(0.8 cm); pair A, B, C, D, E, F, I; real angleC = aCos(3/4); C = (0,0); A = 6*dir(270 - angleC/2); B = 5*dir(270 + angleC/2); I = incenter(A,B,C); D = extension(A, I, B, C); E = extension((A + D)/2, (A + D)/2 + rotate(90)*(A - D), B, I); F = extension((A + D)/2, (A + D)/2 + rotate(90)*(A - D), C, I); draw(A--B--C--cycle); draw(A--interp(A,D,1.4)); draw(A--F--D--E--cycle); draw(E--F); draw(circumcircle(A,B,D)); draw(B--interp(B,E,1.5)); draw(C--interp(C,F,1.5)); dot(\"$A$\", A, SW); dot(\"$B$\", B, dir(0)); dot(\"$C$\", C, N); dot(\"$D$\", D, dir(0)); dot(\"$E$\", E, N); dot(\"$F$\", F, dir(0)); [/asy] Applying the Law of Sines to $\\triangle BED$ and $\\triangle CFD$ gives\\[DE = BD\\cdot\\frac{\\sin y}{\\sin x}\\text{~ and ~} DF = CD\\cdot\\frac{\\sin z}{\\sin x}.\\]By the Angle Bisector Theorem, $BD = 2$ and $CD = 3$. Combining the above information yields \\[[\\triangle AEF] = \\frac{3\\sin y\\cdot\\sin z\\cdot\\cos x}{\\sin^2 x}.\\]Applying the Law of Cosines to $\\triangle ABC$ gives $\\cos 2x = \\frac9{16}$, $\\cos 2y = \\frac1{8}$, and $\\cos 2z = \\frac34$. By the Half Angle Formulas,\\[\\sin^2x = \\frac7{32},~~ \\cos x = \\sqrt{\\frac{25}{32}},~~ \\sin y = \\sqrt{\\frac7{16}}, \\text{~ and ~} \\sin z = \\sqrt{\\frac18}.\\]Therefore \\[[\\triangle AEF] = \\frac{3\\cdot\\sqrt{\\frac{7}{16}}\\cdot\\sqrt{\\frac{1}{8}}\\cdot\\sqrt{\\frac{25}{32}}} {\\frac{7}{32}} = \\frac{15\\sqrt{7}}{14}.\\]The requested sum is $15+7+14 = 36$.",
"Let $x = \\angle BAD = \\angle CAD$, $y = \\angle CBE = \\angle ABE$, and $z = \\angle BCF = \\angle ACF$. Notice that $x+y+z = 90^\\circ$. In $\\triangle ABD$, segment $\\overline{BE}$ is the bisector of $\\angle ABD$, and $E$ lies on the perpendicular bisector of side $\\overline{AD}$. Therefore $E$ is the midpoint of arc $\\stackrel{\\textstyle\\frown}{AD}$ on the circumcircle of $\\triangle ABD$. It follows that $\\angle BED = \\angle BAD = x$ and $\\angle EDA = \\angle EBA = y$. Likewise, $ACDF$ is cyclic, $\\angle CFD = \\angle CAD = x$, and $\\angle FDA = \\angle FCA = z$. Because $\\overline{EF}$ is the perpendicular bisector of $\\overline{AD}$, triangles $AEF$ and $DEF$ are congruent, implying that \\begin{align*} [\\triangle AEF] = [\\triangle DEF] &= \\frac{DE\\cdot DF\\cdot\\sin(\\angle EDF)}{2}\\\\ &= \\frac{DE\\cdot DF\\cdot\\sin(y+z)}{2} = \\frac{DE\\cdot DF\\cdot\\cos x}{2}. \\end{align*} [asy] unitsize(0.8 cm); pair A, B, C, D, E, F, I; real angleC = aCos(3/4); C = (0,0); A = 6*dir(270 - angleC/2); B = 5*dir(270 + angleC/2); I = incenter(A,B,C); D = extension(A, I, B, C); E = extension((A + D)/2, (A + D)/2 + rotate(90)*(A - D), B, I); F = extension((A + D)/2, (A + D)/2 + rotate(90)*(A - D), C, I); draw(A--B--C--cycle); draw(A--interp(A,D,1.4)); draw(A--F--D--E--cycle); draw(E--F); draw(circumcircle(A,B,D)); draw(B--interp(B,E,1.5)); draw(C--interp(C,F,1.5)); dot(\"$A$\", A, SW); dot(\"$B$\", B, dir(0)); dot(\"$C$\", C, N); dot(\"$D$\", D, dir(0)); dot(\"$E$\", E, N); dot(\"$F$\", F, dir(0)); [/asy] Applying the Law of Sines to $\\triangle BED$ and $\\triangle CFD$ gives\\[DE = BD\\cdot\\frac{\\sin y}{\\sin x}\\text{~ and ~} DF = CD\\cdot\\frac{\\sin z}{\\sin x}.\\]By the Angle Bisector Theorem, $BD = 2$ and $CD = 3$. Combining the above information yields \\[[\\triangle AEF] = \\frac{3\\sin y\\cdot\\sin z\\cdot\\cos x}{\\sin^2 x}.\\]Applying the Law of Cosines to $\\triangle ABC$ gives $\\cos 2x = \\frac9{16}$, $\\cos 2y = \\frac1{8}$, and $\\cos 2z = \\frac34$. By the Half Angle Formulas,\\[\\sin^2x = \\frac7{32},~~ \\cos x = \\sqrt{\\frac{25}{32}},~~ \\sin y = \\sqrt{\\frac7{16}}, \\text{~ and ~} \\sin z = \\sqrt{\\frac18}.\\]Therefore \\[[\\triangle AEF] = \\frac{3\\cdot\\sqrt{\\frac{7}{16}}\\cdot\\sqrt{\\frac{1}{8}}\\cdot\\sqrt{\\frac{25}{32}}} {\\frac{7}{32}} = \\frac{15\\sqrt{7}}{14}.\\]The requested sum is $15+7+14 = 36$.",
"Let the point $M$ be the midpoint of $\\overline{AD}$, let $I$ be the incenter of $\\triangle ABC$ which is the common point of lines $AD$, $BE$, and $CF$, and let $r$ be the inradius of $\\triangle ABC$. The semiperimeter of $\\triangle ABC$ is\\[s = \\frac{AB + BC + CA}2 = \\frac{15}2,\\]and Heron's Formula gives the area of $\\triangle ABC$ as\\[\\sqrt{s(s-AB)(s-BC)(s-CA)} = \\frac{15\\sqrt7}4.\\]This area is also $rs$ implying that $r = \\frac{\\sqrt7}2$. Stewart's Theorem gives $AD =3\\sqrt2$. Because the ratio of the areas of $\\triangle IBC$ and $\\triangle ABC$ is $\\frac{ID}{AD},$ it follows that\\[ID = AD\\cdot\\frac{\\frac{r\\cdot BC}2}{\\frac{15\\sqrt7}4} = \\sqrt2.\\]Thus $IM = MD - ID = \\frac{\\sqrt2}2$. [asy] unitsize(0.8 cm); pair A, B, C, D, E, F, I, M; real angleC = aCos(3/4); C = (0,0); A = 6*dir(270 - angleC/2); B = 5*dir(270 + angleC/2); I = incenter(A,B,C); D = extension(A, I, B, C); E = extension((A + D)/2, (A + D)/2 + rotate(90)*(A - D), B, I); F = extension((A + D)/2, (A + D)/2 + rotate(90)*(A - D), C, I); M = (A + D)/2; draw(A--B--C--cycle); draw(A--interp(A,D,1.4)); draw(A--F--D--E--cycle); draw(E--F); draw(B--interp(B,E,1.5)); draw(C--interp(C,F,1.5)); dot(\"$A$\", A, SW); dot(\"$B$\", B, dir(0)); dot(\"$C$\", C, N); dot(\"$D$\", D, dir(0)); dot(\"$E$\", E, N); dot(\"$F$\", F, dir(0)); dot(\"$I$\", I, dir(120)); dot(\"$M$\", M, W); [/asy] Note that $\\angle EFI = 90^{\\circ} - \\angle FIA = 90^{\\circ} - \\angle CID = 90^{\\circ} - \\frac{\\angle A}{2} - \\frac{\\angle C}{2} = \\frac{\\angle B}{2} = \\angle IBC$. Thus $\\triangle IBC \\sim \\triangle IFE$. The height of $\\triangle IBC$ to $I$ is $r=\\frac{\\sqrt7}2$, and the height of $\\triangle IFE$ to $I$ is $IM=\\frac{\\sqrt2}2$, so $EF = BC\\cdot \\frac{IM}r = \\frac{5\\sqrt{14}}{7}$. The needed area of $\\triangle AEF$ is $\\frac12\\cdot EF\\cdot \\frac{AD}2 = \\frac{15\\sqrt7}{14}$, as above.",
"Let the point $M$ be the midpoint of $\\overline{AD}$, let $I$ be the incenter of $\\triangle ABC$ which is the common point of lines $AD$, $BE$, and $CF$, and let $r$ be the inradius of $\\triangle ABC$. The semiperimeter of $\\triangle ABC$ is\\[s = \\frac{AB + BC + CA}2 = \\frac{15}2,\\]and Heron's Formula gives the area of $\\triangle ABC$ as\\[\\sqrt{s(s-AB)(s-BC)(s-CA)} = \\frac{15\\sqrt7}4.\\]This area is also $rs$ implying that $r = \\frac{\\sqrt7}2$. Stewart's Theorem gives $AD =3\\sqrt2$. Because the ratio of the areas of $\\triangle IBC$ and $\\triangle ABC$ is $\\frac{ID}{AD},$ it follows that\\[ID = AD\\cdot\\frac{\\frac{r\\cdot BC}2}{\\frac{15\\sqrt7}4} = \\sqrt2.\\]Thus $IM = MD - ID = \\frac{\\sqrt2}2$. [asy] unitsize(0.8 cm); pair A, B, C, D, E, F, I, M; real angleC = aCos(3/4); C = (0,0); A = 6*dir(270 - angleC/2); B = 5*dir(270 + angleC/2); I = incenter(A,B,C); D = extension(A, I, B, C); E = extension((A + D)/2, (A + D)/2 + rotate(90)*(A - D), B, I); F = extension((A + D)/2, (A + D)/2 + rotate(90)*(A - D), C, I); M = (A + D)/2; draw(A--B--C--cycle); draw(A--interp(A,D,1.4)); draw(A--F--D--E--cycle); draw(E--F); draw(B--interp(B,E,1.5)); draw(C--interp(C,F,1.5)); dot(\"$A$\", A, SW); dot(\"$B$\", B, dir(0)); dot(\"$C$\", C, N); dot(\"$D$\", D, dir(0)); dot(\"$E$\", E, N); dot(\"$F$\", F, dir(0)); dot(\"$I$\", I, dir(120)); dot(\"$M$\", M, W); [/asy] Note that $\\angle EFI = 90^{\\circ} - \\angle FIA = 90^{\\circ} - \\angle CID = 90^{\\circ} - \\frac{\\angle A}{2} - \\frac{\\angle C}{2} = \\frac{\\angle B}{2} = \\angle IBC$. Thus $\\triangle IBC \\sim \\triangle IFE$. The height of $\\triangle IBC$ to $I$ is $r=\\frac{\\sqrt7}2$, and the height of $\\triangle IFE$ to $I$ is $IM=\\frac{\\sqrt2}2$, so $EF = BC\\cdot \\frac{IM}r = \\frac{5\\sqrt{14}}{7}$. The needed area of $\\triangle AEF$ is $\\frac12\\cdot EF\\cdot \\frac{AD}2 = \\frac{15\\sqrt7}{14}$, as above.",
"Firstly, it is easy to find $BD=2,CD=3$ with angle bisector theorem. Using LOC and some trig formulas we get all those values: $cos\\angle{B}=\\frac{1}{8},sin\\angle{B}=\\frac{3\\sqrt{7}}{8},cos\\angle{\\frac{B}{2}}=\\frac{3}{4},cos\\angle{\\frac{{ACB}}{2}}=\\frac{\\sqrt{14}}{4}$ Now we find the coordinates of points $A,E,F$ and we apply shoelace theorem later. Point A's coordinates is $(4*\\frac{1}{8},4*\\frac{3\\sqrt{7}}{8})=(\\frac{1}{2},\\frac{3\\sqrt{7}}{2})$, Let $AJ$ is perpendicular to $BC$, $tan\\angle{JAD}=\\frac{2-\\frac{1}{2}}{\\frac{3\\sqrt{7}}{2}}=\\frac{\\sqrt{7}}{7}$, which means the slope of $FE$ is $\\frac{\\sqrt{7}}{7}$. Find the coordinate of $M$, it is easy, $(\\frac{5}{4},\\frac{3\\sqrt{7}}{4})$, the function $EF$ is $y=\\frac{\\sqrt{7}x}{7}+\\frac{4\\sqrt{7}}{7}$. Now find the intersection of $EF,EB$, $\\frac{\\sqrt{7}x}{7}+\\frac{4\\sqrt{7}}{7}=\\frac{\\sqrt{7}x}{3}$, getting that $x=3,E(3,\\sqrt{7})$ Now we look at line segment $CF$, since $cos\\angle{\\frac{ACB}{2}}=\\frac{\\sqrt{14}}{4},tan\\angle{\\frac{ACB}{2}}=\\frac{\\sqrt{7}}{7}$. Since the line passes $(5,0)$, we can set the equation to get the $CF:y=-\\frac{\\sqrt{7}}{7}+\\frac{5\\sqrt{7}}{7}$, find the intersection of $CF,EF$,$-\\frac{\\sqrt{7}x}{7}+\\frac{5\\sqrt{7}}{7}=\\frac{\\sqrt{7}x}{7}+\\frac{4\\sqrt{7}}{7}$, getting that $F:(\\frac{1}{2},\\frac{9\\sqrt{7}}{14})$ and in the end we use shoelace theorem with coordinates of $A,F,E$ getting the area $\\frac{15\\sqrt{7}}{14}$ leads to the final answer $36 ~bluesoul",
"Using the Claim (below) we get $I$ is orthocenter of $\\triangle DEF,\\angle EDG = \\beta,$ $\\angle FDG = \\gamma.$ So area of $\\triangle DEF$ is \\[[\\triangle DEF] =\\frac {DG \\cdot FE}{2} = \\frac {AD^2}{8} (\\tan \\beta + \\tan \\gamma).\\] Semiperimeter of $\\triangle ABC \\hspace{10mm} s = \\frac{15}{2},$ so the bisector \\[AD= \\frac{2}{AB + AC}\\sqrt{s(s-BC) \\cdot AB\\cdot AC} = 3\\sqrt{2}.\\] We get the inradius by applying Heron's formula \\[r = \\sqrt{\\frac {(s-AB)(s-BC)(s-AC)}{s}} = \\sqrt {\\frac {3\\cdot 5 \\cdot 7}{15 \\cdot 4}}=\\frac {\\sqrt {7}}{2}.\\] We use formulas for inradius and get \\[\\tan \\beta = \\frac{r}{s – AC} = \\frac {\\sqrt {7}}{3} , \\hspace{10mm} \\tan \\gamma = \\frac{r}{s – AB} = \\frac {\\sqrt {7}}{7}.\\] The area $\\hspace{30mm} [\\triangle DEF] = \\frac{15 \\sqrt 7}{14}.$ Claim Let $I$ be incenter of $\\triangle ABC.$ Then bisector $\\overline{BI},$ perpendicular bisector of $\\overline{AD},$ and perpendicular dropped to bisector $\\overline{CI}$ from point $D$ are concurrent. Proof Denote $\\angle BAC = 2 \\alpha, \\angle ABC = 2\\beta,\\angle ACB = 2 \\gamma.$ Then $\\alpha + \\beta + \\gamma = 90^\\circ.$ Denote $P$ the intersection point of $BC$ and the tangent line to the circumcircle at point $A.$ WLOC,$\\hspace{20mm}\\gamma > \\beta$ (case $\\gamma = \\beta$ is trivial). $\\angle PAC = \\angle ABC = 2 \\beta$ (this angles are measured by half the arc $\\overset{\\Large\\frown} {AC}$ of the circumcircle). \\[\\angle APC = \\angle ACD - \\angle PAC = 2(\\gamma - \\beta),\\] \\[\\angle ADP = 180^o – \\angle DAC - \\angle ACD = 180^o – \\alpha – 2 \\gamma = \\alpha + 2 \\beta = \\angle DAP \\implies AP = DP.\\] Therefore bisector of angle P coincite with the perpendicular bisector of $\\overline{AD}$. By applying the Law of Sines to $\\triangle ABP$ we get $\\hspace{20mm}\\frac {BP}{AP} = \\frac {\\sin 2 \\gamma}{\\sin 2 \\beta.}$ Let $E$ be crosspoint of $PG$ and bisector $BI.$ By applying the Law of Sines to $\\triangle BEP$ we get \\[\\frac {EP}{BP} = \\frac {\\sin \\beta}{\\sin(180^o – \\beta – (\\gamma – \\beta))}= \\frac {\\sin \\beta}{\\sin \\gamma}.\\] Let $E'$ be crosspoint of $PG$ and the perpendicular dropped to bisector $\\overline{CI}$ from point $D.$ \\[\\frac {E'P}{DP} = \\frac {\\sin (90^\\circ – \\gamma)}{\\sin(180^o - (90^o – \\gamma) – (\\gamma - \\beta))}= \\frac {\\cos\\gamma}{\\cos \\beta}.\\] $\\hspace{50mm}\\frac {E'P} {EP} = \\frac {E'P}{DP} \\cdot \\frac {AP}{BP} \\cdot \\frac {BP}{EP} = \\frac {\\cos\\gamma}{\\cos \\beta} \\cdot \\frac {\\sin 2\\beta}{\\sin 2\\gamma}\\cdot \\frac {\\sin\\gamma}{\\sin \\beta} = 1 \\implies E$ coincide with $E'.$ vladimir.shelomovskii@gmail.com, vvsss",
"Using the Claim (below) we get $I$ is orthocenter of $\\triangle DEF,\\angle EDG = \\beta,$ $\\angle FDG = \\gamma.$ So area of $\\triangle DEF$ is \\[[\\triangle DEF] =\\frac {DG \\cdot FE}{2} = \\frac {AD^2}{8} (\\tan \\beta + \\tan \\gamma).\\] Semiperimeter of $\\triangle ABC \\hspace{10mm} s = \\frac{15}{2},$ so the bisector \\[AD= \\frac{2}{AB + AC}\\sqrt{s(s-BC) \\cdot AB\\cdot AC} = 3\\sqrt{2}.\\] We get the inradius by applying Heron's formula \\[r = \\sqrt{\\frac {(s-AB)(s-BC)(s-AC)}{s}} = \\sqrt {\\frac {3\\cdot 5 \\cdot 7}{15 \\cdot 4}}=\\frac {\\sqrt {7}}{2}.\\] We use formulas for inradius and get \\[\\tan \\beta = \\frac{r}{s – AC} = \\frac {\\sqrt {7}}{3} , \\hspace{10mm} \\tan \\gamma = \\frac{r}{s – AB} = \\frac {\\sqrt {7}}{7}.\\] The area $\\hspace{30mm} [\\triangle DEF] = \\frac{15 \\sqrt 7}{14}.$ Claim Let $I$ be incenter of $\\triangle ABC.$ Then bisector $\\overline{BI},$ perpendicular bisector of $\\overline{AD},$ and perpendicular dropped to bisector $\\overline{CI}$ from point $D$ are concurrent. Proof Denote $\\angle BAC = 2 \\alpha, \\angle ABC = 2\\beta,\\angle ACB = 2 \\gamma.$ Then $\\alpha + \\beta + \\gamma = 90^\\circ.$ Denote $P$ the intersection point of $BC$ and the tangent line to the circumcircle at point $A.$ WLOC,$\\hspace{20mm}\\gamma > \\beta$ (case $\\gamma = \\beta$ is trivial). $\\angle PAC = \\angle ABC = 2 \\beta$ (this angles are measured by half the arc $\\overset{\\Large\\frown} {AC}$ of the circumcircle). \\[\\angle APC = \\angle ACD - \\angle PAC = 2(\\gamma - \\beta),\\] \\[\\angle ADP = 180^o – \\angle DAC - \\angle ACD = 180^o – \\alpha – 2 \\gamma = \\alpha + 2 \\beta = \\angle DAP \\implies AP = DP.\\] Therefore bisector of angle P coincite with the perpendicular bisector of $\\overline{AD}$. By applying the Law of Sines to $\\triangle ABP$ we get $\\hspace{20mm}\\frac {BP}{AP} = \\frac {\\sin 2 \\gamma}{\\sin 2 \\beta.}$ Let $E$ be crosspoint of $PG$ and bisector $BI.$ By applying the Law of Sines to $\\triangle BEP$ we get \\[\\frac {EP}{BP} = \\frac {\\sin \\beta}{\\sin(180^o – \\beta – (\\gamma – \\beta))}= \\frac {\\sin \\beta}{\\sin \\gamma}.\\] Let $E'$ be crosspoint of $PG$ and the perpendicular dropped to bisector $\\overline{CI}$ from point $D.$ \\[\\frac {E'P}{DP} = \\frac {\\sin (90^\\circ – \\gamma)}{\\sin(180^o - (90^o – \\gamma) – (\\gamma - \\beta))}= \\frac {\\cos\\gamma}{\\cos \\beta}.\\] $\\hspace{50mm}\\frac {E'P} {EP} = \\frac {E'P}{DP} \\cdot \\frac {AP}{BP} \\cdot \\frac {BP}{EP} = \\frac {\\cos\\gamma}{\\cos \\beta} \\cdot \\frac {\\sin 2\\beta}{\\sin 2\\gamma}\\cdot \\frac {\\sin\\gamma}{\\sin \\beta} = 1 \\implies E$ coincide with $E'.$ vladimir.shelomovskii@gmail.com, vvsss",
"Denote $I$ as the incenter of $\\triangle ABC$ and $M$ as the intersecting point of $AD$ and $EF$. We solve this question based on this equation: $[\\triangle AEF]=\\dfrac{1}{2}\\times AM\\times (\\dfrac{IM}{\\tan{\\angle EFI}}+\\dfrac{IM}{\\tan{\\angle FEI}})$. Firstly, let's compute the trig value of $\\angle EFI$ and $\\angle FEI$. Using the Cosine Law and half-angle formula, we obtain: \\[\\cos{\\angle B}=\\dfrac{1}{8}, \\sin{\\dfrac{\\angle B}{2}}=\\dfrac{\\sqrt{7}}{4}, \\cos{\\dfrac{\\angle B}{2}}=\\dfrac{3}{4}\\] \\[\\cos{\\angle C}=\\dfrac{3}{4}, \\sin{\\dfrac{\\angle C}{2}}=\\dfrac{1}{\\sqrt{8}}, \\cos{\\dfrac{\\angle C}{2}}=\\dfrac{\\sqrt{7}}{\\sqrt{8}}\\] Noticing that by angle chasing, $\\angle EFI= 90^{\\circ}-\\angle FIA=90^{\\circ}-\\angle CID=90^{\\circ}-(180^{\\circ}-\\dfrac{\\angle A}{2}-\\angle B-\\dfrac{\\angle C}{2})= \\dfrac{\\angle B}{2}$. Similarly, $\\angle FEI=\\dfrac{\\angle C}{2}$. Thus, $\\tan{\\angle EFI}=\\dfrac{\\sqrt{7}}{3}$ and $\\tan{\\angle FEI}=\\dfrac{1}{\\sqrt{7}}$ Then we find $AM$: by Stewart's theorem, $AM=\\dfrac{AD}{2}=\\dfrac{3\\sqrt{2}}{2}$. For $MI$, first noticing that $MI=MD-ID=\\dfrac{3\\sqrt{2}}{2}-ID$. Then, by Angel Bisector's theorem, we have $BD=2, CD=3$, and applying Sine Law: \\[\\dfrac{ID}{\\sin{\\dfrac{\\angle C}{2}}}=\\dfrac{CD}{\\sin{\\angle CID}}\\] \\[\\dfrac{ID}{\\sin{\\dfrac{\\angle C}{2}}}=\\dfrac{CD}{\\cos{\\dfrac{\\angle B}{2}}}\\] Substituting the known values, we get $ID=\\sqrt{2}$. Thus, $MI=\\dfrac{\\sqrt{2}}{2}$ Now, we already find all the unknown values in the very beginning equation. Substituting all the values, we obtain $[\\triangle AEF]=\\dfrac{15\\sqrt{7}}{14}$, which gives a final answer of $036. ~Ericcc",
"Denote $I$ as the incenter of $\\triangle ABC$ and $M$ as the intersecting point of $AD$ and $EF$. We solve this question based on this equation: $[\\triangle AEF]=\\dfrac{1}{2}\\times AM\\times (\\dfrac{IM}{\\tan{\\angle EFI}}+\\dfrac{IM}{\\tan{\\angle FEI}})$. Firstly, let's compute the trig value of $\\angle EFI$ and $\\angle FEI$. Using the Cosine Law and half-angle formula, we obtain: \\[\\cos{\\angle B}=\\dfrac{1}{8}, \\sin{\\dfrac{\\angle B}{2}}=\\dfrac{\\sqrt{7}}{4}, \\cos{\\dfrac{\\angle B}{2}}=\\dfrac{3}{4}\\] \\[\\cos{\\angle C}=\\dfrac{3}{4}, \\sin{\\dfrac{\\angle C}{2}}=\\dfrac{1}{\\sqrt{8}}, \\cos{\\dfrac{\\angle C}{2}}=\\dfrac{\\sqrt{7}}{\\sqrt{8}}\\] Noticing that by angle chasing, $\\angle EFI= 90^{\\circ}-\\angle FIA=90^{\\circ}-\\angle CID=90^{\\circ}-(180^{\\circ}-\\dfrac{\\angle A}{2}-\\angle B-\\dfrac{\\angle C}{2})= \\dfrac{\\angle B}{2}$. Similarly, $\\angle FEI=\\dfrac{\\angle C}{2}$. Thus, $\\tan{\\angle EFI}=\\dfrac{\\sqrt{7}}{3}$ and $\\tan{\\angle FEI}=\\dfrac{1}{\\sqrt{7}}$ Then we find $AM$: by Stewart's theorem, $AM=\\dfrac{AD}{2}=\\dfrac{3\\sqrt{2}}{2}$. For $MI$, first noticing that $MI=MD-ID=\\dfrac{3\\sqrt{2}}{2}-ID$. Then, by Angel Bisector's theorem, we have $BD=2, CD=3$, and applying Sine Law: \\[\\dfrac{ID}{\\sin{\\dfrac{\\angle C}{2}}}=\\dfrac{CD}{\\sin{\\angle CID}}\\] \\[\\dfrac{ID}{\\sin{\\dfrac{\\angle C}{2}}}=\\dfrac{CD}{\\cos{\\dfrac{\\angle B}{2}}}\\] Substituting the known values, we get $ID=\\sqrt{2}$. Thus, $MI=\\dfrac{\\sqrt{2}}{2}$ Now, we already find all the unknown values in the very beginning equation. Substituting all the values, we obtain $[\\triangle AEF]=\\dfrac{15\\sqrt{7}}{14}$, which gives a final answer of $036. ~Ericcc"
] |
2020-I-14
| 2,020
| 14
|
Let $P(x)$ be a quadratic polynomial with complex coefficients whose $x^2$ coefficient is $1.$ Suppose the equation $P(P(x))=0$ has four distinct solutions, $x=3,4,a,b.$ Find the sum of all possible values of $(a+b)^2.$
| 85
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I
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[
"Either $P(3) = P(4)$ or not. We first see that if $P(3) = P(4)$ it's easy to obtain by Vieta's that $(a+b)^2 = 49$. Now, take $P(3) \\neq P(4)$ and WLOG $P(3) = P(a), P(4) = P(b)$. Now, consider the parabola formed by the graph of $P$. It has vertex $\\frac{3+a}{2}$. Now, say that $P(x) = x^2 - (3+a)x + c$. We note $P(3)P(4) = c = P(3)\\left(4 - 4a + \\frac{8a - 1}{2}\\right) \\implies a = \\frac{7P(3) + 1}{8}$. Now, we note $P(4) = \\frac{7}{2}$ by plugging in again. Now, it's easy to find that $a = -2.5, b = -3.5$, yielding a value of $36$. Finally, we add $49 + 36 = 085.",
"Let the roots of $P(x)$ be $m$ and $n$, then we can write \\[P(x)=x^2-(m+n)x+mn\\] The fact that $P(P(x))=0$ has solutions $x=3,4,a,b$ implies that some combination of $2$ of these are the solution to $P(x)=m$, and the other $2$ are the solution to $P(x)=n$. It's fairly easy to see there are only $2$ possible such groupings: $P(3)=P(4)=m$ and $P(a)=P(b)=n$, or $P(3)=P(a)=m$ and $P(4)=P(b)=n$ (Note that $a,b$ are interchangeable, and so are $m$ and $n$). We now casework: If $P(3)=P(4)=m$, then \\[9-3(m+n)+mn=16-4(m+n)+mn=m \\implies m+n=7\\] \\[a^2-a(m+n)+mn=b^2-b(m+n)+mn=n \\implies a+b=m+n=7\\] so this gives $(a+b)^2=7^2=49$. Next, if $P(3)=P(a)=m$, then \\[9-3(m+n)+mn=a^2-a(m+n)+mn=m \\implies a+3=m+n\\] \\[16-4(m+n)+mn=b^2-b(m+n)+mn=n \\implies b+4=m+n\\] Subtracting the first part of the first equation from the first part of the second equation gives \\[7-(m+n)=n-m \\implies 2n=7 \\implies n=\\frac{7}{2} \\implies m=-3\\] Hence, $a+b=2(m+n)-7=2\\cdot \\frac{1}{2}-7=-6$, and so $(a+b)^2=(-6)^2=36$. Therefore, the solution is $49+36=085 ~ktong",
"Write $P(x) = x^2+wx+z$. Split the problem into two cases: $P(3)\\ne P(4)$ and $P(3) = P(4)$. Case 1: We have $P(3) \\ne P(4)$. We must have \\[w=-P(3)-P(4) = -(9+3w+z)-(16+4w+z) = -25-7w-2z.\\] Rearrange and divide through by $8$ to obtain \\[w = \\frac{-25-2z}{8}.\\] Now, note that \\[z = P(3)P(4) = (9+3w+z)(16+4w+z) = \\left(9 + 3\\cdot \\frac{-25-2z}{8} + z\\right)\\left(16 + 4 \\cdot \\frac{-25-2z}{8} + z\\right) =\\] \\[\\left(-\\frac{3}{8} + \\frac{z}{4}\\right)\\left(\\frac{7}{2}\\right) = -\\frac{21}{16} + \\frac{7z}{8}.\\] Now, rearrange to get \\[\\frac{z}{8} = -\\frac{21}{16}\\] and thus \\[z = -\\frac{21}{2}.\\] Substituting this into our equation for $w$ yields $w = -\\frac{1}{2}$. Then, it is clear that $P$ does not have a double root at $P(3)$, so we must have $P(a) = P(3)$ and $P(b) = P(4)$ or vice versa. This gives $3+a = \\frac{1}{2}$ and $4+b = \\frac{1}{2}$ or vice versa, implying that $a+b = 1-3-4 = -6$ and $(a+b)^2 = 36$. Case 2: We have $P(3) = P(4)$. Then, we must have $w = -7$. It is clear that $P(a) = P(b)$ (we would otherwise get $P(a)=P(3)=P(4)$ implying $a \\in \\{3,4\\}$ or vice versa), so $a+b=-w=7$ and $(a+b)^2 = 49$. Thus, our final answer is $49+36=085. ~GeronimoStilton",
"Let $P(x)=(x-r)(x-s)$. There are two cases: in the first case, $(3-r)(3-s)=(4-r)(4-s)$ equals $r$ (without loss of generality), and thus $(a-r)(a-s)=(b-r)(b-s)=s$. By Vieta's formulas $a+b=r+s=3+4=7$. In the second case, say without loss of generality $(3-r)(3-s)=r$ and $(4-r)(4-s)=s$. Subtracting gives $-7+r+s=r-s$, so $s=7/2$. From this, we have $r=-3$. Note $r+s=1/2$, so by Vieta's, we have $\\{a,b\\}=\\{1/2-3,1/2-4\\}=\\{-5/2,-7/2\\}$. In this case, $a+b=-6$. The requested sum is $36+49=85$.~TheUltimate123",
"Note that because $P\\big(P(3)\\big)=P\\big(P(4)\\big)= 0$, $P(3)$ and $P(4)$ are roots of $P(x)$. There are two cases. CASE 1: $P(3) = P(4)$. Then $P(x)$ is symmetric about $x=\\tfrac72$; that is to say, $P(r) = P(7-r)$ for all $r$. Thus the remaining two roots must sum to $7$. Indeed, the polynomials $P(x) = \\left(x-\\frac72\\right)^2 + \\frac{11}4 \\pm i\\sqrt3$ satisfy the conditions. CASE 2: $P(3)\\neq P(4)$. Then $P(3)$ and $P(4)$ are the two distinct roots of $P(x)$, so\\[P(x) = \\big(x-P(3)\\big)\\big(x-P(4)\\big)\\]for all $x$. Note that any solution to $P\\big(P(x)\\big) = 0$ must satisfy either $P(x) = P(3)$ or $P(x) = P(4)$. Because $P(x)$ is quadratic, the polynomials $P(x) - P(3)$ and $P(x) - P(4)$ each have the same sum of roots as the polynomial $P(x)$, which is $P(3) + P(4)$. Thus the answer in this case is $2\\big(P(3) + P(4)\\big)-7$, and so it suffices to compute the value of $P(3)+P(4)$. Let $P(3)=u$ and $P(4) = v$. Substituting $x=3$ and $x=4$ into the above quadratic polynomial yields the system of equations \\begin{align*} u &= (3-u)(3-v) = 9 - 3u - 3v + uv\\\\ v &= (4-u)(4-v) = 16 - 4u - 4v + uv. \\end{align*}Subtracting the first equation from the second gives $v - u = 7 - u - v$, yielding $v = \\frac72.$ Substituting this value into the second equation gives\\[\\dfrac72 = \\left(4 - u\\right)\\left(4 - \\dfrac72\\right),\\]yielding $u = -3.$ The sum of the two solutions is $2\\left(\\tfrac72-3\\right)-7 = -6$. In this case, $P(x)= (x+3)\\left(x-\\frac72\\right)$. The requested sum of squares is $7^2+(-6)^2 = 085.",
"Note that because $P\\big(P(3)\\big)=P\\big(P(4)\\big)= 0$, $P(3)$ and $P(4)$ are roots of $P(x)$. There are two cases. CASE 1: $P(3) = P(4)$. Then $P(x)$ is symmetric about $x=\\tfrac72$; that is to say, $P(r) = P(7-r)$ for all $r$. Thus the remaining two roots must sum to $7$. Indeed, the polynomials $P(x) = \\left(x-\\frac72\\right)^2 + \\frac{11}4 \\pm i\\sqrt3$ satisfy the conditions. CASE 2: $P(3)\\neq P(4)$. Then $P(3)$ and $P(4)$ are the two distinct roots of $P(x)$, so\\[P(x) = \\big(x-P(3)\\big)\\big(x-P(4)\\big)\\]for all $x$. Note that any solution to $P\\big(P(x)\\big) = 0$ must satisfy either $P(x) = P(3)$ or $P(x) = P(4)$. Because $P(x)$ is quadratic, the polynomials $P(x) - P(3)$ and $P(x) - P(4)$ each have the same sum of roots as the polynomial $P(x)$, which is $P(3) + P(4)$. Thus the answer in this case is $2\\big(P(3) + P(4)\\big)-7$, and so it suffices to compute the value of $P(3)+P(4)$. Let $P(3)=u$ and $P(4) = v$. Substituting $x=3$ and $x=4$ into the above quadratic polynomial yields the system of equations \\begin{align*} u &= (3-u)(3-v) = 9 - 3u - 3v + uv\\\\ v &= (4-u)(4-v) = 16 - 4u - 4v + uv. \\end{align*}Subtracting the first equation from the second gives $v - u = 7 - u - v$, yielding $v = \\frac72.$ Substituting this value into the second equation gives\\[\\dfrac72 = \\left(4 - u\\right)\\left(4 - \\dfrac72\\right),\\]yielding $u = -3.$ The sum of the two solutions is $2\\left(\\tfrac72-3\\right)-7 = -6$. In this case, $P(x)= (x+3)\\left(x-\\frac72\\right)$. The requested sum of squares is $7^2+(-6)^2 = 085.",
"Let $P(x) = (x-c)^2 - d$ for some $c$, $d$. Then, we can write $P(P(x)) = ((x-c)^2 - d - c)^2 - d$. Setting the expression equal to $0$ and solving for $x$ gives: $x = \\pm \\sqrt{ \\pm \\sqrt{d} + d + c} + c$ Therefore, we have that $x$ takes on the four values $\\sqrt{\\sqrt{d} + d + c} + c$, $-\\sqrt{\\sqrt{d} + d + c} + c$, $\\sqrt{-\\sqrt{d} + d + c} + c$, and $-\\sqrt{-\\sqrt{d} + d + c} + c$. Two of these values are $3$ and $4$, and the other two are $a$ and $b$. We can split these four values into two \"groups\" based on the radicand in the expression - for example, the first group consists of the first two values listed above, and the second group consists of the other two values. $\\textbf{Case 1}$: Both the 3 and 4 values are from the same group. In this case, the $a$ and $b$ values are both from the other group. The sum of this is just $2c$ because the radical cancels out. Because of this, we can see that $c$ is just the average of $3$ and $4$, so we have $2c = 3 + 4 = 7$, so $(a+b)^2 = 7^2 = 49$. $\\textbf{Case 2}$: The 3 and 4 values come from different groups. It is easy to see that all possibilities in this case are basically symmetric and yield the same value for $(a+b)^2$. Without loss of generality, assume that $\\sqrt{\\sqrt{d} + d + c} + c = 4$ and $\\sqrt{-\\sqrt{d} + d + c} + c = 3$. Note that we can't switch the values of these two expressions since the first one is guaranteed to be larger. We can write $\\sqrt{\\sqrt{d} + d + c} + c = 1 + \\sqrt{-\\sqrt{d} + d + c} + c$. Moving most terms to the left side and simplifying gives $\\sqrt{\\sqrt{d} + d + c} - \\sqrt{-\\sqrt{d} + d + c} = 1$. We can square both sides and simplify: $\\sqrt{d} + d + c - \\sqrt{d} + d + c - 2\\sqrt{(d + c + \\sqrt{d})(d + c - \\sqrt{d})} = 1$ $2d + 2c - 2\\sqrt{(d + c + \\sqrt{d})(d + c - \\sqrt{d})} = 1$ $\\sqrt{(d + c + \\sqrt{d})(d + c - \\sqrt{d})} = (d+c) - \\frac{1}{2}$ $\\sqrt{(d+c)^2 - (\\sqrt{d})^2} = (d+c) - \\frac{1}{2}$ $\\sqrt{d^2 + 2dc + c^2 - d} = (d+c) - \\frac{1}{2}$ Squaring both sides again gives the following: $d^2 + 2dc + c^2 - d = d^2 + 2dc + c^2 - d - c + \\frac{1}{4}$ Nearly all terms cancel out, yielding $c = \\frac{1}{4}$. By substituting this back in, we obtain $\\sqrt{\\sqrt{d} + d + c} = \\frac{15}{4}$ and $\\sqrt{-\\sqrt{d} + d + c} = \\frac{11}{4}$. The sum of $a$ and $b$ is equal to $-\\sqrt{\\sqrt{d} + d + c} - \\sqrt{-\\sqrt{d} + d + c} + 2c = -\\frac{15}{4} - \\frac{11}{4} + \\frac{1}{2} = -6$, so $(a+b)^2 = 36$. Adding up both values gives $49 + 36 = 085 as our final answer."
] |
2020-I-15
| 2,020
| 15
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Let $\triangle ABC$ be an acute triangle with circumcircle $\omega,$ and let $H$ be the intersection of the altitudes of $\triangle ABC.$ Suppose the tangent to the circumcircle of $\triangle HBC$ at $H$ intersects $\omega$ at points $X$ and $Y$ with $HA=3,HX=2,$ and $HY=6.$ The area of $\triangle ABC$ can be written in the form $m\sqrt{n},$ where $m$ and $n$ are positive integers, and $n$ is not divisible by the square of any prime. Find $m+n.$
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|
[
"The following is a power of a point solution to this menace of a problem: [asy] defaultpen(fontsize(12)+0.6); size(250); pen p=fontsize(10)+gray+0.4; var phi=75.5, theta=130, r=4.8; pair A=r*dir(270-phi-57), C=r*dir(270+phi-57), B=r*dir(theta-57), H=extension(A,foot(A,B,C),B,foot(B,C,A)); path omega=circumcircle(A,B,C), c=circumcircle(H,B,C); pair O=circumcenter(H,B,C), Op=2*H-O, Z=bisectorpoint(O,Op), X=IP(omega,L(H,Z,0,50)), Y=IP(omega,L(H,Z,50,0)), D=2*H-A, K=extension(B,C,X,Y), E=extension(A,origin,X,Y), L=foot(H,B,C); draw(K--C^^omega^^c^^L(A,D,0,1)); draw(Y--K^^L(A,origin,0,1.5)^^L(X,D,0.7,3.5),fuchsia+0.4); draw(CR(H,length(H-L)),royalblue); draw(C--K,p); dot(\"$A$\",A,dir(120)); dot(\"$B$\",B,down); dot(\"$C$\",C,down); dot(\"$H$\",H,down); dot(\"$X$\",X,up); dot(\"$Y$\",Y,down); dot(\"$O$\",origin,down); dot(\"$O'$\",O,down); dot(\"$K$\",K,up); dot(\"$L$\",L,dir(H-X)); dot(\"$D$\",D,down); dot(\"$E$\",E,down); //draw(O--origin,p); //draw(origin--4*dir(57),fuchsia); //draw(Arc(H,3,160,200),royalblue); //draw(Arc(H,2,30,70),heavygreen); //draw(Arc(H,6,220,240),fuchsia); [/asy] Let points be what they appear as in the diagram below. Note that $3HX = HY$ is not insignificant; from here, we set $XH = HE = \\frac{1}{2} EY = HL = 2$ by PoP and trivial construction. Now, $D$ is the reflection of $A$ over $H$. Note $AO \\perp XY$, and therefore by Pythagorean theorem we have $AE = XD = \\sqrt{5}$. Consider $HD = 3$. We have that $\\triangle HXD \\cong HLK$, and therefore we are ready to PoP with respect to $(BHC)$. Setting $BL = x, LC = y$, we obtain $xy = 10$ by PoP on $(ABC)$, and furthermore, we have $KH^2 = 9 = (KL - x)(KL + y) = (\\sqrt{5} - x)(\\sqrt{5} + y)$. Now, we get $4 = \\sqrt{5}(y - x) - xy$, and from $xy = 10$ we take \\[\\frac{14}{\\sqrt{5}} = y - x.\\] However, squaring and manipulating with $xy = 10$ yields that $(x + y)^2 = \\frac{396}{5}$ and from here, since $AL = 5$ we get the area to be $3\\sqrt{55} \\implies 058. ~awang11's sol",
"As in the diagram, let ray $AH$ extended hits BC at L and the circumcircle at say $P$. By power of the point at H, we have $HX \\cdot HY = AH \\cdot HP$. The three values we are given tells us that $HP=\\frac{2\\cdot 6}{3}=4$. L is the midpoint of $HP$(see here: https://www.cut-the-knot.org/Curriculum/Geometry/AltitudeAndCircumcircle.shtml ), so $HL=LP=2$. As in the diagram provided, let K be the intersection of $BC$ and $XY$. By power of a point on the circumcircle of triangle $HBC$, $KH^{2}=KB \\cdot KC$. By power of a point on the circumcircle of triangle $ABC$, $KB \\cdot KC=KX \\cdot KY$, thus $KH^{2}=(KH-2)(KH+6)$. Solving gives $4KH=12$ or $KH=3$. By the Pythagorean Theorem on triangle $HKL$, $KL=\\sqrt{5}$. Now continue with solution 1.",
"[asy] size(10cm); pair A, B, C, D, H, K, O, P, L, M, X, Y; A = (-15, 27); B = (-24, 0); C = (24, 0); D = (-8.28, 18.04); O = (0, 7); P = (0, -7); H = (-15, 13); K = (-15, -13); M = (0, 0); L = (-15, 0); X = (-24.9569, 5.53234); Y = (8.39688, 30.5477); draw(circle(O, 25)); draw(circle(P, 25)); draw(A--B--C--cycle); draw(H -- K); draw(A -- O -- P -- H -- cycle); draw(X -- Y); draw(O -- X, dashed); draw(O -- Y, dashed); draw(O -- B, dashed); draw(O -- C, dashed); label(\"$O$\", O, ENE); label(\"$A$\", A, NW); label(\"$B$\", B, W); label(\"$C$\", C, E); label(\"$H$\", H, E); label(\"$H'$\", K, NE); label(\"$X$\", X, W); label(\"$Y$\", Y, NE); label(\"$O'$\", P, E); label(\"$M$\", M, NE); label(\"$L$\", L, NE); label(\"$D$\", D, NNE); label(\"$2$\", X -- H, NW); label(\"$3$\", H -- A, SW); label(\"$6$\", H -- Y, NW); label(\"$R$\", O -- Y, E); dot(O); dot(P); dot(D); dot(H); [/asy] Diagram not to scale. We first observe that $H'$, the image of the reflection of $H$ over line $BC$, lies on circle $O$. This is because $\\angle HBC = 90 - \\angle C = \\angle H'AC = \\angle H'BC$. This is a well known lemma. The result of this observation is that circle $O'$, the circumcircle of $\\triangle BHC$ is the image of circle $O$ over line $BC$, which in turn implies that $\\overline{AH} = \\overline{OO'}$ and thus $AHO'O$ is a parallelogram. That $AHO'O$ is a parallelogram implies that $AO$ is perpendicular to $\\overline{XY}$, and thus divides segment $\\overline{XY}$ in two equal pieces, $\\overline{XD}$ and $\\overline{DY}$, of length $4$. Using Power of a Point, \\[\\overline{AH} \\cdot \\overline{HH'} = \\overline{XH} \\cdot \\overline{HY} \\Longrightarrow 3 \\cdot \\overline{HH'} = 2 \\cdot 6 \\Longrightarrow \\overline{HH'} = 4\\] This means that $\\overline{HL} = \\frac12 \\cdot 4 = 2$ and $\\overline{AL} = 2 + 3 = 5$, where $L$ is the foot of the altitude from $A$ onto $BC$. All that remains to be found is the length of segment $\\overline{BC}$. Looking at right triangle $\\triangle AHD$, we find that \\[\\overline{AD} = \\sqrt{\\overline{AH}^2 - \\overline{HD}^2} = \\sqrt{3^2 - 2^2} = \\sqrt{5}\\] Looking at right triangle $\\triangle ODY$, we get the equation \\[\\overline{OY}^2 - \\overline{DY}^2 = \\overline{OD}^2 = \\left(\\overline{AO} - \\overline{AD}\\right)^2\\] Plugging in known values, and letting $R$ be the radius of the circle, we find that \\[R^2 - 16 = (R - \\sqrt{5})^2 = R^2 - 2\\sqrt5 R + 5 \\Longrightarrow R = \\frac{21\\sqrt5}{10}\\] Recall that $AHO'O$ is a parallelogram, so $\\overline{AH} = \\overline{OO'} = 3$. So, $\\overline{OM} = \\frac32$, where $M$ is the midpoint of $\\overline{BC}$. This means that \\[\\overline{BC} = 2\\overline{BM} = 2\\sqrt{R^2 - \\left(\\frac32\\right)^2} = 2\\sqrt{\\frac{441}{20} - \\frac{9}{4}} = \\frac{6\\sqrt{55}}{5}\\] Thus, the area of triangle $\\triangle ABC$ is \\[\\frac{\\overline{AL} \\cdot \\overline{BC}}{2} = \\frac{5 \\cdot \\frac{6\\sqrt{55}}{5}}{2} = {3\\sqrt{55}}\\] The answer is $3 + 55 = 058.",
"Extend $\\overline{AH}$ to intersect $\\omega$ again at $P$. The Power of a Point Theorem yields $HP = \\tfrac{HX \\cdot HY}{HA} = 4$. Because $\\angle CAP=\\angle CBP$, and $\\angle CAP$ and $\\angle CBH$ are both complements to $\\angle C$, it follows that $\\angle CBP = \\angle CBH$, implying that $\\overline{BC}$ bisects $\\overline{HP}$, so the length of the altitude from $A$ to $\\overline{BC}$ is $h_a = AH + \\tfrac12 HP = 5$. Let the circumcircle of $\\triangle BCH$ be $\\omega'$. Because $\\triangle BCH \\cong \\triangle BCP$, the two triangles must have the same circumradius. Because the circumcircle of $\\triangle BCP$ is $\\omega$, the circles $\\omega$ and $\\omega'$ have the same radius $R$. Denote the centers of $\\omega$ and $\\omega'$ by $O$ and $O'$, respectively, and let $M$ be the midpoint of $\\overline{XY}$. Note that trapezoid $HMOO'$ has $\\angle H = \\angle M = 90^\\circ$. Also $HM = XM - XH = \\frac12\\cdot XY - HX = 2$ and $HO' = R$. Because $\\omega$ is a translation of $\\omega'$ in the direction of $\\overline{AH}$, it follows that $OO' = AH = 3$. Finally, the Pythagorean Theorem applied to $\\triangle XMO$ yields $MO = \\sqrt{R^2-16}$. Let $T$ be the projection of $O$ onto $\\overline{HO'}$. Then $TO' = R-MO$, so the Pythagorean Theorem applied to $\\triangle TOO'$ yields \\[R - \\sqrt{R^2-16} = \\sqrt{3^2 - 2^2} = \\sqrt{5}.\\]Solving for $R$ gives $R = \\tfrac{21}{2\\sqrt5}$. It follows from properties of the orthocenter $H$ that\\[\\cos\\angle A = \\dfrac{AH}{2R} = \\dfrac{\\sqrt5}{7},\\]so\\[\\sin\\angle A = \\sqrt{1 - \\cos^2\\angle A} = \\dfrac{2\\sqrt{11}}{7}.\\]Therefore by the Extended Law of Sines\\[a = BC = 2R \\sin\\angle A = \\dfrac{6\\sqrt{11}}{\\sqrt5},\\]so \\[[\\triangle ABC] = \\frac12 a h_a = \\frac12 \\cdot \\frac{6\\sqrt{11}}{\\sqrt{5}} \\cdot 5 = 3\\sqrt{55}.\\]The requested sum is $3+55 = 58$. [asy] unitsize(0.6 cm); pair A, B, C, D, H, M, O, Op, P, T, X, Y, Z; real R = 21/(2*sqrt(5)); A = (7/sqrt(5),7/2); O = (0,0); B = intersectionpoint((0,-3/2)--(R,-3/2),Circle(O,R)); C = intersectionpoint((0,-3/2)--(-R,-3/2),Circle(O,R)); H = A + B + C; P = reflect(B,C)*(H); D = (H + P)/2; Op = reflect(B,C)*(O); X = intersectionpoint(H--(H + scale(2)*rotate(90)*(Op - H)), Circle(O,R)); Y = intersectionpoint(H--(H + scale(2)*rotate(90)*(H - Op)), Circle(O,R)); Z = extension(X, Y, B, C); M = (X + Y)/2; T = H + O - M; draw(Circle(O,R)); draw(Circle(Op,R)); draw(A--B--C--cycle); draw(A--P); draw(B--Z--Y); draw(H--Op--O--M); draw(O--T); draw(O--X); dot(\"$A$\", A, NE); dot(\"$B$\", B, SW); dot(\"$C$\", C, W); dot(\"$D$\", D, SW); dot(\"$H$\", H, NE); dot(\"$M$\", M, NE); dot(\"$O$\", O, W); dot(\"$O'$\", Op, W); dot(\"$P$\", P, SE); dot(\"$T$\", T, N); dot(\"$X$\", X, E); dot(\"$Y$\", Y, NW); dot(\"$Z$\", Z, E); label(\"$\\omega$\", R*dir(140), dir(140)); label(\"$\\omega'$\", Op + R*dir(220), dir(220)); [/asy]",
"Extend $\\overline{AH}$ to intersect $\\omega$ again at $P$. The Power of a Point Theorem yields $HP = \\tfrac{HX \\cdot HY}{HA} = 4$. Because $\\angle CAP=\\angle CBP$, and $\\angle CAP$ and $\\angle CBH$ are both complements to $\\angle C$, it follows that $\\angle CBP = \\angle CBH$, implying that $\\overline{BC}$ bisects $\\overline{HP}$, so the length of the altitude from $A$ to $\\overline{BC}$ is $h_a = AH + \\tfrac12 HP = 5$. Let the circumcircle of $\\triangle BCH$ be $\\omega'$. Because $\\triangle BCH \\cong \\triangle BCP$, the two triangles must have the same circumradius. Because the circumcircle of $\\triangle BCP$ is $\\omega$, the circles $\\omega$ and $\\omega'$ have the same radius $R$. Denote the centers of $\\omega$ and $\\omega'$ by $O$ and $O'$, respectively, and let $M$ be the midpoint of $\\overline{XY}$. Note that trapezoid $HMOO'$ has $\\angle H = \\angle M = 90^\\circ$. Also $HM = XM - XH = \\frac12\\cdot XY - HX = 2$ and $HO' = R$. Because $\\omega$ is a translation of $\\omega'$ in the direction of $\\overline{AH}$, it follows that $OO' = AH = 3$. Finally, the Pythagorean Theorem applied to $\\triangle XMO$ yields $MO = \\sqrt{R^2-16}$. Let $T$ be the projection of $O$ onto $\\overline{HO'}$. Then $TO' = R-MO$, so the Pythagorean Theorem applied to $\\triangle TOO'$ yields \\[R - \\sqrt{R^2-16} = \\sqrt{3^2 - 2^2} = \\sqrt{5}.\\]Solving for $R$ gives $R = \\tfrac{21}{2\\sqrt5}$. It follows from properties of the orthocenter $H$ that\\[\\cos\\angle A = \\dfrac{AH}{2R} = \\dfrac{\\sqrt5}{7},\\]so\\[\\sin\\angle A = \\sqrt{1 - \\cos^2\\angle A} = \\dfrac{2\\sqrt{11}}{7}.\\]Therefore by the Extended Law of Sines\\[a = BC = 2R \\sin\\angle A = \\dfrac{6\\sqrt{11}}{\\sqrt5},\\]so \\[[\\triangle ABC] = \\frac12 a h_a = \\frac12 \\cdot \\frac{6\\sqrt{11}}{\\sqrt{5}} \\cdot 5 = 3\\sqrt{55}.\\]The requested sum is $3+55 = 58$. [asy] unitsize(0.6 cm); pair A, B, C, D, H, M, O, Op, P, T, X, Y, Z; real R = 21/(2*sqrt(5)); A = (7/sqrt(5),7/2); O = (0,0); B = intersectionpoint((0,-3/2)--(R,-3/2),Circle(O,R)); C = intersectionpoint((0,-3/2)--(-R,-3/2),Circle(O,R)); H = A + B + C; P = reflect(B,C)*(H); D = (H + P)/2; Op = reflect(B,C)*(O); X = intersectionpoint(H--(H + scale(2)*rotate(90)*(Op - H)), Circle(O,R)); Y = intersectionpoint(H--(H + scale(2)*rotate(90)*(H - Op)), Circle(O,R)); Z = extension(X, Y, B, C); M = (X + Y)/2; T = H + O - M; draw(Circle(O,R)); draw(Circle(Op,R)); draw(A--B--C--cycle); draw(A--P); draw(B--Z--Y); draw(H--Op--O--M); draw(O--T); draw(O--X); dot(\"$A$\", A, NE); dot(\"$B$\", B, SW); dot(\"$C$\", C, W); dot(\"$D$\", D, SW); dot(\"$H$\", H, NE); dot(\"$M$\", M, NE); dot(\"$O$\", O, W); dot(\"$O'$\", Op, W); dot(\"$P$\", P, SE); dot(\"$T$\", T, N); dot(\"$X$\", X, E); dot(\"$Y$\", Y, NW); dot(\"$Z$\", Z, E); label(\"$\\omega$\", R*dir(140), dir(140)); label(\"$\\omega'$\", Op + R*dir(220), dir(220)); [/asy]",
"Note that $\\overline{BC}$ bisects $\\overline{OO'}$. Using the same method from Solution 3 we find $R = \\frac{21}{2\\sqrt{5}}$. Let the midpoint of $\\overline{BC}$ be $N$, then by the Pythagorean Theorem we have $BN^2 = OB^2 - ON^2 = R^2 - \\frac{3}{2}^2$, so $BN = \\frac{3\\sqrt{11}}{\\sqrt{5}}$. Since $h_a = 5$ we have that the area of ABC is $3\\sqrt{55}$ so the answer is $3 + 55 = 58 ~bobjoebilly",
"Note that $\\overline{BC}$ bisects $\\overline{OO'}$. Using the same method from Solution 3 we find $R = \\frac{21}{2\\sqrt{5}}$. Let the midpoint of $\\overline{BC}$ be $N$, then by the Pythagorean Theorem we have $BN^2 = OB^2 - ON^2 = R^2 - \\frac{3}{2}^2$, so $BN = \\frac{3\\sqrt{11}}{\\sqrt{5}}$. Since $h_a = 5$ we have that the area of ABC is $3\\sqrt{55}$ so the answer is $3 + 55 = 58 ~bobjoebilly",
"Let $D$ be the intersection point of line $AH$ and $\\overline{BC}$, noting that $\\overline{AD}\\perp\\overline{BC}$. Because the area of $\\triangle ABC$ is $\\tfrac12\\cdot AD\\cdot BC$, it suffices to compute $AD$ and $BC$ separately. As in the previous solution, $AD = 5$. The value of $BC$ can be found using the following lemma. Lemma: Triangle $AXY$ is isosceles with base $\\overline{XY}$. Proof: Because the circumcircle of $\\triangle BCH$, $\\omega'$, and $\\omega$ have the same radius, there exists a translation $\\Phi$ sending the former to the latter. Because $\\overline{AH}$ is parallel to the line connecting the centers of the two circles, $\\Phi$ must send $H$ to $A$, meaning $\\Phi$ also sends $\\overline{XY}$ to the tangent to $\\omega$ at $A$. But this means that this tangent is parallel to $\\overline{XY}$, which implies the conclusion. Applying Stewart's Theorem to $\\triangle AXY$ yields\\[AX^2 = AH^2 + HX\\cdot HY = 3^2 + 2\\cdot 6 = 21,\\]implying $AX = AY= \\sqrt{21}.$ By the Law of Cosines \\[\\cos \\angle XAY = \\frac{21 + 21 - 64}{2 \\cdot 21} = -\\frac{11}{21},\\] so\\[\\sin \\angle XAY = \\dfrac{4\\sqrt{20}}{21}.\\] Let $R$ be the radius of $\\omega$. By the Extended Law of Sines\\[R = \\frac{XY}{2 \\cdot \\sin \\angle XAY} = \\dfrac{21}{\\sqrt{20}}.\\] Then the solution proceeds as in Solution 3.",
"Let $D$ be the intersection point of line $AH$ and $\\overline{BC}$, noting that $\\overline{AD}\\perp\\overline{BC}$. Because the area of $\\triangle ABC$ is $\\tfrac12\\cdot AD\\cdot BC$, it suffices to compute $AD$ and $BC$ separately. As in the previous solution, $AD = 5$. The value of $BC$ can be found using the following lemma. Lemma: Triangle $AXY$ is isosceles with base $\\overline{XY}$. Proof: Because the circumcircle of $\\triangle BCH$, $\\omega'$, and $\\omega$ have the same radius, there exists a translation $\\Phi$ sending the former to the latter. Because $\\overline{AH}$ is parallel to the line connecting the centers of the two circles, $\\Phi$ must send $H$ to $A$, meaning $\\Phi$ also sends $\\overline{XY}$ to the tangent to $\\omega$ at $A$. But this means that this tangent is parallel to $\\overline{XY}$, which implies the conclusion. Applying Stewart's Theorem to $\\triangle AXY$ yields\\[AX^2 = AH^2 + HX\\cdot HY = 3^2 + 2\\cdot 6 = 21,\\]implying $AX = AY= \\sqrt{21}.$ By the Law of Cosines \\[\\cos \\angle XAY = \\frac{21 + 21 - 64}{2 \\cdot 21} = -\\frac{11}{21},\\] so\\[\\sin \\angle XAY = \\dfrac{4\\sqrt{20}}{21}.\\] Let $R$ be the radius of $\\omega$. By the Extended Law of Sines\\[R = \\frac{XY}{2 \\cdot \\sin \\angle XAY} = \\dfrac{21}{\\sqrt{20}}.\\] Then the solution proceeds as in Solution 3.",
"Define points $D$ and $P$ as above, and note that $AD=5$ and $DH=PD=AD - AH = 2$. Let the circumcircle of $\\triangle BCH$ be $\\omega'$. Extend $\\overline{XY}$ past $X$ until it intersects line $BC$ at point $Z$. Because line $BC$ is a radical axis of $\\omega$ and $\\omega'$, it follows from the Power of a Point Theorem that \\[ZX \\cdot ZY = ZX \\cdot(ZX + 8) = ZH^2 = (ZX+2)^2,\\]from which $ZX=1$. By Pythagorean Theorem $ZD=\\sqrt{ZH^2 - DH^2} = \\sqrt{5}$. Let $m=CD$ and $n=BD$. By the Power of a Point Theorem \\[mn= AD\\cdot PD = 10.\\]On the other hand, \\[ZH^2 = 9 = ZB \\cdot ZC = (\\sqrt{5} - n)(\\sqrt{5} + m) = 5 + \\sqrt{5}(m-n) - mn,\\]from which $m-n = \\frac{14}{\\sqrt{5}}$. Therefore \\[(m+n)^2 = (m-n)^2 + 4mn = \\frac{196}{5} + 40 = \\frac{396}{5}.\\] Thus $BC = m+n=\\sqrt{\\frac{396}{5}} = 6\\sqrt{\\frac{11}{5}}$. Therefore $[\\triangle ABC] = \\frac{1}{2}BC \\cdot AD = 3\\sqrt{55},$ as above.",
"Define points $D$ and $P$ as above, and note that $AD=5$ and $DH=PD=AD - AH = 2$. Let the circumcircle of $\\triangle BCH$ be $\\omega'$. Extend $\\overline{XY}$ past $X$ until it intersects line $BC$ at point $Z$. Because line $BC$ is a radical axis of $\\omega$ and $\\omega'$, it follows from the Power of a Point Theorem that \\[ZX \\cdot ZY = ZX \\cdot(ZX + 8) = ZH^2 = (ZX+2)^2,\\]from which $ZX=1$. By Pythagorean Theorem $ZD=\\sqrt{ZH^2 - DH^2} = \\sqrt{5}$. Let $m=CD$ and $n=BD$. By the Power of a Point Theorem \\[mn= AD\\cdot PD = 10.\\]On the other hand, \\[ZH^2 = 9 = ZB \\cdot ZC = (\\sqrt{5} - n)(\\sqrt{5} + m) = 5 + \\sqrt{5}(m-n) - mn,\\]from which $m-n = \\frac{14}{\\sqrt{5}}$. Therefore \\[(m+n)^2 = (m-n)^2 + 4mn = \\frac{196}{5} + 40 = \\frac{396}{5}.\\] Thus $BC = m+n=\\sqrt{\\frac{396}{5}} = 6\\sqrt{\\frac{11}{5}}$. Therefore $[\\triangle ABC] = \\frac{1}{2}BC \\cdot AD = 3\\sqrt{55},$ as above.",
"Let $O$ be circumcenter of $ABC,$ let $R$ be circumradius of $ABC,$ let $\\omega'$ be the image of circle $\\omega$ over line $BC$ (the circumcircle of $HBC$). Let $P$ be the image of the reflection of $H$ over line $BC, P$ lies on circle $\\omega.$ Let $M$ be the midpoint of $XY.$ Then $P$ lies on $\\omega, OA = O'H, OA || O'H.$ (see here: https://brilliant.org/wiki/triangles-orthocenter/ or https://en.wikipedia.org/wiki/Altitude_(triangle) Russian) $P$ lies on $\\omega \\implies OA = OP = R,$ $OA = O'H, OA || O'H \\implies M$ lies on $OA.$ We use properties of crossing chords and get \\[AH \\cdot HP = XH \\cdot HY = 2 \\cdot 6 \\implies HP = 4, AP = AH + HP = 7.\\] We use properties of radius perpendicular chord and get \\[MH = \\frac{XH + HY}{2} – HY = 2.\\] We find \\[\\sin OAH =\\frac{MH}{AH} = \\frac{2}{3} \\implies \\cos OAH = \\frac{\\sqrt{5}}{3}.\\] We use properties of isosceles $\\triangle OAP$ and find $\\hspace{5mm}R = \\frac{AP}{2\\cos OAP} = \\frac{7}{2\\frac {\\sqrt{5}}{3}} = \\frac{21}{2\\sqrt{5}}.$ We use $OM' = \\frac{AH}{2} = \\frac {3}{2}$ and find $\\hspace{25mm} \\frac{BC}{2} = \\sqrt{R^2 – OM'^2} = 3 \\sqrt {\\frac {11}{5}}.$ The area of $ABC$ \\[[ABC]=\\frac{BC}{2} \\cdot (AH + HD) = 3\\cdot \\sqrt{55} \\implies 3+55 = \\boldsymbol{058}.\\] vladimir.shelomovskii@gmail.com, vvsss"
] |
2020-II-1
| 2,020
| 1
|
Find the number of ordered pairs of positive integers $(m,n)$ such that ${m^2n = 20 ^{20}}$ .
| 231
|
II
|
[
"In this problem, we want to find the number of ordered pairs $(m, n)$ such that $m^2n = 20^{20}$. Let $x = m^2$. Therefore, we want two numbers, $x$ and $n$, such that their product is $20^{20}$ and $x$ is a perfect square. Note that there is exactly one valid $n$ for a unique $x$, which is $\\tfrac{20^{20}}{x}$. This reduces the problem to finding the number of unique perfect square factors of $20^{20}$. $20^{20} = 2^{40} \\cdot 5^{20} = \\left(2^2\\right)^{20}\\cdot\\left(5^2\\right)^{10}.$ Therefore, the answer is $21 \\cdot 11 = 231 ~superagh ~TheBeast5520",
"Because $20^{20}=2^{40}5^{20}$, if $m^2n = 20^{20}$, there must be nonnegative integers $a$, $b$, $c$, and $d$ such that $m = 2^a5^b$ and $n = 2^c5^d$. Then \\[2a + c = 40\\] and \\[2b+d = 20\\] The first equation has $21$ solutions corresponding to $a = 0,1,2,\\dots,20$, and the second equation has $11$ solutions corresponding to $b = 0,1,2,\\dots,10$. Therefore there are a total of $21\\cdot11 = 231.",
"Because $20^{20}=2^{40}5^{20}$, if $m^2n = 20^{20}$, there must be nonnegative integers $a$, $b$, $c$, and $d$ such that $m = 2^a5^b$ and $n = 2^c5^d$. Then \\[2a + c = 40\\] and \\[2b+d = 20\\] The first equation has $21$ solutions corresponding to $a = 0,1,2,\\dots,20$, and the second equation has $11$ solutions corresponding to $b = 0,1,2,\\dots,10$. Therefore there are a total of $21\\cdot11 = 231."
] |
2020-II-2
| 2,020
| 2
|
Let $P$ be a point chosen uniformly at random in the interior of the unit square with vertices at $(0,0), (1,0), (1,1)$ , and $(0,1)$ . The probability that the slope of the line determined by $P$ and the point $\left(\frac58, \frac38 \right)$ is greater than or equal to $\frac12$ can be written as $\frac{m}{n}$ , where $m$ and $n$ are relatively prime positive integers. Find $m+n$ .
| 171
|
II
|
[
"The areas bounded by the unit square and alternately bounded by the lines through $\\left(\\frac{5}{8},\\frac{3}{8}\\right)$ that are vertical or have a slope of $1/2$ show where $P$ can be placed to satisfy the condition. One of the areas is a trapezoid with bases $1/16$ and $3/8$ and height $5/8$. The other area is a trapezoid with bases $7/16$ and $5/8$ and height $3/8$. Then, \\[\\frac{\\frac{1}{16}+\\frac{3}{8}}{2}\\cdot\\frac{5}{8}+\\frac{\\frac{7}{16}+\\frac{5}{8}}{2}\\cdot\\frac{3}{8}=\\frac{86}{256}=\\frac{43}{128}\\implies43+128=171\\] ~mn28407",
"The line through the fixed point $\\left(\\frac58,\\frac38\\right)$ with slope $\\frac12$ has equation $y=\\frac12 x + \\frac1{16}$. The slope between $P$ and the fixed point exceeds $\\frac12$ if $P$ falls within the shaded region in the diagram below consisting of two trapezoids with area \\[\\frac{\\frac1{16}+\\frac38}2\\cdot\\frac58 + \\frac{\\frac58+\\frac7{16}}2\\cdot\\frac38 = \\frac{43}{128}.\\]Because the entire square has area $1,$ the required probability is $\\frac{43}{128}$. The requested sum is $43+128 = 171$. [asy] defaultpen(fontsize(8pt)); unitsize(4cm); pair A = (0,0); pair B = (1,0); pair C = (1,1); pair D = (0,1); pair F = (0, 1/16); pair G = (1, 9/16); pair H = (5/8, 0); pair J = (5/8, 1); pair K = IP(H--J, F--G); pair P = (13/16, 12/16); pair Q = extension(P,K,A,B); pair R = extension(K,P,C,D); draw(A--B--C--D--cycle); label(\"$(0,0)$\", A, SW); label(\"$(1,0)$\", B, SE); label(\"$(1,1)$\", C, E); label(\"$(0,1)$\", D, W); filldraw(A--H--K--F--cycle, lightgray); filldraw(K--G--C--J--cycle, lightgray); dot(K); dot(\"$P$\", P, W); draw(Q -- R, dashed); label(\"$\\frac 38$\", H--K, E); label(\"$\\frac 58$\", K--J, W); label(\"$\\frac 7{16}$\", G--C, E); label(\"$\\frac 38$\", C--J, N); label(\"$\\frac 1{16}$\", A--F, dir(160)); [/asy]",
"The line through the fixed point $\\left(\\frac58,\\frac38\\right)$ with slope $\\frac12$ has equation $y=\\frac12 x + \\frac1{16}$. The slope between $P$ and the fixed point exceeds $\\frac12$ if $P$ falls within the shaded region in the diagram below consisting of two trapezoids with area \\[\\frac{\\frac1{16}+\\frac38}2\\cdot\\frac58 + \\frac{\\frac58+\\frac7{16}}2\\cdot\\frac38 = \\frac{43}{128}.\\]Because the entire square has area $1,$ the required probability is $\\frac{43}{128}$. The requested sum is $43+128 = 171$. [asy] defaultpen(fontsize(8pt)); unitsize(4cm); pair A = (0,0); pair B = (1,0); pair C = (1,1); pair D = (0,1); pair F = (0, 1/16); pair G = (1, 9/16); pair H = (5/8, 0); pair J = (5/8, 1); pair K = IP(H--J, F--G); pair P = (13/16, 12/16); pair Q = extension(P,K,A,B); pair R = extension(K,P,C,D); draw(A--B--C--D--cycle); label(\"$(0,0)$\", A, SW); label(\"$(1,0)$\", B, SE); label(\"$(1,1)$\", C, E); label(\"$(0,1)$\", D, W); filldraw(A--H--K--F--cycle, lightgray); filldraw(K--G--C--J--cycle, lightgray); dot(K); dot(\"$P$\", P, W); draw(Q -- R, dashed); label(\"$\\frac 38$\", H--K, E); label(\"$\\frac 58$\", K--J, W); label(\"$\\frac 7{16}$\", G--C, E); label(\"$\\frac 38$\", C--J, N); label(\"$\\frac 1{16}$\", A--F, dir(160)); [/asy]"
] |
2020-II-3
| 2,020
| 3
|
The value of $x$ that satisfies $\log_{2^x} 3^{20} = \log_{2^{x+3}} 3^{2020}$ can be written as $\frac{m}{n}$ , where $m$ and $n$ are relatively prime positive integers. Find $m+n$ .
| 103
|
II
|
[
"Let $\\log _{2^x}3^{20}=\\log _{2^{x+3}}3^{2020}=n$. Based on the equation, we get $(2^x)^n=3^{20}$ and $(2^{x+3})^n=3^{2020}$. Expanding the second equation, we get $8^n\\cdot2^{xn}=3^{2020}$. Substituting the first equation in, we get $8^n\\cdot3^{20}=3^{2020}$, so $8^n=3^{2000}$. Taking the 100th root, we get $8^{\\frac{n}{100}}=3^{20}$. Therefore, $(2^{\\frac{3}{100}})^n=3^{20}$, and using the our first equation($2^{xn}=3^{20}$), we get $x=\\frac{3}{100}$ and the answer is $103. ~rayfish",
"Because $\\log_a{b^c}=c\\log_a{b},$ we have that $20\\log_{2^x} 3 = 2020\\log_{2^{x+3}} 3,$ or $\\log_{2^x} 3 = 101\\log_{2^{x+3}} 3.$ Since $\\log_a{b}=\\dfrac{1}{\\log_b{a}},$ $\\log_{2^x} 3=\\dfrac{1}{\\log_{3} 2^x},$ and $101\\log_{2^{x+3}} 3=101\\dfrac{1}{\\log_{3}2^{x+3}},$ thus resulting in $\\log_{3}2^{x+3}=101\\log_{3} 2^x,$ or $\\log_{3}2^{x+3}=\\log_{3} 2^{101x}.$ We remove the base 3 logarithm and the power of 2 to yield $x+3=101x,$ or $x=\\dfrac{3}{100}.$ Our answer is $3+100=103 ~ OreoChocolate",
"Using the Change of Base Formula to convert the logarithms in the given equation to base $2$ yields \\[\\frac{\\log_2 3^{20}}{\\log_2 2^x} = \\frac{\\log_2 3^{2020}}{\\log_2 2^{x+3}}, \\text{~ and then ~} \\frac{20\\log_2 3}{x\\cdot\\log_2 2} = \\frac{2020\\log_2 3}{(x+3)\\log_2 2}.\\]Canceling the logarithm factors then yields\\[\\frac{20}x = \\frac{2020}{x+3},\\]which has solution $x = \\frac3{100}.$ The requested sum is $3 + 100 = 103$.",
"Using the Change of Base Formula to convert the logarithms in the given equation to base $2$ yields \\[\\frac{\\log_2 3^{20}}{\\log_2 2^x} = \\frac{\\log_2 3^{2020}}{\\log_2 2^{x+3}}, \\text{~ and then ~} \\frac{20\\log_2 3}{x\\cdot\\log_2 2} = \\frac{2020\\log_2 3}{(x+3)\\log_2 2}.\\]Canceling the logarithm factors then yields\\[\\frac{20}x = \\frac{2020}{x+3},\\]which has solution $x = \\frac3{100}.$ The requested sum is $3 + 100 = 103$.",
"$\\log_{2^x} 3^{20} = 2^{xy} = 3^{20}$ $\\log_{2^{x+3}} 3^{2020} = (2^{x+3})^y = 3^{2020}$ $3^{2020} = (3^{20})^{101}$ $(2^{xy})^{101} = (2^{x+3})^y = 3^{2020}$ $(2^{xy})^{101} = (2^{x+3})^y \\Rightarrow 2^{101xy} = 2^{xy+3y} \\Rightarrow 101xy = xy + 3y \\Rightarrow 101xy = y(x+3)$ $101x = x + 3$ $100x = 3$ $x = \\frac{3}{100}$ $100 + 3 = 103 ~Airplanes2007"
] |
2020-II-4
| 2,020
| 4
|
Triangles $\triangle ABC$ and $\triangle A'B'C'$ lie in the coordinate plane with vertices $A(0,0)$ , $B(0,12)$ , $C(16,0)$ , $A'(24,18)$ , $B'(36,18)$ , $C'(24,2)$ . A rotation of $m$ degrees clockwise around the point $(x,y)$ where $0<m<180$ , will transform $\triangle ABC$ to $\triangle A'B'C'$ . Find $m+x+y$ .
| 108
|
II
|
[
"After sketching, it is clear a $90^{\\circ}$ rotation is done about $(x,y)$. Looking between $A$ and $A'$, $x+y=18$. Thus $90+18=108. ~mn28407",
"Because the rotation sends the vertical segment $\\overline{AB}$ to the horizontal segment $\\overline{A'B'}$, the angle of rotation is $90^\\circ$ degrees clockwise. For any point $(x,y)$ not at the origin, the line segments from $(0,0)$ to $(x,y)$ and from $(x,y)$ to $(x-y,y+x)$ are perpendicular and are the same length. Thus a $90^\\circ$ clockwise rotation around the point $(x,y)$ sends the point $A(0,0)$ to the point $(x-y,y+x) = A'(24,18)$. This has the solution $(x,y) = (21,-3)$. The requested sum is $90+21-3=108$.",
"Because the rotation sends the vertical segment $\\overline{AB}$ to the horizontal segment $\\overline{A'B'}$, the angle of rotation is $90^\\circ$ degrees clockwise. For any point $(x,y)$ not at the origin, the line segments from $(0,0)$ to $(x,y)$ and from $(x,y)$ to $(x-y,y+x)$ are perpendicular and are the same length. Thus a $90^\\circ$ clockwise rotation around the point $(x,y)$ sends the point $A(0,0)$ to the point $(x-y,y+x) = A'(24,18)$. This has the solution $(x,y) = (21,-3)$. The requested sum is $90+21-3=108$.",
"[asy] /* Geogebra to Asymptote conversion by samrocksnature, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ real labelscalefactor = 0.5; /* changes label-to-point distance */ pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */ pen dotstyle = black; /* point style */ real xmin = -8.0451801958033, xmax = 47.246151591238494, ymin = -10.271454747548662, ymax = 21.426040258770957; /* image dimensions */ pen zzttqq = rgb(0.6,0.2,0); pen qqwuqq = rgb(0,0.39215686274509803,0); pen qqffff = rgb(0,1,1); draw((16,0)--(0,0)--(0,12)--cycle, linewidth(2) + zzttqq); draw((24,2)--(24,18)--(36,18)--cycle, linewidth(2) + blue); draw((16,0)--(21,-3)--(24,2)--cycle, linewidth(2) + qqwuqq); draw((21.39134584768662,-2.3477569205223032)--(20.73910276820892,-1.9564110728356852)--(20.347756920522304,-2.608654152313382)--(21,-3)--cycle, linewidth(2) + qqffff); /* draw figures */ draw((16,0)--(0,0), linewidth(2) + zzttqq); draw((0,0)--(0,12), linewidth(2) + zzttqq); draw((0,12)--(16,0), linewidth(2) + zzttqq); draw((24,2)--(24,18), linewidth(2) + blue); draw((24,18)--(36,18), linewidth(2) + blue); draw((36,18)--(24,2), linewidth(2) + blue); draw((16,0)--(24,2), linewidth(2)); draw((16,0)--(21,-3), linewidth(2) + qqwuqq); draw((21,-3)--(24,2), linewidth(2) + qqwuqq); draw((24,2)--(16,0), linewidth(2) + qqwuqq); draw((21,-3)--(20,1), linewidth(2.8) + qqffff); /* dots and labels */ dot((0,0),linewidth(4pt) + dotstyle); label(\"$A$\", (-0.6228029714727868,0.12704474547474198), NE * labelscalefactor); dot((0,12),dotstyle); label(\"$B$\", (0.1301918194013232,12.354245873478124), NE * labelscalefactor); dot((16,0),dotstyle); label(\"$C$\", (16.15822379657881,0.34218611429591583), NE * labelscalefactor); dot((24,18),dotstyle); label(\"$A'$\", (24.154311337765787,18.342347305667463), NE * labelscalefactor); dot((24,2),dotstyle); label(\"$C'$\", (23.186175178070503,1.95574638045472), NE * labelscalefactor); dot((36,18),dotstyle); label(\"$B'$\", (36.13051420214449,18.342347305667463), NE * labelscalefactor); dot((21,-3),dotstyle); label(\"$P$\", (21.35747354309052,-3.458644734878156), NE * labelscalefactor); dot((20,1),linewidth(4pt) + dotstyle); label(\"$D$\", (20.13833911977053,1.2744653791876692), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy] We first draw a diagram with the correct Cartesian coordinates and a center of rotation $P$. Note that $PC=PC'$ because $P$ lies on the perpendicular bisector of $CC'$ (it must be equidistant from $C$ and $C'$ by properties of a rotation). Since $AB$ is vertical while $A'B'$ is horizontal, we have that the angle of rotation must be $90^{\\circ}$, and therefore $\\angle P = 90^{\\circ}$. Therefore, $CPC'$ is a 45-45-90 right triangle, and $CD=DP$. We calculate $D$ to be $(20,1)$. Since we translate $4$ right and $1$ up to get from point $C$ to point $D$, we must translate $1$ right and $4$ down to get to point $P$. This gives us $P(21,-3)$. Our answer is then $90+21-3=108. ~Lopkiloinm & samrocksnature",
"For the above reasons, the transformation is simply a $90^\\circ$ rotation. Proceed with complex numbers on the points $C$ and $C'$. Let $(x, y)$ be the origin. Thus, $C \\rightarrow (16-x)+(-y)i$ and $C' \\rightarrow (24-x)+(2-y)i$. The transformation from $C'$ to $C$ is a multiplication of $i$, which yields $(16-x)+(-y)i=(y-2)+(24-x)i$. Equating the real and complex terms results in the equations $16-x=y-2$ and $-y=24-x$. Solving, $(x, y) : (21, -3) \\rightarrow 90+21-3=108 ~beastgert",
"We know that the rotation point $P$ has to be equidistant from both $A$ and $A'$ so it has to lie on the line that is on the midpoint of the segment $AA'$ and also the line has to be perpendicular to $AA'$. Solving, we get the line is $y=\\frac{-4}{3}x+25$. Doing the same for $B$ and $B'$, we get that $y=-6x+123$. Since the point $P$ of rotation must lie on both of these lines, we set them equal, solve and get: $x=21$,$y=-3$. We can also easily see that the degree of rotation is $90$ since $AB$ is initially vertical, and now it is horizontal. Also, we can just sketch this on a coordinate plane and easily realize the same. Hence, the answer is $21-3+90 = 108",
"We make transformation of line $AB$ into line $A'B'$ using axes symmetry. Point $D(0,18)$ is the crosspoint of this lines. Equation of line $DO$ is \\[x + y = 18.\\] $\\triangle ABC$ maps into $\\triangle A''B''C''$ where \\[A''(18,18), B''(6,18), C''(18,2).\\] We make transform of the line $A''C''$ into line $A'C'$ using axes symmetry with respect to line \\[x = \\frac {A'' + A'}{2} = \\frac {24 + 18}{2} = 21.\\] The composition of two axial symmetries is a rotation through an angle twice as large as the angle between the axes $(45^o)$ around the point of their intersection $O(21, – 3).$ \\[m + x + y = 90 + 21 – 3 = 108\\]. vladimir.shelomovskii@gmail.com, vvsss",
"For a matrix to rotate a figure on a coordinate plane by $m$ degrees, it is written as: $\\left[ {\\begin{array}{cc} cos(m^{\\circ}) & sin(m^{\\circ}) \\\\ -sin(m^{\\circ}) & cos(m^{\\circ}) \\\\ \\end{array} } \\right]$ We can translate the whole figure so that the centre of rotation is at $(0,0)$, which is equivalent to subtracting $x$ and $y$ from all $x$-coordinates and the $y$-coordinates respectively of the given points. We then record all the points $A$, $B$, $C$ in a matrix as follows: $\\left[ {\\begin{array}{ccc} 0-x & 0-x & 16-x \\\\ 0-y & 12-y & 0-y \\\\ \\end{array} } \\right]$ and all the points $A'$, $B'$, $C'$ in a matrix as follows: $\\left[ {\\begin{array}{ccc} 24-x & 36-x & 24-x \\\\ 18-y & 18-y & 2-y \\\\ \\end{array} } \\right]$ Since $\\triangle A'B'C'$ is a rotation around $(x,y)$ of $\\triangle ABC$ by $m^{\\circ}$, by the left multiplication rule, we can equate that: $\\left[ {\\begin{array}{cc} cos(m^{\\circ}) & sin(m^{\\circ}) \\\\ -sin(m^{\\circ}) & cos(m^{\\circ}) \\\\ \\end{array} } \\right]$ $\\left[ {\\begin{array}{ccc} 0-x & 0-x & 16-x \\\\ 0-y & 12-y & 0-y \\\\ \\end{array} } \\right]$ $=$ $\\left[ {\\begin{array}{ccc} 24-x & 36-x & 24-x \\\\ 18-y & 18-y & 2-y \\\\ \\end{array} } \\right]$ We can obtain the follow equations: $\\begin{cases} -xcos(m^{\\circ})-ysin(m^{\\circ})=24-x \\\\ -xcos(m^{\\circ})-ysin(m^{\\circ})+12sin(m^{\\circ})=36-x \\\\ xsin(m^{\\circ})-ycos(m^{\\circ})=24-x \\end{cases}$ From the first 2 equations, we get $m=90$, substituting into the 3rd equation, we get $x+y=18$. Therefore, $m+x+y=90+18=108 ~VitalsBat",
"For a matrix to rotate a figure on a coordinate plane by $m$ degrees, it is written as: $\\left[ {\\begin{array}{cc} cos(m^{\\circ}) & sin(m^{\\circ}) \\\\ -sin(m^{\\circ}) & cos(m^{\\circ}) \\\\ \\end{array} } \\right]$ We can translate the whole figure so that the centre of rotation is at $(0,0)$, which is equivalent to subtracting $x$ and $y$ from all $x$-coordinates and the $y$-coordinates respectively of the given points. We then record all the points $A$, $B$, $C$ in a matrix as follows: $\\left[ {\\begin{array}{ccc} 0-x & 0-x & 16-x \\\\ 0-y & 12-y & 0-y \\\\ \\end{array} } \\right]$ and all the points $A'$, $B'$, $C'$ in a matrix as follows: $\\left[ {\\begin{array}{ccc} 24-x & 36-x & 24-x \\\\ 18-y & 18-y & 2-y \\\\ \\end{array} } \\right]$ Since $\\triangle A'B'C'$ is a rotation around $(x,y)$ of $\\triangle ABC$ by $m^{\\circ}$, by the left multiplication rule, we can equate that: $\\left[ {\\begin{array}{cc} cos(m^{\\circ}) & sin(m^{\\circ}) \\\\ -sin(m^{\\circ}) & cos(m^{\\circ}) \\\\ \\end{array} } \\right]$ $\\left[ {\\begin{array}{ccc} 0-x & 0-x & 16-x \\\\ 0-y & 12-y & 0-y \\\\ \\end{array} } \\right]$ $=$ $\\left[ {\\begin{array}{ccc} 24-x & 36-x & 24-x \\\\ 18-y & 18-y & 2-y \\\\ \\end{array} } \\right]$ We can obtain the follow equations: $\\begin{cases} -xcos(m^{\\circ})-ysin(m^{\\circ})=24-x \\\\ -xcos(m^{\\circ})-ysin(m^{\\circ})+12sin(m^{\\circ})=36-x \\\\ xsin(m^{\\circ})-ycos(m^{\\circ})=24-x \\end{cases}$ From the first 2 equations, we get $m=90$, substituting into the 3rd equation, we get $x+y=18$. Therefore, $m+x+y=90+18=108 ~VitalsBat",
"It is clear that $\\bigtriangleup CPC'$ is a $45-45-90$ right triangle so $m=90$. We use the $\\tan$ angle formula, $\\tan{(a-b)}=\\frac{\\tan(a)-\\tan(b)}{1+\\tan(a)\\tan(b)}$ to find the slope of line $CP$. We know that line $CC'$ has slope $\\frac{1}{4}$ and let $b=-45^{\\circ}$, then plugging both values into the formula, we find that the slope of $CP$ is $\\frac{-3}{5}$. Also, $CC'$ has length $\\sqrt{34}$. Create a right triangle $KCP$ where $KP$ is parallel to the $x$ axis and $CP$ is the hypotenuse. Then $CK=3x$ and $KP=5x$ and doing Pythagorean on $\\bigtriangleup KCP$ gives $x=1$. Therefore, we know that $P$ is a translation 3 units down and 5 units right from $C(16,0)$, from which we obtain $P(21,-3)$. Adding the three variables, we obtain $90+21-3=108 ~Magnetoninja"
] |
2020-II-5
| 2,020
| 5
|
For each positive integer $n$ , let $f(n)$ be the sum of the digits in the base-four representation of $n$ and let $g(n)$ be the sum of the digits in the base-eight representation of $f(n)$ . For example, $f(2020) = f(133210_{\text{4}}) = 10 = 12_{\text{8}}$ , and $g(2020) = \text{the digit sum of }12_{\text{8}} = 3$ . Let $N$ be the least value of $n$ such that the base-sixteen representation of $g(n)$ cannot be expressed using only the digits $0$ through $9$ . Find the remainder when $N$ is divided by $1000$ .
| 151
|
II
|
[
"Let's work backwards. The minimum base-sixteen representation of $g(n)$ that cannot be expressed using only the digits $0$ through $9$ is $A_{16}$, which is equal to $10$ in base 10. Thus, the sum of the digits of the base-eight representation of the sum of the digits of $f(n)$ is $10$. The minimum value for which this is achieved is $37_8$. We have that $37_8 = 31$. Thus, the sum of the digits of the base-four representation of $n$ is $31$. The minimum value for which this is achieved is $13,333,333,333_4$. We just need this value in base 10 modulo 1000. We get $13,333,333,333_4 = 3(1 + 4 + 4^2 + \\dots + 4^8 + 4^9) + 4^{10} = 3\\left(\\dfrac{4^{10} - 1}{3}\\right) + 4^{10} = 2*4^{10} - 1$. Taking this value modulo $1000$, we get the final answer of $151) ~ TopNotchMath",
"First note that if $h_b(s)$ is the least positive integer whose digit sum, in some fixed base $b$, is $s$, then $h_b$ is a strictly increasing function. This together with the fact that $g(N) \\ge 10$ shows that $f(N)$ is the least positive integer whose base-eight digit sum is 10. Thus $f(N) = 37_\\text{eight} = 31$, and $N$ is the least positive integer whose base-four digit sum is $31.$ Therefore \\[N = 13333333333_\\text{four} = 2\\cdot4^{10} - 1 = 2\\cdot1024^2 - 1\\] \\[\\equiv 2\\cdot24^2 - 1 \\equiv 151 \\pmod{1000}.\\]",
"First note that if $h_b(s)$ is the least positive integer whose digit sum, in some fixed base $b$, is $s$, then $h_b$ is a strictly increasing function. This together with the fact that $g(N) \\ge 10$ shows that $f(N)$ is the least positive integer whose base-eight digit sum is 10. Thus $f(N) = 37_\\text{eight} = 31$, and $N$ is the least positive integer whose base-four digit sum is $31.$ Therefore \\[N = 13333333333_\\text{four} = 2\\cdot4^{10} - 1 = 2\\cdot1024^2 - 1\\] \\[\\equiv 2\\cdot24^2 - 1 \\equiv 151 \\pmod{1000}.\\]"
] |
2020-II-6
| 2,020
| 6
|
Define a sequence recursively by $t_1 = 20$ , $t_2 = 21$ , and \[t_n = \frac{5t_{n-1}+1}{25t_{n-2}}\] for all $n \ge 3$ . Then $t_{2020}$ can be written as $\frac{p}{q}$ , where $p$ and $q$ are relatively prime positive integers. Find $p+q$ .
| 626
|
II
|
[
"Let $t_n=\\frac{s_n}{5}$. Then, we have $s_n=\\frac{s_{n-1}+1}{s_{n-2}}$ where $s_1 = 100$ and $s_2 = 105$. By substitution, we find $s_3 = \\frac{53}{50}$, $s_4=\\frac{103}{105\\cdot50}$, $s_5=\\frac{101}{105}$, $s_6=100$, and $s_7=105$. So $s_n$ has a period of $5$. Thus $s_{2020}=s_5=\\frac{101}{105}$. So, $\\frac{101}{105\\cdot 5}\\implies 101+525=626. ~mn28407",
"More generally, let the first two terms be $a$ and $b$ and replace $5$ and $25$ in the recursive formula by $k$ and $k^2$, respectively. Then some algebraic calculation shows that \\[t_3 = \\frac{b\\,k+1}{a\\, k^2},~~t_4 = \\frac{a\\, k + b\\,k+1}{a\\,b\\, k^3},~~ t_5 = \\frac{a\\,k+1}{b\\, k^2},~~ t_6 = a, \\text{~ and ~}t_7 =b,\\]so the sequence is periodic with period $5$. Therefore \\[t_{2020} = t_{5} = \\frac{20\\cdot 5 +1}{21\\cdot 25} = \\frac{101}{525}.\\]The requested sum is $101+525=626$.",
"More generally, let the first two terms be $a$ and $b$ and replace $5$ and $25$ in the recursive formula by $k$ and $k^2$, respectively. Then some algebraic calculation shows that \\[t_3 = \\frac{b\\,k+1}{a\\, k^2},~~t_4 = \\frac{a\\, k + b\\,k+1}{a\\,b\\, k^3},~~ t_5 = \\frac{a\\,k+1}{b\\, k^2},~~ t_6 = a, \\text{~ and ~}t_7 =b,\\]so the sequence is periodic with period $5$. Therefore \\[t_{2020} = t_{5} = \\frac{20\\cdot 5 +1}{21\\cdot 25} = \\frac{101}{525}.\\]The requested sum is $101+525=626$.",
"Let us examine the first few terms of this sequence and see if we can find a pattern. We are obviously given $t_1 = 20$ and $t_2 = 21$, so now we are able to determine the numerical value of $t_3$ using this information: \\[t_3 = \\frac{5t_{3-1}+1}{25t_{3-2}} = \\frac{5t_{2}+1}{25t_{1}} = \\frac{5(21) + 1}{25(20)} = \\frac{105 + 1}{500}t_3 = \\frac{106}{500} = \\frac{53}{250}\\] \\[t_4 = \\frac{5t_{4-1}+1}{25t_{4-2}} = \\frac{5t_{3}+1}{25t_{2}} = \\frac{5(\\frac{53}{250}) + 1}{25(21)} = \\frac{\\frac{53}{50} + 1}{525} = \\frac{\\frac{103}{50}}{525} = \\frac{103}{26250}\\] \\[t_5 = \\frac{5t_{5-1}+1}{25t_{5-2}} = \\frac{5t_{4}+1}{25t_{3}} = \\frac{5(\\frac{103}{26250}) + 1}{25(\\frac{53}{250})} = \\frac{\\frac{103}{5250} + 1}{\\frac{53}{10}} = \\frac{\\frac{5353}{5250}}{\\frac{53}{10}} = \\frac{101}{525}\\] \\[t_6 = \\frac{5t_{6-1}+1}{25t_{6-2}} = \\frac{5t_{5}+1}{25t_{4}} = \\frac{5(\\frac{101}{525}) + 1}{25(\\frac{103}{26250})} = \\frac{\\frac{101}{105} + 1}{\\frac{103}{1050}} = \\frac{\\frac{206}{105}}{\\frac{103}{1050}} \\implies t_6 = 20\\] Alas, we have figured this sequence is period 5! But since $2020 \\equiv 5 \\pmod 5$, we can state that $t_{2020} = t_5 = \\frac{101}{525}$. According to the original problem statement, our answer is $626. ~ nikenissan"
] |
2020-II-7
| 2,020
| 7
|
Two congruent right circular cones each with base radius $3$ and height $8$ have axes of symmetry that intersect at right angles at a point in the interior of the cones a distance $3$ from the base of each cone. A sphere with radius $r$ lies within both cones. The maximum possible value of $r^2$ is $\frac{m}{n}$ , where $m$ and $n$ are relatively prime positive integers. Find $m+n$ .
| 298
|
II
|
[
"Consider the cross section of the cones and sphere by a plane that contains the two axes of symmetry of the cones as shown below. The sphere with maximum radius will be tangent to the sides of each of the cones. The center of that sphere must be on the axis of symmetry of each of the cones and thus must be at the intersection of their axes of symmetry. Let $A$ be the point in the cross section where the bases of the cones meet, and let $C$ be the center of the sphere. Let the axis of symmetry of one of the cones extend from its vertex, $B$, to the center of its base, $D$. Let the sphere be tangent to $\\overline{AB}$ at $E$. The right triangles $\\triangle ABD$ and $\\triangle CBE$ are similar, implying that the radius of the sphere is\\[CE = AD \\cdot\\frac{BC}{AB} = AD \\cdot\\frac{BD-CD}{AB} =3\\cdot\\frac5{\\sqrt{8^2+3^2}} = \\frac{15}{\\sqrt{73}}=\\sqrt{\\frac{225}{73}}.\\]The requested sum is $225+73=298$. [asy] unitsize(0.6cm); pair A = (0,0); pair TriangleOneLeft = (-6,0); pair TriangleOneDown = (-3,-8); pair TriangleOneMid = (-3,0); pair D = (0,-3); pair TriangleTwoDown = (0,-6); pair B = (-8,-3); pair C = IP(TriangleOneMid -- TriangleOneDown, B--D); pair EE = foot(C, A, B); real radius = arclength(C--EE); path circ = Circle(C, radius); draw(A--B--TriangleTwoDown--cycle); draw(B--D); draw(A--TriangleOneLeft--TriangleOneDown--cycle); draw(circ); draw(C--EE); draw(TriangleOneMid -- TriangleOneDown, gray); dot(\"$B$\", B, W); dot(\"$E$\", EE, NW); dot(\"$A$\", A, NE); dot(\"$D$\", D, E); dot(\"$C$\", C, SE); [/asy] Not part of MAA's solution, but this: https://www.geogebra.org/calculator/xv4nm97a is a good visual of the cones in GeoGebra.",
"Consider the cross section of the cones and sphere by a plane that contains the two axes of symmetry of the cones as shown below. The sphere with maximum radius will be tangent to the sides of each of the cones. The center of that sphere must be on the axis of symmetry of each of the cones and thus must be at the intersection of their axes of symmetry. Let $A$ be the point in the cross section where the bases of the cones meet, and let $C$ be the center of the sphere. Let the axis of symmetry of one of the cones extend from its vertex, $B$, to the center of its base, $D$. Let the sphere be tangent to $\\overline{AB}$ at $E$. The right triangles $\\triangle ABD$ and $\\triangle CBE$ are similar, implying that the radius of the sphere is\\[CE = AD \\cdot\\frac{BC}{AB} = AD \\cdot\\frac{BD-CD}{AB} =3\\cdot\\frac5{\\sqrt{8^2+3^2}} = \\frac{15}{\\sqrt{73}}=\\sqrt{\\frac{225}{73}}.\\]The requested sum is $225+73=298$. [asy] unitsize(0.6cm); pair A = (0,0); pair TriangleOneLeft = (-6,0); pair TriangleOneDown = (-3,-8); pair TriangleOneMid = (-3,0); pair D = (0,-3); pair TriangleTwoDown = (0,-6); pair B = (-8,-3); pair C = IP(TriangleOneMid -- TriangleOneDown, B--D); pair EE = foot(C, A, B); real radius = arclength(C--EE); path circ = Circle(C, radius); draw(A--B--TriangleTwoDown--cycle); draw(B--D); draw(A--TriangleOneLeft--TriangleOneDown--cycle); draw(circ); draw(C--EE); draw(TriangleOneMid -- TriangleOneDown, gray); dot(\"$B$\", B, W); dot(\"$E$\", EE, NW); dot(\"$A$\", A, NE); dot(\"$D$\", D, E); dot(\"$C$\", C, SE); [/asy] Not part of MAA's solution, but this: https://www.geogebra.org/calculator/xv4nm97a is a good visual of the cones in GeoGebra.",
"[asy] unitsize(0.6cm); // Coordinates pair A = (0,0); pair B = (6,0); pair C = (0,6); // Calculate point C pair D = (3,8); pair E = (8,3); pair F = (144/73,384/73); // Draw triangles (cones) draw(A--B--D--cycle); draw(A--C--E--cycle); draw(incircle(A,E,F)); pair EE = foot(C, A, B); real radius = arclength(C--EE); path circ = Circle(C, radius); // Label points dot(\"$A$\", A, NW); dot(\"$B$\", B, SW); dot(\"$C$\", C, NE); dot(\"$D$\", D, SE); dot(\"$E$\", E, SE); dot(\"$F$\", F, NE); [/asy] Let $A$ be the origin in the above diagram. Then $B$ is $(6,0)$, $C$ is $(0,6)$, $D$ is $(3,8)$, and $E$ is $(8,3)$. Also, it is easy to see that the inscribed sphere is simply the inscribed circle of $AEF$. Then we want to find the intersection of $AD$ and $CE$ to determine the coordinates of point $F$. Note that $AD$ is line $y=\\frac{8}{3}x$, and $CE$ is line $y=-\\frac{3}{8}x+6$. Then, you can see that these lines are perpendicular, indicating that $AEF$ is a right triangle with right angle at $F$. Finding the intersection of the lines by solving the system, we get that $F$ is the point $(\\frac{144}{73},\\frac{384}{73})$ in this plane. Then, we can find the distances $EF$ and $AF$ by the distance formula, which are $\\frac{55}{\\sqrt{73}}$ and $\\frac{48}{\\sqrt{73}}$ respectively. Also, $AE=\\sqrt{73}$. Then, to find the radius of this triangle's incircle, we use the formula $a=rs$ from which we get that $r=\\frac{15}{\\sqrt{73}}$ and $r^2=\\frac{225}{73} \\implies 298 is the answer. ~SirAppel",
"[asy] unitsize(0.6cm); // Coordinates pair A = (0,0); pair B = (6,0); pair C = (0,6); // Calculate point C pair D = (3,8); pair E = (8,3); pair F = (144/73,384/73); // Draw triangles (cones) draw(A--B--D--cycle); draw(A--C--E--cycle); draw(incircle(A,E,F)); pair EE = foot(C, A, B); real radius = arclength(C--EE); path circ = Circle(C, radius); // Label points dot(\"$A$\", A, NW); dot(\"$B$\", B, SW); dot(\"$C$\", C, NE); dot(\"$D$\", D, SE); dot(\"$E$\", E, SE); dot(\"$F$\", F, NE); [/asy] Let $A$ be the origin in the above diagram. Then $B$ is $(6,0)$, $C$ is $(0,6)$, $D$ is $(3,8)$, and $E$ is $(8,3)$. Also, it is easy to see that the inscribed sphere is simply the inscribed circle of $AEF$. Then we want to find the intersection of $AD$ and $CE$ to determine the coordinates of point $F$. Note that $AD$ is line $y=\\frac{8}{3}x$, and $CE$ is line $y=-\\frac{3}{8}x+6$. Then, you can see that these lines are perpendicular, indicating that $AEF$ is a right triangle with right angle at $F$. Finding the intersection of the lines by solving the system, we get that $F$ is the point $(\\frac{144}{73},\\frac{384}{73})$ in this plane. Then, we can find the distances $EF$ and $AF$ by the distance formula, which are $\\frac{55}{\\sqrt{73}}$ and $\\frac{48}{\\sqrt{73}}$ respectively. Also, $AE=\\sqrt{73}$. Then, to find the radius of this triangle's incircle, we use the formula $a=rs$ from which we get that $r=\\frac{15}{\\sqrt{73}}$ and $r^2=\\frac{225}{73} \\implies 298 is the answer. ~SirAppel",
"Using the diagram above, we notice that the desired length is simply the distance between the point $C$ and $\\overline{AB}$. We can mark $C$ as $(3,3)$ since it is $3$ units away from each of the bases. Point $B$ is $(8,3)$. Thus, line $\\overline{AB}$ is $y = \\frac{3}{8}x \\Rightarrow 3x + 8y = 0$. We can use the distance from point to line formula $\\frac{|Ax_0 + By_0 + C|}{\\sqrt{A^2 + B^2}}$, where $x_0$ and $y_0$ are the coordinates of the point, and A, B, and C are the coefficients of the line in form $Ax + By + C = 0$. Plugging everything in, we get \\[\\frac{|3(3) - 8(3)|}{\\sqrt{8^2 + 3^2}} = \\frac{15}{\\sqrt{73}} \\Rightarrow \\frac{225}{73} \\Rightarrow 298\\].",
"Using the diagram above, we notice that the desired length is simply the distance between the point $C$ and $\\overline{AB}$. We can mark $C$ as $(3,3)$ since it is $3$ units away from each of the bases. Point $B$ is $(8,3)$. Thus, line $\\overline{AB}$ is $y = \\frac{3}{8}x \\Rightarrow 3x + 8y = 0$. We can use the distance from point to line formula $\\frac{|Ax_0 + By_0 + C|}{\\sqrt{A^2 + B^2}}$, where $x_0$ and $y_0$ are the coordinates of the point, and A, B, and C are the coefficients of the line in form $Ax + By + C = 0$. Plugging everything in, we get \\[\\frac{|3(3) - 8(3)|}{\\sqrt{8^2 + 3^2}} = \\frac{15}{\\sqrt{73}} \\Rightarrow \\frac{225}{73} \\Rightarrow 298\\].",
"We graph this on graph paper, with the scale of $\\sqrt{2}:1$. So, we can find $OT$ then divide by $\\sqrt{2}$ to convert to our desired units, then square the result. With 5 minutes' worth of coordbashing, we finally arrive at $298$. [asy] /* Geogebra to Asymptote conversion by samrocksnature, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ real labelscalefactor = 0.5; /* changes label-to-point distance */ pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */ pen dotstyle = black; /* point style */ real xmin = -8.325025958411356, xmax = 8, ymin = -0.6033105644019334, ymax = 12.237120576121757; /* image dimensions */ pen zzttqq = rgb(0.6,0.2,0); pen qqwuqq = rgb(0,0.39215686274509803,0); pen cqcqcq = rgb(0.7529411764705882,0.7529411764705882,0.7529411764705882); draw((5,11)--(0,0)--(-6,6)--cycle, linewidth(1) + zzttqq); draw((6,6)--(-5,11)--(0,0)--cycle, linewidth(1) + qqwuqq); draw((0.2328977836854361,5.767102216314564)--(0.4657955673708722,6)--(0.2328977836854361,6.232897783685436)--(0,6)--cycle, linewidth(1) + qqwuqq); /* draw grid of horizontal/vertical lines */ pen gridstyle = linewidth(0.7) + cqcqcq; real gridx = 1, gridy = 1; /* grid intervals */ for(real i = ceil(xmin/gridx)*gridx; i <= floor(xmax/gridx)*gridx; i += gridx) draw((i,ymin)--(i,ymax), gridstyle); for(real i = ceil(ymin/gridy)*gridy; i <= floor(ymax/gridy)*gridy; i += gridy) draw((xmin,i)--(xmax,i), gridstyle); /* end grid */ Label laxis; laxis.p = fontsize(10); xaxis(xmin, xmax, Ticks(laxis, Step = 1, Size = 2, NoZero),EndArrow(6), above = true); yaxis(ymin, ymax, Ticks(laxis, Step = 1, Size = 2, NoZero),EndArrow(6), above = true); /* draws axes; NoZero hides '0' label */ /* draw figures */ draw((5,11)--(0,0), linewidth(1) + zzttqq); draw((0,0)--(-6,6), linewidth(1) + zzttqq); draw((-6,6)--(5,11), linewidth(1) + zzttqq); draw((6,6)--(-5,11), linewidth(1) + qqwuqq); draw((-5,11)--(0,0), linewidth(1) + qqwuqq); draw((0,0)--(6,6), linewidth(1) + qqwuqq); draw((-3,3)--(5,11), linewidth(1)); draw((-5,11)--(3,3), linewidth(1)); draw(circle((0,6), 2.482817665807104), linewidth(1)); draw((0,6)--(2.2602739726027394,4.972602739726027), linewidth(1)); /* dots and labels */ dot((0,0),linewidth(1pt) + dotstyle); dot((3,3),dotstyle); dot((-3,3),dotstyle); dot((6,6),dotstyle); dot((-6,6),dotstyle); dot((5,11),dotstyle); dot((-5,11),dotstyle); dot((0,6),linewidth(1pt) + dotstyle); label(\"$O$\", (0.059294254264342997,6.119672124650978), NE * labelscalefactor); dot((2.2602739726027394,4.972602739726027),linewidth(1pt) + dotstyle); label(\"$T$\", (2.326166015469254,5.094921876435061), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy] ~samrocksnature",
"Using the graph drawn above in MAA solution, we see that $\\sqrt{AC^2-r^2}$ + $\\sqrt{BC^2-r^2}$ = $AB$. $AB$ = $\\sqrt{8^2+3^2}$ = $\\sqrt{73}$, $AC$ = 3$\\sqrt{2}$, $AB$ = 5. Plugging in and solve for $r^2$ gives us $r^2$ = $\\frac{225}{73}$. ~S17209"
] |
2020-II-8
| 2,020
| 8
|
Define a sequence recursively by $f_1(x)=|x-1|$ and $f_n(x)=f_{n-1}(|x-n|)$ for integers $n>1$ . Find the least value of $n$ such that the sum of the zeros of $f_n$ exceeds $500,000$ .
| 101
|
II
|
[
"First it will be shown by induction that the zeros of $f_n$ are the integers $a, {a+2,} {a+4,} \\dots, {a + n(n-1)}$, where $a = n - \\frac{n(n-1)}2.$ This is certainly true for $n=1$. Suppose that it is true for $n = m-1 \\ge 1$, and note that the zeros of $f_m$ are the solutions of $|x - m| = k$, where $k$ is a nonnegative zero of $f_{m-1}$. Because the zeros of $f_{m-1}$ form an arithmetic sequence with common difference $2,$ so do the zeros of $f_m$. The greatest zero of $f_{m-1}$ is\\[m-1+\\frac{(m-1)(m-2)}2 =\\frac{m(m-1)}2,\\]so the greatest zero of $f_m$ is $m+\\frac{m(m-1)}2$ and the least is $m-\\frac{m(m-1)}2$. It follows that the number of zeros of $f_n$ is $\\frac{n(n-1)}2+1=\\frac{n^2-n+2}2$, and their average value is $n$. The sum of the zeros of $f_n$ is\\[\\frac{n^3-n^2+2n}2.\\]Let $S(n)=n^3-n^2+2n$, so the sum of the zeros exceeds $500{,}000$ if and only if $S(n) > 1{,}000{,}000 = 100^3\\!.$ Because $S(n)$ is increasing for $n > 2$, the values $S(100) = 1{,}000{,}000 - 10{,}000 + 200 = 990{,}200$ and $S(101)=1{,}030{,}301 - 10{,}201 + 202 = 1{,}020{,}302$ show that the requested value of $n$ is $101.",
"First it will be shown by induction that the zeros of $f_n$ are the integers $a, {a+2,} {a+4,} \\dots, {a + n(n-1)}$, where $a = n - \\frac{n(n-1)}2.$ This is certainly true for $n=1$. Suppose that it is true for $n = m-1 \\ge 1$, and note that the zeros of $f_m$ are the solutions of $|x - m| = k$, where $k$ is a nonnegative zero of $f_{m-1}$. Because the zeros of $f_{m-1}$ form an arithmetic sequence with common difference $2,$ so do the zeros of $f_m$. The greatest zero of $f_{m-1}$ is\\[m-1+\\frac{(m-1)(m-2)}2 =\\frac{m(m-1)}2,\\]so the greatest zero of $f_m$ is $m+\\frac{m(m-1)}2$ and the least is $m-\\frac{m(m-1)}2$. It follows that the number of zeros of $f_n$ is $\\frac{n(n-1)}2+1=\\frac{n^2-n+2}2$, and their average value is $n$. The sum of the zeros of $f_n$ is\\[\\frac{n^3-n^2+2n}2.\\]Let $S(n)=n^3-n^2+2n$, so the sum of the zeros exceeds $500{,}000$ if and only if $S(n) > 1{,}000{,}000 = 100^3\\!.$ Because $S(n)$ is increasing for $n > 2$, the values $S(100) = 1{,}000{,}000 - 10{,}000 + 200 = 990{,}200$ and $S(101)=1{,}030{,}301 - 10{,}201 + 202 = 1{,}020{,}302$ show that the requested value of $n$ is $101.",
"Starting from $f_1(x)=|x-1|$, we can track the solutions, the number of solutions, and their sum: \\[\\begin{array}{c|c|c|c} n&Solutions&number&sum\\\\ 1&1&1&1\\\\ 2&1,3&2&4\\\\ 3&0,2,4,6&4&12\\\\ 4&-2,0,2...10&7&28\\\\ 5&-5,-3,-1...15&11&55\\\\ \\end{array}\\] It is clear that the solutions form arithmetic sequences with a difference of 2, and the negative solutions cancel out all but $n$ of the $1+\\frac{n(n-1)}{2}$ solutions. Thus, the sum of the solutions is $n \\cdot [1+\\frac{n(n-1)}{2}]$, which is a cubic function. (Side Note: Gergor-Newton Interpolation Formula is applicable here) $n \\cdot [1+\\frac{n(n-1)}{2}]>500,000$ Multiplying both sides by $2$, $n \\cdot [2+n(n-1)]>1,000,000$ 1 million is $10^6=100^3$, so the solution should be close to $100$. 100 is slightly too small, so $101 works. ~ dragnin",
"Starting from $f_1(x)=|x-1|$, we can track the solutions, the number of solutions, and their sum: \\[\\begin{array}{c|c|c|c} n&Solutions&number&sum\\\\ 1&1&1&1\\\\ 2&1,3&2&4\\\\ 3&0,2,4,6&4&12\\\\ 4&-2,0,2...10&7&28\\\\ 5&-5,-3,-1...15&11&55\\\\ \\end{array}\\] It is clear that the solutions form arithmetic sequences with a difference of 2, and the negative solutions cancel out all but $n$ of the $1+\\frac{n(n-1)}{2}$ solutions. Thus, the sum of the solutions is $n \\cdot [1+\\frac{n(n-1)}{2}]$, which is a cubic function. (Side Note: Gergor-Newton Interpolation Formula is applicable here) $n \\cdot [1+\\frac{n(n-1)}{2}]>500,000$ Multiplying both sides by $2$, $n \\cdot [2+n(n-1)]>1,000,000$ 1 million is $10^6=100^3$, so the solution should be close to $100$. 100 is slightly too small, so $101 works. ~ dragnin"
] |
2020-II-9
| 2,020
| 9
|
While watching a show, Ayako, Billy, Carlos, Dahlia, Ehuang, and Frank sat in that order in a row of six chairs. During the break, they went to the kitchen for a snack. When they came back, they sat on those six chairs in such a way that if two of them sat next to each other before the break, then they did not sit next to each other after the break. Find the number of possible seating orders they could have chosen after the break.
| 90
|
II
|
[
"There are $2^{5}-1$ intersections that we must consider if we are to perform a PIE bash on this problem. Since we don't really want to think that hard, and bashing does not take that long for this problem, we can write down half of all permutations that satisfy the conditions presented in the problem in \"lexicographically next\" order to keep track easily. We do this for all cases such that the first \"person\" is $A-C$, and multiply by two, since the number of working permutations with $D-F$ as the first person is the same as if it were $A-C$, hence, after doing such a bash, we get $45\\times2=90$ permutations that result in no consecutive letters being adjacent to each other. ~afatperson",
"There are $2^{5}-1$ intersections that we must consider if we are to perform a PIE bash on this problem. Since we don't really want to think that hard, and bashing does not take that long for this problem, we can write down half of all permutations that satisfy the conditions presented in the problem in \"lexicographically next\" order to keep track easily. We do this for all cases such that the first \"person\" is $A-C$, and multiply by two, since the number of working permutations with $D-F$ as the first person is the same as if it were $A-C$, hence, after doing such a bash, we get $45\\times2=90$ permutations that result in no consecutive letters being adjacent to each other. ~afatperson",
"Ayako ($A$), Billy $(B)$, Carlos $(C)$, Dahlia $(D)$, Ehuang $(E)$, and Frank $(F)$ originally sat in the order $ABCDEF$. Let $T(XY)$ denote the set of seatings where $X$ and $Y$ sit next to each other after the break. Then the required number of seating orders is given by the Inclusion-Exclusion Principle as \\[6!-\\big(|T(AB)|+|T(BC)|+|T(CD)|+|T(DE)|+|T(EF)|\\big)+\\]\\[\\big(|T(AB)\\cap T(BC)|+|T(AB)\\cap T(CD)|+\\cdots\\big) - \\cdots.\\]Each term can be calculated separately. (a) $|T(AB)|=|T(BC)|=|T(CD)|=|T(DE)|=|T(EF)|=2\\cdot5!=240.$ Because there are $5$ terms, the sum is $5\\cdot240=1200$. (b) For $|T(XY)\\cap T(ZW)|$, if $Y=Z$, then $XYW$ must sit consecutively, so $|T(XY)\\cap T(ZW)|=2\\cdot4!=48$. There are $4$ terms that satisfy $Y=Z$, so the sum is $4\\cdot 48=192$. If $XY$ and $ZW$ are pairwise disjoint, then $|T(XY)\\cap T(ZW)|=2^2\\cdot4!=96$. There are $6$ terms, so the sum is $6\\cdot96=576$. (c) If there are at least three pairs that sit next to each other, consider these three subcases: If the three pairs are consecutive, the sum is $3\\cdot2\\cdot3!=36$. If exactly two of the pairs are consecutive, the sum is $6\\cdot2^2\\cdot3!=144$. If none of the three pairs is consecutive, the sum is $1\\cdot2^3\\cdot3!=48.$ (d) If there are at least four pairs that sit next to each other, then if the pairs are consecutive, the sum is $2\\cdot2\\cdot2!=8$. If the pairs are not consecutive, then the sum is $3\\cdot2^2\\cdot2!=24$. (e) If all five pairs sit next to each other, the number is $1\\cdot2\\cdot1!=2$. Therefore the required number of seating orders is\\[6!-1200+(192+576)-{(36+144+48)+(8+24)-2}=90.\\] Note: See the A002464 of the On-Line Encyclopedia of Integer Sequences for equivalent formulations.",
"Ayako ($A$), Billy $(B)$, Carlos $(C)$, Dahlia $(D)$, Ehuang $(E)$, and Frank $(F)$ originally sat in the order $ABCDEF$. Let $T(XY)$ denote the set of seatings where $X$ and $Y$ sit next to each other after the break. Then the required number of seating orders is given by the Inclusion-Exclusion Principle as \\[6!-\\big(|T(AB)|+|T(BC)|+|T(CD)|+|T(DE)|+|T(EF)|\\big)+\\]\\[\\big(|T(AB)\\cap T(BC)|+|T(AB)\\cap T(CD)|+\\cdots\\big) - \\cdots.\\]Each term can be calculated separately. (a) $|T(AB)|=|T(BC)|=|T(CD)|=|T(DE)|=|T(EF)|=2\\cdot5!=240.$ Because there are $5$ terms, the sum is $5\\cdot240=1200$. (b) For $|T(XY)\\cap T(ZW)|$, if $Y=Z$, then $XYW$ must sit consecutively, so $|T(XY)\\cap T(ZW)|=2\\cdot4!=48$. There are $4$ terms that satisfy $Y=Z$, so the sum is $4\\cdot 48=192$. If $XY$ and $ZW$ are pairwise disjoint, then $|T(XY)\\cap T(ZW)|=2^2\\cdot4!=96$. There are $6$ terms, so the sum is $6\\cdot96=576$. (c) If there are at least three pairs that sit next to each other, consider these three subcases: If the three pairs are consecutive, the sum is $3\\cdot2\\cdot3!=36$. If exactly two of the pairs are consecutive, the sum is $6\\cdot2^2\\cdot3!=144$. If none of the three pairs is consecutive, the sum is $1\\cdot2^3\\cdot3!=48.$ (d) If there are at least four pairs that sit next to each other, then if the pairs are consecutive, the sum is $2\\cdot2\\cdot2!=8$. If the pairs are not consecutive, then the sum is $3\\cdot2^2\\cdot2!=24$. (e) If all five pairs sit next to each other, the number is $1\\cdot2\\cdot1!=2$. Therefore the required number of seating orders is\\[6!-1200+(192+576)-{(36+144+48)+(8+24)-2}=90.\\] Note: See the A002464 of the On-Line Encyclopedia of Integer Sequences for equivalent formulations.",
"We proceed with casework based on the person who sits first after the break. $\\textbf{Case 1:}$ A is first. Then the first three people in the row can be ACE, ACF, ADB, ADF, AEB, AEC, AFB, AFC, or AFD, which yield 2, 1, 2, 2, 1, 2, 0, 1, and 1 possible configurations, respectively, implying 2 + 1 + 2 + 2 + 1 + 2 + 0 + 1 + 1 = 12 possible configurations in this case. $\\textbf{Case 2:}$ B is first. Then the first three people in the row can be BDA, BDF, BEA, BEC, BFA, BFC, or BFD, which yield 2, 4, 2, 4, 0, 1, and 2 possible configurations, respectively, implying 2 + 4 + 2 + 4 + 0 + 1 + 2 = 15 possible configurations in this case. $\\textbf{Case 3:}$ C is first. Then the first three people in the row can be CAD, CAE, CAF, CEA, CEB, CFA, CFB, or CFD, which yield 1, 2, 1, 4, 4, 2, 2, and 2 possible configurations, respectively, implying 1 + 2 + 1 + 4 + 4 + 2 + 2 + 2 = 18 possible configurations in this case. Finally, the number of valid configurations for A and F, B and E, and C and D are equal by symmetry, so our final answer is 2(12 + 15 + 18), which computes to be $090 ~peace09",
"We determine the order of A, B, C, relative to each other. Then, we will insert D, E, F into the alignment and calculate the total number of possibilities. This solution can be visualised as standing in lines rather than sitting on chairs. There are 6 possible alignments for A, B, and C, but we only evaluate 3 because the other 3 cases can be mirrored to overlap these 3 cases. $\\textbf{Case 1: A B C}$ In this case, there must be a person standing between A and B and also between B and C. Also, D cannot be adjacent to C. There are 9 possibilities. $\\textbf{Case 2: A C B}$ In this case, there must be a person standing between B and C. Also, D cannot be adjacent to C. There are 12 possibilities. $\\textbf{Case 3: C A B}$ In this case, there must be a person standing between A and B. Also, D cannot be adjacent to C. There are 24 possibilities. So the total number of cases is 2(9+12+24)=$090. -Superdolphin",
"Consider arrangements of numbers $1, 2, 3, ..., n$. Let $f(n,k)$ be the number of arrangements in which $1$ and $2$, $2$ and $3$, $3$ and $4$, ..., $k-1$ and $k$ aren't together. We need to find $f(6,6)$. Let $d(n,k)$ be the number of arrangements in which $1$ and $2$, $2$ and $3$, $3$ and $4$, ..., $k-2$ and $k-1$ aren't together, but $(k-1)$ and $k$ are together. Then, \\[f(n,k+1) = f(n,k) - d(n,k+1) ......(1)\\] \\[d(n,k+1) = 2f(n-1,k) + d(n-1,k) ......(2)\\] Hence, \\[f(n,k+1)=f(n,k)-f(n-1,k)-f(n-1,k-1)\\] And because $f(n,0) = f(n,1) = n!$, it's easy to get $f(6,6) = 090, which is the second equation. ~Shawn",
"Consider arrangements of numbers $1, 2, 3, ..., n$. Let $f(n,k)$ be the number of arrangements in which $1$ and $2$, $2$ and $3$, $3$ and $4$, ..., $k-1$ and $k$ aren't together. We need to find $f(6,6)$. Let $d(n,k)$ be the number of arrangements in which $1$ and $2$, $2$ and $3$, $3$ and $4$, ..., $k-2$ and $k-1$ aren't together, but $(k-1)$ and $k$ are together. Then, \\[f(n,k+1) = f(n,k) - d(n,k+1) ......(1)\\] \\[d(n,k+1) = 2f(n-1,k) + d(n-1,k) ......(2)\\] Hence, \\[f(n,k+1)=f(n,k)-f(n-1,k)-f(n-1,k-1)\\] And because $f(n,0) = f(n,1) = n!$, it's easy to get $f(6,6) = 090, which is the second equation. ~Shawn",
"We will split up our cases based on the positions of $A$, $B$, and $C$ relative to each other. Case $1$: $A$ $B$ $C$: $9$ cases Case $2$: $A$ $C$ $B$: $16$ cases Case $3$: $B$ $A$ $C$: $20$ cases Since we can reverse the first case to make $C$ $B$ $A$, reverse the second case to make $B$ $C$ $A$, and reverse the third case to make $C$ $A$ $B$, by symmetry we have $2(9+16+20) = 090 ways in total. ~xHypotenuse",
"We will split up our cases based on the positions of $A$, $B$, and $C$ relative to each other. Case $1$: $A$ $B$ $C$: $9$ cases Case $2$: $A$ $C$ $B$: $16$ cases Case $3$: $B$ $A$ $C$: $20$ cases Since we can reverse the first case to make $C$ $B$ $A$, reverse the second case to make $B$ $C$ $A$, and reverse the third case to make $C$ $A$ $B$, by symmetry we have $2(9+16+20) = 090 ways in total. ~xHypotenuse"
] |
2020-II-10
| 2,020
| 10
|
Find the sum of all positive integers $n$ such that when $1^3+2^3+3^3+\cdots +n^3$ is divided by $n+5$ , the remainder is $17$ .
| 239
|
II
|
[
"The formula for the sum of cubes, also known as Nicomachus's Theorem, is as follows: \\[1^3+2^3+3^3+\\dots+k^3=(1+2+3+\\dots+k)^2=\\left(\\frac{k(k+1)}{2}\\right)^2\\] for any positive integer $k$. So let's apply this to this problem. Let $m=n+5$. Then we have \\begin{align*} 1^3+2^3+3^3+\\dots+(m-5)^3&\\equiv 17 \\mod m \\\\ \\left(\\frac{(m-5)(m-4)}{2}\\right)^2&\\equiv 17 \\mod m \\\\ \\left(\\dfrac{m(m-9)+20}2\\right)^2&\\equiv 17\\mod m \\\\ \\left(m(m-9)+20\\right)^2&\\equiv 4\\cdot 17\\mod m \\\\ \\left(20\\right)^2&\\equiv 68\\mod m \\\\ 332 &\\equiv 0 \\mod m \\\\ \\end{align*} So, $m\\in\\{83,166,332\\}$. Testing the cases, only $332$ fails. This leaves $78+161=239 and formatting adjustments and intermediate steps for clarification by Technodoggo.",
"The sum of the cubes from 1 to $n$ is \\[1^3+2^3+\\cdots+n^3=\\frac{n^2(n+1)^2}{4}.\\]For this to be equal to $(n+5)q+17$ for some integer $q$, it must be that\\[n^2(n+1)^2=4(n+5)q+4\\cdot 17,\\]so\\[n^2(n+1)^2 \\equiv 4 \\cdot 17= 68\\hskip-.2cm \\pmod{n+5}.\\]But $n^2(n+1)^2 \\equiv (-5)^2(-4)^2 = 400 \\pmod{n+5}.$ Thus $n^2(n+1)^2$ is congruent to both $68$ and $400,$ which implies that $n+5$ divides $400-68 = 332=2^2 \\cdot 83$. Because $n+5 > 17$, the only choices for $n+5$ are $83, 166,$ and $332.$ Checking all three cases verifies that $n=78$ and $n=161$ work, but $n=327$ does not. The requested sum is $78+161 = 239$.",
"The sum of the cubes from 1 to $n$ is \\[1^3+2^3+\\cdots+n^3=\\frac{n^2(n+1)^2}{4}.\\]For this to be equal to $(n+5)q+17$ for some integer $q$, it must be that\\[n^2(n+1)^2=4(n+5)q+4\\cdot 17,\\]so\\[n^2(n+1)^2 \\equiv 4 \\cdot 17= 68\\hskip-.2cm \\pmod{n+5}.\\]But $n^2(n+1)^2 \\equiv (-5)^2(-4)^2 = 400 \\pmod{n+5}.$ Thus $n^2(n+1)^2$ is congruent to both $68$ and $400,$ which implies that $n+5$ divides $400-68 = 332=2^2 \\cdot 83$. Because $n+5 > 17$, the only choices for $n+5$ are $83, 166,$ and $332.$ Checking all three cases verifies that $n=78$ and $n=161$ work, but $n=327$ does not. The requested sum is $78+161 = 239$.",
"The sum of the cubes of the integers from $1$ through $n$ is\\[1^3+2^3+\\cdots+n^3=\\frac{n^2(n+1)^2}{4},\\]which, when divided by $n+5$, has quotient\\[Q=\\frac14n^3 -\\frac34n^2+4n-20 = \\frac{n^2(n-3)}4+4n-20\\]with remainder $100.$ If $n$ is not congruent to $1\\pmod4$, then $Q$ is an integer, and\\[\\frac{n^2(n+1)^2}{4} = (n+5)Q + 100 \\equiv 17\\pmod{n+5},\\]so $n+5$ divides $100 - 17 =83$, and $n = 78$. If $n \\equiv 1 \\pmod4$, then $Q$ is half of an integer, and letting $n = 4k+1$ for some integer $k$ gives\\[\\frac{n^2(n+1)^2}{4} = 2(2k+3)Q + 100 \\equiv 17\\pmod{n+5}.\\]Thus $2k+3$ divides $100-17 = 83$. It follows that $k=40$, and $n = 161$. The requested sum is $161 + 78 = 239$.",
"The sum of the cubes of the integers from $1$ through $n$ is\\[1^3+2^3+\\cdots+n^3=\\frac{n^2(n+1)^2}{4},\\]which, when divided by $n+5$, has quotient\\[Q=\\frac14n^3 -\\frac34n^2+4n-20 = \\frac{n^2(n-3)}4+4n-20\\]with remainder $100.$ If $n$ is not congruent to $1\\pmod4$, then $Q$ is an integer, and\\[\\frac{n^2(n+1)^2}{4} = (n+5)Q + 100 \\equiv 17\\pmod{n+5},\\]so $n+5$ divides $100 - 17 =83$, and $n = 78$. If $n \\equiv 1 \\pmod4$, then $Q$ is half of an integer, and letting $n = 4k+1$ for some integer $k$ gives\\[\\frac{n^2(n+1)^2}{4} = 2(2k+3)Q + 100 \\equiv 17\\pmod{n+5}.\\]Thus $2k+3$ divides $100-17 = 83$. It follows that $k=40$, and $n = 161$. The requested sum is $161 + 78 = 239$.",
"Using the formula for $\\sum_{k=1}^n k^3$, \\[1^3 + 2^3 + 3^3 + ... + n^3 = \\frac{n^2(n+1)^2}{4}\\] Since $1^3 + 2^3 + 3^3 + ... + n^3$ divided by $n + 5$ has a remainder of $17$, \\[\\frac{n^2(n+1)^2}{4} \\equiv 17\\pmod {n + 5}\\] Using the rules of modular arithmetic, \\[n^2(n+1)^2 \\equiv 68\\pmod {n + 5}\\]\\[n^2(n+1)^2 - 68\\equiv 0\\pmod {n + 5}\\] Expanding the left hand side, \\[n^4 + 2 n^3 + n^2 - 68\\equiv 0\\pmod {n + 5}\\] This means that $n^4 + 2 n^3 + n^2 - 68$ is divisible by ${n + 5}$. \\[(n + 5) | (n^4 + 2 n^3 + n^2 - 68)\\] Dividing polynomials, \\[\\frac{n^4 + 2 n^3 + n^2 - 68}{n + 5}\\] \\[= n^3 - 3 n^2 + 16n - 80 + \\frac{332}{(n + 5)}\\] $(n + 5)$ $|$ $(n^4 + 2 n^3 + n^2 - 68)$ $\\iff$ $\\frac{332}{(n + 5)}$ $\\in$ $\\mathbb{Z}$ $\\frac{332}{(n + 5)}$ $\\in$ $\\mathbb{Z}$ $\\iff$ $(n + 5) = \\pm 1, \\pm 2, \\pm 4, \\pm 83, \\pm 166, \\pm 332$ Note that $n$ $\\in$ $\\mathbb{N}$ and $n + 5 > 17$ (because the remainder when dividing by $n + 5$ is $17$, so $n + 5$ must be greater than $17$), so all options $\\leq 17$ can be eliminated. \\[(n + 5) = 83, 166, 332\\] \\[n = 78, 161, 327\\] Checking all 3 cases, $n = 78$ and $n = 161$ work; $n = 327$ fails. Therefore, the answer is $78 + 161 = 239. ~ {TSun} ~",
"As before, we note that $(5+a)^3 + (n-a)^3 \\equiv (5+a)^3 - (n+5 - (n-a))^3 \\equiv 0 \\pmod {n+5}.$ Thus, we can pair up the terms from $5^3$ to $n^3$ and cancel them. We have to deal with two cases: If $n$ is even, then $5^3+6^3 + \\cdots + n^3 \\equiv 0 \\pmod {n+5},$ as there are an even number of terms and they pair and cancel. We thus get $1^2+2^3+3^3+4^3 = 100 \\equiv 17 \\pmod {n+5},$ or $(n+5) | 83,$ which yields $n=78.$ If $n$ is odd, then $1^3+2^3+\\cdots + n^3 \\equiv 1^3+2^3+3^3+4^3+\\left( \\frac{n+5}{2} \\right)^3 \\equiv 17 \\pmod {n+5}.$ Letting $k = \\frac{n+5}{2}$ yields $k^2 + 83 \\equiv 0 \\pmod {2k}.$ However, this means that $83$ is divisible by $k,$ so $k=1,83.$ Plugging this back into $n$ yields $n=2(83)-5 = 161$ in the latter case. Thus, the sum of all possible $n$ is just $78+161 = 239 - ccx09",
"As before, we note that $(5+a)^3 + (n-a)^3 \\equiv (5+a)^3 - (n+5 - (n-a))^3 \\equiv 0 \\pmod {n+5}.$ Thus, we can pair up the terms from $5^3$ to $n^3$ and cancel them. We have to deal with two cases: If $n$ is even, then $5^3+6^3 + \\cdots + n^3 \\equiv 0 \\pmod {n+5},$ as there are an even number of terms and they pair and cancel. We thus get $1^2+2^3+3^3+4^3 = 100 \\equiv 17 \\pmod {n+5},$ or $(n+5) | 83,$ which yields $n=78.$ If $n$ is odd, then $1^3+2^3+\\cdots + n^3 \\equiv 1^3+2^3+3^3+4^3+\\left( \\frac{n+5}{2} \\right)^3 \\equiv 17 \\pmod {n+5}.$ Letting $k = \\frac{n+5}{2}$ yields $k^2 + 83 \\equiv 0 \\pmod {2k}.$ However, this means that $83$ is divisible by $k,$ so $k=1,83.$ Plugging this back into $n$ yields $n=2(83)-5 = 161$ in the latter case. Thus, the sum of all possible $n$ is just $78+161 = 239 - ccx09"
] |
2020-II-11
| 2,020
| 11
|
Let $P(x) = x^2 - 3x - 7$ , and let $Q(x)$ and $R(x)$ be two quadratic polynomials also with the coefficient of $x^2$ equal to $1$ . David computes each of the three sums $P + Q$ , $P + R$ , and $Q + R$ and is surprised to find that each pair of these sums has a common root, and these three common roots are distinct. If $Q(0) = 2$ , then $R(0) = \frac{m}{n}$ , where $m$ and $n$ are relatively prime positive integers. Find $m + n$ .
| 71
|
II
|
[
"Let $Q(x) = x^2 + ax + 2$ and $R(x) = x^2 + bx + c$. We can write the following: \\[P + Q = 2x^2 + (a - 3)x - 5\\] \\[P + R = 2x^2 + (b - 3)x + (c - 7)\\] \\[Q + R = 2x^2 + (a + b)x + (c + 2)\\] Let the common root of $P+Q,P+R$ be $r$; $P+R,Q+R$ be $s$; and $P+Q,Q+R$ be $t$. We then have that the roots of $P+Q$ are $r,t$, the roots of $P + R$ are $r, s$, and the roots of $Q + R$ are $s,t$. By Vieta's, we have: \\[r + t = \\dfrac{3 - a}{2}\\tag{1}\\] \\[r + s = \\dfrac{3 - b}{2}\\tag{2}\\] \\[s + t = \\dfrac{-a - b}{2}\\tag{3}\\] \\[rt = \\dfrac{-5}{2}\\tag{4}\\] \\[rs = \\dfrac{c - 7}{2}\\tag{5}\\] \\[st = \\dfrac{c + 2}{2}\\tag{6}\\] Subtracting $(3)$ from $(1)$, we get $r - s = \\dfrac{3 + b}{2}$. Adding this to $(2)$, we get $2r = 3 \\implies r = \\dfrac{3}{2}$. This gives us that $t = \\dfrac{-5}{3}$ from $(4)$. Substituting these values into $(5)$ and $(6)$, we get $s = \\dfrac{c-7}{3}$ and $s = \\dfrac{-3c - 6}{10}$. Equating these values, we get $\\dfrac{c-7}{3} = \\dfrac{-3c-6}{10} \\implies c = \\dfrac{52}{19} = R(0)$. Thus, our answer is $52 + 19 = 071. ~ TopNotchMath",
"We know that $P(x)=x^2-3x-7$. Since $Q(0)=2$, the constant term in $Q(x)$ is $2$. Let $Q(x)=x^2+ax+2$. Finally, let $R(x)=x^2+bx+c$. $P(x)+Q(x)=2x^2+(a-3)x-5$. Let its roots be $p$ and $q$. $P(x)+R(x)=2x^2+(b-3)x+(c-7)$ Let its roots be $p$ and $r$. $Q(x)+R(x)=2x^2+(a+b)x+(c+2)$. Let its roots be $q$ and $r$. By vietas, $p+q=\\frac{3-a}{2}, p+r=\\frac{3-b}{2}, q+r=\\frac{-(a+b)}{2}$ We could work out the system of equations, but it's pretty easy to see that $p=\\frac32, q=-\\frac{a}{2}, r=-\\frac{b}{2}$. $\\text{Again, by vietas, }pq=-\\frac52\\text{, } pr=\\frac{c-7}{2}\\text{, } qr=\\frac{c+2}{2}\\text{, } \\text{multiplying everything together a}\\text{nd taking the sqrt of both sides,}$ \\[(pqr)^2=\\left(-\\frac52\\right)\\left(\\frac{c-7}{2}\\right)\\left(\\frac{c+2}{2}\\right)\\] \\[pqr=\\sqrt{\\left(-\\frac52\\right)\\left(\\frac{c-7}{2}\\right)\\left(\\frac{c+2}{2}\\right)}\\] $\\text{Dividing this }\\text{equation by }qr=\\frac{c+2}{2}$ \\[\\frac{pqr}{qr}=\\frac{\\sqrt{\\left(-\\frac52\\right)\\left(\\frac{c-7}{2}\\right)\\left(\\frac{c+2}{2}\\right)}}{\\frac{c+2}{2}}\\] \\[p = \\frac{\\sqrt{\\left(-\\frac52\\right)\\left(\\frac{c-7}{2}\\right)}}{\\sqrt{\\frac{c+2}{2}}}\\] $\\text{Recall th}\\text{at }p=\\frac32 \\text{ and square both sides}$ \\[\\frac94=\\frac{\\left(-\\frac52\\right)\\left(\\frac{c-7}{2}\\right)}{\\frac{c+2}{2}}\\] $\\text{Solving gives } c=\\frac{52}{19}, \\text{ so our answer is }071 ~quacker88",
"Let the common root of $P+Q$ and $P+R$ be $p$, the common root of $P+Q$ and $Q+R$ be $q$, and the common root of $Q+R$ and $P+R$ be $r$. Because $p$ and $q$ are both roots of $P+Q$ and $P+Q$ has leading coefficient $2$, it follows that $P(x) + Q(x) = 2(x-p)(x-q).$ Similarly, $P(x) + R(x) = 2(x-p)(x-r)$ and $Q(x) + R(x) = 2(x-q)(x-r)$. Adding these three equations together and dividing by $2$ yields\\[P(x) + Q(x) + R(x) = (x-p)(x-q) + (x-p)(x-r) + (x-q)(x-r),\\]so \\[P(x) = (P(x) + Q(x) + R(x)) - (Q(x) + R(x))\\] \\[= (x-p)(x-q) + (x-p)(x-r) - (x-q)(x-r)\\] \\[= x^2 - 2px + (pq + pr - qr).\\] Similarly, \\[Q(x) = x^2 - 2qx + (pq + qr - pr) \\text{~ and}\\] \\[R(x) = x^2 - 2rx + (pr + qr - pq).\\] Comparing the $x$ coefficients yields $p = \\tfrac32$, and comparing the constant coefficients yields $-7 = pq + pr - qr = \\tfrac32(q+r) - qr$. The fact that $Q(0) = 2$ implies that $\\tfrac32(q-r) + qr = 2$. Adding these two equations yields $q = -\\tfrac53$, and so substituting back in to solve for $r$ gives $r=-\\tfrac{27}{19}$. Finally,\\[R(0) = pr + qr - pq = \\left(-\\frac{27}{19}\\right)\\left(\\frac32-\\frac53\\right) + \\frac52 = \\frac{9}{38} + \\frac52 = \\frac{52}{19}.\\]The requested sum is $52 + 19 = 71$. Note that $Q(x) = x^2 + \\frac{10}3x + 2$ and $R(x) = x^2 + \\frac{54}{19}x + \\frac{52}{19}$.",
"Let the common root of $P+Q$ and $P+R$ be $p$, the common root of $P+Q$ and $Q+R$ be $q$, and the common root of $Q+R$ and $P+R$ be $r$. Because $p$ and $q$ are both roots of $P+Q$ and $P+Q$ has leading coefficient $2$, it follows that $P(x) + Q(x) = 2(x-p)(x-q).$ Similarly, $P(x) + R(x) = 2(x-p)(x-r)$ and $Q(x) + R(x) = 2(x-q)(x-r)$. Adding these three equations together and dividing by $2$ yields\\[P(x) + Q(x) + R(x) = (x-p)(x-q) + (x-p)(x-r) + (x-q)(x-r),\\]so \\[P(x) = (P(x) + Q(x) + R(x)) - (Q(x) + R(x))\\] \\[= (x-p)(x-q) + (x-p)(x-r) - (x-q)(x-r)\\] \\[= x^2 - 2px + (pq + pr - qr).\\] Similarly, \\[Q(x) = x^2 - 2qx + (pq + qr - pr) \\text{~ and}\\] \\[R(x) = x^2 - 2rx + (pr + qr - pq).\\] Comparing the $x$ coefficients yields $p = \\tfrac32$, and comparing the constant coefficients yields $-7 = pq + pr - qr = \\tfrac32(q+r) - qr$. The fact that $Q(0) = 2$ implies that $\\tfrac32(q-r) + qr = 2$. Adding these two equations yields $q = -\\tfrac53$, and so substituting back in to solve for $r$ gives $r=-\\tfrac{27}{19}$. Finally,\\[R(0) = pr + qr - pq = \\left(-\\frac{27}{19}\\right)\\left(\\frac32-\\frac53\\right) + \\frac52 = \\frac{9}{38} + \\frac52 = \\frac{52}{19}.\\]The requested sum is $52 + 19 = 71$. Note that $Q(x) = x^2 + \\frac{10}3x + 2$ and $R(x) = x^2 + \\frac{54}{19}x + \\frac{52}{19}$."
] |
2020-II-12
| 2,020
| 12
|
Let $m$ and $n$ be odd integers greater than $1.$ An $m\times n$ rectangle is made up of unit squares where the squares in the top row are numbered left to right with the integers $1$ through $n$ , those in the second row are numbered left to right with the integers $n + 1$ through $2n$ , and so on. Square $200$ is in the top row, and square $2000$ is in the bottom row. Find the number of ordered pairs $(m,n)$ of odd integers greater than $1$ with the property that, in the $m\times n$ rectangle, the line through the centers of squares $200$ and $2000$ intersects the interior of square $1099$ .
| 248
|
II
|
[
"Let us take some cases. Since $m$ and $n$ are odds, and $200$ is in the top row and $2000$ in the bottom, $m$ has to be $3$, $5$, $7$, or $9$. Also, taking a look at the diagram, the slope of the line connecting those centers has to have an absolute value of $< 1$. Therefore, $m < 1800 \\mod n < 1800-m$. If $m=3$, $n$ can range from $667$ to $999$. However, $900$ divides $1800$, so looking at mods, we can easily eliminate $899$ and $901$. Now, counting these odd integers, we get $167 - 2 = 165$. Similarly, let $m=5$. Then $n$ can range from $401$ to $499$. However, $450|1800$, so one can remove $449$ and $451$. Counting odd integers, we get $50 - 2 = 48$. Take $m=7$. Then, $n$ can range from $287$ to $333$. However, $300|1800$, so one can verify and eliminate $299$ and $301$. Counting odd integers, we get $24 - 2 = 22$. Let $m = 9$. Then $n$ can vary from $223$ to $249$. However, $225|1800$. Checking that value and the values around it, we can eliminate $225$. Counting odd integers, we get $14 - 1 = 13$. Add all of our cases to get \\[165+48+22+13 = 248\\] -Solution by thanosaops",
"Because square $2000$ is in the bottom row, it follows that $\\frac{2000}m \\le n < \\frac{2000}{m-1}$. Moreover, because square $200$ is in the top row, and square $2000$ is not in the top row, $1 < m \\le 10$. In particular, because the number of rows in the rectangle must be odd, $m$ must be one of $3, 5, 7,$ or $9.$ For each possible choice of $m$ and $n$, let $\\ell_{m,n}$ denote the line through the centers of squares $200$ and $2000.$ Note that for odd values of $m$, the line $\\ell_{m,n}$ passes through the center of square $1100.$ Thus $\\ell_{m,n}$ intersects the interior of cell $1099$ exactly when its slope is strictly between $-1$ and $1$. The line $\\ell_{m,n}$ is vertical whenever square $2000$ is the $200$th square in the bottom row of the rectangle. This would happen for $m = 3, 5, 7, 9$ when $n = 900, 450, 300, 225$, respectively. When $n$ is 1 greater than or 1 less than these numbers, the slope of $\\ell_{m,n}$ is $1$ or $-1$, respectively. In all other cases the slope is strictly between $-1$ and $1.$ The admissible values for $n$ for each possible value of $m$ are given in the following table. \\[\\begin{tabular}{|c|c|c|c|c|}\\hline m & minimum n & maximum n & avoided n & number of odd n\\\\\\hline 3&667&999&899, 900, 901&165\\\\\\hline 5&400&499&449, 450, 451&48\\\\\\hline 7&286&333&299, 300, 301&22\\\\\\hline 9&223&249&224, 225, 226&13\\\\\\hline \\end{tabular}\\] This accounts for $165 + 48 + 22 + 13 = 248$ rectangles.",
"Because square $2000$ is in the bottom row, it follows that $\\frac{2000}m \\le n < \\frac{2000}{m-1}$. Moreover, because square $200$ is in the top row, and square $2000$ is not in the top row, $1 < m \\le 10$. In particular, because the number of rows in the rectangle must be odd, $m$ must be one of $3, 5, 7,$ or $9.$ For each possible choice of $m$ and $n$, let $\\ell_{m,n}$ denote the line through the centers of squares $200$ and $2000.$ Note that for odd values of $m$, the line $\\ell_{m,n}$ passes through the center of square $1100.$ Thus $\\ell_{m,n}$ intersects the interior of cell $1099$ exactly when its slope is strictly between $-1$ and $1$. The line $\\ell_{m,n}$ is vertical whenever square $2000$ is the $200$th square in the bottom row of the rectangle. This would happen for $m = 3, 5, 7, 9$ when $n = 900, 450, 300, 225$, respectively. When $n$ is 1 greater than or 1 less than these numbers, the slope of $\\ell_{m,n}$ is $1$ or $-1$, respectively. In all other cases the slope is strictly between $-1$ and $1.$ The admissible values for $n$ for each possible value of $m$ are given in the following table. \\[\\begin{tabular}{|c|c|c|c|c|}\\hline m & minimum n & maximum n & avoided n & number of odd n\\\\\\hline 3&667&999&899, 900, 901&165\\\\\\hline 5&400&499&449, 450, 451&48\\\\\\hline 7&286&333&299, 300, 301&22\\\\\\hline 9&223&249&224, 225, 226&13\\\\\\hline \\end{tabular}\\] This accounts for $165 + 48 + 22 + 13 = 248$ rectangles."
] |
2020-II-13
| 2,020
| 13
|
Convex pentagon $ABCDE$ has side lengths $AB=5$ , $BC=CD=DE=6$ , and $EA=7$ . Moreover, the pentagon has an inscribed circle (a circle tangent to each side of the pentagon). Find the area of $ABCDE$ .
| 60
|
II
|
[
"Assume the incircle touches $AB$, $BC$, $CD$, $DE$, $EA$ at $P,Q,R,S,T$ respectively. Then let $PB=x=BQ=RD=SD$, $ET=y=ES=CR=CQ$, $AP=AT=z$. So we have $x+y=6$, $x+z=5$ and $y+z$=7, solve it we have $x=2$, $z=3$, $y=4$. Let the center of the incircle be $I$, by SAS we can proof triangle $BIQ$ is congruent to triangle $DIS$, and triangle $CIR$ is congruent to triangle $SIE$. Then we have $\\angle AED=\\angle BCD$, $\\angle ABC=\\angle CDE$. Extend $CD$, cross ray $AB$ at $M$, ray $AE$ at $N$, then by AAS we have triangle $END$ is congruent to triangle $BMC$. Thus $\\angle M=\\angle N$. Let $EN=MC=a$, then $BM=DN=a+2$. So by law of cosine in triangle $END$ and triangle $ANM$ we can obtain \\[\\frac{2a+8}{2(a+7)}=\\cos N=\\frac{a^2+(a+2)^2-36}{2a(a+2)}\\], solved it gives us $a=8$, which yield triangle $ANM$ to be a triangle with side length 15, 15, 24, draw a height from $A$ to $NM$ divides it into two triangles with side lengths 9, 12, 15, so the area of triangle $ANM$ is 108. Triangle $END$ is a triangle with side lengths 6, 8, 10, so the area of two of them is 48, so the area of pentagon is $108-48=60. -Fanyuchen20020715",
"Suppose that the circle intersects $\\overline{AB}$, $\\overline{BC}$, $\\overline{CD}$, $\\overline{DE}$, and $\\overline{EA}$ at $P$, $Q$, $R$, $S$, and $T$ respectively. Then $AT = AP = a$, $BP = BQ = b$, $CQ = CR = c$, $DR = DS = d$, and $ES = ET = e$. So $a + b = 5$, $b + c = 6$, $c + d = 6$, $d + e = 6$, and $e + a = 7$. Then $2a + 2b + 2c + 2d + 2e = 30$, so $a + b + c + d + e= 15$. Then we can solve for each individually. $a = 3$, $b = 2$, $c = 4$, $d = 2$, and $e = 4$. To find the radius, we notice that $4 \\arctan(\\frac{2}{r}) + 4 \\arctan(\\frac{4}{r}) + 2 \\arctan (\\frac{3}{r}) = 360 ^ \\circ$, or $2 \\arctan(\\frac{2}{r}) + 2 \\arctan(\\frac{4}{r}) + \\arctan (\\frac{3}{r}) = 180 ^ \\circ$. Each of these angles in this could be represented by complex numbers. When two complex numbers are multiplied, their angles add up to create the angle of the resulting complex number. Thus, $(r + 2i)^2 \\cdot (r + 4i)^2 \\cdot (r + 3i)$ is real. Expanding, we get: \\[(r^2 + 4ir - 4)(r^2 + 8ir -16)(r + 3i)\\] \\[(r^4 + 12ir^3 - 52r^2 - 96ir + 64)(r + 3i)\\] On the last expanding, we only multiply the reals with the imaginaries and vice versa, because we only care that the imaginary component equals 0. \\[15ir^4 - 252ir^2 + 192i = 0\\] \\[5r^4 - 84r^2 + 64 = 0\\] \\[(5r^2 - 4)(r^2 - 16) = 0\\] $r$ must equal 4, as r cannot be negative or be approximately equal to 1. Thus, the area of $ABCDE$ is $4 \\cdot (a + b + c + d + e) = 4 \\cdot 15 = 60 -nihao4112",
"Suppose that the circle intersects $\\overline{AB}$, $\\overline{BC}$, $\\overline{CD}$, $\\overline{DE}$, and $\\overline{EA}$ at $P$, $Q$, $R$, $S$, and $T$ respectively. Then $AT = AP = a$, $BP = BQ = b$, $CQ = CR = c$, $DR = DS = d$, and $ES = ET = e$. So $a + b = 5$, $b + c = 6$, $c + d = 6$, $d + e = 6$, and $e + a = 7$. Then $2a + 2b + 2c + 2d + 2e = 30$, so $a + b + c + d + e= 15$. Then we can solve for each individually. $a = 3$, $b = 2$, $c = 4$, $d = 2$, and $e = 4$. To find the radius, we notice that $4 \\arctan(\\frac{2}{r}) + 4 \\arctan(\\frac{4}{r}) + 2 \\arctan (\\frac{3}{r}) = 360 ^ \\circ$, or $2 \\arctan(\\frac{2}{r}) + 2 \\arctan(\\frac{4}{r}) + \\arctan (\\frac{3}{r}) = 180 ^ \\circ$. Each of these angles in this could be represented by complex numbers. When two complex numbers are multiplied, their angles add up to create the angle of the resulting complex number. Thus, $(r + 2i)^2 \\cdot (r + 4i)^2 \\cdot (r + 3i)$ is real. Expanding, we get: \\[(r^2 + 4ir - 4)(r^2 + 8ir -16)(r + 3i)\\] \\[(r^4 + 12ir^3 - 52r^2 - 96ir + 64)(r + 3i)\\] On the last expanding, we only multiply the reals with the imaginaries and vice versa, because we only care that the imaginary component equals 0. \\[15ir^4 - 252ir^2 + 192i = 0\\] \\[5r^4 - 84r^2 + 64 = 0\\] \\[(5r^2 - 4)(r^2 - 16) = 0\\] $r$ must equal 4, as r cannot be negative or be approximately equal to 1. Thus, the area of $ABCDE$ is $4 \\cdot (a + b + c + d + e) = 4 \\cdot 15 = 60 -nihao4112",
"This pentagon is very close to a regular pentagon with side lengths $6$. The area of a regular pentagon with side lengths $s$ is $\\frac{5s^2}{4\\sqrt{5-2\\sqrt{5}}}$. $5-2\\sqrt{5}$ is slightly greater than $\\frac{1}{2}$ given that $2\\sqrt{5}$ is slightly less than $\\frac{9}{2}$. $4\\sqrt{5-2\\sqrt{5}}$ is then slightly greater than $2\\sqrt{2}$. We will approximate that to be $2.9$. The area is now roughly $\\frac{180}{2.9}$, but because the actual pentagon is not regular, but has the same perimeter of the regular one that we are comparing to we can say that this is an overestimate on the area and turn the $2.9$ into $3$ thus turning the area into $\\frac{180}{3}$ which is $60$ and since $60$ is a multiple of the semiperimeter $15$, we can safely say that the answer is most likely $60. ~Lopkiloinm",
"This pentagon is very close to a regular pentagon with side lengths $6$. The area of a regular pentagon with side lengths $s$ is $\\frac{5s^2}{4\\sqrt{5-2\\sqrt{5}}}$. $5-2\\sqrt{5}$ is slightly greater than $\\frac{1}{2}$ given that $2\\sqrt{5}$ is slightly less than $\\frac{9}{2}$. $4\\sqrt{5-2\\sqrt{5}}$ is then slightly greater than $2\\sqrt{2}$. We will approximate that to be $2.9$. The area is now roughly $\\frac{180}{2.9}$, but because the actual pentagon is not regular, but has the same perimeter of the regular one that we are comparing to we can say that this is an overestimate on the area and turn the $2.9$ into $3$ thus turning the area into $\\frac{180}{3}$ which is $60$ and since $60$ is a multiple of the semiperimeter $15$, we can safely say that the answer is most likely $60. ~Lopkiloinm",
"Because the AIME answers have to be a whole number it would meant the radius of the circle have to be a whole number, thus by drawing the diagram and experimenting, we can safely say the radius is 4 and the answer is 60 (Edit: While the guess would be technically correct, the assumption that the radius would have to be a whole number for the ans to be a whole number is wrong) By EtherealMidnight (Edit: I think that will actually work because the area of $ABCDE$ is equal to the semi-perimeter times the radius. By a simple calculation, we know that the semi-perimeter is an integer so the radius should also be an integer) By YBSuburbanTea ...the radius could be a fraction with denominator 3, 5, or 15, and the area of the pentagon would still be an integer. - GeometryJake",
"Because the AIME answers have to be a whole number it would meant the radius of the circle have to be a whole number, thus by drawing the diagram and experimenting, we can safely say the radius is 4 and the answer is 60 (Edit: While the guess would be technically correct, the assumption that the radius would have to be a whole number for the ans to be a whole number is wrong) By EtherealMidnight (Edit: I think that will actually work because the area of $ABCDE$ is equal to the semi-perimeter times the radius. By a simple calculation, we know that the semi-perimeter is an integer so the radius should also be an integer) By YBSuburbanTea ...the radius could be a fraction with denominator 3, 5, or 15, and the area of the pentagon would still be an integer. - GeometryJake",
"Let $\\omega$ be the inscribed circle, $I$ be its center, and $r$ be its radius. The area of $ABCDE$ is equal to its semiperimeter, $15,$ times $r$, so the problem is reduced to finding $r$. Let $a$ be the length of the tangent segment from $A$ to $\\omega$, and analogously define $b$, $c$, $d$, and $e$. Then $a+b=5$, $b+c= c+d=d+e=6$, and $e+a=7$, with a total of $a+b+c+d+e=15$. Hence $a=3$, $b=d=2$, and $c=e=4$. It follows that $\\angle B= \\angle D$ and $\\angle C= \\angle E$. Let $Q$ be the point where $\\omega$ is tangent to $\\overline{CD}$. Then $\\angle IAE = \\angle IAB =\\frac{1}{2}\\angle A$. Now we claim that points $A, I, Q$ are collinear, which can be proved if $\\angle{AIQ}=\\angle{QIA}=180^{\\circ}$. The sum of the internal angles in polygons $ABCQI$ and $AIQDE$ are equal, so $\\angle IAE + \\angle AIQ + \\angle IQD + \\angle D + \\angle E = \\angle IAB + \\angle B + \\angle C + \\angle CQI + \\angle QIA$, which implies that $\\angle AIQ$ must be $180^\\circ$. Therefore points $A$, $I$, and $Q$ are collinear. [asy] defaultpen(fontsize(8pt)); unitsize(0.025cm); pair[] vertices = {(0,0), (5,0), (8.6,4.8), (3.8,8.4), (-1.96, 6.72)}; string[] labels = {\"$A$\", \"$B$\", \"$C$\", \"$D$\", \"$E$\"}; pair[] dirs = {SW, SE,E, N, NW}; string[] smallLabels = {\"$a$\", \"$b$\", \"$c$\", \"$d$\", \"$e$\"}; pair I = (3,4); real rad = 4; pair Q = foot(I, vertices[2], vertices[3]); pair[] interpoints = {}; for(int i =0; i<vertices.length; ++i){ interpoints.push(foot(I, vertices[i], vertices[(i+1)%vertices.length])); } for(int i = 0; i< vertices.length; ++i){ draw(vertices[i]--vertices[(i+1)%vertices.length]); dot(labels[i],vertices[i],dirs[i]); draw(I--vertices[i]); } draw(Circle(I, rad)); dot(\"$I$\", I, dir(200)); draw(I--Q); dot(\"$Q$\", Q, NE); for(int i = 0; i < vertices.length; ++i){ label(smallLabels[i], vertices[i] --interpoints[i]); //dot(interpoints[i], blue); label(smallLabels[i], interpoints[(i-1)%vertices.length] -- vertices[i]); } [/asy] Because $\\overline{AQ} \\perp \\overline{CD}$, it follows that\\[AC^2-AD^2=CQ^2-DQ^2=c^2-d^2=12.\\]Another expression for $AC^2-AD^2$ can be found as follows. Note that $\\tan \\left(\\frac{\\angle B}{2}\\right) = \\frac{r}{2}$ and $\\tan \\left(\\frac{\\angle E}{2}\\right) = \\frac{r}{4}$, so \\[\\cos (\\angle B) =\\frac{1-\\tan^2 \\left(\\frac{\\angle B}{2}\\right)}{1+\\tan^2 \\left(\\frac{\\angle B}{2}\\right)} = \\frac{4-r^2}{4+r^2}\\]and \\[\\cos (\\angle E) = \\frac{1-\\tan^2 \\left(\\frac{\\angle E}{2}\\right)}{1+\\tan^2 \\left(\\frac{\\angle E}{2}\\right)}= \\frac{16-r^2}{16+r^2}.\\]Applying the Law of Cosines to $\\triangle ABC$ and $\\triangle AED$ gives \\[AC^2=AB^2+BC^2-2\\cdot AB\\cdot BC\\cdot \\cos (\\angle B) = 5^2+6^2-2 \\cdot 5 \\cdot 6 \\cdot \\frac{4-r^2}{4+r^2}\\] and \\[AD^2=AE^2+DE^2-2 \\cdot AE \\cdot DE \\cdot \\cos(\\angle E) = 7^2+6^2-2 \\cdot 7 \\cdot 6 \\cdot \\frac{16-r^2}{16+r^2}.\\] Hence \\[12=AC^2- AD^2= 5^2-2\\cdot 5 \\cdot 6\\cdot \\frac{4-r^2}{4+r^2} -7^2+2\\cdot 7 \\cdot 6 \\cdot \\frac{16-r^2}{16+r^2},\\] yielding \\[2\\cdot 7 \\cdot 6 \\cdot \\frac{16-r^2}{16+r^2}- 2\\cdot 5 \\cdot 6\\cdot \\frac{4-r^2}{4+r^2}= 36;\\] equivalently \\[7(16-r^2)(4+r^2)-5(4-r^2)(16+r^2) = 3(16+r^2)(4+r^2).\\] Substituting $x=r^2$ gives the quadratic equation $5x^2-84x+64=0$, with solutions $\\frac{42 - 38}{5}=\\frac45$, and $\\frac{42 + 38}{5}= 16$. The solution $r^2=\\frac45$ corresponds to a five-pointed star, which is not convex. Indeed, if $r<3$, then $\\tan \\left(\\frac{\\angle A}{2}\\right)$, $\\tan \\left(\\frac{\\angle C}{2}\\right)$, and $\\tan \\left(\\frac{\\angle E}{2}\\right)$ are less than $1,$ implying that $\\angle A$, $\\angle C$, and $\\angle E$ are acute, which cannot happen in a convex pentagon. Thus $r^2=16$ and $r=4$. The requested area is $15\\cdot4 = 60.",
"Let $\\omega$ be the inscribed circle, $I$ be its center, and $r$ be its radius. The area of $ABCDE$ is equal to its semiperimeter, $15,$ times $r$, so the problem is reduced to finding $r$. Let $a$ be the length of the tangent segment from $A$ to $\\omega$, and analogously define $b$, $c$, $d$, and $e$. Then $a+b=5$, $b+c= c+d=d+e=6$, and $e+a=7$, with a total of $a+b+c+d+e=15$. Hence $a=3$, $b=d=2$, and $c=e=4$. It follows that $\\angle B= \\angle D$ and $\\angle C= \\angle E$. Let $Q$ be the point where $\\omega$ is tangent to $\\overline{CD}$. Then $\\angle IAE = \\angle IAB =\\frac{1}{2}\\angle A$. Now we claim that points $A, I, Q$ are collinear, which can be proved if $\\angle{AIQ}=\\angle{QIA}=180^{\\circ}$. The sum of the internal angles in polygons $ABCQI$ and $AIQDE$ are equal, so $\\angle IAE + \\angle AIQ + \\angle IQD + \\angle D + \\angle E = \\angle IAB + \\angle B + \\angle C + \\angle CQI + \\angle QIA$, which implies that $\\angle AIQ$ must be $180^\\circ$. Therefore points $A$, $I$, and $Q$ are collinear. [asy] defaultpen(fontsize(8pt)); unitsize(0.025cm); pair[] vertices = {(0,0), (5,0), (8.6,4.8), (3.8,8.4), (-1.96, 6.72)}; string[] labels = {\"$A$\", \"$B$\", \"$C$\", \"$D$\", \"$E$\"}; pair[] dirs = {SW, SE,E, N, NW}; string[] smallLabels = {\"$a$\", \"$b$\", \"$c$\", \"$d$\", \"$e$\"}; pair I = (3,4); real rad = 4; pair Q = foot(I, vertices[2], vertices[3]); pair[] interpoints = {}; for(int i =0; i<vertices.length; ++i){ interpoints.push(foot(I, vertices[i], vertices[(i+1)%vertices.length])); } for(int i = 0; i< vertices.length; ++i){ draw(vertices[i]--vertices[(i+1)%vertices.length]); dot(labels[i],vertices[i],dirs[i]); draw(I--vertices[i]); } draw(Circle(I, rad)); dot(\"$I$\", I, dir(200)); draw(I--Q); dot(\"$Q$\", Q, NE); for(int i = 0; i < vertices.length; ++i){ label(smallLabels[i], vertices[i] --interpoints[i]); //dot(interpoints[i], blue); label(smallLabels[i], interpoints[(i-1)%vertices.length] -- vertices[i]); } [/asy] Because $\\overline{AQ} \\perp \\overline{CD}$, it follows that\\[AC^2-AD^2=CQ^2-DQ^2=c^2-d^2=12.\\]Another expression for $AC^2-AD^2$ can be found as follows. Note that $\\tan \\left(\\frac{\\angle B}{2}\\right) = \\frac{r}{2}$ and $\\tan \\left(\\frac{\\angle E}{2}\\right) = \\frac{r}{4}$, so \\[\\cos (\\angle B) =\\frac{1-\\tan^2 \\left(\\frac{\\angle B}{2}\\right)}{1+\\tan^2 \\left(\\frac{\\angle B}{2}\\right)} = \\frac{4-r^2}{4+r^2}\\]and \\[\\cos (\\angle E) = \\frac{1-\\tan^2 \\left(\\frac{\\angle E}{2}\\right)}{1+\\tan^2 \\left(\\frac{\\angle E}{2}\\right)}= \\frac{16-r^2}{16+r^2}.\\]Applying the Law of Cosines to $\\triangle ABC$ and $\\triangle AED$ gives \\[AC^2=AB^2+BC^2-2\\cdot AB\\cdot BC\\cdot \\cos (\\angle B) = 5^2+6^2-2 \\cdot 5 \\cdot 6 \\cdot \\frac{4-r^2}{4+r^2}\\] and \\[AD^2=AE^2+DE^2-2 \\cdot AE \\cdot DE \\cdot \\cos(\\angle E) = 7^2+6^2-2 \\cdot 7 \\cdot 6 \\cdot \\frac{16-r^2}{16+r^2}.\\] Hence \\[12=AC^2- AD^2= 5^2-2\\cdot 5 \\cdot 6\\cdot \\frac{4-r^2}{4+r^2} -7^2+2\\cdot 7 \\cdot 6 \\cdot \\frac{16-r^2}{16+r^2},\\] yielding \\[2\\cdot 7 \\cdot 6 \\cdot \\frac{16-r^2}{16+r^2}- 2\\cdot 5 \\cdot 6\\cdot \\frac{4-r^2}{4+r^2}= 36;\\] equivalently \\[7(16-r^2)(4+r^2)-5(4-r^2)(16+r^2) = 3(16+r^2)(4+r^2).\\] Substituting $x=r^2$ gives the quadratic equation $5x^2-84x+64=0$, with solutions $\\frac{42 - 38}{5}=\\frac45$, and $\\frac{42 + 38}{5}= 16$. The solution $r^2=\\frac45$ corresponds to a five-pointed star, which is not convex. Indeed, if $r<3$, then $\\tan \\left(\\frac{\\angle A}{2}\\right)$, $\\tan \\left(\\frac{\\angle C}{2}\\right)$, and $\\tan \\left(\\frac{\\angle E}{2}\\right)$ are less than $1,$ implying that $\\angle A$, $\\angle C$, and $\\angle E$ are acute, which cannot happen in a convex pentagon. Thus $r^2=16$ and $r=4$. The requested area is $15\\cdot4 = 60.",
"Define $a$, $b$, $c$, $d$, $e$, and $r$ as in Solution 5. Then, as in Solution 5, $a=3$, $b=d=2$, $c=e=4$, $\\angle B= \\angle D$, and $\\angle C= \\angle E$. Let $\\alpha =\\frac{\\angle A}{2}$, $\\beta = \\frac{\\angle B}{2}$, and $\\gamma=\\frac{\\angle C}{2}$. It follows that $540^{\\circ} = 2\\alpha + 4 \\beta + 4 \\gamma$, so $270^{\\circ} = \\alpha + 2\\beta + 2 \\gamma$. Thus \\[\\tan(2\\beta + 2 \\gamma) = \\frac{1}{\\tan \\alpha},\\] $\\tan(\\beta) = \\frac{r}{2}$, $\\tan(\\gamma) = \\frac{r}{4}$, and $\\tan(\\alpha) = \\frac {r}{3}$. By the Tangent Addition Formula, \\[\\tan(\\beta +\\gamma) = \\frac{6r}{8-r^2}\\] and \\[\\tan(2\\beta + 2\\gamma) = \\frac{\\frac{12r}{8-r^2}}{1-\\frac{36r^2}{(8-r^2)^2}} = \\frac{12r(8-r^2)}{(8-r^2)^2-36r^2}.\\] Therefore \\[\\frac{12r(8-r^2)}{(8-r^2)^2-36r^2} = \\frac{3}{r},\\] which simplifies to $5r^4 - 84r^2 + 64 = 0$. Then the solution proceeds as in Solution 5.",
"Define $a$, $b$, $c$, $d$, $e$, and $r$ as in Solution 5. Then, as in Solution 5, $a=3$, $b=d=2$, $c=e=4$, $\\angle B= \\angle D$, and $\\angle C= \\angle E$. Let $\\alpha =\\frac{\\angle A}{2}$, $\\beta = \\frac{\\angle B}{2}$, and $\\gamma=\\frac{\\angle C}{2}$. It follows that $540^{\\circ} = 2\\alpha + 4 \\beta + 4 \\gamma$, so $270^{\\circ} = \\alpha + 2\\beta + 2 \\gamma$. Thus \\[\\tan(2\\beta + 2 \\gamma) = \\frac{1}{\\tan \\alpha},\\] $\\tan(\\beta) = \\frac{r}{2}$, $\\tan(\\gamma) = \\frac{r}{4}$, and $\\tan(\\alpha) = \\frac {r}{3}$. By the Tangent Addition Formula, \\[\\tan(\\beta +\\gamma) = \\frac{6r}{8-r^2}\\] and \\[\\tan(2\\beta + 2\\gamma) = \\frac{\\frac{12r}{8-r^2}}{1-\\frac{36r^2}{(8-r^2)^2}} = \\frac{12r(8-r^2)}{(8-r^2)^2-36r^2}.\\] Therefore \\[\\frac{12r(8-r^2)}{(8-r^2)^2-36r^2} = \\frac{3}{r},\\] which simplifies to $5r^4 - 84r^2 + 64 = 0$. Then the solution proceeds as in Solution 5.",
"Define $a$, $b$, $c$, $d$, $e$, and $r$ as in Solution 5. Note that \\[\\arctan\\left(\\frac{a}{r}\\right) + \\arctan\\left(\\frac{b}{r}\\right) + \\arctan\\left(\\frac{c}{r}\\right) + \\arctan\\left(\\frac{d}{r}\\right) + \\arctan\\left(\\frac{e}{r}\\right) = 180^{\\circ}.\\] Hence \\[\\operatorname{Arg}(r + 3i) + 2\\cdot \\operatorname{Arg}(r + 2i) + 2\\cdot \\operatorname{Arg}(r + 4i) = 180^{\\circ}.\\] Therefore \\[\\operatorname{Im} \\big( (r + 3i)(r+2i)^2(r+4i)^2 \\big) = 0.\\] Simplifying this equation gives the same quadratic equation in $r^2$ as in Solution 5.",
"Define $a$, $b$, $c$, $d$, $e$, and $r$ as in Solution 5. Note that \\[\\arctan\\left(\\frac{a}{r}\\right) + \\arctan\\left(\\frac{b}{r}\\right) + \\arctan\\left(\\frac{c}{r}\\right) + \\arctan\\left(\\frac{d}{r}\\right) + \\arctan\\left(\\frac{e}{r}\\right) = 180^{\\circ}.\\] Hence \\[\\operatorname{Arg}(r + 3i) + 2\\cdot \\operatorname{Arg}(r + 2i) + 2\\cdot \\operatorname{Arg}(r + 4i) = 180^{\\circ}.\\] Therefore \\[\\operatorname{Im} \\big( (r + 3i)(r+2i)^2(r+4i)^2 \\big) = 0.\\] Simplifying this equation gives the same quadratic equation in $r^2$ as in Solution 5.",
"Notation shown on diagram. As in solution 5, we get $\\overline{AQ} \\perp \\overline{CD}, AG = 3, GB = 2, CQ = 4$ and so on. Let $\\overline{AB}$ cross $\\overline{CD}$ at $F, \\overline{AE}$ cross $\\overline{CD}$ at $F', CF = x.$ $FQ = FG \\implies FB = x+2.$ $\\angle BAQ = \\angle EAQ \\implies DF' = x + 2, EF' = x.$ Triangle $\\triangle AFF'$ has semiperimeter $s = 2x + 11.$ The radius of incircle $\\omega$ is $r =\\sqrt{\\frac{s-FF’}{s}}(s-AF) = \\sqrt{\\frac{3}{2x +11}}(x+4).$ Triangle $\\triangle BCF$ has semiperimeter $s = x + 4.$ The radius of excircle $\\omega$ is $r = \\sqrt{\\frac{s(s-BF)(s-CF)}{s-BC}} = \\sqrt{ \\frac{(x+4)\\cdot 2 \\cdot 4}{x - 2}}.$ It is the same radius, therefore \\[\\sqrt{\\frac{3}{2x +11}}(x+4) = \\sqrt{\\frac{8(x+4)}{x – 2}} \\implies \\frac {3(x+4)}{2x+11} = \\frac {8}{x-2} \\implies (x-8)(3x + 14) = 0 \\implies x = 8, r = 4.\\] Then the solution proceeds as in Solution 5. vladimir.shelomovskii@gmail.com, vvsss",
"Notation shown on diagram. As in solution 5, we get $\\overline{AQ} \\perp \\overline{CD}, AG = 3, GB = 2, CQ = 4$ and so on. Let $\\overline{AB}$ cross $\\overline{CD}$ at $F, \\overline{AE}$ cross $\\overline{CD}$ at $F', CF = x.$ $FQ = FG \\implies FB = x+2.$ $\\angle BAQ = \\angle EAQ \\implies DF' = x + 2, EF' = x.$ Triangle $\\triangle AFF'$ has semiperimeter $s = 2x + 11.$ The radius of incircle $\\omega$ is $r =\\sqrt{\\frac{s-FF’}{s}}(s-AF) = \\sqrt{\\frac{3}{2x +11}}(x+4).$ Triangle $\\triangle BCF$ has semiperimeter $s = x + 4.$ The radius of excircle $\\omega$ is $r = \\sqrt{\\frac{s(s-BF)(s-CF)}{s-BC}} = \\sqrt{ \\frac{(x+4)\\cdot 2 \\cdot 4}{x - 2}}.$ It is the same radius, therefore \\[\\sqrt{\\frac{3}{2x +11}}(x+4) = \\sqrt{\\frac{8(x+4)}{x – 2}} \\implies \\frac {3(x+4)}{2x+11} = \\frac {8}{x-2} \\implies (x-8)(3x + 14) = 0 \\implies x = 8, r = 4.\\] Then the solution proceeds as in Solution 5. vladimir.shelomovskii@gmail.com, vvsss"
] |
2020-II-14
| 2,020
| 14
|
For real number $x$ let $\lfloor x\rfloor$ be the greatest integer less than or equal to $x$ , and define $\{x\} = x - \lfloor x \rfloor$ to be the fractional part of $x$ . For example, $\{3\} = 0$ and $\{4.56\} = 0.56$ . Define $f(x)=x\{x\}$ , and let $N$ be the number of real-valued solutions to the equation $f(f(f(x)))=17$ for $0\leq x\leq 2020$ . Find the remainder when $N$ is divided by $1000$ .
| 10
|
II
|
[
"Note that the upper bound for our sum is $2019,$ and not $2020,$ because if it were $2020$ then the function composition cannot equal to $17.$ From there, it's not too hard to see that, by observing the function composition from right to left, $N$ is (note that the summation starts from the right to the left): \\[\\sum_{x=17}^{2019} \\sum_{y=x}^{2019} \\sum_{z=y}^{2019} 1 .\\] One can see an easy combinatorical argument exists which is the official solution, but I will present another solution here for the sake of variety. Applying algebraic manipulation and the hockey-stick identity $3$ times gives \\[\\sum_{x=17}^{2019} \\sum_{y=x}^{2019} \\sum_{z=y}^{2019} 1\\] \\[=\\sum_{x=17}^{2019} \\sum_{y=x}^{2019} \\sum_{z=y}^{2019} \\binom{z-y}{0}\\] \\[=\\sum_{x=17}^{2019} \\sum_{y=x}^{2019} \\binom{2020-y}{1}\\] \\[=\\sum_{x=17}^{2019} \\binom{2021-x}{2}\\] \\[=\\binom{2005}{3}\\] Hence, \\[N = \\frac{2005 \\cdot 2004 \\cdot 2003}{3 \\cdot 2\\cdot 1} \\equiv 010 (\\mathrm{mod} \\hskip .2cm 1000)\\]",
"To solve $f(f(f(x)))=17$, we need to solve $f(x) = y$ where $f(f(y))=17$, and to solve that we need to solve $f(y) = z$ where $f(z) = 17$. It is clear to see for some integer $a \\geq 17$ there is exactly one value of $z$ in the interval $[a, a+1)$ where $f(z) = 17$. To understand this, imagine the graph of $f(z)$ on the interval $[a, a+1)$ The graph starts at $0$, is continuous and increasing, and approaches $a+1$. So as long as $a+1 > 17$, there will be a solution for $z$ in the interval. Using this logic, we can find the number of solutions to $f(x) = y$. For every interval $[a, a+1)$ where $a \\geq \\left \\lfloor{y}\\right \\rfloor$ there will be one solution for $x$ in that interval. However, the question states $0 \\leq x \\leq 2020$, but because $x=2020$ doesn't work we can change it to $0 \\leq x < 2020$. Therefore, $\\left \\lfloor{y}\\right \\rfloor \\leq a \\leq 2019$, and there are $2020 - \\left \\lfloor{y}\\right \\rfloor$ solutions to $f(x) = y$. We can solve $f(y) = z$ similarly. $0 \\leq y < 2020$ to satisfy the bounds of $x$, so there are $2020 - \\left \\lfloor{z}\\right \\rfloor$ solutions to $f(y) = z$, and $0 \\leq z < 2020$ to satisfy the bounds of $y$. Going back to $f(z) = 17$, there is a single solution for z in the interval $[a, a+1)$, where $17 \\leq a \\leq 2019$. (We now have an upper bound for $a$ because we know $z < 2020$.) There are $2003$ solutions for $z$, and the floors of these solutions create the sequence $17, 18, 19, ..., 2018, 2019$ Lets first look at the solution of $z$ where $\\left \\lfloor{z}\\right \\rfloor = 17$. Then $f(y) = z$ would have $2003$ solutions, and the floors of these solutions would also create the sequence $17, 18, 19, ..., 2018, 2019$. If we used the solution of $y$ where $\\left \\lfloor{y}\\right \\rfloor = 17$, there would be $2003$ solutions for $f(x) = y$. If we used the solution of $y$ where $\\left \\lfloor{y}\\right \\rfloor = 18$, there would be $2002$ solutions for $x$, and so on. So for the solution of $z$ where $\\left \\lfloor{z}\\right \\rfloor = 17$, there will be $2003 + 2002 + 2001 + ... + 2 + 1 = \\binom{2004}{2}$ solutions for $x$ If we now look at the solution of $z$ where $\\left \\lfloor{z}\\right \\rfloor = 18$, there would be $\\binom{2003}{2}$ solutions for $x$. If we looked at the solution of $z$ where $\\left \\lfloor{z}\\right \\rfloor = 19$, there would be $\\binom{2002}{2}$ solutions for $x$, and so on. The total number of solutions to $x$ is $\\binom{2004}{2} + \\binom{2003}{2} + \\binom{2002}{2} + ... + \\binom{3}{2} + \\binom{2}{2}$. Using the hockey stick theorem, we see this equals $\\binom{2005}{3}$, and when we take the remainder of that number when divided by $1000$, we get the answer, $10 ~aragornmf",
"For any nonnegative integer $n$, the function $f$ increases on the interval $[n,n+1)$, with $f(n)=0$ and $f(x)<n+1$ for every $x$ in this interval. On this interval $f(x)=x(x-n)$, which takes on every real value in the interval $[0,n+1)$ exactly once. Thus for each nonnegative real number $y$, the equation $f(x) = y$ has exactly one solution $x \\in [n, n+1)$ for every $n \\ge \\lfloor y \\rfloor$. For each integer $a\\geq 17$ there is exactly one $x$ with $\\lfloor x\\rfloor=a$ such that $f(x)=17$; likewise for each integer $b\\geq a\\geq 17$ there is exactly one $x$ with $\\lfloor f(x)\\rfloor=a$ and $\\lfloor x\\rfloor=b$ such that $f(f(x))=17$. Finally, for each integer $c \\ge b \\ge a \\ge 17$ there is exactly one $x$ with $\\lfloor f(f(x)) \\rfloor = a$, $\\lfloor f(x)\\rfloor=b$, and $\\lfloor x\\rfloor=c$ such that $f(f(f(x)))=17$. Thus $f(f(f(x)))=17$ has exactly one solution $x$ with $0\\leq x\\leq 2020$ for each triple of integers $(a,b,c)$ with $17\\leq a\\leq b\\leq c<2020$, noting that $x=2020$ is not a solution. This nondecreasing ordered triple can be identified with a multiset of three elements of the set of $2003$ integers $\\{17,18,19,\\ldots,2019\\}$, which can be selected in $\\binom{2005}3$ ways. Thus \\[N=\\frac{2005\\cdot 2004\\cdot 2003}{6}\\equiv 10\\hskip -.2cm \\pmod{1000}.\\]",
"For any nonnegative integer $n$, the function $f$ increases on the interval $[n,n+1)$, with $f(n)=0$ and $f(x)<n+1$ for every $x$ in this interval. On this interval $f(x)=x(x-n)$, which takes on every real value in the interval $[0,n+1)$ exactly once. Thus for each nonnegative real number $y$, the equation $f(x) = y$ has exactly one solution $x \\in [n, n+1)$ for every $n \\ge \\lfloor y \\rfloor$. For each integer $a\\geq 17$ there is exactly one $x$ with $\\lfloor x\\rfloor=a$ such that $f(x)=17$; likewise for each integer $b\\geq a\\geq 17$ there is exactly one $x$ with $\\lfloor f(x)\\rfloor=a$ and $\\lfloor x\\rfloor=b$ such that $f(f(x))=17$. Finally, for each integer $c \\ge b \\ge a \\ge 17$ there is exactly one $x$ with $\\lfloor f(f(x)) \\rfloor = a$, $\\lfloor f(x)\\rfloor=b$, and $\\lfloor x\\rfloor=c$ such that $f(f(f(x)))=17$. Thus $f(f(f(x)))=17$ has exactly one solution $x$ with $0\\leq x\\leq 2020$ for each triple of integers $(a,b,c)$ with $17\\leq a\\leq b\\leq c<2020$, noting that $x=2020$ is not a solution. This nondecreasing ordered triple can be identified with a multiset of three elements of the set of $2003$ integers $\\{17,18,19,\\ldots,2019\\}$, which can be selected in $\\binom{2005}3$ ways. Thus \\[N=\\frac{2005\\cdot 2004\\cdot 2003}{6}\\equiv 10\\hskip -.2cm \\pmod{1000}.\\]"
] |
2020-II-15
| 2,020
| 15
|
Let $\triangle ABC$ be an acute scalene triangle with circumcircle $\omega$ . The tangents to $\omega$ at $B$ and $C$ intersect at $T$ . Let $X$ and $Y$ be the projections of $T$ onto lines $AB$ and $AC$ , respectively. Suppose $BT = CT = 16$ , $BC = 22$ , and $TX^2 + TY^2 + XY^2 = 1143$ . Find $XY^2$ .
| 717
|
II
|
[
"Let $O$ be the circumcenter of $\\triangle ABC$; say $OT$ intersects $BC$ at $M$; draw segments $XM$, and $YM$. We have $MT=3\\sqrt{15}$. Since $\\angle A=\\angle CBT=\\angle BCT$, we have $\\cos A=\\tfrac{11}{16}$. Notice that $AXTY$ is cyclic, so $\\angle XTY=180^{\\circ}-A$, so $\\cos XTY=-\\cos A$, and the cosine law in $\\triangle TXY$ gives \\[1143-2XY^2=-\\frac{11}{8}\\cdot XT\\cdot YT.\\] Since $\\triangle BMT \\cong \\triangle CMT$, we have $TM\\perp BC$, and therefore quadrilaterals $BXTM$ and $CYTM$ are cyclic. Let $P$ (resp. $Q$) be the midpoint of $BT$ (resp. $CT$). So $P$ (resp. $Q$) is the center of $(BXTM)$ (resp. $CYTM$). Then $\\theta=\\angle ABC=\\angle MTX$ and $\\phi=\\angle ACB=\\angle YTM$. So $\\angle XPM=2\\theta$, so\\[\\frac{\\frac{XM}{2}}{XP}=\\sin \\theta,\\]which yields $XM=2XP\\sin \\theta=BT(=CT)\\sin \\theta=TY$. Similarly we have $YM=XT$. Ptolemy's theorem in $BXTM$ gives \\[16TY=11TX+3\\sqrt{15}BX,\\] while Pythagoras' theorem gives $BX^2+XT^2=16^2$. Similarly, Ptolemy's theorem in $YTMC$ gives\\[16TX=11TY+3\\sqrt{15}CY\\] while Pythagoras' theorem in $\\triangle CYT$ gives $CY^2+YT^2=16^2$. Solve this for $XT$ and $TY$ and substitute into the equation about $\\cos XTY$ to obtain the result $XY^2=717 is a parallelogram, which is an important theorem in Olympiad, and there are some other ways of computation under this observation.) -Fanyuchen20020715",
"Let $M$ denote the midpoint of $\\overline{BC}$. The critical claim is that $M$ is the orthocenter of $\\triangle AXY$, which has the circle with diameter $\\overline{AT}$ as its circumcircle. To see this, note that because $\\angle BXT = \\angle BMT = 90^\\circ$, the quadrilateral $MBXT$ is cyclic, it follows that \\[\\angle MXA = \\angle MXB = \\angle MTB = 90^\\circ - \\angle TBM = 90^\\circ - \\angle A,\\] implying that $\\overline{MX} \\perp \\overline{AC}$. Similarly, $\\overline{MY} \\perp \\overline{AB}$. In particular, $MXTY$ is a parallelogram. [asy] defaultpen(fontsize(8pt)); unitsize(0.8cm); pair A = (0,0); pair B = (-1.26,-4.43); pair C = (-1.26+3.89, -4.43); pair M = (B+C)/2; pair O = circumcenter(A,B,C); pair T = (0.68, -6.49); pair X = foot(T,A,B); pair Y = foot(T,A,C); path omega = circumcircle(A,B,C); real rad = circumradius(A,B,C); filldraw(A--B--C--cycle, 0.2*royalblue+white); label(\"$\\omega$\", O + rad*dir(45), SW); //filldraw(T--Y--M--X--cycle, rgb(150, 247, 254)); filldraw(T--Y--M--X--cycle, 0.2*heavygreen+white); draw(M--T); draw(X--Y); draw(B--T--C); draw(A--X--Y--cycle); draw(omega); dot(\"$X$\", X, W); dot(\"$Y$\", Y, E); dot(\"$O$\", O, W); dot(\"$T$\", T, S); dot(\"$A$\", A, N); dot(\"$B$\", B, W); dot(\"$C$\", C, E); dot(\"$M$\", M, N); [/asy] Hence, by the Parallelogram Law, \\[TM^2 + XY^2 = 2(TX^2 + TY^2) = 2(1143-XY^2).\\] But $TM^2 = TB^2 - BM^2 = 16^2 - 11^2 = 135$. Therefore \\[XY^2 = \\frac13(2 \\cdot 1143-135) = 717.\\]",
"Let $M$ denote the midpoint of $\\overline{BC}$. The critical claim is that $M$ is the orthocenter of $\\triangle AXY$, which has the circle with diameter $\\overline{AT}$ as its circumcircle. To see this, note that because $\\angle BXT = \\angle BMT = 90^\\circ$, the quadrilateral $MBXT$ is cyclic, it follows that \\[\\angle MXA = \\angle MXB = \\angle MTB = 90^\\circ - \\angle TBM = 90^\\circ - \\angle A,\\] implying that $\\overline{MX} \\perp \\overline{AC}$. Similarly, $\\overline{MY} \\perp \\overline{AB}$. In particular, $MXTY$ is a parallelogram. [asy] defaultpen(fontsize(8pt)); unitsize(0.8cm); pair A = (0,0); pair B = (-1.26,-4.43); pair C = (-1.26+3.89, -4.43); pair M = (B+C)/2; pair O = circumcenter(A,B,C); pair T = (0.68, -6.49); pair X = foot(T,A,B); pair Y = foot(T,A,C); path omega = circumcircle(A,B,C); real rad = circumradius(A,B,C); filldraw(A--B--C--cycle, 0.2*royalblue+white); label(\"$\\omega$\", O + rad*dir(45), SW); //filldraw(T--Y--M--X--cycle, rgb(150, 247, 254)); filldraw(T--Y--M--X--cycle, 0.2*heavygreen+white); draw(M--T); draw(X--Y); draw(B--T--C); draw(A--X--Y--cycle); draw(omega); dot(\"$X$\", X, W); dot(\"$Y$\", Y, E); dot(\"$O$\", O, W); dot(\"$T$\", T, S); dot(\"$A$\", A, N); dot(\"$B$\", B, W); dot(\"$C$\", C, E); dot(\"$M$\", M, N); [/asy] Hence, by the Parallelogram Law, \\[TM^2 + XY^2 = 2(TX^2 + TY^2) = 2(1143-XY^2).\\] But $TM^2 = TB^2 - BM^2 = 16^2 - 11^2 = 135$. Therefore \\[XY^2 = \\frac13(2 \\cdot 1143-135) = 717.\\]",
"Let $H$ be the orthocenter of $\\triangle AXY$. Lemma 1: $H$ is the midpoint of $BC$. Proof: Let $H'$ be the midpoint of $BC$, and observe that $XBH'T$ and $TH'CY$ are cyclical. Define $H'Y \\cap BA=E$ and $H'X \\cap AC=F$, then note that: \\[\\angle H'BT=\\angle H'CT=\\angle H'XT=\\angle H'YT=\\angle A.\\] That implies that $\\angle H'XB=\\angle H'YC=90^\\circ-\\angle A$, $\\angle CH'Y=\\angle EH'B=90^\\circ-\\angle B$, and $\\angle BH'Y=\\angle FH'C=90^\\circ-\\angle C$. Thus $YH'\\perp AX$ and $XH' \\perp AY$; $H'$ is indeed the same as $H$, and we have proved lemma 1. Since $AXTY$ is cyclical, $\\angle XTY=\\angle XHY$ and this implies that $XHYT$ is a paralelogram. By the Law of Cosines: \\[XY^2=XT^2+TY^2+2(XT)(TY)\\cdot \\cos(\\angle A)\\] \\[XY^2=XH^2+HY^2+2(XH)(HY) \\cdot \\cos(\\angle A)\\] \\[HT^2=HX^2+XT^2-2(HX)(XT) \\cdot \\cos(\\angle A)\\] \\[HT^2=HY^2+YT^2-2(HY)(YT) \\cdot \\cos(\\angle A).\\] We add all these equations to get: \\[HT^2+XY^2=2(XT^2+TY^2) \\qquad (1).\\] We have that $BH=HC=11$ and $BT=TC=16$ using our midpoints. Note that $HT \\perp BC$, so by the Pythagorean Theorem, it follows that $HT^2=135$. We were also given that $XT^2+TY^2=1143-XY^2$, which we multiply by $2$ to use equation $(1)$. \\[2(XT^2+TY^2)=2286-2 \\cdot XY^2\\] Since $2(XT^2+TY^2)=2(HT^2+TY^2)=HT^2+XY^2$, we have \\[135+XY^2=2286-2 \\cdot XY^2\\] \\[3 \\cdot XY^2=2151.\\] Therefore, $XY^2=717. ~ MathLuis",
"Let $H$ be the orthocenter of $\\triangle AXY$. Lemma 1: $H$ is the midpoint of $BC$. Proof: Let $H'$ be the midpoint of $BC$, and observe that $XBH'T$ and $TH'CY$ are cyclical. Define $H'Y \\cap BA=E$ and $H'X \\cap AC=F$, then note that: \\[\\angle H'BT=\\angle H'CT=\\angle H'XT=\\angle H'YT=\\angle A.\\] That implies that $\\angle H'XB=\\angle H'YC=90^\\circ-\\angle A$, $\\angle CH'Y=\\angle EH'B=90^\\circ-\\angle B$, and $\\angle BH'Y=\\angle FH'C=90^\\circ-\\angle C$. Thus $YH'\\perp AX$ and $XH' \\perp AY$; $H'$ is indeed the same as $H$, and we have proved lemma 1. Since $AXTY$ is cyclical, $\\angle XTY=\\angle XHY$ and this implies that $XHYT$ is a paralelogram. By the Law of Cosines: \\[XY^2=XT^2+TY^2+2(XT)(TY)\\cdot \\cos(\\angle A)\\] \\[XY^2=XH^2+HY^2+2(XH)(HY) \\cdot \\cos(\\angle A)\\] \\[HT^2=HX^2+XT^2-2(HX)(XT) \\cdot \\cos(\\angle A)\\] \\[HT^2=HY^2+YT^2-2(HY)(YT) \\cdot \\cos(\\angle A).\\] We add all these equations to get: \\[HT^2+XY^2=2(XT^2+TY^2) \\qquad (1).\\] We have that $BH=HC=11$ and $BT=TC=16$ using our midpoints. Note that $HT \\perp BC$, so by the Pythagorean Theorem, it follows that $HT^2=135$. We were also given that $XT^2+TY^2=1143-XY^2$, which we multiply by $2$ to use equation $(1)$. \\[2(XT^2+TY^2)=2286-2 \\cdot XY^2\\] Since $2(XT^2+TY^2)=2(HT^2+TY^2)=HT^2+XY^2$, we have \\[135+XY^2=2286-2 \\cdot XY^2\\] \\[3 \\cdot XY^2=2151.\\] Therefore, $XY^2=717. ~ MathLuis",
"Using the Claim (below) we get $\\triangle ABC \\sim \\triangle XTM \\sim \\triangle YMT.$ Corresponding sides of similar $\\triangle XTM \\sim \\triangle YMT$ is $MT,$ so $\\triangle XTM = \\triangle YMT \\implies MY = XT, MX = TY \\implies XMYT$ – parallelogram. \\[4 TD^2 = MT^2 = \\sqrt{BT^2 - BM^2} =\\sqrt{153}.\\] The formula for median $DT$ of triangle $XYT$ is \\[2 DT^2 = XT^2 + TY^2 – \\frac{XY^2}{2},\\] \\[3 \\cdot XY^2 = 2XT^2 + 2TY^2 + 2XY^2 – 4 DT^2,\\] \\[3 \\cdot XY^2 = 2 \\cdot 1143-153 = 2151 \\implies XY^2 = 717 vladimir.shelomovskii@gmail.com, vvsss",
"Using the Claim (below) we get $\\triangle ABC \\sim \\triangle XTM \\sim \\triangle YMT.$ Corresponding sides of similar $\\triangle XTM \\sim \\triangle YMT$ is $MT,$ so $\\triangle XTM = \\triangle YMT \\implies MY = XT, MX = TY \\implies XMYT$ – parallelogram. \\[4 TD^2 = MT^2 = \\sqrt{BT^2 - BM^2} =\\sqrt{153}.\\] The formula for median $DT$ of triangle $XYT$ is \\[2 DT^2 = XT^2 + TY^2 – \\frac{XY^2}{2},\\] \\[3 \\cdot XY^2 = 2XT^2 + 2TY^2 + 2XY^2 – 4 DT^2,\\] \\[3 \\cdot XY^2 = 2 \\cdot 1143-153 = 2151 \\implies XY^2 = 717 vladimir.shelomovskii@gmail.com, vvsss",
"Let $M$ be the midpoint of $BC$. Note that $\\angle XYT = \\angle XAT = \\angle MAC$ because $AT$ is a symmedian. Similarly, $\\angle TXY = \\angle MAB$. $XT = 16\\sin{C}$ and $YT = 16\\sin{C}$. By law of sines on $XYT$, $\\frac{16\\sin{C}}{\\sin{XYT}} = \\frac{XY}{\\sin{A}}$. However by law of sines on $MAC$, $\\frac{\\sin{C}}{\\sin{MAC}} = \\frac{AM}{11}$. Combining these two yields, $\\frac{16}{11} AM = \\frac{XY}{\\sin{A}}$. Since $\\sin{A} = \\frac{\\sqrt{135}}{16}$, we have $XY^2 = \\frac{135}{121} AM^2$. Letting $AB = x$, $AC = y$, and $AM = d$, we have $x^2 + y^2 = 2d^2 + 242$ by Stewarts. Since $x = 2R\\sin{C}$, and $y = 2R\\sin{B}$ by extended law of sines, we can write $\\sin^2{B} + \\sin^2{C} = \\frac{2d^2 + 242}{4R^2}$. By law of cosines on $ABC$, $x^2 + y^2 - 2xy(\\frac{11}{16}) = 484$, $\\frac{11}{8} xy + 484 = 2d^2 + 242$, $xy = \\frac{8}{11} (2d^2 - 242)$. Then similarly as before we can write $\\sin{B}\\sin{C} = \\frac{\\frac{8}{11} (2d^2 - 242)}{4R^2}$. By law of cosines on $XYT$ and using $XT^2 + XY^2 + YT^2 = 1143$, we have $512(\\sin^2{B} + \\sin^2{C}) - 1143 = -512\\sin{B}\\sin{C}(\\frac{11}{16})$. Substituting our previous values here and using $R = \\frac{176}{\\sqrt{135}}$ yields a value for $d^2$, and multiplying by $\\frac{135}{121}$ gives $717. ~sdfgfjh",
"Let $M$ be the midpoint of $BC$. Note that $\\angle XYT = \\angle XAT = \\angle MAC$ because $AT$ is a symmedian. Similarly, $\\angle TXY = \\angle MAB$. $XT = 16\\sin{C}$ and $YT = 16\\sin{C}$. By law of sines on $XYT$, $\\frac{16\\sin{C}}{\\sin{XYT}} = \\frac{XY}{\\sin{A}}$. However by law of sines on $MAC$, $\\frac{\\sin{C}}{\\sin{MAC}} = \\frac{AM}{11}$. Combining these two yields, $\\frac{16}{11} AM = \\frac{XY}{\\sin{A}}$. Since $\\sin{A} = \\frac{\\sqrt{135}}{16}$, we have $XY^2 = \\frac{135}{121} AM^2$. Letting $AB = x$, $AC = y$, and $AM = d$, we have $x^2 + y^2 = 2d^2 + 242$ by Stewarts. Since $x = 2R\\sin{C}$, and $y = 2R\\sin{B}$ by extended law of sines, we can write $\\sin^2{B} + \\sin^2{C} = \\frac{2d^2 + 242}{4R^2}$. By law of cosines on $ABC$, $x^2 + y^2 - 2xy(\\frac{11}{16}) = 484$, $\\frac{11}{8} xy + 484 = 2d^2 + 242$, $xy = \\frac{8}{11} (2d^2 - 242)$. Then similarly as before we can write $\\sin{B}\\sin{C} = \\frac{\\frac{8}{11} (2d^2 - 242)}{4R^2}$. By law of cosines on $XYT$ and using $XT^2 + XY^2 + YT^2 = 1143$, we have $512(\\sin^2{B} + \\sin^2{C}) - 1143 = -512\\sin{B}\\sin{C}(\\frac{11}{16})$. Substituting our previous values here and using $R = \\frac{176}{\\sqrt{135}}$ yields a value for $d^2$, and multiplying by $\\frac{135}{121}$ gives $717. ~sdfgfjh"
] |
2021-I-1
| 2,021
| 1
|
Zou and Chou are practicing their $100$ -meter sprints by running $6$ races against each other. Zou wins the first race, and after that, the probability that one of them wins a race is $\frac23$ if they won the previous race but only $\frac13$ if they lost the previous race. The probability that Zou will win exactly $5$ of the $6$ races is $\frac mn$ , where $m$ and $n$ are relatively prime positive integers. Find $m+n$ .
| 97
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I
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[
"For the next five races, Zou wins four and loses one. Let $W$ and $L$ denote a win and a loss, respectively. There are five possible outcome sequences for Zou: $LWWWW$ $WLWWW$ $WWLWW$ $WWWLW$ $WWWWL$ We proceed with casework: Case (1): Sequences #1-4, in which Zou does not lose the last race. The probability that Zou loses a race is $\\frac13,$ and the probability that Zou wins the next race is $\\frac13.$ For each of the three other races, the probability that Zou wins is $\\frac23.$ There are four sequences in this case. The probability of one such sequence is $\\left(\\frac13\\right)^2\\left(\\frac23\\right)^3.$ Case (2): Sequence #5, in which Zou loses the last race. The probability that Zou loses a race is $\\frac13.$ For each of the four other races, the probability that Zou wins is $\\frac23.$ There is one sequence in this case. The probability is $\\left(\\frac13\\right)^1\\left(\\frac23\\right)^4.$ Answer The requested probability is \\[4\\left(\\frac13\\right)^2\\left(\\frac23\\right)^3+\\left(\\frac13\\right)^1\\left(\\frac23\\right)^4=\\frac{32}{243}+\\frac{16}{243}=\\frac{48}{243}=\\frac{16}{81},\\] from which the answer is $16+81=097 ~MRENTHUSIASM",
"For the next five races, Zou wins four and loses one. Let $W$ and $L$ denote a win and a loss, respectively. There are five possible outcome sequences for Zou: $LWWWW$ $WLWWW$ $WWLWW$ $WWWLW$ $WWWWL$ We proceed with casework: Case (1): Sequences #1-4, in which Zou does not lose the last race. The probability that Zou loses a race is $\\frac13,$ and the probability that Zou wins the next race is $\\frac13.$ For each of the three other races, the probability that Zou wins is $\\frac23.$ There are four sequences in this case. The probability of one such sequence is $\\left(\\frac13\\right)^2\\left(\\frac23\\right)^3.$ Case (2): Sequence #5, in which Zou loses the last race. The probability that Zou loses a race is $\\frac13.$ For each of the four other races, the probability that Zou wins is $\\frac23.$ There is one sequence in this case. The probability is $\\left(\\frac13\\right)^1\\left(\\frac23\\right)^4.$ Answer The requested probability is \\[4\\left(\\frac13\\right)^2\\left(\\frac23\\right)^3+\\left(\\frac13\\right)^1\\left(\\frac23\\right)^4=\\frac{32}{243}+\\frac{16}{243}=\\frac{48}{243}=\\frac{16}{81},\\] from which the answer is $16+81=097 ~MRENTHUSIASM",
"We have $5$ cases, depending on which race Zou lost. Let $\\text{W}$ denote a won race, and $\\text{L}$ denote a lost race for Zou. The possible cases are $\\text{WWWWL, WWWLW, WWLWW, WLWWW, LWWWW}$. The first case has probability $\\left(\\frac{2}{3} \\right)^4 \\cdot \\frac{1}{3} = \\frac{16}{3^5}$. The second case has probability $\\left( \\frac{2}{3} \\right)^3 \\cdot \\frac{1}{3} \\cdot \\frac{1}{3} = \\frac{8}{3^5}$. The third has probability $\\left( \\frac{2}{3} \\right)^2 \\cdot \\frac{1}{3} \\cdot \\frac{1}{3} \\cdot \\frac{2}{3} = \\frac{8}{3^5}$. The fourth has probability $\\frac{2}{3} \\cdot \\frac{1}{3} \\cdot \\frac{1}{3} \\cdot \\left( \\frac{2}{3} \\right)^2 = \\frac{8}{3^5}$. Lastly, the fifth has probability $\\frac{1}{3} \\cdot \\frac{1}{3} \\cdot \\left( \\frac{2}{3} \\right)^3 = \\frac{8}{3^5}$. Adding these up, the total probability is $\\frac{16 + 8 \\cdot 4}{3^5} = \\frac{16 \\cdot 3}{3^5} = \\frac{16}{81}$, so $m+n = 097. ~rocketsri This is for if you're paranoid like me, and like to sometimes write out all of the cases if there are only a few.",
"We have $5$ cases, depending on which race Zou lost. Let $\\text{W}$ denote a won race, and $\\text{L}$ denote a lost race for Zou. The possible cases are $\\text{WWWWL, WWWLW, WWLWW, WLWWW, LWWWW}$. The first case has probability $\\left(\\frac{2}{3} \\right)^4 \\cdot \\frac{1}{3} = \\frac{16}{3^5}$. The second case has probability $\\left( \\frac{2}{3} \\right)^3 \\cdot \\frac{1}{3} \\cdot \\frac{1}{3} = \\frac{8}{3^5}$. The third has probability $\\left( \\frac{2}{3} \\right)^2 \\cdot \\frac{1}{3} \\cdot \\frac{1}{3} \\cdot \\frac{2}{3} = \\frac{8}{3^5}$. The fourth has probability $\\frac{2}{3} \\cdot \\frac{1}{3} \\cdot \\frac{1}{3} \\cdot \\left( \\frac{2}{3} \\right)^2 = \\frac{8}{3^5}$. Lastly, the fifth has probability $\\frac{1}{3} \\cdot \\frac{1}{3} \\cdot \\left( \\frac{2}{3} \\right)^3 = \\frac{8}{3^5}$. Adding these up, the total probability is $\\frac{16 + 8 \\cdot 4}{3^5} = \\frac{16 \\cdot 3}{3^5} = \\frac{16}{81}$, so $m+n = 097. ~rocketsri This is for if you're paranoid like me, and like to sometimes write out all of the cases if there are only a few.",
"Case 1: Zou loses the first race In this case, Zou must win the rest of the races. Thus, our probability is $\\frac{8}{243}$. Case 2: Zou loses the last race There is only one possibility for this, so our probability is $\\frac{16}{243}$. Case 3: Neither happens There are three ways that this happens. Each has one loss that is not the last race. Therefore, the probability that one happens is $\\frac{1}{3} \\cdot \\frac{1}{3} \\cdot \\frac{2}{3} \\cdot \\frac{2}{3} \\cdot \\frac{2}{3} = \\frac{8}{243}$. Thus, the total probability is $\\frac{8}{243} \\cdot 3 = \\frac{24}{243}$. Adding these up, we get $\\frac{48}{243} = \\frac{16}{81}$, so $16+81=097. ~mathboy100",
"Case 1: Zou loses the first race In this case, Zou must win the rest of the races. Thus, our probability is $\\frac{8}{243}$. Case 2: Zou loses the last race There is only one possibility for this, so our probability is $\\frac{16}{243}$. Case 3: Neither happens There are three ways that this happens. Each has one loss that is not the last race. Therefore, the probability that one happens is $\\frac{1}{3} \\cdot \\frac{1}{3} \\cdot \\frac{2}{3} \\cdot \\frac{2}{3} \\cdot \\frac{2}{3} = \\frac{8}{243}$. Thus, the total probability is $\\frac{8}{243} \\cdot 3 = \\frac{24}{243}$. Adding these up, we get $\\frac{48}{243} = \\frac{16}{81}$, so $16+81=097. ~mathboy100",
"Note that Zou wins one race. The probability that he wins the last race is $\\left(\\frac{2}{3}\\right)^4\\left(\\frac{1}{3}\\right)=\\frac{16}{243}.$ Now, if he doesn't win the last race, then there must be two races where the winner of the previous race loses. We can choose any $4$ of the middle races for Zou to win. So the probability for this case is $4\\left(\\frac{2}{3}\\right)^3\\left(\\frac{1}{3}\\right)^2=\\frac{32}{243}.$ Thus, the answer is $\\frac{16}{243}+\\frac{32}{243}=\\frac{16}{81}\\implies097 ~pinkpig",
"Note that Zou wins one race. The probability that he wins the last race is $\\left(\\frac{2}{3}\\right)^4\\left(\\frac{1}{3}\\right)=\\frac{16}{243}.$ Now, if he doesn't win the last race, then there must be two races where the winner of the previous race loses. We can choose any $4$ of the middle races for Zou to win. So the probability for this case is $4\\left(\\frac{2}{3}\\right)^3\\left(\\frac{1}{3}\\right)^2=\\frac{32}{243}.$ Thus, the answer is $\\frac{16}{243}+\\frac{32}{243}=\\frac{16}{81}\\implies097 ~pinkpig"
] |
2021-I-2
| 2,021
| 2
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In the diagram below, $ABCD$ is a rectangle with side lengths $AB=3$ and $BC=11$ , and $AECF$ is a rectangle with side lengths $AF=7$ and $FC=9,$ as shown. The area of the shaded region common to the interiors of both rectangles is $\frac mn$ , where $m$ and $n$ are relatively prime positive integers. Find $m+n$ . [asy] pair A, B, C, D, E, F; A = (0,3); B=(0,0); C=(11,0); D=(11,3); E=foot(C, A, (9/4,0)); F=foot(A, C, (35/4,3)); draw(A--B--C--D--cycle); draw(A--E--C--F--cycle); filldraw(A--(9/4,0)--C--(35/4,3)--cycle,gray*0.5+0.5*lightgray); dot(A^^B^^C^^D^^E^^F); label("$A$", A, W); label("$B$", B, W); label("$C$", C, (1,0)); label("$D$", D, (1,0)); label("$F$", F, N); label("$E$", E, S); [/asy]
| 109
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I
|
[
"Let $G$ be the intersection of $AD$ and $FC$. From vertical angles, we know that $\\angle FGA= \\angle DGC$. Also, because we are given that $ABCD$ and $AFCE$ are rectangles, we know that $\\angle AFG= \\angle CDG=90 ^{\\circ}$. Therefore, by AA similarity, we know that $\\triangle AFG\\sim\\triangle CDG$. Let $AG=x$. Then, we have $DG=11-x$. By similar triangles, we know that $FG=\\frac{7}{3}(11-x)$ and $CG=\\frac{3}{7}x$. We have $\\frac{7}{3}(11-x)+\\frac{3}{7}x=FC=9$. Solving for $x$, we have $x=\\frac{35}{4}$. The area of the shaded region is just $3\\cdot \\frac{35}{4}=\\frac{105}{4}$. Thus, the answer is $105+4=109. ~yuanyuanC",
"Let $G$ be the intersection of $AD$ and $FC$. From vertical angles, we know that $\\angle FGA= \\angle DGC$. Also, because we are given that $ABCD$ and $AFCE$ are rectangles, we know that $\\angle AFG= \\angle CDG=90 ^{\\circ}$. Therefore, by AA similarity, we know that $\\triangle AFG\\sim\\triangle CDG$. Let $AG=x$. Then, we have $DG=11-x$. By similar triangles, we know that $FG=\\frac{7}{3}(11-x)$ and $CG=\\frac{3}{7}x$. We have $\\frac{7}{3}(11-x)+\\frac{3}{7}x=FC=9$. Solving for $x$, we have $x=\\frac{35}{4}$. The area of the shaded region is just $3\\cdot \\frac{35}{4}=\\frac{105}{4}$. Thus, the answer is $105+4=109. ~yuanyuanC",
"Again, let the intersection of $AE$ and $BC$ be $G$. By AA similarity, $\\triangle AFG \\sim \\triangle CDG$ with a $\\frac{7}{3}$ ratio. Define $x$ as $\\frac{[CDG]}{9}$. Because of similar triangles, $[AFG] = 49x$. Using $ABCD$, the area of the parallelogram is $33-18x$. Using $AECF$, the area of the parallelogram is $63-98x$. These equations are equal, so we can solve for $x$ and obtain $x = \\frac{3}{8}$. Thus, $18x = \\frac{27}{4}$, so the area of the parallelogram is $33 - \\frac{27}{4} = \\frac{105}{4}$. Finally, the answer is $105+4=109. ~mathboy100",
"Again, let the intersection of $AE$ and $BC$ be $G$. By AA similarity, $\\triangle AFG \\sim \\triangle CDG$ with a $\\frac{7}{3}$ ratio. Define $x$ as $\\frac{[CDG]}{9}$. Because of similar triangles, $[AFG] = 49x$. Using $ABCD$, the area of the parallelogram is $33-18x$. Using $AECF$, the area of the parallelogram is $63-98x$. These equations are equal, so we can solve for $x$ and obtain $x = \\frac{3}{8}$. Thus, $18x = \\frac{27}{4}$, so the area of the parallelogram is $33 - \\frac{27}{4} = \\frac{105}{4}$. Finally, the answer is $105+4=109. ~mathboy100",
"Let the intersection of $AE$ and $BC$ be $G$, and let $BG=x$, so $CG=11-x$. By the Pythagorean theorem, ${AG}^2={AB}^2+{BG}^2$, so $AG=\\sqrt{x^2+9}$, and thus $EG=9-\\sqrt{x^2+9}$. By the Pythagorean theorem again, ${CG}^2={EG}^2+{CE}^2$: \\[11-x=\\sqrt{7^2+(9-\\sqrt{x^2+9})^2}.\\] Solving, we get $x=\\frac{9}{4}$, so the area of the parallelogram is $3\\cdot\\left(11-\\frac{9}{4}\\right)=\\frac{105}{4}$, and $105+4=109. ~JulianaL25",
"Let the intersection of $AE$ and $BC$ be $G$, and let $BG=x$, so $CG=11-x$. By the Pythagorean theorem, ${AG}^2={AB}^2+{BG}^2$, so $AG=\\sqrt{x^2+9}$, and thus $EG=9-\\sqrt{x^2+9}$. By the Pythagorean theorem again, ${CG}^2={EG}^2+{CE}^2$: \\[11-x=\\sqrt{7^2+(9-\\sqrt{x^2+9})^2}.\\] Solving, we get $x=\\frac{9}{4}$, so the area of the parallelogram is $3\\cdot\\left(11-\\frac{9}{4}\\right)=\\frac{105}{4}$, and $105+4=109. ~JulianaL25",
"Let $P = AD \\cap FC$, and $K = AE \\cap BC$. Also let $AP = x$. $CK$ also has to be $x$ by parallelogram properties. Then $PD$ and $BK$ must be $11-x$ because the sum of the segments has to be $11$. We can easily solve for $PC$ by the Pythagorean Theorem: \\begin{align*} DC^2 + PD^2 &= PC^2\\\\ 9 + (11-x)^2 &= PC^2 \\end{align*} It follows shortly that $PC = \\sqrt{x^2-22x+30}$. Also, $FC = 9$, and $FP + PC = 9$. We can then say that $PC = \\sqrt{x^2-22x+30}$, so $FP = 9 - \\sqrt{x^2-22x+30}$. Now we can apply the Pythagorean Theorem to $\\triangle AFP$. \\begin{align*} AF^2 + FP^2 = AP^2\\\\ 49 + \\left(9 - \\sqrt{x^2-22x+30}\\right)^2 = x^2 \\end{align*} This simplifies (not-as-shortly) to $x = \\dfrac{35}{4}$. Now we have to solve for the area of $APCK$. We know that the height is $3$ because the height of the parallelogram is the same as the height of the smaller rectangle. From the area of a parallelogram (we know that the base is $\\dfrac{35}{4}$ and the height is $3$), it is clear that the area is $\\dfrac{105}{4}$, giving an answer of $109)",
"Let $P = AD \\cap FC$, and $K = AE \\cap BC$. Also let $AP = x$. $CK$ also has to be $x$ by parallelogram properties. Then $PD$ and $BK$ must be $11-x$ because the sum of the segments has to be $11$. We can easily solve for $PC$ by the Pythagorean Theorem: \\begin{align*} DC^2 + PD^2 &= PC^2\\\\ 9 + (11-x)^2 &= PC^2 \\end{align*} It follows shortly that $PC = \\sqrt{x^2-22x+30}$. Also, $FC = 9$, and $FP + PC = 9$. We can then say that $PC = \\sqrt{x^2-22x+30}$, so $FP = 9 - \\sqrt{x^2-22x+30}$. Now we can apply the Pythagorean Theorem to $\\triangle AFP$. \\begin{align*} AF^2 + FP^2 = AP^2\\\\ 49 + \\left(9 - \\sqrt{x^2-22x+30}\\right)^2 = x^2 \\end{align*} This simplifies (not-as-shortly) to $x = \\dfrac{35}{4}$. Now we have to solve for the area of $APCK$. We know that the height is $3$ because the height of the parallelogram is the same as the height of the smaller rectangle. From the area of a parallelogram (we know that the base is $\\dfrac{35}{4}$ and the height is $3$), it is clear that the area is $\\dfrac{105}{4}$, giving an answer of $109)",
"Suppose $B=(0,0).$ It follows that $A=(0,3),C=(11,0),$ and $D=(11,3).$ Since $AECF$ is a rectangle, we have $AE=FC=9$ and $EC=AF=7.$ The equation of the circle with center $A$ and radius $\\overline{AE}$ is $x^2+(y-3)^2=81,$ and the equation of the circle with center $C$ and radius $\\overline{CE}$ is $(x-11)^2+y^2=49.$ We now have a system of two equations with two variables. Expanding and rearranging respectively give \\begin{align*} x^2+y^2-6y&=72, &(1) \\\\ x^2+y^2-22x&=-72. &(2) \\end{align*} Subtracting $(2)$ from $(1),$ we obtain $22x-6y=144.$ Simplifying and rearranging produce \\[x=\\frac{3y+72}{11}. \\hspace{34.5mm} (*)\\] Substituting $(*)$ into $(1)$ gives \\[\\left(\\frac{3y+72}{11}\\right)^2+y^2-6y=72,\\] which is a quadratic of $y.$ We clear fractions by multiplying both sides by $11^2=121,$ then solve by factoring: \\begin{align*} \\left(3y+72\\right)^2+121y^2-726y&=8712 \\\\ \\left(9y^2+432y+5184\\right)+121y^2-726y&=8712 \\\\ 130y^2-294y-3528&=0 \\\\ 2(5y+21)(13y-84)&=0 \\\\ y&=-\\frac{21}{5},\\frac{84}{13}. \\end{align*} Since $E$ is in Quadrant IV, we have $E=\\left(\\frac{3\\left(-\\frac{21}{5}\\right)+72}{11},-\\frac{21}{5}\\right)=\\left(\\frac{27}{5},-\\frac{21}{5}\\right).$ It follows that the equation of $\\overleftrightarrow{AE}$ is $y=-\\frac{4}{3}x+3.$ Let $G$ be the intersection of $\\overline{AD}$ and $\\overline{FC},$ and $H$ be the intersection of $\\overline{AE}$ and $\\overline{BC}.$ Since $H$ is the $x$-intercept of $\\overleftrightarrow{AE},$ we get $H=\\left(\\frac94,0\\right).$ By symmetry, quadrilateral $AGCH$ is a parallelogram. Its area is $HC\\cdot AB=\\left(11-\\frac94\\right)\\cdot3=\\frac{105}{4},$ from which the requested sum is $105+4=109 ~MRENTHUSIASM",
"Suppose $B=(0,0).$ It follows that $A=(0,3),C=(11,0),$ and $D=(11,3).$ Since $AECF$ is a rectangle, we have $AE=FC=9$ and $EC=AF=7.$ The equation of the circle with center $A$ and radius $\\overline{AE}$ is $x^2+(y-3)^2=81,$ and the equation of the circle with center $C$ and radius $\\overline{CE}$ is $(x-11)^2+y^2=49.$ We now have a system of two equations with two variables. Expanding and rearranging respectively give \\begin{align*} x^2+y^2-6y&=72, &(1) \\\\ x^2+y^2-22x&=-72. &(2) \\end{align*} Subtracting $(2)$ from $(1),$ we obtain $22x-6y=144.$ Simplifying and rearranging produce \\[x=\\frac{3y+72}{11}. \\hspace{34.5mm} (*)\\] Substituting $(*)$ into $(1)$ gives \\[\\left(\\frac{3y+72}{11}\\right)^2+y^2-6y=72,\\] which is a quadratic of $y.$ We clear fractions by multiplying both sides by $11^2=121,$ then solve by factoring: \\begin{align*} \\left(3y+72\\right)^2+121y^2-726y&=8712 \\\\ \\left(9y^2+432y+5184\\right)+121y^2-726y&=8712 \\\\ 130y^2-294y-3528&=0 \\\\ 2(5y+21)(13y-84)&=0 \\\\ y&=-\\frac{21}{5},\\frac{84}{13}. \\end{align*} Since $E$ is in Quadrant IV, we have $E=\\left(\\frac{3\\left(-\\frac{21}{5}\\right)+72}{11},-\\frac{21}{5}\\right)=\\left(\\frac{27}{5},-\\frac{21}{5}\\right).$ It follows that the equation of $\\overleftrightarrow{AE}$ is $y=-\\frac{4}{3}x+3.$ Let $G$ be the intersection of $\\overline{AD}$ and $\\overline{FC},$ and $H$ be the intersection of $\\overline{AE}$ and $\\overline{BC}.$ Since $H$ is the $x$-intercept of $\\overleftrightarrow{AE},$ we get $H=\\left(\\frac94,0\\right).$ By symmetry, quadrilateral $AGCH$ is a parallelogram. Its area is $HC\\cdot AB=\\left(11-\\frac94\\right)\\cdot3=\\frac{105}{4},$ from which the requested sum is $105+4=109 ~MRENTHUSIASM",
"Let the intersection of $AE$ and $BC$ be $G$. It is useful to find $\\tan(\\angle DAE)$, because $\\tan(\\angle DAE)=\\frac{3}{BG}$ and $\\frac{3}{\\tan(\\angle DAE)}=BG$. From there, subtracting the areas of the two triangles from the larger rectangle, we get Area = $33-3BG=33-\\frac{9}{\\tan(\\angle DAE)}$. let $\\angle CAD = \\alpha$. Let $\\angle CAE = \\beta$. Note, $\\alpha+\\beta=\\angle DAE$. $\\alpha=\\tan^{-1}\\left(\\frac{3}{11}\\right)$ $\\beta=\\tan^{-1}\\left(\\frac{7}{9}\\right)$ $\\tan(\\angle DAE) = \\tan\\left(\\tan^{-1}\\left(\\frac{3}{11}\\right)+\\tan^{-1}\\left(\\frac{7}{9}\\right)\\right) = \\frac{\\frac{3}{11}+\\frac{7}{9}}{1-\\frac{3}{11}\\cdot\\frac{7}{9}} = \\frac{\\frac{104}{99}}{\\frac{78}{99}} = \\frac{4}{3}$ $\\mathrm{Area}=33-\\frac{9}{\\frac{4}{3}} = 33-\\frac{27}{4 } = \\frac{105}{4}$. The answer is $105+4=109. ~twotothetenthis1024",
"Let the intersection of $AE$ and $BC$ be $G$. It is useful to find $\\tan(\\angle DAE)$, because $\\tan(\\angle DAE)=\\frac{3}{BG}$ and $\\frac{3}{\\tan(\\angle DAE)}=BG$. From there, subtracting the areas of the two triangles from the larger rectangle, we get Area = $33-3BG=33-\\frac{9}{\\tan(\\angle DAE)}$. let $\\angle CAD = \\alpha$. Let $\\angle CAE = \\beta$. Note, $\\alpha+\\beta=\\angle DAE$. $\\alpha=\\tan^{-1}\\left(\\frac{3}{11}\\right)$ $\\beta=\\tan^{-1}\\left(\\frac{7}{9}\\right)$ $\\tan(\\angle DAE) = \\tan\\left(\\tan^{-1}\\left(\\frac{3}{11}\\right)+\\tan^{-1}\\left(\\frac{7}{9}\\right)\\right) = \\frac{\\frac{3}{11}+\\frac{7}{9}}{1-\\frac{3}{11}\\cdot\\frac{7}{9}} = \\frac{\\frac{104}{99}}{\\frac{78}{99}} = \\frac{4}{3}$ $\\mathrm{Area}=33-\\frac{9}{\\frac{4}{3}} = 33-\\frac{27}{4 } = \\frac{105}{4}$. The answer is $105+4=109. ~twotothetenthis1024"
] |
2021-I-3
| 2,021
| 3
|
Find the number of positive integers less than $1000$ that can be expressed as the difference of two integral powers of $2.$
| 50
|
I
|
[
"We want to find the number of positive integers $n<1000$ which can be written in the form $n = 2^a - 2^b$ for some non-negative integers $a > b \\ge 0$ (note that if $a=b$, then $2^a-2^b = 0$). We first observe $a$ must be at most 10; if $a \\ge 11$, then $2^a - 2^b \\ge 2^{10} > 1000$. As $2^{10} = 1024 \\approx 1000$, we can first choose two different numbers $a > b$ from the set $\\{0,1,2,\\ldots,10\\}$ in $\\binom{11}{2}=55$ ways. This includes $(a,b) = (10,0)$, $(10,1)$, $(10,2)$, $(10,3)$, $(10,4)$ which are invalid as $2^a - 2^b > 1000$ in this case. For all other choices $a$ and $b$, the value of $2^a - 2^b$ is less than 1000. We claim that for all other choices of $a$ and $b$, the values of $2^a - 2^b$ are pairwise distinct. More specifically, if $(a_1,b_1) \\neq (a_2,b_2)$ where $10 \\ge a_1 > b_1 \\ge 0$ and $10 \\ge a_2 > b_2 \\ge 0$, we must show that $2^{a_1}-2^{b_1} \\neq 2^{a_2} - 2^{b_2}$. Suppose otherwise for sake of contradiction; rearranging yields $2^{a_1}+2^{b_2} = 2^{a_2}+2^{b_1}$. We use the fact that every positive integer has a unique binary representation: If $a_1 \\neq b_2$ then $\\{a_1,b_2\\} = \\{a_2,b_1\\}$; from here we can deduce either $a_1=a_2$ and $b_1=b_2$ (contradicting the assumption that $(a_1,b_1) \\neq (a_2,b_2)$, or $a_1=b_1$ and $a_2=b_2$ (contradicting the assumption $a_1>b_1$ and $a_2>b_2$). If $a_1 = b_2$ then $2^{a_1}+2^{b_2} = 2 \\times 2^{a_1}$, and it follows that $a_1=a_2=b_1=b_2$, also contradicting the assumption $(a_1,b_1) \\neq (a_2,b_2)$. Hence we obtain contradiction.* Then there are $\\binom{11}{2}-5$ choices for $(a,b)$ for which $2^a - 2^b$ is a positive integer less than 1000; by the above claim, each choice of $(a,b)$ results in a different positive integer $n$. Then there are $55-5 = 050, contradiction. Note by Ross Gao",
"Case 1: When our answer is in the form $2^n-2^i$, where $i$ is an integer such that $0\\le i\\le 4$. We start with the subcase where it is $2^n-2^0$, for some integer $n$ where $n>0$ (this is because the case where $n=0$ yields $2^0-2^0=0$, which doesn't work because it must be a positive integer.) Note that $2^{10}=1024$, and $2^9=512$. Our answer needs to be less than $1000$, so the maximum possible result (in this case) is $2^9-2^0$. Our lowest result is $2^1-2^0$. All the positive powers of two less than $1024$ work, so we have $9$ possibilities for this subcase. For subcases $i=1, i=2, i=3,$ and $i=4$, we have $8, 7, 6,$ and $5$ possibilities, respectively. Case 2: When our answer is in the form of $2^n-2^j$, where $j$ is an integer such that $5\\le j\\le 9$. We can start with the subcase where $j=5$. We notice that $2^5=32$, and $2^{10}-2^5=992$ which is less than $1000$, so the greatest result in this subcase is actually $2^{10}-2^5$, and the lowest is $2^6-2^5$. Thus, we have $5$ possibilities. For the other four subcases, we have $4, 3, 2,$ and $1$ possibilities, respectively. Answer: We note that these are our only cases, as numbers in the form of $2^n-2^{10}$ and beyond are greater than $1000$. Thus, our result is $9+8+7+6+5+5+4+3+2+1=(9+8+7+6+5+4+3+2+1)+5=50. ~jehu26",
"Case 1: When our answer is in the form $2^n-2^i$, where $i$ is an integer such that $0\\le i\\le 4$. We start with the subcase where it is $2^n-2^0$, for some integer $n$ where $n>0$ (this is because the case where $n=0$ yields $2^0-2^0=0$, which doesn't work because it must be a positive integer.) Note that $2^{10}=1024$, and $2^9=512$. Our answer needs to be less than $1000$, so the maximum possible result (in this case) is $2^9-2^0$. Our lowest result is $2^1-2^0$. All the positive powers of two less than $1024$ work, so we have $9$ possibilities for this subcase. For subcases $i=1, i=2, i=3,$ and $i=4$, we have $8, 7, 6,$ and $5$ possibilities, respectively. Case 2: When our answer is in the form of $2^n-2^j$, where $j$ is an integer such that $5\\le j\\le 9$. We can start with the subcase where $j=5$. We notice that $2^5=32$, and $2^{10}-2^5=992$ which is less than $1000$, so the greatest result in this subcase is actually $2^{10}-2^5$, and the lowest is $2^6-2^5$. Thus, we have $5$ possibilities. For the other four subcases, we have $4, 3, 2,$ and $1$ possibilities, respectively. Answer: We note that these are our only cases, as numbers in the form of $2^n-2^{10}$ and beyond are greater than $1000$. Thus, our result is $9+8+7+6+5+5+4+3+2+1=(9+8+7+6+5+4+3+2+1)+5=50. ~jehu26",
"We look for all positive integers of the form $2^a-2^b<1000,$ where $0\\leq b<a.$ Performing casework on $a,$ we can enumerate all possibilities in the table below: \\[\\begin{array}{c|c} & \\\\ [-2.25ex] \\boldsymbol{a} & \\boldsymbol{b} \\\\ \\hline & \\\\ [-2ex] 1 & 0 \\\\ 2 & 0,1 \\\\ 3 & 0,1,2 \\\\ 4 & 0,1,2,3 \\\\ 5 & 0,1,2,3,4 \\\\ 6 & 0,1,2,3,4,5 \\\\ 7 & 0,1,2,3,4,5,6 \\\\ 8 & 0,1,2,3,4,5,6,7 \\\\ 9 & 0,1,2,3,4,5,6,7,8 \\\\ 10 & \\xcancel{0},\\xcancel{1},\\xcancel{2},\\xcancel{3},\\xcancel{4},5,6,7,8,9 \\\\ [0.5ex] \\end{array}\\] As indicated by the X-marks, the ordered pairs $(a,b)=(10,0),(10,1),(10,2),(10,3),(10,4)$ generate $2^a-2^b>1000,$ which are invalid. Note that each of the remaining ordered pairs generates one unique desired positive integer. We prove this statement as follows: The positive integers generated for each value of $a$ are clearly different. For all integers $k$ such that $1\\leq k\\leq9,$ the largest positive integer generated for $a=k$ is $1$ less than the smallest positive integer generated for $a=k+1.$ Together, we have justified our claim in bold. The answer is \\[1+2+3+4+5+6+7+8+9+5=050.\\] ~MRENTHUSIASM",
"We look for all positive integers of the form $2^a-2^b<1000,$ where $0\\leq b<a.$ Performing casework on $a,$ we can enumerate all possibilities in the table below: \\[\\begin{array}{c|c} & \\\\ [-2.25ex] \\boldsymbol{a} & \\boldsymbol{b} \\\\ \\hline & \\\\ [-2ex] 1 & 0 \\\\ 2 & 0,1 \\\\ 3 & 0,1,2 \\\\ 4 & 0,1,2,3 \\\\ 5 & 0,1,2,3,4 \\\\ 6 & 0,1,2,3,4,5 \\\\ 7 & 0,1,2,3,4,5,6 \\\\ 8 & 0,1,2,3,4,5,6,7 \\\\ 9 & 0,1,2,3,4,5,6,7,8 \\\\ 10 & \\xcancel{0},\\xcancel{1},\\xcancel{2},\\xcancel{3},\\xcancel{4},5,6,7,8,9 \\\\ [0.5ex] \\end{array}\\] As indicated by the X-marks, the ordered pairs $(a,b)=(10,0),(10,1),(10,2),(10,3),(10,4)$ generate $2^a-2^b>1000,$ which are invalid. Note that each of the remaining ordered pairs generates one unique desired positive integer. We prove this statement as follows: The positive integers generated for each value of $a$ are clearly different. For all integers $k$ such that $1\\leq k\\leq9,$ the largest positive integer generated for $a=k$ is $1$ less than the smallest positive integer generated for $a=k+1.$ Together, we have justified our claim in bold. The answer is \\[1+2+3+4+5+6+7+8+9+5=050.\\] ~MRENTHUSIASM",
"Because the difference is less than $1000$, we can simply list out all numbers that satisfy $2^n < 1000$. We get $0 \\le n < 10$, where n is an integer. Because the sequence $2^n$ is geometric, the difference of any two terms will be unique. $\\binom{10}{2}$ will be the number of differences for $0\\le n < 10$. However, we also need to consider the case in which $n=10$. With simple counting, we find that $5$ numbers: $(32, 64, 128, 256, 512)$ could be subtracted from $1024$, which makes another 5 cases. There is no need to check for higher exponents since the lowest difference would be $2^{11} - 2^{10} = 1024$, which exceeds $1000$. Thus, the final answer is $\\binom{10}{2} + 5 = 050 ~TOMYANG",
"Because the difference is less than $1000$, we can simply list out all numbers that satisfy $2^n < 1000$. We get $0 \\le n < 10$, where n is an integer. Because the sequence $2^n$ is geometric, the difference of any two terms will be unique. $\\binom{10}{2}$ will be the number of differences for $0\\le n < 10$. However, we also need to consider the case in which $n=10$. With simple counting, we find that $5$ numbers: $(32, 64, 128, 256, 512)$ could be subtracted from $1024$, which makes another 5 cases. There is no need to check for higher exponents since the lowest difference would be $2^{11} - 2^{10} = 1024$, which exceeds $1000$. Thus, the final answer is $\\binom{10}{2} + 5 = 050 ~TOMYANG"
] |
2021-I-4
| 2,021
| 4
|
Find the number of ways $66$ identical coins can be separated into three nonempty piles so that there are fewer coins in the first pile than in the second pile and fewer coins in the second pile than in the third pile.
| 331
|
I
|
[
"Suppose we have $1$ coin in the first pile. Then $(1, 2, 63), (1, 3, 62), \\ldots, (1, 32, 33)$ all work for a total of $31$ piles. Suppose we have $2$ coins in the first pile, then $(2, 3, 61), (2, 4, 60), \\ldots, (2, 31, 33)$ all work, for a total of $29$. Continuing this pattern until $21$ coins in the first pile, we have the sum \\begin{align*} 31+29+28+26+25+\\cdots+4+2+1 &= (31+28+25+22+\\cdots+1)+(29+26+23+\\cdots+2) \\\\ &= 176+155 \\\\ &= 331. \\end{align*}",
"We make an equation: $a+b+c=66,$ where $a<b<c.$ We don't have a clear solution, so we'll try complementary counting. First, let's find where $a\\geq b\\geq c.$ By stars and bars, we have $\\dbinom{65}{2}=2080$ to assign positive integer solutions to $a + b + c = 66.$ Now we need to subtract off the cases where it doesn't satisfy the condition. We start with $a = b.$ We can write that as $2b + c = 66.$ We can find there are 32 integer solutions to this equation. There are $32$ solutions for $b=c$ and $a = c$ by symmetry. We also need to add back $2$ because we subtracted $(a,b,c)=(22,22,22)$ $3$ times. We then have to divide by $6$ because there are $3!=6$ ways to order $a, b,$ and $c.$ Therefore, we have $\\dfrac{\\dbinom{65}{2}-96+2}{6} = \\dfrac{1986}{6} = 331 ~Arcticturn",
"Let the piles have $a, b$ and $c$ coins, with $0 < a < b < c$. Then, let $b = a + k_1$, and $c = b + k_2$, such that each $k_i \\geq 1$. The sum is then $a + a+k_1 + a+k_1+k_2 = 66 \\implies 3a+2k_1 + k_2 = 66$. This is simply the number of positive solutions to the equation $3x+2y+z = 66$. Now, we take cases on $a$. If $a = 1$, then $2k_1 + k_2 = 63 \\implies 1 \\leq k_1 \\leq 31$. Each value of $k_1$ corresponds to a unique value of $k_2$, so there are $31$ solutions in this case. Similarly, if $a = 2$, then $2k_1 + k_2 = 60 \\implies 1 \\leq k_1 \\leq 29$, for a total of $29$ solutions in this case. If $a = 3$, then $2k_1 + k_2 = 57 \\implies 1 \\leq k_1 \\leq 28$, for a total of $28$ solutions. In general, the number of solutions is just all the numbers that aren't a multiple of $3$, that are less than or equal to $31$. We then add our cases to get \\begin{align*} 1 + 2 + 4 + 5 + \\cdots + 31 &= 1 + 2 + 3 + \\cdots + 31 - 3(1 + 2 + 3 + \\cdots + 10) \\\\ &= \\frac{31(32)}{2} - 3(55) \\\\ &= 31 \\cdot 16 - 165 \\\\ &= 496 - 165 \\\\ &= 331 \\end{align*} as our answer.",
"Let the first pile have $a$ coins, the second $a+b$ coins, and the third $a+b+c$ coins, where $a$, $b$, and $c$ are strictly positive integers. Thus the total number of coins is $3a+2b+c=66$. Perform the substitution $x=a-1$, $y=b-1$, and $z=c-1$ to yield the equation $3x+2y+z=60$, where $x$, $y$, and $z$ are instead nonnegative integers. From here we can set up the generating function $(x^0+x^3+\\cdots+x^{60})(x^0+x^2+\\cdots+x^{60})(x^0+x^1+\\cdots+x^{60})$. We need to find the coefficient of $x^{60}$. Multiplying the second and third polynomials with clever reasoning returns $(x^0+x^3+\\cdots+x^{60})(x^0+x^1+2x^2+2x^3+\\cdots+31x^{60}+\\cdots+2x^{117}+2x^{118}+x^{119}+x^{120})$ where in the second polynomial, for every two terms, the coefficient increases or decreases by one (depending on which side of the polynomial the term resides). One can notice here that for every term in the first polynomial there exists one and only one term in the second polynomial that, when multiplied, yield $x^{60}$. Furthermore, we need only consider the coefficients of the second polynomial. The corresponding coefficient for $x^{60}$ is $1$, for $x^{57}$ is $2$, and for $x^{54}$ is $4$. We notice the pattern: increase by one, increase by two, and so on. When does this pattern stop? For $x^0$, the corresponding coefficient is $31$, and we notice that $31\\equiv1\\mod3$. As a result, we know that the pattern has $21$ terms, and we can take advantage of the first $20$ by symmetry. The answer is simply $10(1+29)+31=331. ~eevee9406 Solution 4"
] |
2021-I-5
| 2,021
| 5
|
Call a three-term strictly increasing arithmetic sequence of integers special if the sum of the squares of the three terms equals the product of the middle term and the square of the common difference. Find the sum of the third terms of all special sequences.
| 31
|
I
|
[
"Let the terms be $a-b$, $a$, and $a+b$. Then we want $(a-b)^2+a^2+(a+b)^2=ab^2$, or $3a^2+2b^2=ab^2$. Rearranging, we get $b^2=\\frac{3a^2}{a-2}$. Simplifying further, $b^2=3a+6+\\frac{12}{a-2}$. Looking at this second equation, since the right side must be an integer, $a-2$ must equal $\\pm1, 2, 3, 4, 6, 12$. Looking at the first equation, we see $a>2$ since $b^2$ is positive. This means we must test $a=3, 4, 5, 6, 8, 14$. After testing these, we see that only $a=5$ and $a=14$ work which give $b=5$ and $b=7$ respectively. Thus the answer is $10+21=031.",
"Let the common difference be $d$ and let the middle term be $x$. Then, we have that the sequence is \\[x-d,~x,~x+d.\\] This means that the sum of the squares of the 3 terms of the sequence is \\[(x-d)^2+x^2+(x+d)^2=x^2-2xd+d^2+x^2+x^2+2xd+d^2=3x^2+2d^2.\\] We know that this must be equal to $xd^2,$ so we can write that \\[3x^2+2d^2=xd^2,\\] and it follows that \\[3x^2-xd^2+2d^2=3x^2-\\left(d^2\\right)x+2d^2=0.\\] Now, we can treat $d$ as a constant and use the quadratic formula to get \\[x=\\frac{d^2\\pm \\sqrt{d^4-4(3)(2d^2)}}{6}.\\] We can factor pull $d^2$ out of the square root to get \\[x=\\frac{d^2\\pm d\\sqrt{d^2-24}}{6}.\\] Here, it is easy to figure out the values of $d$. Let $\\sqrt{d^2-24} = k$, then $d^2-k^2=24$ which is $(d+k)(d-k)=24,$ note that $d$, $k$ are integers. Examining the parity, we find that $d+k$ and $d-k$ are of the same parity. Now, we solve by factoring. We can find that $d=5$ and $d=7$ are the only positive integer values of $d$ that make $\\sqrt{d^2-24}$ a positive integer.$^{*}$ $d=5$ gives $x=5$ and $x=\\frac{10}{3}$, but we can ignore the latter. $d=7$ gives $x=14$, as well as a fraction which we can ignore. Since $d=5,~x=5$ and $d=7, x=14$ are the only two solutions and we want the sum of the third terms, our answer is $(5+5)+(7+14)=10+21=031 are integers see if you can figure it out. -PureSwag",
"Proceed as in solution 2, until we reach \\[3x^2+2d^2=xd^2,\\] Write $d^2=\\frac{3x^2}{x-2}$, it follows that $x-2=3k^2$ for some (positive) integer k and $k \\mid x$. Taking both sides modulo $k$, $-2 \\equiv 0 \\pmod{k}$, so $k \\mid 2 \\rightarrow k=1,2$. When $k=1$, we have $x=5$ and $d=5$. When $k=2$, we have $x=14$ and $d=7$. Summing the two cases, we have $10+21=031. -Ross Gao",
"As in Solution 1, write the three integers in the sequence as $a-d$, $a$, and $a+d$. Then the sum of the squares of the three integers is $(a-d)^2+a^2+(a+d)^2 = 3a^2+2d^2$. Setting this equal to the middle term times the common difference squared, which is $ad^2$, and solving for $d^2$ we get: $3a^2+2d^2 = ad^2 \\implies ad^2-2d^2 = 3a^2 \\implies d^2(a-2) = 3a^2 \\implies d^2 = \\frac{3a^2}{a-2}$ The numerator has to be positive, so the denominator has to be positive too for the sequence to be strictly increasing; that is, $a>2$. For $\\frac{3a^2}{a-2}$ to be a perfect square, $\\frac{3}{a-2}$ must be a perfect square as well. This means that $a-2$ is divisible by 3, and whatever left over is a perfect square. We can express this as an equation: let the perfect square left over be $n^2$. Then: $3n^2 = a-2$. Now when you divide the numerator and denominator by 3, you are left with $d^2 = \\frac{a^2}{n^2} \\implies d = \\frac{a}{n}$. Because the sequence is of integers, d must also be an integer, which means that $n$ must divide $a$. Taking the above equation we can solve for $a$: $3n^2 = a-2 \\implies a = 3n^2+2$. This means that $3n^2+2$ is divisible by $n$. $3n^2$ is automatically divisible by $n$, so $2$ must be divisible by $n$. Then $n$ must be either of $\\{1,2\\}$. Plugging back into the equation, $n = 1 \\implies a = 5 \\implies d = 5$, so $a+d = 5+5 = 10$. $n = 2 \\implies a = 14 \\implies d = 7$, so $a+d = 14+7 = 21$. Finally, $10+21 = 031 -KingRavi",
"As in Solution 1, write the three integers in the sequence as $a-d$, $a$, and $a+d$. Then the sum of the squares of the three integers is $(a-d)^2+a^2+(a+d)^2 = 3a^2+2d^2$. Setting this equal to the middle term times the common difference squared, which is $ad^2$, and solving for $d^2$ we get: $3a^2+2d^2 = ad^2 \\implies ad^2-2d^2 = 3a^2 \\implies d^2(a-2) = 3a^2 \\implies d^2 = \\frac{3a^2}{a-2}$ The numerator has to be positive, so the denominator has to be positive too for the sequence to be strictly increasing; that is, $a>2$. For $\\frac{3a^2}{a-2}$ to be a perfect square, $\\frac{3}{a-2}$ must be a perfect square as well. This means that $a-2$ is divisible by 3, and whatever left over is a perfect square. We can express this as an equation: let the perfect square left over be $n^2$. Then: $3n^2 = a-2$. Now when you divide the numerator and denominator by 3, you are left with $d^2 = \\frac{a^2}{n^2} \\implies d = \\frac{a}{n}$. Because the sequence is of integers, d must also be an integer, which means that $n$ must divide $a$. Taking the above equation we can solve for $a$: $3n^2 = a-2 \\implies a = 3n^2+2$. This means that $3n^2+2$ is divisible by $n$. $3n^2$ is automatically divisible by $n$, so $2$ must be divisible by $n$. Then $n$ must be either of $\\{1,2\\}$. Plugging back into the equation, $n = 1 \\implies a = 5 \\implies d = 5$, so $a+d = 5+5 = 10$. $n = 2 \\implies a = 14 \\implies d = 7$, so $a+d = 14+7 = 21$. Finally, $10+21 = 031 -KingRavi",
"Following from previous solutions, we derive $3x^2+2a^2=xa^2.$ We divide both sides to get $3\\left(\\frac{x}{a}\\right)^2+2=x.$ Since $x$ is an integer, $\\frac{x}{a}$ must also be an integer, so we have $x=pa$, for some factor $p$. We then get $3p^2+2=pa.$ We then take this to modulo $p$, getting $2\\equiv 0 \\pmod p.$ The only possibilities for $p$ are therefore 1 and 2. We plug these into $3p^2+2=pa$, for $a=5$ and $x=5$, giving us the sequence $0,5,10$, or $2a=14$ and $x=14$, for the sequence $7,14,21.$ $10+21 = 031 -RYang2"
] |
2021-I-6
| 2,021
| 6
|
Segments $\overline{AB}, \overline{AC},$ and $\overline{AD}$ are edges of a cube and $\overline{AG}$ is a diagonal through the center of the cube. Point $P$ satisfies $BP=60\sqrt{10}$ , $CP=60\sqrt{5}$ , $DP=120\sqrt{2}$ , and $GP=36\sqrt{7}$ . Find $AP.$
| 192
|
I
|
[
"First scale down the whole cube by $12$. Let point $P$ have coordinates $(x, y, z)$, point $A$ have coordinates $(0, 0, 0)$, and $s$ be the side length. Then we have the equations \\begin{align*} (s-x)^2+y^2+z^2&=\\left(5\\sqrt{10}\\right)^2, \\\\ x^2+(s-y)^2+z^2&=\\left(5\\sqrt{5}\\right)^2, \\\\ x^2+y^2+(s-z)^2&=\\left(10\\sqrt{2}\\right)^2, \\\\ (s-x)^2+(s-y)^2+(s-z)^2&=\\left(3\\sqrt{7}\\right)^2. \\end{align*} These simplify into \\begin{align*} s^2+x^2+y^2+z^2-2sx&=250, \\\\ s^2+x^2+y^2+z^2-2sy&=125, \\\\ s^2+x^2+y^2+z^2-2sz&=200, \\\\ 3s^2-2s(x+y+z)+x^2+y^2+z^2&=63. \\end{align*} Adding the first three equations together, we get $3s^2-2s(x+y+z)+3(x^2+y^2+z^2)=575$. Subtracting this from the fourth equation, we get $2(x^2+y^2+z^2)=512$, so $x^2+y^2+z^2=256$. This means $PA=16$. However, we scaled down everything by $12$ so our answer is $16*12=192. ~JHawk0224",
"Once the equations for the distance between point P and the vertices of the cube have been written, we can add the first, second, and third to receive, \\[2(x^2 + y^2 + z^2) + (s-x)^2 + (s-y)^2 + (s-z)^2 = 250 + 125 + 200.\\] Subtracting the fourth equation gives \\begin{align*} 2(x^2 + y^2 + z^2) &= 575 - 63 \\\\ x^2 + y^2 + z^2 &= 256 \\\\ \\sqrt{x^2 + y^2 + z^2} &= 16. \\end{align*} Since point $A = (0,0,0), PA = 16$, and since we scaled the answer is $16 \\cdot 12 = 192. ~Aaryabhatta1",
"Once the equations for the distance between point P and the vertices of the cube have been written, we can add the first, second, and third to receive, \\[2(x^2 + y^2 + z^2) + (s-x)^2 + (s-y)^2 + (s-z)^2 = 250 + 125 + 200.\\] Subtracting the fourth equation gives \\begin{align*} 2(x^2 + y^2 + z^2) &= 575 - 63 \\\\ x^2 + y^2 + z^2 &= 256 \\\\ \\sqrt{x^2 + y^2 + z^2} &= 16. \\end{align*} Since point $A = (0,0,0), PA = 16$, and since we scaled the answer is $16 \\cdot 12 = 192. ~Aaryabhatta1",
"Let $E$ be the vertex of the cube such that $ABED$ is a square. Using the British Flag Theorem, we can easily show that \\[PA^2 + PE^2 = PB^2 + PD^2\\] and \\[PA^2 + PG^2 = PC^2 + PE^2\\] Hence, by adding the two equations together, we get $2PA^2 + PG^2 = PB^2 + PC^2 + PD^2$. Substituting in the values we know, we get $2PA^2 + 7\\cdot 36^2 =10\\cdot60^2 + 5\\cdot 60^2 + 2\\cdot 120^2$. Thus, we can solve for $PA$, which ends up being $192. (Lokman GÖKÇE)",
"For all points $X$ in space, define the function $f:\\mathbb{R}^{3}\\rightarrow\\mathbb{R}$ by $f(X)=PX^{2}-GX^{2}$. Then $f$ is linear; let $O=\\frac{2A+G}{3}$ be the center of $\\triangle BCD$. Then since $f$ is linear, \\begin{align*} 3f(O)=f(B)+f(C)+f(D)&=2f(A)+f(G) \\\\ \\left(PB^{2}-GB^{2}\\right)+\\left(PC^{2}-GC^{2}\\right)+\\left(PD^{2}-GD^{2}\\right)&=2\\left(PA^{2}-GA^{2}\\right)+PG^{2} \\\\ \\left(60\\sqrt{10}\\right)^{2}-2x^{2}+\\left(60\\sqrt{5}\\right)^{2}-2x^{2}+\\left(120\\sqrt{2}\\right)^{2}-2x^{2}&=2PA^{2}-2\\cdot 3x^{2}+\\left(36\\sqrt{7}\\right)^{2}, \\end{align*} where $x$ denotes the side length of the cube. Thus \\begin{align*} 36\\text{,}000+18\\text{,}000+28\\text{,}800-6x^{2}&=2PA^{2}-6x^{2}+9072 \\\\ 82\\text{,}800-6x^{2}&=2PA^{2}-6x^{2}+9072 \\\\ 73\\text{,}728&=2PA^{2} \\\\ 36\\text{,}864&=PA^{2} \\\\ PA&=192. \\end{align*}"
] |
2021-I-7
| 2,021
| 7
|
Find the number of pairs $(m,n)$ of positive integers with $1\le m<n\le 30$ such that there exists a real number $x$ satisfying \[\sin(mx)+\sin(nx)=2.\]
| 63
|
I
|
[
"It is trivial that the maximum value of $\\sin \\theta$ is $1$, is achieved at $\\theta = \\frac{\\pi}{2}+2k\\pi$ for some integer $k$. This implies that $\\sin(mx) = \\sin(nx) = 1$, and that $mx = \\frac{\\pi}{2}+2a\\pi$ and $nx = \\frac{\\pi}{2}+2b\\pi$, for integers $a, b$. Taking their ratio, we have \\[\\frac{mx}{nx} = \\frac{\\frac{\\pi}{2}+2a\\pi}{\\frac{\\pi}{2}+2b\\pi} \\implies \\frac{m}{n} = \\frac{4a + 1}{4b + 1} \\implies \\frac{m}{4a + 1} = \\frac{n}{4b + 1} = k.\\] It remains to find all $m, n$ that satisfy this equation. If $k = 1$, then $m \\equiv n \\equiv 1 \\pmod 4$. This corresponds to choosing two elements from the set $\\{1, 5, 9, 13, 17, 21, 25, 29\\}$. There are $\\binom 82$ ways to do so. If $k < 1$, by multiplying $m$ and $n$ by the same constant $c = \\frac{1}{k}$, we have that $mc \\equiv nc \\equiv 1 \\pmod 4$. Then either $m \\equiv n \\equiv 1 \\pmod 4$, or $m \\equiv n \\equiv 3 \\pmod 4$. But the first case was already counted, so we don't need to consider that case. The other case corresponds to choosing two numbers from the set $\\{3, 7, 11, 15, 19, 23, 27\\}$. There are $\\binom 72$ ways here. (This argument seems to have a logical flaw *check note at bottom*) Finally, if $k > 1$, note that $k$ must be an integer. This means that $m, n$ belong to the set $\\{k, 5k, 9k, \\dots\\}$, or $\\{3k, 7k, 11k, \\dots\\}$. Taking casework on $k$, we get the sets $\\{2, 10, 18, 26\\}, \\{6, 14, 22, 30\\}, \\{4, 20\\}, \\{12, 28\\}$. Some sets have been omitted; this is because they were counted in the other cases already. This sums to $\\binom 42 + \\binom 42 + \\binom 22 + \\binom 22$. In total, there are $\\binom 82 + \\binom 72 + \\binom 42 + \\binom 42 + \\binom 22 + \\binom 22 = 063. This solution was brought to you by ~Leonard_my_dude~",
"In order for $\\sin(mx) + \\sin(nx) = 2$, $\\sin(mx) = \\sin(nx) = 1$. This happens when $mx \\equiv nx \\equiv \\frac{\\pi}{2} ($mod $2\\pi).$ This means that $mx = \\frac{\\pi}{2} + 2\\pi\\alpha$ and $nx = \\frac{\\pi}{2} + 2\\pi\\beta$ for any integers $\\alpha$ and $\\beta$. As in Solution 1, take the ratio of the two equations: \\[\\frac{mx}{nx} = \\frac{\\frac{\\pi}{2}+2\\pi\\alpha}{\\frac{\\pi}{2}+2\\pi\\beta} \\implies \\frac{m}{n} = \\frac{\\frac{1}{2}+2\\alpha}{\\frac{1}{2}+2\\beta} \\implies \\frac{m}{n} = \\frac{4\\alpha+1}{4\\beta+1}\\] Now notice that the numerator and denominator of $\\frac{4\\alpha+1}{4\\beta+1}$ are both odd, which means that $m$ and $n$ have the same power of two (the powers of 2 cancel out). Let the common power be $p$: then $m = 2^p\\cdot a$, and $n = 2^p\\cdot b$ where $a$ and $b$ are integers between 1 and 30. We can now rewrite the equation: \\[\\frac{2^p\\cdot a}{2^p\\cdot b} = \\frac{4\\alpha+1}{4\\beta+1} \\implies \\frac{a}{b} = \\frac{4\\alpha+1}{4\\beta+1}\\] Now it is easy to tell that $a \\equiv 1 ($mod $4)$ and $b \\equiv 1 ($mod $4)$. However, there is another case: that $a \\equiv 3 ($mod $4)$ and $b \\equiv 3 ($mod $4)$. This is because multiplying both $4\\alpha+1$ and $4\\beta+1$ by $-1$ will not change the fraction, but each congruence will be changed to $-1 ($mod $4) \\equiv 3 ($mod $4)$. From the first set of congruences, we find that $a$ and $b$ can be two of $\\{1, 5, 9, \\ldots, 29\\}$. From the second set of congruences, we find that $a$ and $b$ can be two of $\\{3, 7, 11, \\ldots, 27\\}$. Now all we have to do is multiply by $2^p$ to get back to $m$ and $n$. Let’s organize the solutions in order of increasing values of $p$, keeping in mind that $m$ and $n$ are bounded between 1 and 30. For $p = 0$ we get $\\{1, 5, 9, \\ldots, 29\\}, \\{3, 7, 11, \\ldots, 27\\}$. For $p = 1$ we get $\\{2, 10, 18, 26\\}, \\{6, 14, 22, 30\\}$ For $p = 2$ we get $\\{4, 20\\}, \\{12, 28\\}$ Note that $16\\mid{a}$ since $m$ will cancel out a factor of 4 from $a$, and $\\frac{a}{m}$ must contain a factor of 4. Again, $1-4X$ will never contribute a factor of 2. Simply inspecting, we see two feasible values for $a$ and $m$ such that $a+m\\leq30$. If we increase the value of $p$ more, there will be less than two integers in our sets, so we are done there. There are 8 numbers in the first set, 7 in the second, 4 in the third, 4 in the fourth, 2 in the fifth, and 2 in the sixth. In each of these sets we can choose 2 numbers to be $m$ and $n$ and then assign them in increasing order. Thus there are: \\[\\dbinom{8}{2}+\\dbinom{7}{2}+\\dbinom{4}{2}+\\dbinom{4}{2}+\\dbinom{2}{2}+\\dbinom{2}{2} = 28+21+6+6+1+1 = 063 that satisfy the conditions. -KingRavi",
"We know that the range of sine is between $-1$ and $1$, inclusive. Thus, the only way for the sum to be $2$ is for $\\sin(mx)=\\sin(nx)=1$. Note that $\\sin(90+360k)=1$. Assuming $mx$ and $nx$ are both positive, $m$ and $n$ could be $1,5,9,13,17,21,25,29$. There are $8$ ways, so $\\dbinom{8}{2}$. If both are negative, $m$ and $n$ could be $3,7,11,15,19,23,27$. There are $7$ ways, so $\\dbinom{7}{2}$. However, the pair $(1,5)$ could also be $(2, 10)$ and so on. The same goes for some other pairs. In total there are $14$ of these extra pairs. The answer is $28+21+14 = 063.",
"The equation implies that $\\sin(mx)=\\sin(nx)=1$. Therefore, we can write $mx$ as $2{\\pi}k_1+\\frac{\\pi}{2}$ and $nx$ as $2{\\pi}k_2+\\frac{\\pi}{2}$ for integers $k_1$ and $k_2$. Then, $\\frac{mx}{nx}=\\frac{m}{n}=\\frac{2k_1+\\frac{1}{2}}{2k_2+\\frac{1}{2}}$. Cross multiplying, we get $m\\cdot{(2k_2+\\frac{1}{2})}=n\\cdot{(2k_1+\\frac{1}{2})} \\Longrightarrow 4k_2m-4k_1n=n-m$. Let $n-m=a$ so the equation becomes $4(m(k_2-k_1)+k_1a)=a$. Let $k_2-k_1=X$ and $k_1=Y$, then the equation becomes $a=4Ym+4Xa \\Longrightarrow \\frac{a(1-4X)}{m}=4Y$. Note that $X$ and $Y$ can vary accordingly, and $4\\mid{a}$. Next, we do casework on $m\\pmod{4}$: If $m\\equiv 1\\pmod{4}$: Once $a$ and $m$ are determined, $n$ is determined, so $a+m\\leq30$. $a\\in \\{4,8,12,\\dots,28\\}$ and $m\\in \\{1,5,9,\\dots,29\\}$. Therefore, there are $\\sum_{i=1}^{7}{i}=28$ ways for this case such that $a+m\\leq30$. If $m\\equiv 3\\pmod{4}$: $a\\in \\{4,8,12,\\dots,28\\}$ and $m\\in \\{3,7,11,\\dots,27\\}$. Therefore, there are $\\sum_{i=1}^{6}{i}=21$ ways such that $a+m\\leq30$. If $m\\equiv 2\\pmod{4}$: Note that $8\\mid{a}$ since $m$ in this case will have a factor of 2, which will cancel out a factor of 2 in $a$, and we need the left hand side to divide 4. Also, $1-4X\\equiv 1\\pmod{4}$ so it is odd and will therefore never contribute a factor of 2. $a\\in \\{8,16,24\\}$ and $m\\in \\{2,6,10,\\dots,30\\}$. Following the condition $a+m\\leq30$, we conclude that there are $6+4+2=12$ ways for this case. If $m\\equiv 0\\pmod{4}$: Adding all the cases up, we obtain $28+21+12+2=063 ~Magnetoninja"
] |
2021-I-8
| 2,021
| 8
|
Find the number of integers $c$ such that the equation \[\left||20|x|-x^2|-c\right|=21\] has $12$ distinct real solutions.
| 57
|
I
|
[
"We take cases for the outermost absolute value, then rearrange: \\[\\left|20|x|-x^2\\right|=c\\pm21.\\] Let $f(x)=\\left|20|x|-x^2\\right|.$ We rewrite $f(x)$ as a piecewise function without using absolute values: \\[f(x) = \\begin{cases} \\left|-20x-x^2\\right| & \\mathrm{if} \\ x \\le 0 \\begin{cases} 20x+x^2 & \\mathrm{if} \\ x\\le-20 \\\\ -20x-x^2 & \\mathrm{if} \\ -20<x\\leq0 \\end{cases} \\\\ \\left|20x-x^2\\right| & \\mathrm{if} \\ x > 0 \\begin{cases} 20x-x^2 & \\mathrm{if} \\ 0<x\\leq20 \\\\ -20x+x^2 & \\mathrm{if} \\ x>20 \\end{cases} \\end{cases}.\\] We graph $y=f(x)$ with all extremum points labeled, as shown below. The fact that $f(x)$ is an even function ($f(x)=f(-x)$ holds for all real numbers $x,$ so the graph of $y=f(x)$ is symmetric about the $y$-axis) should facilitate the process of graphing. [asy] /* Made by MRENTHUSIASM */ size(1200,300); real xMin = -65; real xMax = 65; real yMin = -50; real yMax = 125; draw((xMin,0)--(xMax,0),black+linewidth(1.5),EndArrow(5)); draw((0,yMin)--(0,yMax),black+linewidth(1.5),EndArrow(5)); label(\"$x$\",(xMax,0),(2,0)); label(\"$y$\",(0,yMax),(0,2)); real f(real x) { return abs(20*abs(x)-x^2); } real g(real x) { return 21; } real h(real x) { return -21; } draw(graph(f,-25,25),red,\"$y=\\left|20|x|-x^2\\right|$\"); draw(graph(g,-65,65),blue,\"$y=\\pm21$\"); draw(graph(h,-65,65),blue); pair A[]; A[0] = (-20,0); A[1] = (-10,100); A[2] = (0,0); A[3] = (10,100); A[4] = (20,0); for(int i = 0; i <= 4; ++i) { dot(A[i],red+linewidth(4.5)); } label(\"$(-20,0)$\",A[0],(-1.5,-1.5),red,UnFill); label(\"$(-10,100)$\",A[1],(-1.5,1.5),red); label(\"$(0,0)$\",A[2],(0,-1.5),red,UnFill); label(\"$(10,100)$\",A[3],(1.5,1.5),red); label(\"$(20,0)$\",A[4],(1.5,-1.5),red,UnFill); add(legend(),point(E),40E,UnFill); [/asy] Since $f(x)=c\\pm21$ has $12$ distinct real solutions, it is clear that each case has $6$ distinct real solutions geometrically. We shift the graphs of $y=\\pm21$ up $c$ units, where $c\\geq0:$ For $f(x)=c+21$ to have $6$ distinct real solutions, we need $0\\leq c<79.$ For $f(x)=c-21$ to have $6$ distinct real solutions, we need $21<c<121.$ Taking the intersection of these two cases gives $21<c<79,$ from which there are $79-21-1=057 ~MRENTHUSIASM",
"We take cases for the outermost absolute value, then rearrange: \\[\\left|20|x|-x^2\\right|=c\\pm21.\\] Let $f(x)=\\left|20|x|-x^2\\right|.$ We rewrite $f(x)$ as a piecewise function without using absolute values: \\[f(x) = \\begin{cases} \\left|-20x-x^2\\right| & \\mathrm{if} \\ x \\le 0 \\begin{cases} 20x+x^2 & \\mathrm{if} \\ x\\le-20 \\\\ -20x-x^2 & \\mathrm{if} \\ -20<x\\leq0 \\end{cases} \\\\ \\left|20x-x^2\\right| & \\mathrm{if} \\ x > 0 \\begin{cases} 20x-x^2 & \\mathrm{if} \\ 0<x\\leq20 \\\\ -20x+x^2 & \\mathrm{if} \\ x>20 \\end{cases} \\end{cases}.\\] We graph $y=f(x)$ with all extremum points labeled, as shown below. The fact that $f(x)$ is an even function ($f(x)=f(-x)$ holds for all real numbers $x,$ so the graph of $y=f(x)$ is symmetric about the $y$-axis) should facilitate the process of graphing. [asy] /* Made by MRENTHUSIASM */ size(1200,300); real xMin = -65; real xMax = 65; real yMin = -50; real yMax = 125; draw((xMin,0)--(xMax,0),black+linewidth(1.5),EndArrow(5)); draw((0,yMin)--(0,yMax),black+linewidth(1.5),EndArrow(5)); label(\"$x$\",(xMax,0),(2,0)); label(\"$y$\",(0,yMax),(0,2)); real f(real x) { return abs(20*abs(x)-x^2); } real g(real x) { return 21; } real h(real x) { return -21; } draw(graph(f,-25,25),red,\"$y=\\left|20|x|-x^2\\right|$\"); draw(graph(g,-65,65),blue,\"$y=\\pm21$\"); draw(graph(h,-65,65),blue); pair A[]; A[0] = (-20,0); A[1] = (-10,100); A[2] = (0,0); A[3] = (10,100); A[4] = (20,0); for(int i = 0; i <= 4; ++i) { dot(A[i],red+linewidth(4.5)); } label(\"$(-20,0)$\",A[0],(-1.5,-1.5),red,UnFill); label(\"$(-10,100)$\",A[1],(-1.5,1.5),red); label(\"$(0,0)$\",A[2],(0,-1.5),red,UnFill); label(\"$(10,100)$\",A[3],(1.5,1.5),red); label(\"$(20,0)$\",A[4],(1.5,-1.5),red,UnFill); add(legend(),point(E),40E,UnFill); [/asy] Since $f(x)=c\\pm21$ has $12$ distinct real solutions, it is clear that each case has $6$ distinct real solutions geometrically. We shift the graphs of $y=\\pm21$ up $c$ units, where $c\\geq0:$ For $f(x)=c+21$ to have $6$ distinct real solutions, we need $0\\leq c<79.$ For $f(x)=c-21$ to have $6$ distinct real solutions, we need $21<c<121.$ Taking the intersection of these two cases gives $21<c<79,$ from which there are $79-21-1=057 ~MRENTHUSIASM",
"Graph $y=|20|x|-x^2|$ (If you are having trouble, look at the description in the next two lines and/or the diagram in Solution 1). Notice that we want this to be equal to $c-21$ and $c+21$. We see that from left to right, the graph first dips from very positive to $0$ at $x=-20$, then rebounds up to $100$ at $x=-10$, then falls back down to $0$ at $x=0$. The positive $x$ are symmetric, so the graph re-ascends to $100$ at $x=10$, falls back to $0$ at $x=10$, and rises to arbitrarily large values afterwards. Now we analyze the $y$ (varied by $c$) values. At $y=k<0$, we will have no solutions, as the line $y=k$ will have no intersections with our graph. At $y=0$, we will have exactly $3$ solutions for the three zeroes. At $y=n$ for any $n$ strictly between $0$ and $100$, we will have exactly $6$ solutions. At $y=100$, we will have $4$ solutions, because local maxima are reached at $x= \\pm 10$. At $y=m>100$, we will have exactly $2$ solutions. To get $12$ distinct solutions for $y=|20|x|-x^2|=c \\pm 21$, both $c +21$ and $c-21$ must produce $6$ solutions. Thus $0<c-21$ and $c+21<100$, so $c \\in \\{ 22, 23, \\dots , 77, 78 \\}$ is required. It is easy to verify that all of these choices of $c$ produce $12$ distinct solutions (none overlap), so our answer is $057.",
"Graph $y=|20|x|-x^2|$ (If you are having trouble, look at the description in the next two lines and/or the diagram in Solution 1). Notice that we want this to be equal to $c-21$ and $c+21$. We see that from left to right, the graph first dips from very positive to $0$ at $x=-20$, then rebounds up to $100$ at $x=-10$, then falls back down to $0$ at $x=0$. The positive $x$ are symmetric, so the graph re-ascends to $100$ at $x=10$, falls back to $0$ at $x=10$, and rises to arbitrarily large values afterwards. Now we analyze the $y$ (varied by $c$) values. At $y=k<0$, we will have no solutions, as the line $y=k$ will have no intersections with our graph. At $y=0$, we will have exactly $3$ solutions for the three zeroes. At $y=n$ for any $n$ strictly between $0$ and $100$, we will have exactly $6$ solutions. At $y=100$, we will have $4$ solutions, because local maxima are reached at $x= \\pm 10$. At $y=m>100$, we will have exactly $2$ solutions. To get $12$ distinct solutions for $y=|20|x|-x^2|=c \\pm 21$, both $c +21$ and $c-21$ must produce $6$ solutions. Thus $0<c-21$ and $c+21<100$, so $c \\in \\{ 22, 23, \\dots , 77, 78 \\}$ is required. It is easy to verify that all of these choices of $c$ produce $12$ distinct solutions (none overlap), so our answer is $057.",
"Let $y = |x|.$ Then the equation becomes $\\left|\\left|20y-y^2\\right|-c\\right| = 21$, or $\\left|y^2-20y\\right| = c \\pm 21$. Note that since $y = |x|$, $y$ is nonnegative, so we only care about nonnegative solutions in $y$. Notice that each positive solution in $y$ gives two solutions in $x$ ($x = \\pm y$), whereas if $y = 0$ is a solution, this only gives one solution in $x$, $x = 0$. Since the total number of solutions in $x$ is even, $y = 0$ must not be a solution. Hence, we require that $\\left|y^2-20y\\right| = c \\pm 21$ has exactly $6$ positive solutions and is not solved by $y = 0.$ If $c < 21$, then $c - 21$ is negative, and therefore cannot be the absolute value of $y^2 - 20y$. This means the equation's only solutions are in $\\left|y^2-20y\\right| = c + 21$. There is no way for this equation to have $6$ solutions, since the quadratic $y^2-20y$ can only take on each of the two values $\\pm(c + 21)$ at most twice, yielding at most $4$ solutions. Hence, $c \\ge 21$. $c$ also can't equal $21$, since this would mean $y = 0$ would solve the equation. Hence, $c > 21.$ At this point, the equation $y^2-20y = c \\pm 21$ will always have exactly $2$ positive solutions, since $y^2-20y$ takes on each positive value exactly once when $y$ is restricted to positive values (graph it to see this), and $c \\pm 21$ are both positive. Therefore, we just need $y^2-20y = -(c \\pm 21)$ to have the remaining $4$ solutions exactly. This means the horizontal lines at $-(c \\pm 21)$ each intersect the parabola $y^2 - 20y$ in two places. This occurs when the two lines are above the parabola's vertex $(10,-100)$. Hence we have \\begin{align*} -(c + 21) &> -100 \\\\ c + 21 &< 100 \\\\ c &< 79. \\end{align*} Hence, the integers $c$ satisfying the conditions are those satisfying $21 < c < 79.$ There are $057.",
"Let $y = |x|.$ Then the equation becomes $\\left|\\left|20y-y^2\\right|-c\\right| = 21$, or $\\left|y^2-20y\\right| = c \\pm 21$. Note that since $y = |x|$, $y$ is nonnegative, so we only care about nonnegative solutions in $y$. Notice that each positive solution in $y$ gives two solutions in $x$ ($x = \\pm y$), whereas if $y = 0$ is a solution, this only gives one solution in $x$, $x = 0$. Since the total number of solutions in $x$ is even, $y = 0$ must not be a solution. Hence, we require that $\\left|y^2-20y\\right| = c \\pm 21$ has exactly $6$ positive solutions and is not solved by $y = 0.$ If $c < 21$, then $c - 21$ is negative, and therefore cannot be the absolute value of $y^2 - 20y$. This means the equation's only solutions are in $\\left|y^2-20y\\right| = c + 21$. There is no way for this equation to have $6$ solutions, since the quadratic $y^2-20y$ can only take on each of the two values $\\pm(c + 21)$ at most twice, yielding at most $4$ solutions. Hence, $c \\ge 21$. $c$ also can't equal $21$, since this would mean $y = 0$ would solve the equation. Hence, $c > 21.$ At this point, the equation $y^2-20y = c \\pm 21$ will always have exactly $2$ positive solutions, since $y^2-20y$ takes on each positive value exactly once when $y$ is restricted to positive values (graph it to see this), and $c \\pm 21$ are both positive. Therefore, we just need $y^2-20y = -(c \\pm 21)$ to have the remaining $4$ solutions exactly. This means the horizontal lines at $-(c \\pm 21)$ each intersect the parabola $y^2 - 20y$ in two places. This occurs when the two lines are above the parabola's vertex $(10,-100)$. Hence we have \\begin{align*} -(c + 21) &> -100 \\\\ c + 21 &< 100 \\\\ c &< 79. \\end{align*} Hence, the integers $c$ satisfying the conditions are those satisfying $21 < c < 79.$ There are $057.",
"Removing the absolute value bars from the equation successively, we get \\begin{align*} \\left|\\left|20|x|-x^2\\right|-c\\right|&=21 \\\\ \\left|20|x|-x^2\\right|&= c \\pm21 \\\\ 20|x|-x^2 &= \\pm c \\pm 21 \\\\ x^2 \\pm 20x \\pm c \\pm21 &= 0. \\end{align*} The discriminant of this equation is \\[\\sqrt{400-4(\\pm c \\pm 21)}.\\] Equating the discriminant to $0$, we see that there will be two distinct solutions to each of the possible quadratics above only in the interval $-79 < c < 79$. However, the number of zeros the equation $ax^2+b|x|+k$ has is determined by where $ax^2+bx+k$ and $ax^2-bx+k$ intersect, namely at $(0,k)$. When $k>0$, $a>0$, $ax^2+b|x|+k$ will have only $2$ solutions, and when $k<0$, $a>0$, then there will be $4$ real solutions, if they exist at all. In order to have $12$ solutions here, we thus need to ensure $-c+21<0$, so that exactly $2$ out of the $4$ possible equations of the form $ax^2+b|x|+k$ given above have y-intercepts below $0$ and only $2$ real solutions, while the remaining $2$ equations have $4$ solutions. This occurs when $c>21$, so our final bounds are $21<c<79$, giving us $057.",
"Removing the absolute value bars from the equation successively, we get \\begin{align*} \\left|\\left|20|x|-x^2\\right|-c\\right|&=21 \\\\ \\left|20|x|-x^2\\right|&= c \\pm21 \\\\ 20|x|-x^2 &= \\pm c \\pm 21 \\\\ x^2 \\pm 20x \\pm c \\pm21 &= 0. \\end{align*} The discriminant of this equation is \\[\\sqrt{400-4(\\pm c \\pm 21)}.\\] Equating the discriminant to $0$, we see that there will be two distinct solutions to each of the possible quadratics above only in the interval $-79 < c < 79$. However, the number of zeros the equation $ax^2+b|x|+k$ has is determined by where $ax^2+bx+k$ and $ax^2-bx+k$ intersect, namely at $(0,k)$. When $k>0$, $a>0$, $ax^2+b|x|+k$ will have only $2$ solutions, and when $k<0$, $a>0$, then there will be $4$ real solutions, if they exist at all. In order to have $12$ solutions here, we thus need to ensure $-c+21<0$, so that exactly $2$ out of the $4$ possible equations of the form $ax^2+b|x|+k$ given above have y-intercepts below $0$ and only $2$ real solutions, while the remaining $2$ equations have $4$ solutions. This occurs when $c>21$, so our final bounds are $21<c<79$, giving us $057."
] |
2021-I-9
| 2,021
| 9
|
Let $ABCD$ be an isosceles trapezoid with $AD=BC$ and $AB<CD.$ Suppose that the distances from $A$ to the lines $BC,CD,$ and $BD$ are $15,18,$ and $10,$ respectively. Let $K$ be the area of $ABCD.$ Find $\sqrt2 \cdot K.$
| 567
|
I
|
[
"Let $\\overline{AE}, \\overline{AF},$ and $\\overline{AG}$ be the perpendiculars from $A$ to $\\overleftrightarrow{BC}, \\overleftrightarrow{CD},$ and $\\overleftrightarrow{BD},$ respectively. Next, let $H$ be the intersection of $\\overline{AF}$ and $\\overline{BD}.$ We set $AB=x$ and $AH=y,$ as shown below. [asy] /* Made by MRENTHUSIASM */ size(250); pair A, B, C, D, E, F, G, H; A = (-45sqrt(2)/8,18); B = (45sqrt(2)/8,18); C = (81sqrt(2)/8,0); D = (-81sqrt(2)/8,0); E = foot(A,C,B); F = foot(A,C,D); G = foot(A,B,D); H = intersectionpoint(A--F,B--D); markscalefactor=0.1; draw(rightanglemark(A,E,B),red); draw(rightanglemark(A,F,C),red); draw(rightanglemark(A,G,D),red); dot(\"$A$\",A,1.5*NW,linewidth(4)); dot(\"$B$\",B,1.5*NE,linewidth(4)); dot(\"$C$\",C,1.5*SE,linewidth(4)); dot(\"$D$\",D,1.5*SW,linewidth(4)); dot(\"$E$\",E,1.5*dir(E),linewidth(4)); dot(\"$F$\",F,1.5*S,linewidth(4)); dot(\"$G$\",G,SE,linewidth(4)); dot(\"$H$\",H,SE,linewidth(4)); draw(A--B--C--D--cycle^^B--D^^B--E); draw(A--E^^A--F^^A--G,dashed); label(\"$10$\",midpoint(A--G),1.5*(1,0)); label(\"$15$\",midpoint(A--E),1.5*N); Label L = Label(\"$18$\", align=(0,0), position=MidPoint, filltype=Fill(0,3,white)); draw(C+(5,0)--(81sqrt(2)/8,18)+(5,0), L=L, arrow=Arrows(),bar=Bars(15)); label(\"$x$\",midpoint(A--B),N); label(\"$y$\",midpoint(A--H),W); [/asy] From here, we obtain $HF=18-y$ by segment subtraction, and $BG=\\sqrt{x^2-10^2}$ and $HG=\\sqrt{y^2-10^2}$ by the Pythagorean Theorem. Since $\\angle ABG$ and $\\angle HAG$ are both complementary to $\\angle AHB,$ we have $\\angle ABG = \\angle HAG,$ from which $\\triangle ABG \\sim \\triangle HAG$ by AA. It follows that $\\frac{BG}{AG}=\\frac{AG}{HG},$ so $BG\\cdot HG=AG^2,$ or \\[\\sqrt{x^2-10^2}\\cdot\\sqrt{y^2-10^2}=10^2. \\hspace{10mm} (1)\\] Since $\\angle AHB = \\angle FHD$ by vertical angles, we have $\\triangle AHB \\sim \\triangle FHD$ by AA, with the ratio of similitude $\\frac{AH}{FH}=\\frac{BA}{DF}.$ It follows that $DF=BA\\cdot\\frac{FH}{AH}=x\\cdot\\frac{18-y}{y}.$ Since $\\angle EBA = \\angle ECD = \\angle FDA$ by angle chasing, we have $\\triangle EBA \\sim \\triangle FDA$ by AA, with the ratio of similitude $\\frac{EA}{FA}=\\frac{BA}{DA}.$ It follows that $DA=BA\\cdot\\frac{FA}{EA}=x\\cdot\\frac{18}{15}=\\frac{6}{5}x.$ By the Pythagorean Theorem on right $\\triangle ADF,$ we have $DF^2+AF^2=AD^2,$ or \\[\\left(x\\cdot\\frac{18-y}{y}\\right)^2+18^2=\\left(\\frac{6}{5}x\\right)^2. \\hspace{7mm} (2)\\] Solving this system of equations ($(1)$ and $(2)$), we get $x=\\frac{45\\sqrt2}{4}$ and $y=\\frac{90}{7},$ so $AB=x=\\frac{45\\sqrt2}{4}$ and $CD=AB+2DF=x+2\\left(x\\cdot\\frac{18-y}{y}\\right)=\\frac{81\\sqrt2}{4}.$ Finally, the area of $ABCD$ is \\[K=\\frac{AB+CD}{2}\\cdot AF=\\frac{567\\sqrt2}{2},\\] from which $\\sqrt2 \\cdot K=567 ~Chupdogs",
"Let $\\overline{AE}, \\overline{AF},$ and $\\overline{AG}$ be the perpendiculars from $A$ to $\\overleftrightarrow{BC}, \\overleftrightarrow{CD},$ and $\\overleftrightarrow{BD},$ respectively. Next, let $H$ be the intersection of $\\overline{AF}$ and $\\overline{BD}.$ We set $AB=x$ and $AH=y,$ as shown below. [asy] /* Made by MRENTHUSIASM */ size(250); pair A, B, C, D, E, F, G, H; A = (-45sqrt(2)/8,18); B = (45sqrt(2)/8,18); C = (81sqrt(2)/8,0); D = (-81sqrt(2)/8,0); E = foot(A,C,B); F = foot(A,C,D); G = foot(A,B,D); H = intersectionpoint(A--F,B--D); markscalefactor=0.1; draw(rightanglemark(A,E,B),red); draw(rightanglemark(A,F,C),red); draw(rightanglemark(A,G,D),red); dot(\"$A$\",A,1.5*NW,linewidth(4)); dot(\"$B$\",B,1.5*NE,linewidth(4)); dot(\"$C$\",C,1.5*SE,linewidth(4)); dot(\"$D$\",D,1.5*SW,linewidth(4)); dot(\"$E$\",E,1.5*dir(E),linewidth(4)); dot(\"$F$\",F,1.5*S,linewidth(4)); dot(\"$G$\",G,SE,linewidth(4)); dot(\"$H$\",H,SE,linewidth(4)); draw(A--B--C--D--cycle^^B--D^^B--E); draw(A--E^^A--F^^A--G,dashed); label(\"$10$\",midpoint(A--G),1.5*(1,0)); label(\"$15$\",midpoint(A--E),1.5*N); Label L = Label(\"$18$\", align=(0,0), position=MidPoint, filltype=Fill(0,3,white)); draw(C+(5,0)--(81sqrt(2)/8,18)+(5,0), L=L, arrow=Arrows(),bar=Bars(15)); label(\"$x$\",midpoint(A--B),N); label(\"$y$\",midpoint(A--H),W); [/asy] From here, we obtain $HF=18-y$ by segment subtraction, and $BG=\\sqrt{x^2-10^2}$ and $HG=\\sqrt{y^2-10^2}$ by the Pythagorean Theorem. Since $\\angle ABG$ and $\\angle HAG$ are both complementary to $\\angle AHB,$ we have $\\angle ABG = \\angle HAG,$ from which $\\triangle ABG \\sim \\triangle HAG$ by AA. It follows that $\\frac{BG}{AG}=\\frac{AG}{HG},$ so $BG\\cdot HG=AG^2,$ or \\[\\sqrt{x^2-10^2}\\cdot\\sqrt{y^2-10^2}=10^2. \\hspace{10mm} (1)\\] Since $\\angle AHB = \\angle FHD$ by vertical angles, we have $\\triangle AHB \\sim \\triangle FHD$ by AA, with the ratio of similitude $\\frac{AH}{FH}=\\frac{BA}{DF}.$ It follows that $DF=BA\\cdot\\frac{FH}{AH}=x\\cdot\\frac{18-y}{y}.$ Since $\\angle EBA = \\angle ECD = \\angle FDA$ by angle chasing, we have $\\triangle EBA \\sim \\triangle FDA$ by AA, with the ratio of similitude $\\frac{EA}{FA}=\\frac{BA}{DA}.$ It follows that $DA=BA\\cdot\\frac{FA}{EA}=x\\cdot\\frac{18}{15}=\\frac{6}{5}x.$ By the Pythagorean Theorem on right $\\triangle ADF,$ we have $DF^2+AF^2=AD^2,$ or \\[\\left(x\\cdot\\frac{18-y}{y}\\right)^2+18^2=\\left(\\frac{6}{5}x\\right)^2. \\hspace{7mm} (2)\\] Solving this system of equations ($(1)$ and $(2)$), we get $x=\\frac{45\\sqrt2}{4}$ and $y=\\frac{90}{7},$ so $AB=x=\\frac{45\\sqrt2}{4}$ and $CD=AB+2DF=x+2\\left(x\\cdot\\frac{18-y}{y}\\right)=\\frac{81\\sqrt2}{4}.$ Finally, the area of $ABCD$ is \\[K=\\frac{AB+CD}{2}\\cdot AF=\\frac{567\\sqrt2}{2},\\] from which $\\sqrt2 \\cdot K=567 ~Chupdogs",
"First, draw the diagram. Then, notice that since $ABCD$ is isosceles, $\\Delta ABD \\cong \\Delta BAC$, and the length of the altitude from $B$ to $AC$ is also $10$. Let the foot of this altitude be $F$, and let the foot of the altitude from $A$ to $BC$ be denoted as $E$. Then, $\\Delta BCF \\sim \\Delta ACE$. So, $\\frac{BC}{AC} = \\frac{BF}{AE} = \\frac{2}{3}$. Now, notice that $[ABC] = \\frac{10 \\cdot AC} {2} = \\frac{AB \\cdot 18}{2} \\implies AC = \\frac{9 \\cdot AB}{5}$, where $[ABC]$ denotes the area of triangle $ABC$. Letting $AB = x$, this equality becomes $AC = \\frac{9x}{5}$. Also, from $\\frac{BC}{AC} = \\frac{2}{3}$, we have $BC = \\frac{6x}{5}$. Now, by the Pythagorean theorem on triangles $ABF$ and $CBF$, we have $AF = \\sqrt{x^{2}-100}$ and $CF = \\sqrt{ \\left( \\frac{6x}{5} \\right) ^{2}-100}$. Notice that $AC = AF + CF$, so $\\frac{9x}{5} = \\sqrt{x^{2}-100} + \\sqrt{ \\left( \\frac{6x}{5} \\right) ^{2}-100}$. Squaring both sides of the equation once, moving $x^{2}-100$ and $\\left( \\frac{6x}{5} \\right) ^{2}-100$ to the right, dividing both sides by $2$, and squaring the equation once more, we are left with $\\frac{32x^{4}}{25} = 324x^{2}$. Dividing both sides by $x^{2}$ (since we know $x$ is positive), we are left with $\\frac{32x^{2}}{25} = 324$. Solving for $x$ gives us $x = \\frac{45}{2\\sqrt{2}}$. Now, let the foot of the perpendicular from $A$ to $CD$ be $G$. Then let $DG = y$. Let the foot of the perpendicular from $B$ to $CD$ be $H$. Then, $CH$ is also equal to $y$. Notice that $ABHG$ is a rectangle, so $GH = x$. Now, we have $CG = GH + CH = x + y$. By the Pythagorean theorem applied to $\\Delta AGC$, we have $(x+y)^{2}+18^{2}= \\left( \\frac{9x}{5} \\right) ^{2}$. We know that $\\frac{9x}{5} = \\frac{9}{5} \\cdot \\frac{45}{2\\sqrt{2}} = \\frac{81}{2\\sqrt{2}}$, so we can plug this into this equation. Solving for $x+y$, we get $x+y=\\frac{63}{2\\sqrt{2}}$. Finally, to find $[ABCD]$, we use the formula for the area of a trapezoid: $K = [ABCD] = \\frac{b_{1}+b_{2}}{2} \\cdot h = \\frac{AB+CD}{2} \\cdot 18 = \\frac{x+(CG+DG)}{2} \\cdot 18 = \\frac{2x+2y}{2} \\cdot 18 = (x+y) \\cdot 18 = \\frac{63}{2\\sqrt{2}} \\cdot 18 = \\frac{567}{\\sqrt{2}}$. The problem asks us for $K \\cdot \\sqrt{2}$, which comes out to be $567. ~advanture",
"First, draw the diagram. Then, notice that since $ABCD$ is isosceles, $\\Delta ABD \\cong \\Delta BAC$, and the length of the altitude from $B$ to $AC$ is also $10$. Let the foot of this altitude be $F$, and let the foot of the altitude from $A$ to $BC$ be denoted as $E$. Then, $\\Delta BCF \\sim \\Delta ACE$. So, $\\frac{BC}{AC} = \\frac{BF}{AE} = \\frac{2}{3}$. Now, notice that $[ABC] = \\frac{10 \\cdot AC} {2} = \\frac{AB \\cdot 18}{2} \\implies AC = \\frac{9 \\cdot AB}{5}$, where $[ABC]$ denotes the area of triangle $ABC$. Letting $AB = x$, this equality becomes $AC = \\frac{9x}{5}$. Also, from $\\frac{BC}{AC} = \\frac{2}{3}$, we have $BC = \\frac{6x}{5}$. Now, by the Pythagorean theorem on triangles $ABF$ and $CBF$, we have $AF = \\sqrt{x^{2}-100}$ and $CF = \\sqrt{ \\left( \\frac{6x}{5} \\right) ^{2}-100}$. Notice that $AC = AF + CF$, so $\\frac{9x}{5} = \\sqrt{x^{2}-100} + \\sqrt{ \\left( \\frac{6x}{5} \\right) ^{2}-100}$. Squaring both sides of the equation once, moving $x^{2}-100$ and $\\left( \\frac{6x}{5} \\right) ^{2}-100$ to the right, dividing both sides by $2$, and squaring the equation once more, we are left with $\\frac{32x^{4}}{25} = 324x^{2}$. Dividing both sides by $x^{2}$ (since we know $x$ is positive), we are left with $\\frac{32x^{2}}{25} = 324$. Solving for $x$ gives us $x = \\frac{45}{2\\sqrt{2}}$. Now, let the foot of the perpendicular from $A$ to $CD$ be $G$. Then let $DG = y$. Let the foot of the perpendicular from $B$ to $CD$ be $H$. Then, $CH$ is also equal to $y$. Notice that $ABHG$ is a rectangle, so $GH = x$. Now, we have $CG = GH + CH = x + y$. By the Pythagorean theorem applied to $\\Delta AGC$, we have $(x+y)^{2}+18^{2}= \\left( \\frac{9x}{5} \\right) ^{2}$. We know that $\\frac{9x}{5} = \\frac{9}{5} \\cdot \\frac{45}{2\\sqrt{2}} = \\frac{81}{2\\sqrt{2}}$, so we can plug this into this equation. Solving for $x+y$, we get $x+y=\\frac{63}{2\\sqrt{2}}$. Finally, to find $[ABCD]$, we use the formula for the area of a trapezoid: $K = [ABCD] = \\frac{b_{1}+b_{2}}{2} \\cdot h = \\frac{AB+CD}{2} \\cdot 18 = \\frac{x+(CG+DG)}{2} \\cdot 18 = \\frac{2x+2y}{2} \\cdot 18 = (x+y) \\cdot 18 = \\frac{63}{2\\sqrt{2}} \\cdot 18 = \\frac{567}{\\sqrt{2}}$. The problem asks us for $K \\cdot \\sqrt{2}$, which comes out to be $567. ~advanture",
"Make $AE$ perpendicular to $BC$; $AG$ perpendicular to $BD$; $AF$ perpendicular $DC$. It's obvious that $\\triangle{AEB} \\sim \\triangle{AFD}$. Let $EB=5x; AB=5y; DF=6x; AD=6y$. Then make $BQ$ perpendicular to $DC$, it's easy to get $BQ=18$. Since $AB$ parallel to $DC$, $\\angle{ABG}=\\angle{BDQ}$, so $\\triangle{ABG} \\sim \\triangle{BDQ}$. After drawing the altitude, it's obvious that $FQ=AB=5y$, so $DQ=5y+6x$. According to the property of similar triangles, $AG/BQ=BG/DQ$. So, $\\frac{5}{9}=\\frac{GB}{(6x+5y)}$, or $GB=\\frac{(30x+25y)}{9}$. Now, we see the $\\triangle AEB$, pretty easy to find that $15^2+(5x)^2=(5y)^2$, then we get $x^2+9=y^2$, then express $y$ into $x$ form that $y=\\sqrt{x^2+9}$ we put the length of $BG$ back to $\\triangle AGB$: $BG^2+100=AB^2$. So, \\[\\frac{[30x+25\\sqrt{(x^2+9)}]^2}{81}+100=(5\\sqrt{x^2+9})^2.\\] After calculating, we can have a final equation of $x^2+9=\\sqrt{x^2+9}\\cdot3x$. It's easy to find $x=\\frac{3\\sqrt{2}}{4}$ then $y=\\frac{9\\sqrt{2}}{4}$. So, \\[\\sqrt{2}\\cdot K = \\sqrt{2}\\cdot(5y+5y+6x+6x)\\cdot9=567.\\] ~bluesoul",
"Make $AE$ perpendicular to $BC$; $AG$ perpendicular to $BD$; $AF$ perpendicular $DC$. It's obvious that $\\triangle{AEB} \\sim \\triangle{AFD}$. Let $EB=5x; AB=5y; DF=6x; AD=6y$. Then make $BQ$ perpendicular to $DC$, it's easy to get $BQ=18$. Since $AB$ parallel to $DC$, $\\angle{ABG}=\\angle{BDQ}$, so $\\triangle{ABG} \\sim \\triangle{BDQ}$. After drawing the altitude, it's obvious that $FQ=AB=5y$, so $DQ=5y+6x$. According to the property of similar triangles, $AG/BQ=BG/DQ$. So, $\\frac{5}{9}=\\frac{GB}{(6x+5y)}$, or $GB=\\frac{(30x+25y)}{9}$. Now, we see the $\\triangle AEB$, pretty easy to find that $15^2+(5x)^2=(5y)^2$, then we get $x^2+9=y^2$, then express $y$ into $x$ form that $y=\\sqrt{x^2+9}$ we put the length of $BG$ back to $\\triangle AGB$: $BG^2+100=AB^2$. So, \\[\\frac{[30x+25\\sqrt{(x^2+9)}]^2}{81}+100=(5\\sqrt{x^2+9})^2.\\] After calculating, we can have a final equation of $x^2+9=\\sqrt{x^2+9}\\cdot3x$. It's easy to find $x=\\frac{3\\sqrt{2}}{4}$ then $y=\\frac{9\\sqrt{2}}{4}$. So, \\[\\sqrt{2}\\cdot K = \\sqrt{2}\\cdot(5y+5y+6x+6x)\\cdot9=567.\\] ~bluesoul",
"Let the foot of the altitude from $A$ to $BC$ be $P$, to $CD$ be $Q$, and to $BD$ be $R$. Note that all isosceles trapezoids are cyclic quadrilaterals; thus, $A$ is on the circumcircle of $\\triangle BCD$ and we have that $PRQ$ is the Simson Line from $A$. As $\\angle QAB = 90^\\circ$, we have that $\\angle QAR = 90^\\circ - \\angle RAB =\\angle ABR = \\angle APR = \\angle APQ$, with the last equality coming from cyclic quadrilateral $APBR$. Thus, $\\triangle QAR \\sim \\triangle QPA$ and we have that $\\frac{AQ}{AR} = \\frac{PQ}{PA}$ or that $\\frac{18}{10} = \\frac{QP}{15}$, which we can see gives us that $QP = 27$. Further ratios using the same similar triangles gives that $QR = 12$ and $RP = 15$. We also see that quadrilaterals $APBR$ and $ARQD$ are both cyclic,it is clear that $\\angle AQR =\\angle ADR$ and $\\angle APR =\\angle ABR$,which shows $\\triangle ABD \\sim \\triangle APQ$. we wish to find the ratio of similitude between the two triangles. To do this, we use the one number we have for $\\triangle ABD$: we know that the altitude from $A$ to $BD$ has length $10$. As the two triangles are similar, if we can find the height from $A$ to $PQ$, we can take the ratio of the two heights as the ratio of similitude. To do this, we once again note that $QP = 27$. Using this, we can drop the altitude from $A$ to $QP$ and let it intersect $QP$ at $H$. Then, let $QH = x$ and thus $HP=27-x$. We then have by the Pythagorean Theorem on $\\triangle AQH$ and $\\triangle APH$: \\begin{align*} 18^2 - x^2 &= 15^2 - (27-x)^2 \\\\ 324 - x^2 &= 225 - (x^2-54x+729) \\\\ 54x &= 828 \\\\ x &= \\frac{46}{3}. \\end{align*} Then, $RH = QH - QR = \\frac{46}{3} - 12 = \\frac{10}{3}$. This gives us then from right triangle $\\triangle ARH$ that $AH = \\frac{20\\sqrt{2}}{3}$ and thus the ratio of $\\triangle APQ$ to $\\triangle ABD$ is $\\frac{3\\sqrt{2}}{4}$. From this, we see then that \\[AB = AP \\cdot \\frac{3\\sqrt{2}}{4} = 15 \\cdot \\frac{3\\sqrt{2}}{4} = \\frac{45\\sqrt{2}}{4}\\] and \\[AD = AQ \\cdot \\frac{3\\sqrt{2}}{4} = 18 \\cdot \\frac{3\\sqrt{2}}{4} = \\frac{27\\sqrt{2}}{2}.\\] The Pythagorean Theorem on $\\triangle AQD$ then gives that \\[QD = \\sqrt{AD^2 - AQ^2} = \\sqrt{\\left(\\frac{27\\sqrt{2}}{2}\\right)^2 - 18^2} = \\sqrt{\\frac{81}{2}} = \\frac{9\\sqrt{2}}{2}.\\] Then, we have the height of trapezoid $ABCD$ is $AQ = 18$, the top base is $AB = \\frac{45\\sqrt{2}}{4}$, and the bottom base is $CD = \\frac{45\\sqrt{2}}{4} + 2\\cdot\\frac{9\\sqrt{2}}{2}$. From the equation of a trapezoid, $K = \\frac{b_1+b_2}{2} \\cdot h = \\frac{63\\sqrt{2}}{4} \\cdot 18 = \\frac{567\\sqrt{2}}{2}$, so the answer is $K\\sqrt{2} = 567. ~lvmath ~Corrected by dongjiu0728(Former Ans made the wrong QH and wrong QR、RP but somehow came to the right answer,which trapped me for a weekend)",
"Let the foot of the altitude from $A$ to $BC$ be $P$, to $CD$ be $Q$, and to $BD$ be $R$. Note that all isosceles trapezoids are cyclic quadrilaterals; thus, $A$ is on the circumcircle of $\\triangle BCD$ and we have that $PRQ$ is the Simson Line from $A$. As $\\angle QAB = 90^\\circ$, we have that $\\angle QAR = 90^\\circ - \\angle RAB =\\angle ABR = \\angle APR = \\angle APQ$, with the last equality coming from cyclic quadrilateral $APBR$. Thus, $\\triangle QAR \\sim \\triangle QPA$ and we have that $\\frac{AQ}{AR} = \\frac{PQ}{PA}$ or that $\\frac{18}{10} = \\frac{QP}{15}$, which we can see gives us that $QP = 27$. Further ratios using the same similar triangles gives that $QR = 12$ and $RP = 15$. We also see that quadrilaterals $APBR$ and $ARQD$ are both cyclic,it is clear that $\\angle AQR =\\angle ADR$ and $\\angle APR =\\angle ABR$,which shows $\\triangle ABD \\sim \\triangle APQ$. we wish to find the ratio of similitude between the two triangles. To do this, we use the one number we have for $\\triangle ABD$: we know that the altitude from $A$ to $BD$ has length $10$. As the two triangles are similar, if we can find the height from $A$ to $PQ$, we can take the ratio of the two heights as the ratio of similitude. To do this, we once again note that $QP = 27$. Using this, we can drop the altitude from $A$ to $QP$ and let it intersect $QP$ at $H$. Then, let $QH = x$ and thus $HP=27-x$. We then have by the Pythagorean Theorem on $\\triangle AQH$ and $\\triangle APH$: \\begin{align*} 18^2 - x^2 &= 15^2 - (27-x)^2 \\\\ 324 - x^2 &= 225 - (x^2-54x+729) \\\\ 54x &= 828 \\\\ x &= \\frac{46}{3}. \\end{align*} Then, $RH = QH - QR = \\frac{46}{3} - 12 = \\frac{10}{3}$. This gives us then from right triangle $\\triangle ARH$ that $AH = \\frac{20\\sqrt{2}}{3}$ and thus the ratio of $\\triangle APQ$ to $\\triangle ABD$ is $\\frac{3\\sqrt{2}}{4}$. From this, we see then that \\[AB = AP \\cdot \\frac{3\\sqrt{2}}{4} = 15 \\cdot \\frac{3\\sqrt{2}}{4} = \\frac{45\\sqrt{2}}{4}\\] and \\[AD = AQ \\cdot \\frac{3\\sqrt{2}}{4} = 18 \\cdot \\frac{3\\sqrt{2}}{4} = \\frac{27\\sqrt{2}}{2}.\\] The Pythagorean Theorem on $\\triangle AQD$ then gives that \\[QD = \\sqrt{AD^2 - AQ^2} = \\sqrt{\\left(\\frac{27\\sqrt{2}}{2}\\right)^2 - 18^2} = \\sqrt{\\frac{81}{2}} = \\frac{9\\sqrt{2}}{2}.\\] Then, we have the height of trapezoid $ABCD$ is $AQ = 18$, the top base is $AB = \\frac{45\\sqrt{2}}{4}$, and the bottom base is $CD = \\frac{45\\sqrt{2}}{4} + 2\\cdot\\frac{9\\sqrt{2}}{2}$. From the equation of a trapezoid, $K = \\frac{b_1+b_2}{2} \\cdot h = \\frac{63\\sqrt{2}}{4} \\cdot 18 = \\frac{567\\sqrt{2}}{2}$, so the answer is $K\\sqrt{2} = 567. ~lvmath ~Corrected by dongjiu0728(Former Ans made the wrong QH and wrong QR、RP but somehow came to the right answer,which trapped me for a weekend)",
"Let $E,F,$ and $G$ be the feet of the altitudes from $A$ to $BC,CD,$ and $DB$, respectively. Claim: We have $2$ pairs of similar right triangles: $\\triangle AEB \\sim \\triangle AFD$ and $\\triangle AGD \\sim \\triangle AEC$. Proof: Note that $ABCD$ is cyclic. We need one more angle, and we get this from this cyclic quadrilateral: \\begin{align*} \\angle ABE &= 180^\\circ - \\angle ABC =\\angle ADC = \\angle ADG, \\\\ \\angle ADG &= \\angle ADB =\\angle ACB = \\angle ACE. \\hspace{20mm} \\square \\end{align*} Let $AD=a$. We obtain from the similarities $AB = \\frac{5a}{6}$ and $AC=BD=\\frac{3a}{2}$. By Ptolemy, $\\left(\\frac{3a}{2}\\right)^2 = a^2 + \\frac{5a}{6} \\cdot CD$, so $\\frac{5a^2}{4} = \\frac{5a}{6} \\cdot CD$. We obtain $CD=\\frac{3a}{2}$, so $DF=\\frac{CD-AB}{2}=\\frac{a}{3}$. Applying the Pythagorean theorem on $\\triangle ADF$, we get $324=a^2 - \\frac{a^2}{9}=\\frac{8a^2}{9}$. Thus, $a=\\frac{27}{\\sqrt{2}}$, and $[ABCD]=\\frac{AB+CD}{2} \\cdot 18 = \\frac{\\frac{5a}{6} +\\frac{9a}{6}}{2} \\cdot 18 = 18 \\cdot \\frac{7}{6} \\cdot \\frac{27}{\\sqrt{2}} = \\frac{567}{\\sqrt{2}}$, yielding $\\sqrt2\\cdot[ABCD]=567.",
"Let $E,F,$ and $G$ be the feet of the altitudes from $A$ to $BC,CD,$ and $DB$, respectively. Claim: We have $2$ pairs of similar right triangles: $\\triangle AEB \\sim \\triangle AFD$ and $\\triangle AGD \\sim \\triangle AEC$. Proof: Note that $ABCD$ is cyclic. We need one more angle, and we get this from this cyclic quadrilateral: \\begin{align*} \\angle ABE &= 180^\\circ - \\angle ABC =\\angle ADC = \\angle ADG, \\\\ \\angle ADG &= \\angle ADB =\\angle ACB = \\angle ACE. \\hspace{20mm} \\square \\end{align*} Let $AD=a$. We obtain from the similarities $AB = \\frac{5a}{6}$ and $AC=BD=\\frac{3a}{2}$. By Ptolemy, $\\left(\\frac{3a}{2}\\right)^2 = a^2 + \\frac{5a}{6} \\cdot CD$, so $\\frac{5a^2}{4} = \\frac{5a}{6} \\cdot CD$. We obtain $CD=\\frac{3a}{2}$, so $DF=\\frac{CD-AB}{2}=\\frac{a}{3}$. Applying the Pythagorean theorem on $\\triangle ADF$, we get $324=a^2 - \\frac{a^2}{9}=\\frac{8a^2}{9}$. Thus, $a=\\frac{27}{\\sqrt{2}}$, and $[ABCD]=\\frac{AB+CD}{2} \\cdot 18 = \\frac{\\frac{5a}{6} +\\frac{9a}{6}}{2} \\cdot 18 = 18 \\cdot \\frac{7}{6} \\cdot \\frac{27}{\\sqrt{2}} = \\frac{567}{\\sqrt{2}}$, yielding $\\sqrt2\\cdot[ABCD]=567.",
"Let $AD=BC=a$. Draw diagonal $AC$ and let $G$ be the foot of the perpendicular from $B$ to $AC$, $F$ be the foot of the perpendicular from $A$ to line $BC$, and $H$ be the foot of the perpendicular from $A$ to $DC$. Note that $\\triangle CBG\\sim\\triangle CAF$, and we get that $\\frac{10}{15}=\\frac{a}{AC}$. Therefore, $AC=\\frac32 a$. It then follows that $\\triangle ABF\\sim\\triangle ADH$. Using similar triangles, we can then find that $AB=\\frac{5}{6}a$. Using the Law of Cosines on $\\triangle ABC$, We can find that the $\\cos\\angle ABC=-\\frac{1}{3}$. Since $\\angle ABF=\\angle ADH$, and each is supplementary to $\\angle ABC$, we know that the $\\cos\\angle ADH=\\frac{1}{3}$. It then follows that $a=\\frac{27\\sqrt{2}}{2}$. Then it can be found that the area $K$ is $\\frac{567\\sqrt{2}}{2}$. Multiplying this by $\\sqrt{2}$, the answer is $567. ~happykeeper",
"Let $AD=BC=a$. Draw diagonal $AC$ and let $G$ be the foot of the perpendicular from $B$ to $AC$, $F$ be the foot of the perpendicular from $A$ to line $BC$, and $H$ be the foot of the perpendicular from $A$ to $DC$. Note that $\\triangle CBG\\sim\\triangle CAF$, and we get that $\\frac{10}{15}=\\frac{a}{AC}$. Therefore, $AC=\\frac32 a$. It then follows that $\\triangle ABF\\sim\\triangle ADH$. Using similar triangles, we can then find that $AB=\\frac{5}{6}a$. Using the Law of Cosines on $\\triangle ABC$, We can find that the $\\cos\\angle ABC=-\\frac{1}{3}$. Since $\\angle ABF=\\angle ADH$, and each is supplementary to $\\angle ABC$, we know that the $\\cos\\angle ADH=\\frac{1}{3}$. It then follows that $a=\\frac{27\\sqrt{2}}{2}$. Then it can be found that the area $K$ is $\\frac{567\\sqrt{2}}{2}$. Multiplying this by $\\sqrt{2}$, the answer is $567. ~happykeeper",
"Draw the distances in terms of $B$, as shown in the diagram. By similar triangles, $\\triangle{AEC}\\sim\\triangle{BIC}$. As a result, let $AB=u$, then $BC=AD=\\frac{6}{5}u$ and $2AC=3BC$. The triangle $ABC$ is $6-5-9$ which $\\cos(\\angle{ABC})=-\\frac{1}{3}$. By angle subtraction, $\\cos(180-\\theta)=-\\cos\\theta$. Therefore, $AB=\\frac{45}{2\\sqrt{2}}=\\frac{45\\sqrt{2}}{4}$ and $AD=BC=\\frac{27}{\\sqrt{2}}$. By trapezoid area formula, the area of $ABCD$ is equal to $(AB+DF)\\cdot 18=567\\cdot \\frac{\\sqrt{2}}{2}$ which $\\sqrt{2}\\cdot k=567. ~math2718281828459",
"Draw the distances in terms of $B$, as shown in the diagram. By similar triangles, $\\triangle{AEC}\\sim\\triangle{BIC}$. As a result, let $AB=u$, then $BC=AD=\\frac{6}{5}u$ and $2AC=3BC$. The triangle $ABC$ is $6-5-9$ which $\\cos(\\angle{ABC})=-\\frac{1}{3}$. By angle subtraction, $\\cos(180-\\theta)=-\\cos\\theta$. Therefore, $AB=\\frac{45}{2\\sqrt{2}}=\\frac{45\\sqrt{2}}{4}$ and $AD=BC=\\frac{27}{\\sqrt{2}}$. By trapezoid area formula, the area of $ABCD$ is equal to $(AB+DF)\\cdot 18=567\\cdot \\frac{\\sqrt{2}}{2}$ which $\\sqrt{2}\\cdot k=567. ~math2718281828459",
"[asy] size(250); pair A, B, C, D, E, F, G, H; A = (-45sqrt(2)/8,18); B = (45sqrt(2)/8,18); C = (81sqrt(2)/8,0); D = (-81sqrt(2)/8,0); E = foot(A,C,B); F = foot(A,C,D); G = foot(A,B,D); H = intersectionpoint(A--F,B--D); markscalefactor=0.1; draw(rightanglemark(A,E,B),red); draw(rightanglemark(A,F,C),red); draw(rightanglemark(A,G,D),red); filldraw(A--D--F--cycle,yellow,black+linewidth(1.5)); filldraw(A--B--E--cycle,yellow,black+linewidth(1.5)); dot(\"$A$\",A,1.5*NW,linewidth(4)); dot(\"$B$\",B,1.5*NE,linewidth(4)); dot(\"$C$\",C,1.5*SE,linewidth(4)); dot(\"$D$\",D,1.5*SW,linewidth(4)); dot(E,linewidth(4)); dot(F,linewidth(4)); dot(G,linewidth(4)); label(\"$E$\",E,NE); label(\"$F$\",F, S); label(\"$G$\",G,SE); draw(A--B--C--D--cycle^^B--D^^B--E); draw(A--E^^A--F^^A--G,dashed); label(\"$10$\",midpoint(A--G),1.5*(1,0)); label(\"$15$\",midpoint(A--E),1.5*N); label(\"$5x$\",midpoint(A--B),S); label(\"$6x$\",midpoint(A--D),1.5*(-1,0)); Label L = Label(\"$18$\", align=(0,0), position=MidPoint, filltype=Fill(0,3,white)); draw(C+(5,0)--(81sqrt(2)/8,18)+(5,0), L=L, arrow=Arrows(),bar=Bars(15)); [/asy] Let the points formed by dropping altitudes from $A$ to the lines $BC$, $CD$, and $BD$ be $E$, $F$, and $G$, respectively. We have \\[\\triangle ABE \\sim \\triangle ADF \\implies \\frac{AD}{18} = \\frac{AB}{15} \\implies AD = \\frac{6}{5}AB\\] and \\[BD\\cdot10 = 2[ABD] = AB\\cdot18 \\implies BD = \\frac{9}{5}AB.\\] For convenience, let $AB = 5x$. By Heron's formula on $\\triangle ABD$, we have sides $5x,6x,9x$ and semiperimeter $10x$, so \\[\\sqrt{10x\\cdot5x\\cdot4x\\cdot1x} = [ABD] = \\frac{AB\\cdot18}{2} = 45x \\implies 10\\sqrt{2}x^2 = 45x \\implies x= \\frac{45}{10\\sqrt{2}},\\] so $AB = 5x = \\frac{45}{2\\sqrt{2}}$. Then, \\[BE = \\sqrt{AB^2 - CA^2} = \\sqrt{\\left(\\frac{45}{2\\sqrt{2}}\\right)^2 - 15^2} = \\sqrt{\\frac{225}{8}} = \\frac{15}{2\\sqrt{2}}\\] and \\[\\triangle ABE \\sim \\triangle ADF \\implies DF = \\frac{6}{5}BE = \\frac{6}{5}\\cdot\\frac{15}{2\\sqrt{2}} = \\frac{18}{2\\sqrt{2}}.\\] Finally, recalling that $ABCD$ is isosceles, \\[K = [ABCD] = \\frac{18}{2}(AB + (AB + 2DF)) = 18(AB + DF) = 18\\left(\\frac{45}{2\\sqrt{2}} + \\frac{18}{2\\sqrt{2}}\\right) = \\frac{567}{\\sqrt{2}},\\] so $\\sqrt{2}\\cdot K = 567. ~emerald_block",
"[asy] size(250); pair A, B, C, D, E, F, G, H; A = (-45sqrt(2)/8,18); B = (45sqrt(2)/8,18); C = (81sqrt(2)/8,0); D = (-81sqrt(2)/8,0); E = foot(A,C,B); F = foot(A,C,D); G = foot(A,B,D); H = intersectionpoint(A--F,B--D); markscalefactor=0.1; draw(rightanglemark(A,E,B),red); draw(rightanglemark(A,F,C),red); draw(rightanglemark(A,G,D),red); filldraw(A--D--F--cycle,yellow,black+linewidth(1.5)); filldraw(A--B--E--cycle,yellow,black+linewidth(1.5)); dot(\"$A$\",A,1.5*NW,linewidth(4)); dot(\"$B$\",B,1.5*NE,linewidth(4)); dot(\"$C$\",C,1.5*SE,linewidth(4)); dot(\"$D$\",D,1.5*SW,linewidth(4)); dot(E,linewidth(4)); dot(F,linewidth(4)); dot(G,linewidth(4)); label(\"$E$\",E,NE); label(\"$F$\",F, S); label(\"$G$\",G,SE); draw(A--B--C--D--cycle^^B--D^^B--E); draw(A--E^^A--F^^A--G,dashed); label(\"$10$\",midpoint(A--G),1.5*(1,0)); label(\"$15$\",midpoint(A--E),1.5*N); label(\"$5x$\",midpoint(A--B),S); label(\"$6x$\",midpoint(A--D),1.5*(-1,0)); Label L = Label(\"$18$\", align=(0,0), position=MidPoint, filltype=Fill(0,3,white)); draw(C+(5,0)--(81sqrt(2)/8,18)+(5,0), L=L, arrow=Arrows(),bar=Bars(15)); [/asy] Let the points formed by dropping altitudes from $A$ to the lines $BC$, $CD$, and $BD$ be $E$, $F$, and $G$, respectively. We have \\[\\triangle ABE \\sim \\triangle ADF \\implies \\frac{AD}{18} = \\frac{AB}{15} \\implies AD = \\frac{6}{5}AB\\] and \\[BD\\cdot10 = 2[ABD] = AB\\cdot18 \\implies BD = \\frac{9}{5}AB.\\] For convenience, let $AB = 5x$. By Heron's formula on $\\triangle ABD$, we have sides $5x,6x,9x$ and semiperimeter $10x$, so \\[\\sqrt{10x\\cdot5x\\cdot4x\\cdot1x} = [ABD] = \\frac{AB\\cdot18}{2} = 45x \\implies 10\\sqrt{2}x^2 = 45x \\implies x= \\frac{45}{10\\sqrt{2}},\\] so $AB = 5x = \\frac{45}{2\\sqrt{2}}$. Then, \\[BE = \\sqrt{AB^2 - CA^2} = \\sqrt{\\left(\\frac{45}{2\\sqrt{2}}\\right)^2 - 15^2} = \\sqrt{\\frac{225}{8}} = \\frac{15}{2\\sqrt{2}}\\] and \\[\\triangle ABE \\sim \\triangle ADF \\implies DF = \\frac{6}{5}BE = \\frac{6}{5}\\cdot\\frac{15}{2\\sqrt{2}} = \\frac{18}{2\\sqrt{2}}.\\] Finally, recalling that $ABCD$ is isosceles, \\[K = [ABCD] = \\frac{18}{2}(AB + (AB + 2DF)) = 18(AB + DF) = 18\\left(\\frac{45}{2\\sqrt{2}} + \\frac{18}{2\\sqrt{2}}\\right) = \\frac{567}{\\sqrt{2}},\\] so $\\sqrt{2}\\cdot K = 567. ~emerald_block",
"Let $\\overline{AE}, \\overline{AF},$ and $\\overline{AG}$ be the perpendiculars from $A$ to ${BC}, {CD},$ and ${BD},$ respectively. $AE = 15, AF = 18, AG =10$. Denote by $G'$ the base of the perpendicular from $B$ to $AC, H$ be the base of the perpendicular from $C$ to $AB$. Denote $\\theta = \\angle{CBH}.$ It is clear that \\[BG' = AG, CH = AF, \\triangle CBH \\ =\\triangle ADF,\\] the area of $ABCD$ is equal to the area of the rectangle $AFCH.$ The problem is reduced to finding $AH$. In triangle $ABC$ all altitudes are known: \\[AB : BC : AC = \\frac{1}{CH}\\ : \\frac{1}{AE}\\ : \\frac{1}{BG'}\\ =\\] \\[= \\frac{1}{AF}\\ : \\frac{1}{AE}\\ : \\frac{1}{AG}\\ = 5 : 6 : 9.\\] We apply the Law of Cosines to $\\triangle ABC$ and get$:$ \\begin{align*} 2\\cdot AB\\cdot BC \\cdot \\cos\\theta = AC^2 – AB^2 – BC^2, \\end{align*} \\begin{align*} 2\\cdot 5\\cdot 6\\cdot \\cos\\theta = 60 \\cos\\theta = 9^2 – 5^2 – 6^2 = 20, \\cos\\theta =\\frac{1}{3}. \\end{align*} \\begin{align*} BH = BC \\cos\\theta = \\frac{BC}{3}.\\end{align*} We apply the Pythagorean Law to $\\triangle HBC$ and get$:$ \\begin{align*} HC^2 = 18^2 = BC^2 – BH^2 = 9\\cdot BH^2 – BH^2 = 8 BH^2.\\end{align*} \\begin{align*} BH = \\frac{9}{\\sqrt2}, AH = (\\frac{5}{2} + 1)\\cdot BH = \\frac{63}{2\\cdot \\sqrt2}. \\end{align*} Required area is \\begin{align*} K = \\frac{63}{2\\cdot \\sqrt{2}} \\cdot 18 = \\frac{567}{\\sqrt{2}} \\implies \\sqrt{2} K=567. \\end{align*} vladimir.shelomovskii@gmail.com, vvsss",
"Let $\\overline{AE}, \\overline{AF},$ and $\\overline{AG}$ be the perpendiculars from $A$ to ${BC}, {CD},$ and ${BD},$ respectively. $AE = 15, AF = 18, AG =10$. Denote by $G'$ the base of the perpendicular from $B$ to $AC, H$ be the base of the perpendicular from $C$ to $AB$. Denote $\\theta = \\angle{CBH}.$ It is clear that \\[BG' = AG, CH = AF, \\triangle CBH \\ =\\triangle ADF,\\] the area of $ABCD$ is equal to the area of the rectangle $AFCH.$ The problem is reduced to finding $AH$. In triangle $ABC$ all altitudes are known: \\[AB : BC : AC = \\frac{1}{CH}\\ : \\frac{1}{AE}\\ : \\frac{1}{BG'}\\ =\\] \\[= \\frac{1}{AF}\\ : \\frac{1}{AE}\\ : \\frac{1}{AG}\\ = 5 : 6 : 9.\\] We apply the Law of Cosines to $\\triangle ABC$ and get$:$ \\begin{align*} 2\\cdot AB\\cdot BC \\cdot \\cos\\theta = AC^2 – AB^2 – BC^2, \\end{align*} \\begin{align*} 2\\cdot 5\\cdot 6\\cdot \\cos\\theta = 60 \\cos\\theta = 9^2 – 5^2 – 6^2 = 20, \\cos\\theta =\\frac{1}{3}. \\end{align*} \\begin{align*} BH = BC \\cos\\theta = \\frac{BC}{3}.\\end{align*} We apply the Pythagorean Law to $\\triangle HBC$ and get$:$ \\begin{align*} HC^2 = 18^2 = BC^2 – BH^2 = 9\\cdot BH^2 – BH^2 = 8 BH^2.\\end{align*} \\begin{align*} BH = \\frac{9}{\\sqrt2}, AH = (\\frac{5}{2} + 1)\\cdot BH = \\frac{63}{2\\cdot \\sqrt2}. \\end{align*} Required area is \\begin{align*} K = \\frac{63}{2\\cdot \\sqrt{2}} \\cdot 18 = \\frac{567}{\\sqrt{2}} \\implies \\sqrt{2} K=567. \\end{align*} vladimir.shelomovskii@gmail.com, vvsss",
"Let $F$ be on $DC$ such that $AF \\| DC$. Let $G$ be on $BD$ such that $AG \\| BD$. Let $m$ be the length of $AB$. Let $n$ be the length of $AD$. The area of $\\triangle ABD$ can be expressed in three ways: $\\frac{1}{2}(15)(BC) = \\frac{1}{2}(15)(n)$, $\\frac{1}{2}(18)(m)$, and $\\frac{1}{2}(10)(BD)$. \\[\\frac{1}{2}(15)(n) = \\frac{1}{2}(18)(m)\\] \\[15n = 18m\\] \\[5n = 6m\\] \\[n = \\frac{6}{5}m\\] Now, $BD = BG + GD = \\sqrt{m^2-100} + \\sqrt{n^2-100}$. We can substitute in $n = \\frac{6}{5}m$ to get $BD = \\sqrt{m^2-100} + \\sqrt{(\\frac{6}{5}m)^2-100}$. We have \\[\\frac{1}{2}(10)\\left(\\sqrt{m^2-100} + \\sqrt{(\\frac{6}{5}m)^2-100}\\right) = \\frac{1}{2}(18)(m)\\] After a fairly straightforward algebraic bash, we get $m = \\frac{45\\sqrt{2}}{4}$, and $n = (\\frac{6}{5})(\\frac{45\\sqrt{2}}{4}) = \\frac{27\\sqrt{2}}{2}$. By the Pythagorean Theorem on $\\triangle ADF$, $DF^2 = n^2 - 18^2 = \\frac{729}{2} - 324 = \\frac{81}{2}$, and $DF = \\frac{9\\sqrt{2}}{2}$. Thus, $DC = 2DF + AB = 9\\sqrt{2}+\\frac{45\\sqrt{2}}{4} = \\frac{81\\sqrt{2}}{4}$. Therefore, $K = \\frac{1}{2}(\\frac{45\\sqrt{2}}{4}+\\frac{81\\sqrt{2}}{4}) \\cdot 18 = \\frac{63\\sqrt{2}}{2} \\cdot 18 = \\frac{567\\sqrt{2}}{2}$. The requested answer is $K \\cdot \\sqrt{2} = 567. ~ adam_zheng",
"Let $F$ be on $DC$ such that $AF \\| DC$. Let $G$ be on $BD$ such that $AG \\| BD$. Let $m$ be the length of $AB$. Let $n$ be the length of $AD$. The area of $\\triangle ABD$ can be expressed in three ways: $\\frac{1}{2}(15)(BC) = \\frac{1}{2}(15)(n)$, $\\frac{1}{2}(18)(m)$, and $\\frac{1}{2}(10)(BD)$. \\[\\frac{1}{2}(15)(n) = \\frac{1}{2}(18)(m)\\] \\[15n = 18m\\] \\[5n = 6m\\] \\[n = \\frac{6}{5}m\\] Now, $BD = BG + GD = \\sqrt{m^2-100} + \\sqrt{n^2-100}$. We can substitute in $n = \\frac{6}{5}m$ to get $BD = \\sqrt{m^2-100} + \\sqrt{(\\frac{6}{5}m)^2-100}$. We have \\[\\frac{1}{2}(10)\\left(\\sqrt{m^2-100} + \\sqrt{(\\frac{6}{5}m)^2-100}\\right) = \\frac{1}{2}(18)(m)\\] After a fairly straightforward algebraic bash, we get $m = \\frac{45\\sqrt{2}}{4}$, and $n = (\\frac{6}{5})(\\frac{45\\sqrt{2}}{4}) = \\frac{27\\sqrt{2}}{2}$. By the Pythagorean Theorem on $\\triangle ADF$, $DF^2 = n^2 - 18^2 = \\frac{729}{2} - 324 = \\frac{81}{2}$, and $DF = \\frac{9\\sqrt{2}}{2}$. Thus, $DC = 2DF + AB = 9\\sqrt{2}+\\frac{45\\sqrt{2}}{4} = \\frac{81\\sqrt{2}}{4}$. Therefore, $K = \\frac{1}{2}(\\frac{45\\sqrt{2}}{4}+\\frac{81\\sqrt{2}}{4}) \\cdot 18 = \\frac{63\\sqrt{2}}{2} \\cdot 18 = \\frac{567\\sqrt{2}}{2}$. The requested answer is $K \\cdot \\sqrt{2} = 567. ~ adam_zheng"
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2021-I-10
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Consider the sequence $(a_k)_{k\ge 1}$ of positive rational numbers defined by $a_1 = \frac{2020}{2021}$ and for $k\ge 1$ , if $a_k = \frac{m}{n}$ for relatively prime positive integers $m$ and $n$ , then \[a_{k+1} = \frac{m + 18}{n+19}.\] Determine the sum of all positive integers $j$ such that the rational number $a_j$ can be written in the form $\frac{t}{t+1}$ for some positive integer $t$ .
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I
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[
"We know that $a_{1}=\\tfrac{t}{t+1}$ when $t=2020$ so $1$ is a possible value of $j$. Note also that $a_{2}=\\tfrac{2038}{2040}=\\tfrac{1019}{1020}=\\tfrac{t}{t+1}$ for $t=1019$. Then $a_{2+q}=\\tfrac{1019+18q}{1020+19q}$ unless $1019+18q$ and $1020+19q$ are not relatively prime which happens when $q+1$ divides $18q+1019$ (by the Euclidean Algorithm), or $q+1$ divides $1001$. Thus, the least value of $q$ is $6$ and $j=2+6=8$. We know $a_{8}=\\tfrac{1019+108}{1020+114}=\\tfrac{1127}{1134}=\\tfrac{161}{162}$. Now $a_{8+q}=\\tfrac{161+18q}{162+19q}$ unless $18q+161$ and $19q+162$ are not relatively prime which happens the first time $q+1$ divides $18q+161$ or $q+1$ divides $143$ or $q=10$, and $j=8+10=18$. We have $a_{18}=\\tfrac{161+180}{162+190}=\\tfrac{341}{352}=\\tfrac{31}{32}$. Now $a_{18+q}=\\tfrac{31+18q}{32+19q}$ unless $18q+31$ and $19q+32$ are not relatively prime. This happens the first time $q+1$ divides $18q+31$ implying $q+1$ divides $13$, which is prime so $q=12$ and $j=18+12=30$. We have $a_{30}=\\tfrac{31+216}{32+228}=\\tfrac{247}{260}=\\tfrac{19}{20}$. We have $a_{30+q}=\\tfrac{18q+19}{19q+20}$, which is always reduced by EA, so the sum of all $j$ is $1+2+8+18+30=059. This follows from the Euclidean Algorithm. ~Magnetoninja",
"Let $a_{j_1}, a_{j_2}, a_{j_3}, \\ldots, a_{j_u}$ be all terms in the form $\\frac{t}{t+1},$ where $j_1<j_2<j_3<\\cdots<j_u,$ and $t$ is some positive integer. We wish to find $\\sum_{i=1}^{u}{j_i}.$ Suppose $a_{j_i}=\\frac{m}{m+1}$ for some positive integer $m.$ To find $\\boldsymbol{a_{j_{i+1}},}$ we look for the smallest positive integer $\\boldsymbol{k'}$ for which \\[\\boldsymbol{a_{j_{i+1}}=a_{j_i+k'}=\\frac{m+18k'}{m+1+19k'}}\\] is reducible: If $\\frac{m+18k'}{m+1+19k'}$ is reducible, then there exists a common factor $d>1$ for $m+18k'$ and $m+1+19k'.$ By the Euclidean Algorithm, we have \\begin{align*} d\\mid m+18k' \\text{ and } d\\mid m+1+19k' &\\implies d\\mid m+18k' \\text{ and } d\\mid k'+1 \\\\ &\\implies d\\mid m-18 \\text{ and } d\\mid k'+1. \\end{align*} Since $m-18$ and $k'+1$ are not relatively prime, and $m$ is fixed, the smallest value of $k'$ such that $\\frac{m+18k'}{m+1+19k'}$ is reducible occurs when $k'+1$ is the smallest prime factor of $m-18.$ We will prove that for such value of $\\boldsymbol{k',}$ the number $\\boldsymbol{a_{j_{i+1}}}$ can be written in the form $\\boldsymbol{\\frac{t}{t+1}:}$ \\[a_{j_{i+1}}=a_{j_i+k'}=\\frac{m+18k'}{m+1+19k'}=\\frac{(m-18)+18(k'+1)}{(m-18)+19(k'+1)}=\\frac{\\frac{m-18}{k'+1}+18}{\\frac{m-18}{k'+1}+19}, \\hspace{10mm} (*)\\] where $t=\\frac{m-18}{k'+1}+18$ must be a positive integer. We start with $m=2020$ and $a_{j_1}=a_1=\\frac{2020}{2021},$ then find $a_{j_2}, a_{j_3}, \\ldots, a_{j_u}$ by filling out the table below recursively: \\[\\begin{array}{c|c|c|c|c|c} & & & & & \\\\ [-2ex] \\boldsymbol{i} & \\boldsymbol{m} & \\boldsymbol{m-18} & \\boldsymbol{k'+1} & \\boldsymbol{k'} & \\boldsymbol{a_{j_{i+1}} \\left(\\textbf{by } (*)\\right)} \\\\ [0.5ex] \\hline & & & & & \\\\ [-1.5ex] 1 & 2020 & 2002 & 2 & 1 & \\hspace{4.25mm} a_2 = \\frac{1019}{1020} \\\\ [1ex] 2 & 1019 & 1001 & 7 & 6 & \\hspace{2.75mm} a_8 = \\frac{161}{162} \\\\ [1ex] 3 & 161 & 143 & 11 & 10 & a_{18} = \\frac{31}{32} \\\\ [1ex] 4 & 31 & 13 & 13 & 12 & a_{30} = \\frac{19}{20} \\\\ [1ex] 5 & 19 & 1 & \\text{N/A} & \\text{N/A} & \\text{N/A} \\\\ [1ex] \\end{array}\\] As $\\left(j_1,j_2,j_3,j_4,j_5\\right)=(1,2,8,18,30),$ the answer is $\\sum_{i=1}^{5}{j_i}=059 Now we continue with the last two paragraphs of the solution above. ~MRENTHUSIASM",
"Let $a_{j_1}, a_{j_2}, a_{j_3}, \\ldots, a_{j_u}$ be all terms in the form $\\frac{t}{t+1},$ where $j_1<j_2<j_3<\\cdots<j_u,$ and $t$ is some positive integer. We wish to find $\\sum_{i=1}^{u}{j_i}.$ Suppose $a_{j_i}=\\frac{m}{m+1}$ for some positive integer $m.$ To find $\\boldsymbol{a_{j_{i+1}},}$ we look for the smallest positive integer $\\boldsymbol{k'}$ for which \\[\\boldsymbol{a_{j_{i+1}}=a_{j_i+k'}=\\frac{m+18k'}{m+1+19k'}}\\] is reducible: If $\\frac{m+18k'}{m+1+19k'}$ is reducible, then there exists a common factor $d>1$ for $m+18k'$ and $m+1+19k'.$ By the Euclidean Algorithm, we have \\begin{align*} d\\mid m+18k' \\text{ and } d\\mid m+1+19k' &\\implies d\\mid m+18k' \\text{ and } d\\mid k'+1 \\\\ &\\implies d\\mid m-18 \\text{ and } d\\mid k'+1. \\end{align*} Since $m-18$ and $k'+1$ are not relatively prime, and $m$ is fixed, the smallest value of $k'$ such that $\\frac{m+18k'}{m+1+19k'}$ is reducible occurs when $k'+1$ is the smallest prime factor of $m-18.$ We will prove that for such value of $\\boldsymbol{k',}$ the number $\\boldsymbol{a_{j_{i+1}}}$ can be written in the form $\\boldsymbol{\\frac{t}{t+1}:}$ \\[a_{j_{i+1}}=a_{j_i+k'}=\\frac{m+18k'}{m+1+19k'}=\\frac{(m-18)+18(k'+1)}{(m-18)+19(k'+1)}=\\frac{\\frac{m-18}{k'+1}+18}{\\frac{m-18}{k'+1}+19}, \\hspace{10mm} (*)\\] where $t=\\frac{m-18}{k'+1}+18$ must be a positive integer. We start with $m=2020$ and $a_{j_1}=a_1=\\frac{2020}{2021},$ then find $a_{j_2}, a_{j_3}, \\ldots, a_{j_u}$ by filling out the table below recursively: \\[\\begin{array}{c|c|c|c|c|c} & & & & & \\\\ [-2ex] \\boldsymbol{i} & \\boldsymbol{m} & \\boldsymbol{m-18} & \\boldsymbol{k'+1} & \\boldsymbol{k'} & \\boldsymbol{a_{j_{i+1}} \\left(\\textbf{by } (*)\\right)} \\\\ [0.5ex] \\hline & & & & & \\\\ [-1.5ex] 1 & 2020 & 2002 & 2 & 1 & \\hspace{4.25mm} a_2 = \\frac{1019}{1020} \\\\ [1ex] 2 & 1019 & 1001 & 7 & 6 & \\hspace{2.75mm} a_8 = \\frac{161}{162} \\\\ [1ex] 3 & 161 & 143 & 11 & 10 & a_{18} = \\frac{31}{32} \\\\ [1ex] 4 & 31 & 13 & 13 & 12 & a_{30} = \\frac{19}{20} \\\\ [1ex] 5 & 19 & 1 & \\text{N/A} & \\text{N/A} & \\text{N/A} \\\\ [1ex] \\end{array}\\] As $\\left(j_1,j_2,j_3,j_4,j_5\\right)=(1,2,8,18,30),$ the answer is $\\sum_{i=1}^{5}{j_i}=059 Now we continue with the last two paragraphs of the solution above. ~MRENTHUSIASM"
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2021-I-11
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Let $ABCD$ be a cyclic quadrilateral with $AB=4,BC=5,CD=6,$ and $DA=7.$ Let $A_1$ and $C_1$ be the feet of the perpendiculars from $A$ and $C,$ respectively, to line $BD,$ and let $B_1$ and $D_1$ be the feet of the perpendiculars from $B$ and $D,$ respectively, to line $AC.$ The perimeter of $A_1B_1C_1D_1$ is $\frac mn,$ where $m$ and $n$ are relatively prime positive integers. Find $m+n.$
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I
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[
"This solution refers to the Diagram section. By the Converse of the Inscribed Angle Theorem, if distinct points $X$ and $Y$ lie on the same side of $\\overline{PQ}$ (but not on $\\overline{PQ}$ itself) for which $\\angle PXQ=\\angle PYQ,$ then $P,Q,X,$ and $Y$ are cyclic. From the Converse of the Inscribed Angle Theorem, quadrilaterals $ABA_1B_1,BCC_1B_1,CDC_1D_1,$ and $DAA_1D_1$ are all cyclic. Suppose $\\overline{AC}$ and $\\overline{BD}$ intersect at $E,$ and let $\\angle AEB=\\theta.$ It follows that $\\angle CED=\\theta$ and $\\angle BEC=\\angle DEA=180^\\circ-\\theta.$ We obtain the following diagram: [asy] /* Made by MRENTHUSIASM */ size(300); pair A, B, C, D, A1, B1, C1, D1, P, M1, M2; A = origin; C = (sqrt(53041)/31,0); B = intersectionpoints(Circle(A,4),Circle(C,5))[0]; D = intersectionpoints(Circle(A,7),Circle(C,6))[1]; A1 = foot(A,B,D); C1 = foot(C,B,D); B1 = foot(B,A,C); D1 = foot(D,A,C); P = intersectionpoint(A--C,B--D); M1 = midpoint(A--B); M2 = midpoint(C--D); markscalefactor=0.025; draw(rightanglemark(A,A1,B),red); draw(rightanglemark(B,B1,A),red); draw(rightanglemark(C,C1,D),red); draw(rightanglemark(D,D1,C),red); draw(Arc(M1,A,B)^^Arc(M2,C,D),blue); draw(A1--B1--C1--D1--cycle,green); dot(\"$A$\",A,1.5*W,linewidth(4)); dot(\"$B$\",B,1.5*dir(180-aCos(11/59)),linewidth(4)); dot(\"$C$\",C,1.5*E,linewidth(4)); dot(\"$D$\",D,1.5*dir(-aCos(11/59)),linewidth(4)); dot(\"$A_1$\",A1,1.5*dir(A1-A),linewidth(4)); dot(\"$B_1$\",B1,1.5*S,linewidth(4)); dot(\"$C_1$\",C1,1.5*dir(C1-C),linewidth(4)); dot(\"$D_1$\",D1,1.5*N,linewidth(4)); dot(\"$E$\",P,dir((180-aCos(11/59))/2),linewidth(4)); label(\"$\\theta$\",P,dir(180-aCos(11/59)/2),red); draw(A--B--C--D--cycle^^A--C^^B--D^^circumcircle(A,B,C)); draw(A--A1^^B--B1^^C--C1^^D--D1,dashed); [/asy] In every cyclic quadrilateral, each pair of opposite angles is supplementary. So, we have $\\angle EA_1B_1=\\angle EAB$ (both supplementary to $\\angle B_1A_1B$) and $\\angle EB_1A_1=\\angle EBA$ (both supplementary to $\\angle A_1B_1A$), from which $\\triangle A_1B_1E \\sim \\triangle ABE$ by AA, with the ratio of similitude \\[\\frac{A_1B_1}{AB}=\\underbrace{\\frac{A_1E}{AE}}_{\\substack{\\text{right} \\\\ \\triangle A_1AE}}=\\underbrace{\\frac{B_1E}{BE}}_{\\substack{\\text{right} \\\\ \\triangle B_1BE}}=\\cos\\theta. \\hspace{15mm}(1)\\] Similarly, we have $\\angle EC_1D_1=\\angle ECD$ (both supplementary to $\\angle D_1C_1D$) and $\\angle ED_1C_1=\\angle EDC$ (both supplementary to $\\angle C_1D_1C$), from which $\\triangle C_1D_1E \\sim \\triangle CDE$ by AA, with the ratio of similitude \\[\\frac{C_1D_1}{CD}=\\underbrace{\\frac{C_1E}{CE}}_{\\substack{\\text{right} \\\\ \\triangle C_1CE}}=\\underbrace{\\frac{D_1E}{DE}}_{\\substack{\\text{right} \\\\ \\triangle D_1DE}}=\\cos\\theta. \\hspace{14.75mm}(2)\\] We apply the Transitive Property to $(1)$ and $(2):$ We get $\\frac{B_1E}{BE}=\\frac{C_1E}{CE}=\\cos\\theta,$ so $\\triangle B_1C_1E \\sim \\triangle BCE$ by SAS, with the ratio of similitude \\[\\frac{B_1C_1}{BC}=\\frac{B_1E}{BE}=\\frac{C_1E}{CE}=\\cos\\theta. \\hspace{14.75mm}(3)\\] We get $\\frac{D_1E}{DE}=\\frac{A_1E}{AE}=\\cos\\theta,$ so $\\triangle D_1A_1E \\sim \\triangle DAE$ by SAS, with the ratio of similitude \\[\\frac{D_1A_1}{DA}=\\frac{D_1E}{DE}=\\frac{A_1E}{AE}=\\cos\\theta. \\hspace{14mm}(4)\\] From $(1),(2),(3),$ and $(4),$ the perimeter of $A_1B_1C_1D_1$ is \\begin{align*} A_1B_1+B_1C_1+C_1D_1+D_1A_1&=AB\\cos\\theta+BC\\cos\\theta+CD\\cos\\theta+DA\\cos\\theta \\\\ &=(AB+BC+CD+DA)\\cos\\theta \\\\ &=22\\cos\\theta. &&\\hspace{5mm}(\\bigstar) \\end{align*} Two solutions follow from here: Solution 1.1 (Law of Cosines) Note that $\\cos(180^\\circ-\\theta)=-\\cos\\theta$ holds for all $\\theta.$ We apply the Law of Cosines to $\\triangle ABE, \\triangle BCE, \\triangle CDE,$ and $\\triangle DAE,$ respectively: \\begin{alignat*}{12} &&&AE^2+BE^2-2\\cdot AE\\cdot BE\\cdot\\cos\\angle AEB&&=AB^2&&\\quad\\implies\\quad AE^2+BE^2-2\\cdot AE\\cdot BE\\cdot\\cos\\theta&&=4^2, \\hspace{15mm} &(1\\star) \\\\ &&&BE^2+CE^2-2\\cdot BE\\cdot CE\\cdot\\cos\\angle BEC&&=BC^2&&\\quad\\implies\\quad BE^2+CE^2+2\\cdot BE\\cdot CE\\cdot\\cos\\theta&&=5^2, \\hspace{15mm} &(2\\star) \\\\ &&&CE^2+DE^2-2\\cdot CE\\cdot DE\\cdot\\cos\\angle CED&&=CD^2&&\\quad\\implies\\quad CE^2+DE^2-2\\cdot CE\\cdot DE\\cdot\\cos\\theta&&=6^2, \\hspace{15mm} &(3\\star) \\\\ &&&DE^2+AE^2-2\\cdot DE\\cdot AE\\cdot\\cos\\angle DEA&&=DA^2&&\\quad\\implies\\quad DE^2+AE^2+2\\cdot DE\\cdot AE\\cdot\\cos\\theta&&=7^2. \\hspace{15mm} &(4\\star) \\\\ \\end{alignat*} We subtract $(1\\star)+(3\\star)$ from $(2\\star)+(4\\star):$ \\begin{align*} 2\\cdot AE\\cdot BE\\cdot\\cos\\theta+2\\cdot BE\\cdot CE\\cdot\\cos\\theta+2\\cdot CE\\cdot DE\\cdot\\cos\\theta+2\\cdot DE\\cdot AE\\cdot\\cos\\theta&=22 \\\\ 2\\cdot\\cos\\theta\\cdot(\\phantom{ }\\underbrace{AE\\cdot BE+BE\\cdot CE+CE\\cdot DE+DE\\cdot AE}_{\\text{Use the result from }\\textbf{Remark}\\text{.}}\\phantom{ })&=22 \\\\ 2\\cdot\\cos\\theta\\cdot59&=22 \\\\ \\cos\\theta&=\\frac{11}{59}. \\end{align*} Finally, substituting this result into $(\\bigstar)$ gives $22\\cos\\theta=\\frac{242}{59},$ from which the answer is $242+59=301 we have \\begin{align*} AE\\cdot BE+BE\\cdot CE+CE\\cdot DE+DE\\cdot AE &= (AE+CE)(BE+DE) &&\\hspace{10mm}\\text{Factor by Grouping} \\\\ &=AC\\cdot BD &&\\hspace{10mm}\\text{Segment Addition} \\\\ &=AB\\cdot CD+BC\\cdot DA &&\\hspace{10mm}\\text{Ptolemy's Theorem} \\\\ &=59. &&\\hspace{10mm}\\text{Substitution} \\end{align*} ~MRENTHUSIASM",
"This solution refers to the Diagram section. By the Converse of the Inscribed Angle Theorem, if distinct points $X$ and $Y$ lie on the same side of $\\overline{PQ}$ (but not on $\\overline{PQ}$ itself) for which $\\angle PXQ=\\angle PYQ,$ then $P,Q,X,$ and $Y$ are cyclic. From the Converse of the Inscribed Angle Theorem, quadrilaterals $ABA_1B_1,BCC_1B_1,CDC_1D_1,$ and $DAA_1D_1$ are all cyclic. Suppose $\\overline{AC}$ and $\\overline{BD}$ intersect at $E,$ and let $\\angle AEB=\\theta.$ It follows that $\\angle CED=\\theta$ and $\\angle BEC=\\angle DEA=180^\\circ-\\theta.$ We obtain the following diagram: [asy] /* Made by MRENTHUSIASM */ size(300); pair A, B, C, D, A1, B1, C1, D1, P, M1, M2; A = origin; C = (sqrt(53041)/31,0); B = intersectionpoints(Circle(A,4),Circle(C,5))[0]; D = intersectionpoints(Circle(A,7),Circle(C,6))[1]; A1 = foot(A,B,D); C1 = foot(C,B,D); B1 = foot(B,A,C); D1 = foot(D,A,C); P = intersectionpoint(A--C,B--D); M1 = midpoint(A--B); M2 = midpoint(C--D); markscalefactor=0.025; draw(rightanglemark(A,A1,B),red); draw(rightanglemark(B,B1,A),red); draw(rightanglemark(C,C1,D),red); draw(rightanglemark(D,D1,C),red); draw(Arc(M1,A,B)^^Arc(M2,C,D),blue); draw(A1--B1--C1--D1--cycle,green); dot(\"$A$\",A,1.5*W,linewidth(4)); dot(\"$B$\",B,1.5*dir(180-aCos(11/59)),linewidth(4)); dot(\"$C$\",C,1.5*E,linewidth(4)); dot(\"$D$\",D,1.5*dir(-aCos(11/59)),linewidth(4)); dot(\"$A_1$\",A1,1.5*dir(A1-A),linewidth(4)); dot(\"$B_1$\",B1,1.5*S,linewidth(4)); dot(\"$C_1$\",C1,1.5*dir(C1-C),linewidth(4)); dot(\"$D_1$\",D1,1.5*N,linewidth(4)); dot(\"$E$\",P,dir((180-aCos(11/59))/2),linewidth(4)); label(\"$\\theta$\",P,dir(180-aCos(11/59)/2),red); draw(A--B--C--D--cycle^^A--C^^B--D^^circumcircle(A,B,C)); draw(A--A1^^B--B1^^C--C1^^D--D1,dashed); [/asy] In every cyclic quadrilateral, each pair of opposite angles is supplementary. So, we have $\\angle EA_1B_1=\\angle EAB$ (both supplementary to $\\angle B_1A_1B$) and $\\angle EB_1A_1=\\angle EBA$ (both supplementary to $\\angle A_1B_1A$), from which $\\triangle A_1B_1E \\sim \\triangle ABE$ by AA, with the ratio of similitude \\[\\frac{A_1B_1}{AB}=\\underbrace{\\frac{A_1E}{AE}}_{\\substack{\\text{right} \\\\ \\triangle A_1AE}}=\\underbrace{\\frac{B_1E}{BE}}_{\\substack{\\text{right} \\\\ \\triangle B_1BE}}=\\cos\\theta. \\hspace{15mm}(1)\\] Similarly, we have $\\angle EC_1D_1=\\angle ECD$ (both supplementary to $\\angle D_1C_1D$) and $\\angle ED_1C_1=\\angle EDC$ (both supplementary to $\\angle C_1D_1C$), from which $\\triangle C_1D_1E \\sim \\triangle CDE$ by AA, with the ratio of similitude \\[\\frac{C_1D_1}{CD}=\\underbrace{\\frac{C_1E}{CE}}_{\\substack{\\text{right} \\\\ \\triangle C_1CE}}=\\underbrace{\\frac{D_1E}{DE}}_{\\substack{\\text{right} \\\\ \\triangle D_1DE}}=\\cos\\theta. \\hspace{14.75mm}(2)\\] We apply the Transitive Property to $(1)$ and $(2):$ We get $\\frac{B_1E}{BE}=\\frac{C_1E}{CE}=\\cos\\theta,$ so $\\triangle B_1C_1E \\sim \\triangle BCE$ by SAS, with the ratio of similitude \\[\\frac{B_1C_1}{BC}=\\frac{B_1E}{BE}=\\frac{C_1E}{CE}=\\cos\\theta. \\hspace{14.75mm}(3)\\] We get $\\frac{D_1E}{DE}=\\frac{A_1E}{AE}=\\cos\\theta,$ so $\\triangle D_1A_1E \\sim \\triangle DAE$ by SAS, with the ratio of similitude \\[\\frac{D_1A_1}{DA}=\\frac{D_1E}{DE}=\\frac{A_1E}{AE}=\\cos\\theta. \\hspace{14mm}(4)\\] From $(1),(2),(3),$ and $(4),$ the perimeter of $A_1B_1C_1D_1$ is \\begin{align*} A_1B_1+B_1C_1+C_1D_1+D_1A_1&=AB\\cos\\theta+BC\\cos\\theta+CD\\cos\\theta+DA\\cos\\theta \\\\ &=(AB+BC+CD+DA)\\cos\\theta \\\\ &=22\\cos\\theta. &&\\hspace{5mm}(\\bigstar) \\end{align*} Two solutions follow from here: Solution 1.1 (Law of Cosines) Note that $\\cos(180^\\circ-\\theta)=-\\cos\\theta$ holds for all $\\theta.$ We apply the Law of Cosines to $\\triangle ABE, \\triangle BCE, \\triangle CDE,$ and $\\triangle DAE,$ respectively: \\begin{alignat*}{12} &&&AE^2+BE^2-2\\cdot AE\\cdot BE\\cdot\\cos\\angle AEB&&=AB^2&&\\quad\\implies\\quad AE^2+BE^2-2\\cdot AE\\cdot BE\\cdot\\cos\\theta&&=4^2, \\hspace{15mm} &(1\\star) \\\\ &&&BE^2+CE^2-2\\cdot BE\\cdot CE\\cdot\\cos\\angle BEC&&=BC^2&&\\quad\\implies\\quad BE^2+CE^2+2\\cdot BE\\cdot CE\\cdot\\cos\\theta&&=5^2, \\hspace{15mm} &(2\\star) \\\\ &&&CE^2+DE^2-2\\cdot CE\\cdot DE\\cdot\\cos\\angle CED&&=CD^2&&\\quad\\implies\\quad CE^2+DE^2-2\\cdot CE\\cdot DE\\cdot\\cos\\theta&&=6^2, \\hspace{15mm} &(3\\star) \\\\ &&&DE^2+AE^2-2\\cdot DE\\cdot AE\\cdot\\cos\\angle DEA&&=DA^2&&\\quad\\implies\\quad DE^2+AE^2+2\\cdot DE\\cdot AE\\cdot\\cos\\theta&&=7^2. \\hspace{15mm} &(4\\star) \\\\ \\end{alignat*} We subtract $(1\\star)+(3\\star)$ from $(2\\star)+(4\\star):$ \\begin{align*} 2\\cdot AE\\cdot BE\\cdot\\cos\\theta+2\\cdot BE\\cdot CE\\cdot\\cos\\theta+2\\cdot CE\\cdot DE\\cdot\\cos\\theta+2\\cdot DE\\cdot AE\\cdot\\cos\\theta&=22 \\\\ 2\\cdot\\cos\\theta\\cdot(\\phantom{ }\\underbrace{AE\\cdot BE+BE\\cdot CE+CE\\cdot DE+DE\\cdot AE}_{\\text{Use the result from }\\textbf{Remark}\\text{.}}\\phantom{ })&=22 \\\\ 2\\cdot\\cos\\theta\\cdot59&=22 \\\\ \\cos\\theta&=\\frac{11}{59}. \\end{align*} Finally, substituting this result into $(\\bigstar)$ gives $22\\cos\\theta=\\frac{242}{59},$ from which the answer is $242+59=301 we have \\begin{align*} AE\\cdot BE+BE\\cdot CE+CE\\cdot DE+DE\\cdot AE &= (AE+CE)(BE+DE) &&\\hspace{10mm}\\text{Factor by Grouping} \\\\ &=AC\\cdot BD &&\\hspace{10mm}\\text{Segment Addition} \\\\ &=AB\\cdot CD+BC\\cdot DA &&\\hspace{10mm}\\text{Ptolemy's Theorem} \\\\ &=59. &&\\hspace{10mm}\\text{Substitution} \\end{align*} ~MRENTHUSIASM",
"Note that $\\cos(180^\\circ-\\theta)=-\\cos\\theta$ holds for all $\\theta.$ We apply the Law of Cosines to $\\triangle ABE, \\triangle BCE, \\triangle CDE,$ and $\\triangle DAE,$ respectively: \\begin{alignat*}{12} &&&AE^2+BE^2-2\\cdot AE\\cdot BE\\cdot\\cos\\angle AEB&&=AB^2&&\\quad\\implies\\quad AE^2+BE^2-2\\cdot AE\\cdot BE\\cdot\\cos\\theta&&=4^2, \\hspace{15mm} &(1\\star) \\\\ &&&BE^2+CE^2-2\\cdot BE\\cdot CE\\cdot\\cos\\angle BEC&&=BC^2&&\\quad\\implies\\quad BE^2+CE^2+2\\cdot BE\\cdot CE\\cdot\\cos\\theta&&=5^2, \\hspace{15mm} &(2\\star) \\\\ &&&CE^2+DE^2-2\\cdot CE\\cdot DE\\cdot\\cos\\angle CED&&=CD^2&&\\quad\\implies\\quad CE^2+DE^2-2\\cdot CE\\cdot DE\\cdot\\cos\\theta&&=6^2, \\hspace{15mm} &(3\\star) \\\\ &&&DE^2+AE^2-2\\cdot DE\\cdot AE\\cdot\\cos\\angle DEA&&=DA^2&&\\quad\\implies\\quad DE^2+AE^2+2\\cdot DE\\cdot AE\\cdot\\cos\\theta&&=7^2. \\hspace{15mm} &(4\\star) \\\\ \\end{alignat*} We subtract $(1\\star)+(3\\star)$ from $(2\\star)+(4\\star):$ \\begin{align*} 2\\cdot AE\\cdot BE\\cdot\\cos\\theta+2\\cdot BE\\cdot CE\\cdot\\cos\\theta+2\\cdot CE\\cdot DE\\cdot\\cos\\theta+2\\cdot DE\\cdot AE\\cdot\\cos\\theta&=22 \\\\ 2\\cdot\\cos\\theta\\cdot(\\phantom{ }\\underbrace{AE\\cdot BE+BE\\cdot CE+CE\\cdot DE+DE\\cdot AE}_{\\text{Use the result from }\\textbf{Remark}\\text{.}}\\phantom{ })&=22 \\\\ 2\\cdot\\cos\\theta\\cdot59&=22 \\\\ \\cos\\theta&=\\frac{11}{59}. \\end{align*} Finally, substituting this result into $(\\bigstar)$ gives $22\\cos\\theta=\\frac{242}{59},$ from which the answer is $242+59=301 we have \\begin{align*} AE\\cdot BE+BE\\cdot CE+CE\\cdot DE+DE\\cdot AE &= (AE+CE)(BE+DE) &&\\hspace{10mm}\\text{Factor by Grouping} \\\\ &=AC\\cdot BD &&\\hspace{10mm}\\text{Segment Addition} \\\\ &=AB\\cdot CD+BC\\cdot DA &&\\hspace{10mm}\\text{Ptolemy's Theorem} \\\\ &=59. &&\\hspace{10mm}\\text{Substitution} \\end{align*} ~MRENTHUSIASM",
"Note that $\\cos(180^\\circ-\\theta)=-\\cos\\theta$ holds for all $\\theta.$ We apply the Law of Cosines to $\\triangle ABE, \\triangle BCE, \\triangle CDE,$ and $\\triangle DAE,$ respectively: \\begin{alignat*}{12} &&&AE^2+BE^2-2\\cdot AE\\cdot BE\\cdot\\cos\\angle AEB&&=AB^2&&\\quad\\implies\\quad AE^2+BE^2-2\\cdot AE\\cdot BE\\cdot\\cos\\theta&&=4^2, \\hspace{15mm} &(1\\star) \\\\ &&&BE^2+CE^2-2\\cdot BE\\cdot CE\\cdot\\cos\\angle BEC&&=BC^2&&\\quad\\implies\\quad BE^2+CE^2+2\\cdot BE\\cdot CE\\cdot\\cos\\theta&&=5^2, \\hspace{15mm} &(2\\star) \\\\ &&&CE^2+DE^2-2\\cdot CE\\cdot DE\\cdot\\cos\\angle CED&&=CD^2&&\\quad\\implies\\quad CE^2+DE^2-2\\cdot CE\\cdot DE\\cdot\\cos\\theta&&=6^2, \\hspace{15mm} &(3\\star) \\\\ &&&DE^2+AE^2-2\\cdot DE\\cdot AE\\cdot\\cos\\angle DEA&&=DA^2&&\\quad\\implies\\quad DE^2+AE^2+2\\cdot DE\\cdot AE\\cdot\\cos\\theta&&=7^2. \\hspace{15mm} &(4\\star) \\\\ \\end{alignat*} We subtract $(1\\star)+(3\\star)$ from $(2\\star)+(4\\star):$ \\begin{align*} 2\\cdot AE\\cdot BE\\cdot\\cos\\theta+2\\cdot BE\\cdot CE\\cdot\\cos\\theta+2\\cdot CE\\cdot DE\\cdot\\cos\\theta+2\\cdot DE\\cdot AE\\cdot\\cos\\theta&=22 \\\\ 2\\cdot\\cos\\theta\\cdot(\\phantom{ }\\underbrace{AE\\cdot BE+BE\\cdot CE+CE\\cdot DE+DE\\cdot AE}_{\\text{Use the result from }\\textbf{Remark}\\text{.}}\\phantom{ })&=22 \\\\ 2\\cdot\\cos\\theta\\cdot59&=22 \\\\ \\cos\\theta&=\\frac{11}{59}. \\end{align*} Finally, substituting this result into $(\\bigstar)$ gives $22\\cos\\theta=\\frac{242}{59},$ from which the answer is $242+59=301 we have \\begin{align*} AE\\cdot BE+BE\\cdot CE+CE\\cdot DE+DE\\cdot AE &= (AE+CE)(BE+DE) &&\\hspace{10mm}\\text{Factor by Grouping} \\\\ &=AC\\cdot BD &&\\hspace{10mm}\\text{Segment Addition} \\\\ &=AB\\cdot CD+BC\\cdot DA &&\\hspace{10mm}\\text{Ptolemy's Theorem} \\\\ &=59. &&\\hspace{10mm}\\text{Substitution} \\end{align*} ~MRENTHUSIASM",
"Let the brackets denote areas. We find $[ABCD]$ in two different ways: Note that $\\sin(180^\\circ-\\theta)=\\sin\\theta$ holds for all $\\theta.$ By area addition, we get \\begin{align*} [ABCD]&=[ABE]+[BCE]+[CDE]+[DAE] \\\\ &=\\frac12\\cdot AE\\cdot BE\\cdot\\sin\\angle AEB+\\frac12\\cdot BE\\cdot CE\\cdot\\sin\\angle BEC+\\frac12\\cdot CE\\cdot DE\\cdot\\sin\\angle CED+\\frac12\\cdot DE\\cdot AE\\cdot\\sin\\angle DEA \\\\ &=\\frac12\\cdot AE\\cdot BE\\cdot\\sin\\theta+\\frac12\\cdot BE\\cdot CE\\cdot\\sin\\theta+\\frac12\\cdot CE\\cdot DE\\cdot\\sin\\theta+\\frac12\\cdot DE\\cdot AE\\cdot\\sin\\theta \\\\ &=\\frac12\\cdot\\sin\\theta\\cdot(\\phantom{ }\\underbrace{AE\\cdot BE+BE\\cdot CE+CE\\cdot DE+DE\\cdot AE}_{\\text{Use the result from }\\textbf{Remark}\\text{.}}\\phantom{ }) \\\\ &=\\frac12\\cdot\\sin\\theta\\cdot59. \\end{align*} By Brahmagupta's Formula, we get \\[[ABCD]=\\sqrt{(s-AB)(s-BC)(s-CD)(s-DA)}=2\\sqrt{210},\\] where $s=\\frac{AB+BC+CD+DA}{2}=11$ is the semiperimeter of $ABCD.$ Equating the expressions for $[ABCD],$ we have \\[\\frac12\\cdot\\sin\\theta\\cdot59=2\\sqrt{210},\\] so $\\sin\\theta=\\frac{4\\sqrt{210}}{59}.$ Since $0^\\circ<\\theta<90^\\circ,$ we have $\\cos\\theta>0.$ It follows that \\[\\cos\\theta=\\sqrt{1-\\sin^2\\theta}=\\frac{11}{59}.\\] Finally, substituting this result into $(\\bigstar)$ gives $22\\cos\\theta=\\frac{242}{59},$ from which the answer is $242+59=301 we have \\begin{align*} AE\\cdot BE+BE\\cdot CE+CE\\cdot DE+DE\\cdot AE &= (AE+CE)(BE+DE) &&\\hspace{10mm}\\text{Factor by Grouping} \\\\ &=AC\\cdot BD &&\\hspace{10mm}\\text{Segment Addition} \\\\ &=AB\\cdot CD+BC\\cdot DA &&\\hspace{10mm}\\text{Ptolemy's Theorem} \\\\ &=59. &&\\hspace{10mm}\\text{Substitution} \\end{align*} ~MRENTHUSIASM",
"Let the brackets denote areas. We find $[ABCD]$ in two different ways: Note that $\\sin(180^\\circ-\\theta)=\\sin\\theta$ holds for all $\\theta.$ By area addition, we get \\begin{align*} [ABCD]&=[ABE]+[BCE]+[CDE]+[DAE] \\\\ &=\\frac12\\cdot AE\\cdot BE\\cdot\\sin\\angle AEB+\\frac12\\cdot BE\\cdot CE\\cdot\\sin\\angle BEC+\\frac12\\cdot CE\\cdot DE\\cdot\\sin\\angle CED+\\frac12\\cdot DE\\cdot AE\\cdot\\sin\\angle DEA \\\\ &=\\frac12\\cdot AE\\cdot BE\\cdot\\sin\\theta+\\frac12\\cdot BE\\cdot CE\\cdot\\sin\\theta+\\frac12\\cdot CE\\cdot DE\\cdot\\sin\\theta+\\frac12\\cdot DE\\cdot AE\\cdot\\sin\\theta \\\\ &=\\frac12\\cdot\\sin\\theta\\cdot(\\phantom{ }\\underbrace{AE\\cdot BE+BE\\cdot CE+CE\\cdot DE+DE\\cdot AE}_{\\text{Use the result from }\\textbf{Remark}\\text{.}}\\phantom{ }) \\\\ &=\\frac12\\cdot\\sin\\theta\\cdot59. \\end{align*} By Brahmagupta's Formula, we get \\[[ABCD]=\\sqrt{(s-AB)(s-BC)(s-CD)(s-DA)}=2\\sqrt{210},\\] where $s=\\frac{AB+BC+CD+DA}{2}=11$ is the semiperimeter of $ABCD.$ Equating the expressions for $[ABCD],$ we have \\[\\frac12\\cdot\\sin\\theta\\cdot59=2\\sqrt{210},\\] so $\\sin\\theta=\\frac{4\\sqrt{210}}{59}.$ Since $0^\\circ<\\theta<90^\\circ,$ we have $\\cos\\theta>0.$ It follows that \\[\\cos\\theta=\\sqrt{1-\\sin^2\\theta}=\\frac{11}{59}.\\] Finally, substituting this result into $(\\bigstar)$ gives $22\\cos\\theta=\\frac{242}{59},$ from which the answer is $242+59=301 we have \\begin{align*} AE\\cdot BE+BE\\cdot CE+CE\\cdot DE+DE\\cdot AE &= (AE+CE)(BE+DE) &&\\hspace{10mm}\\text{Factor by Grouping} \\\\ &=AC\\cdot BD &&\\hspace{10mm}\\text{Segment Addition} \\\\ &=AB\\cdot CD+BC\\cdot DA &&\\hspace{10mm}\\text{Ptolemy's Theorem} \\\\ &=59. &&\\hspace{10mm}\\text{Substitution} \\end{align*} ~MRENTHUSIASM",
"The angle $\\theta$ between diagonals satisfies \\[\\tan{\\frac{\\theta}{2}}=\\sqrt{\\frac{(s-b)(s-d)}{(s-a)(s-c)}}\\] (see https://en.wikipedia.org/wiki/Cyclic_quadrilateral#Angle_formulas). Thus, \\[\\tan{\\frac{\\theta}{2}}=\\sqrt{\\frac{(11-4)(11-6)}{(11-5)(11-7)}}\\text{ or }\\tan{\\frac{\\theta}{2}}=\\sqrt{\\frac{(11-5)(11-7)}{(11-4)(11-6)}}.\\] That is, $\\tan^2{\\frac{\\theta}{2}}=\\frac{1-\\cos^2{\\frac{\\theta}{2}}}{\\cos^2{\\frac{\\theta}{2}}}=\\frac{24}{35}$ or $\\frac{35}{24}$. Thus, $\\cos^2{\\frac{\\theta}{2}}=\\frac{35}{59}$ or $\\frac{24}{59}$. So, \\[\\cos{\\theta}=2\\cos^2{\\frac{\\theta}{2}}-1=\\pm\\frac{11}{59}.\\] In this context, $\\cos{\\theta}>0$. Thus, $\\cos{\\theta}=\\frac{11}{59}$. The perimeter of $A_1B_1C_1D_1$ is \\[22\\cdot\\cos{\\theta}=22\\cdot\\frac{11}{59}=\\frac{242}{59},\\] and the answer is $m+n=242+59=301. ~y.grace.yu",
"The angle $\\theta$ between diagonals satisfies \\[\\tan{\\frac{\\theta}{2}}=\\sqrt{\\frac{(s-b)(s-d)}{(s-a)(s-c)}}\\] (see https://en.wikipedia.org/wiki/Cyclic_quadrilateral#Angle_formulas). Thus, \\[\\tan{\\frac{\\theta}{2}}=\\sqrt{\\frac{(11-4)(11-6)}{(11-5)(11-7)}}\\text{ or }\\tan{\\frac{\\theta}{2}}=\\sqrt{\\frac{(11-5)(11-7)}{(11-4)(11-6)}}.\\] That is, $\\tan^2{\\frac{\\theta}{2}}=\\frac{1-\\cos^2{\\frac{\\theta}{2}}}{\\cos^2{\\frac{\\theta}{2}}}=\\frac{24}{35}$ or $\\frac{35}{24}$. Thus, $\\cos^2{\\frac{\\theta}{2}}=\\frac{35}{59}$ or $\\frac{24}{59}$. So, \\[\\cos{\\theta}=2\\cos^2{\\frac{\\theta}{2}}-1=\\pm\\frac{11}{59}.\\] In this context, $\\cos{\\theta}>0$. Thus, $\\cos{\\theta}=\\frac{11}{59}$. The perimeter of $A_1B_1C_1D_1$ is \\[22\\cdot\\cos{\\theta}=22\\cdot\\frac{11}{59}=\\frac{242}{59},\\] and the answer is $m+n=242+59=301. ~y.grace.yu",
"We assume that the two quadrilateral mentioned in the problem are similar (due to both of them being cyclic). Note that by Ptolemy’s, one of the diagonals has length $\\sqrt{4 \\cdot 6 + 5 \\cdot 7} = \\sqrt{59}.$ [I don't believe this is correct... are the two diagonals of $ABCD$ necessarily congruent? -peace09]* WLOG we focus on diagonal $BD.$ To find the diagonal of the inner quadrilateral, we drop the altitude from $A$ and $C$ and calculate the length of $A_1C_1.$ Let $x$ be $A_1D$ (Thus $A_1B = \\sqrt{59} - x.$ By Pythagorean theorem, we have \\[49 - x^2 = 16 - \\left(\\sqrt{59} - x\\right)^2 \\implies 92 = 2\\sqrt{59}x \\implies x = \\frac{46}{\\sqrt{59}} = \\frac{46\\sqrt{59}}{59}.\\] Now let $y$ be $C_1D.$ (thus making $C_1B = \\sqrt{59} - y$). Similarly, we have \\[36 - y^2 = 25 - \\left(\\sqrt{59} - y\\right)^2 \\implies 70 = 2\\sqrt{59}y \\implies y = \\frac{35}{\\sqrt{59}} = \\frac{35\\sqrt{59}}{59}.\\] We see that $A_1C_1$, the scaled down diagonal is just $x - y = \\frac{11\\sqrt{59}}{59},$ which is $\\frac{\\frac{11\\sqrt{59}}{59}}{\\sqrt{59}} = \\frac{11}{59}$ times our original diagonal $BD,$ implying a scale factor of $\\frac{11}{59}.$ Thus, due to perimeters scaling linearly, the perimeter of the new quadrilateral is simply $\\frac{11}{59} \\cdot 22 = \\frac{242}{59},$ making our answer $242+59 = 301. ~MATH-TITAN",
"We assume that the two quadrilateral mentioned in the problem are similar (due to both of them being cyclic). Note that by Ptolemy’s, one of the diagonals has length $\\sqrt{4 \\cdot 6 + 5 \\cdot 7} = \\sqrt{59}.$ [I don't believe this is correct... are the two diagonals of $ABCD$ necessarily congruent? -peace09]* WLOG we focus on diagonal $BD.$ To find the diagonal of the inner quadrilateral, we drop the altitude from $A$ and $C$ and calculate the length of $A_1C_1.$ Let $x$ be $A_1D$ (Thus $A_1B = \\sqrt{59} - x.$ By Pythagorean theorem, we have \\[49 - x^2 = 16 - \\left(\\sqrt{59} - x\\right)^2 \\implies 92 = 2\\sqrt{59}x \\implies x = \\frac{46}{\\sqrt{59}} = \\frac{46\\sqrt{59}}{59}.\\] Now let $y$ be $C_1D.$ (thus making $C_1B = \\sqrt{59} - y$). Similarly, we have \\[36 - y^2 = 25 - \\left(\\sqrt{59} - y\\right)^2 \\implies 70 = 2\\sqrt{59}y \\implies y = \\frac{35}{\\sqrt{59}} = \\frac{35\\sqrt{59}}{59}.\\] We see that $A_1C_1$, the scaled down diagonal is just $x - y = \\frac{11\\sqrt{59}}{59},$ which is $\\frac{\\frac{11\\sqrt{59}}{59}}{\\sqrt{59}} = \\frac{11}{59}$ times our original diagonal $BD,$ implying a scale factor of $\\frac{11}{59}.$ Thus, due to perimeters scaling linearly, the perimeter of the new quadrilateral is simply $\\frac{11}{59} \\cdot 22 = \\frac{242}{59},$ making our answer $242+59 = 301. ~MATH-TITAN",
"Solution In accordance with Claim 1, the ratios of pairs of one-color segments are the same and equal to $\\cos \\theta,$ where $\\theta$ is the acute angle between the diagonals. \\begin{align*} s &= A'B' + B'C' + C'D' + D'A' \\\\ &= (AB + BC + CD + DA)\\cos \\theta \\\\ &= (a + b + c + d)\\cos \\theta \\\\ &= 22\\cos \\theta. \\end{align*} In accordance with Claim 2, \\begin{align*} 2(ac + bd)\\cos \\theta = |d^2 – c^2 + b^2 – a^2|.\\end{align*} \\[2 \\cdot 59 \\cos \\theta = |13 + 9|.\\] \\[s = 22\\cos \\theta = \\frac{22 \\cdot 11}{59} = \\frac{242}{59}.\\] Therefore, the answer is $242+59=301: \\begin{align*} B'D^2 = AD^2 + AB'^2 – 2 AD \\cdot AB' \\cos \\theta,\\end{align*} \\[B'D^2 = CD^2 + CB'^2 + 2 CD \\cdot CB' \\cos \\theta,\\] \\[d^2 + b^2 – 2 bd \\cos \\theta = c^2 + a^2 + 2ac \\cos \\theta,\\] \\[2(ac + bd) \\cos \\theta = |d^2 – c^2 + b^2 – a^2|.\\] vladimir.shelomovskii@gmail.com, vvsss",
"Solution In accordance with Claim 1, the ratios of pairs of one-color segments are the same and equal to $\\cos \\theta,$ where $\\theta$ is the acute angle between the diagonals. \\begin{align*} s &= A'B' + B'C' + C'D' + D'A' \\\\ &= (AB + BC + CD + DA)\\cos \\theta \\\\ &= (a + b + c + d)\\cos \\theta \\\\ &= 22\\cos \\theta. \\end{align*} In accordance with Claim 2, \\begin{align*} 2(ac + bd)\\cos \\theta = |d^2 – c^2 + b^2 – a^2|.\\end{align*} \\[2 \\cdot 59 \\cos \\theta = |13 + 9|.\\] \\[s = 22\\cos \\theta = \\frac{22 \\cdot 11}{59} = \\frac{242}{59}.\\] Therefore, the answer is $242+59=301: \\begin{align*} B'D^2 = AD^2 + AB'^2 – 2 AD \\cdot AB' \\cos \\theta,\\end{align*} \\[B'D^2 = CD^2 + CB'^2 + 2 CD \\cdot CB' \\cos \\theta,\\] \\[d^2 + b^2 – 2 bd \\cos \\theta = c^2 + a^2 + 2ac \\cos \\theta,\\] \\[2(ac + bd) \\cos \\theta = |d^2 – c^2 + b^2 – a^2|.\\] vladimir.shelomovskii@gmail.com, vvsss"
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2021-I-12
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Let $A_1A_2A_3\ldots A_{12}$ be a dodecagon ( $12$ -gon). Three frogs initially sit at $A_4,A_8,$ and $A_{12}$ . At the end of each minute, simultaneously, each of the three frogs jumps to one of the two vertices adjacent to its current position, chosen randomly and independently with both choices being equally likely. All three frogs stop jumping as soon as two frogs arrive at the same vertex at the same time. The expected number of minutes until the frogs stop jumping is $\frac mn$ , where $m$ and $n$ are relatively prime positive integers. Find $m+n$ .
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I
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[
"Define the distance between two frogs as the number of sides between them that do not contain the third frog. Let $E(a,b,c)$ denote the expected number of minutes until the frogs stop jumping, such that the distances between the frogs are $a,b,$ and $c$ (in either clockwise or counterclockwise order). Without the loss of generality, assume that $a\\leq b\\leq c.$ We wish to find $E(4,4,4).$ Note that: At any moment before the frogs stop jumping, the only possibilities for $(a,b,c)$ are $(4,4,4),(2,4,6),$ and $(2,2,8).$ $E(a,b,c)$ does not depend on the actual positions of the frogs, but depends on the distances between the frogs. At the end of each minute, each frog has $2$ outcomes. So, there are $2^3=8$ outcomes in total. We have the following system of equations: \\begin{align*} E(4,4,4)&=1+\\frac{2}{8}E(4,4,4)+\\frac{6}{8}E(2,4,6), \\\\ E(2,4,6)&=1+\\frac{4}{8}E(2,4,6)+\\frac{1}{8}E(4,4,4)+\\frac{1}{8}E(2,2,8), \\\\ E(2,2,8)&=1+\\frac{2}{8}E(2,2,8)+\\frac{2}{8}E(2,4,6). \\end{align*} Rearranging and simplifying each equation, we get \\begin{align*} E(4,4,4)&=\\frac{4}{3}+E(2,4,6), &(1) \\\\ E(2,4,6)&=2+\\frac{1}{4}E(4,4,4)+\\frac{1}{4}E(2,2,8), &\\hspace{12.75mm}(2) \\\\ E(2,2,8)&=\\frac{4}{3}+\\frac{1}{3}E(2,4,6). &(3) \\end{align*} Substituting $(1)$ and $(3)$ into $(2),$ we obtain \\[E(2,4,6)=2+\\frac{1}{4}\\left[\\frac{4}{3}+E(2,4,6)\\right]+\\frac{1}{4}\\left[\\frac{4}{3}+\\frac{1}{3}E(2,4,6)\\right],\\] from which $E(2,4,6)=4.$ Substituting this into $(1)$ gives $E(4,4,4)=\\frac{16}{3}.$ Therefore, the answer is $16+3=019 ~Ross Gao ~MRENTHUSIASM",
"We can solve the problem by removing $1$ frog, and calculate the expected time for the remaining $2$ frogs. In the original problem, when the movement stops, $2$ of the $3$ frogs meet. Because the $3$ frogs cannot meet at one vertex, the probability that those two specific frogs meet is $\\tfrac13$. If the expected time for the two frog problem is $E'$, then the expected time for the original problem is $\\tfrac 13 E'$. The distance between the two frogs can only be $0$, $2$, $4$, $6$. We use the distances as the states to draw the following Markov Chain. This Markov Chain is much simpler than that of Solution 1 Supplement in the Remark section. \\begin{align*} E(2) &= 1 + \\frac12 \\cdot E(2) + \\frac14 \\cdot E(4)\\\\ E(4) &= 1 + \\frac14 \\cdot E(2) + \\frac12 \\cdot E(4) + \\frac14 \\cdot E(6)\\\\ E(6) &= 1 + \\frac12 \\cdot E(4) + \\frac12 \\cdot E(6) \\end{align*} By solving the above system of equations, $E(4) = 16$. The answer for the original problem is $\\frac{16}{3}$, $16 + 3 = \\textbf{019} ~isabelchen",
"We can solve the problem by removing $1$ frog, and calculate the expected time for the remaining $2$ frogs. In the original problem, when the movement stops, $2$ of the $3$ frogs meet. Because the $3$ frogs cannot meet at one vertex, the probability that those two specific frogs meet is $\\tfrac13$. If the expected time for the two frog problem is $E'$, then the expected time for the original problem is $\\tfrac 13 E'$. The distance between the two frogs can only be $0$, $2$, $4$, $6$. We use the distances as the states to draw the following Markov Chain. This Markov Chain is much simpler than that of Solution 1 Supplement in the Remark section. \\begin{align*} E(2) &= 1 + \\frac12 \\cdot E(2) + \\frac14 \\cdot E(4)\\\\ E(4) &= 1 + \\frac14 \\cdot E(2) + \\frac12 \\cdot E(4) + \\frac14 \\cdot E(6)\\\\ E(6) &= 1 + \\frac12 \\cdot E(4) + \\frac12 \\cdot E(6) \\end{align*} By solving the above system of equations, $E(4) = 16$. The answer for the original problem is $\\frac{16}{3}$, $16 + 3 = \\textbf{019} ~isabelchen",
"Solution 1 can be represented by the following Markov Chain: From state $(4, 4, 4)$ to state $(4, 4, 4)$: the $3$ frogs must jump in the same direction, $2 \\cdot \\frac18 = \\frac14$. From state $(4, 4, 4)$ to state $(2, 4, 6)$: $2$ frogs must jump in the same direction, and the other must jump in the opposite direction, $\\binom32 \\cdot 2 \\cdot \\frac18 = \\frac34$. From state $(2, 4, 6)$ to state $(4, 4, 4)$: the $2$ frogs with a distance of $4$ in between must jump in the same direction so that they will be further away from the other frog, and the other frog must jump in the opposite direction as those $2$ frogs, $\\frac18$. From state $(2, 4, 6)$ to state $(2, 4, 6)$: the $3$ frogs can all jump in the same direction; or the $2$ frogs with a distance of $2$ in between jumps away from each other and the other frog jumps closer to the closest frog; or the $2$ frogs with a distance of $6$ in between jump closer to each other and away from the third frog, the third frog jumps closer to the closest frog; $(2 + 1 + 1) \\cdot \\frac18 = \\frac12$. From state $(2, 4, 6)$ to state $(2, 2, 8)$: the $2$ frogs with a distance of $2$ in between must jump closer to the other frog and the other frog must jump close to those $2$ frogs, $\\frac18$. From state $(2, 2, 8)$ to state $(2, 4, 6)$: $2$ frogs that have no frogs in between must both jump in the same direction away from the other frog, the other frog must also jump away from those $2$ frogs, $2 \\cdot \\frac18 = \\frac14$. From state $(2, 2, 8)$ to state $(2, 2, 8)$: the $3$ frogs must all jump in the same direction, $2 \\cdot \\frac18 = \\frac14$. From state $(2, 2, 8)$ to state $(0, x, y)$: frogs with a distance of $2$ must jump closer to each other, the other frog can jump in any direction, $\\frac12 \\cdot \\frac12 \\cdot 2 = \\frac12$. From state $(2, 4, 6)$ to state $(0, x, y)$: the frogs with a distance of $2$ must jump closer to each other, the other frog can jump in any direction, $\\frac12 \\cdot \\frac12 = \\frac14$. Because $a + b + c = 12$, we can use $(a, b)$ to represent the states which is simpler than using $(a, b, c)$ in Solution 1. Summary The two above Markov Chains are Absorbing Markov Chain. The state of $2$ frogs meeting is the absorbing state. This problem asks for the Expected Number of Steps before being absorbed in the absorbing state. Let $p_{ij} = P(X_{n+1} = j | X_n = i)$, the probability that state $i$ transits to state $j$ on the next step. Let $e_i$ be the expected number of steps before being absorbed in the absorbing state when starting from transient state $i$. $e_i = \\sum_{j} [p_{ij} \\cdot ( 1 + e_{j})] = \\sum_{j} (p_{ij} + p_{ij} \\cdot e_{j}) = \\sum_{j} p_{ij} + \\sum_{j} (p_{ij} \\cdot e_{j}) = 1 + \\sum_{j} (p_{ij} \\cdot e_{j})$ $e_i$ is $1$ plus the sum of the products of $p_{ij}$ and $s_j$ of all the next state $j$, $\\sum_{j} p_{ij} = 1$. 2014 AMC12B Problem 22 and 2017 AMC12A Problem 22 are similar problem with simpler states, both problems can be solved by Absorbing Markov Chain. ~isabelchen"
] |
2021-I-13
| 2,021
| 13
|
Circles $\omega_1$ and $\omega_2$ with radii $961$ and $625$ , respectively, intersect at distinct points $A$ and $B$ . A third circle $\omega$ is externally tangent to both $\omega_1$ and $\omega_2$ . Suppose line $AB$ intersects $\omega$ at two points $P$ and $Q$ such that the measure of minor arc $\widehat{PQ}$ is $120^{\circ}$ . Find the distance between the centers of $\omega_1$ and $\omega_2$ .
| 672
|
I
|
[
"Let $O_i$ and $r_i$ be the center and radius of $\\omega_i$, and let $O$ and $r$ be the center and radius of $\\omega$. Since $\\overline{AB}$ extends to an arc with arc $120^\\circ$, the distance from $O$ to $\\overline{AB}$ is $r/2$. Let $X=\\overline{AB}\\cap \\overline{O_1O_2}$. Consider $\\triangle OO_1O_2$. The line $\\overline{AB}$ is perpendicular to $\\overline{O_1O_2}$ and passes through $X$. Let $H$ be the foot from $O$ to $\\overline{O_1O_2}$; so $HX=r/2$. We have by tangency $OO_1=r+r_1$ and $OO_2=r+r_2$. Let $O_1O_2=d$. [asy] unitsize(3cm); pointpen=black; pointfontpen=fontsize(9); pair A=dir(110), B=dir(230), C=dir(310); DPA(A--B--C--A); pair H = foot(A, B, C); draw(A--H); pair X = 0.3*B + 0.7*C; pair Y = A+X-H; draw(X--1.3*Y-0.3*X); draw(A--Y, dotted); pair R1 = 1.3*X-0.3*Y; pair R2 = 0.7*X+0.3*Y; draw(R1--X); D(\"O\",A,dir(A)); D(\"O_1\",B,dir(B)); D(\"O_2\",C,dir(C)); D(\"H\",H,dir(270)); D(\"X\",X,dir(225)); D(\"A\",R1,dir(180)); D(\"B\",R2,dir(180)); draw(rightanglemark(Y,X,C,3)); [/asy] Since $X$ is on the radical axis of $\\omega_1$ and $\\omega_2$, it has equal power with respect to both circles, so \\[O_1X^2 - r_1^2 = O_2X^2-r_2^2 \\implies O_1X-O_2X = \\frac{r_1^2-r_2^2}{d}\\]since $O_1X+O_2X=d$. Now we can solve for $O_1X$ and $O_2X$, and in particular, \\begin{align*} O_1H &= O_1X - HX = \\frac{d+\\frac{r_1^2-r_2^2}{d}}{2} - \\frac{r}{2} \\\\ O_2H &= O_2X + HX = \\frac{d-\\frac{r_1^2-r_2^2}{d}}{2} + \\frac{r}{2}. \\end{align*} We want to solve for $d$. By the Pythagorean Theorem (twice): \\begin{align*} &\\qquad -OH^2 = O_2H^2 - (r+r_2)^2 = O_1H^2 - (r+r_1)^2 \\\\ &\\implies \\left(d+r-\\tfrac{r_1^2-r_2^2}{d}\\right)^2 - 4(r+r_2)^2 = \\left(d-r+\\tfrac{r_1^2-r_2^2}{d}\\right)^2 - 4(r+r_1)^2 \\\\ &\\implies 2dr - 2(r_1^2-r_2^2)-8rr_2-4r_2^2 = -2dr+2(r_1^2-r_2^2)-8rr_1-4r_1^2 \\\\ &\\implies 4dr = 8rr_2-8rr_1 \\\\ &\\implies d=2r_2-2r_1 \\end{align*} Therefore, $d=2(r_2-r_1) = 2(961-625)=672.",
"Let $O_i$ and $r_i$ be the center and radius of $\\omega_i$, and let $O$ and $r$ be the center and radius of $\\omega$. Since $\\overline{AB}$ extends to an arc with arc $120^\\circ$, the distance from $O$ to $\\overline{AB}$ is $r/2$. Let $X=\\overline{AB}\\cap \\overline{O_1O_2}$. Consider $\\triangle OO_1O_2$. The line $\\overline{AB}$ is perpendicular to $\\overline{O_1O_2}$ and passes through $X$. Let $H$ be the foot from $O$ to $\\overline{O_1O_2}$; so $HX=r/2$. We have by tangency $OO_1=r+r_1$ and $OO_2=r+r_2$. Let $O_1O_2=d$. [asy] unitsize(3cm); pointpen=black; pointfontpen=fontsize(9); pair A=dir(110), B=dir(230), C=dir(310); DPA(A--B--C--A); pair H = foot(A, B, C); draw(A--H); pair X = 0.3*B + 0.7*C; pair Y = A+X-H; draw(X--1.3*Y-0.3*X); draw(A--Y, dotted); pair R1 = 1.3*X-0.3*Y; pair R2 = 0.7*X+0.3*Y; draw(R1--X); D(\"O\",A,dir(A)); D(\"O_1\",B,dir(B)); D(\"O_2\",C,dir(C)); D(\"H\",H,dir(270)); D(\"X\",X,dir(225)); D(\"A\",R1,dir(180)); D(\"B\",R2,dir(180)); draw(rightanglemark(Y,X,C,3)); [/asy] Since $X$ is on the radical axis of $\\omega_1$ and $\\omega_2$, it has equal power with respect to both circles, so \\[O_1X^2 - r_1^2 = O_2X^2-r_2^2 \\implies O_1X-O_2X = \\frac{r_1^2-r_2^2}{d}\\]since $O_1X+O_2X=d$. Now we can solve for $O_1X$ and $O_2X$, and in particular, \\begin{align*} O_1H &= O_1X - HX = \\frac{d+\\frac{r_1^2-r_2^2}{d}}{2} - \\frac{r}{2} \\\\ O_2H &= O_2X + HX = \\frac{d-\\frac{r_1^2-r_2^2}{d}}{2} + \\frac{r}{2}. \\end{align*} We want to solve for $d$. By the Pythagorean Theorem (twice): \\begin{align*} &\\qquad -OH^2 = O_2H^2 - (r+r_2)^2 = O_1H^2 - (r+r_1)^2 \\\\ &\\implies \\left(d+r-\\tfrac{r_1^2-r_2^2}{d}\\right)^2 - 4(r+r_2)^2 = \\left(d-r+\\tfrac{r_1^2-r_2^2}{d}\\right)^2 - 4(r+r_1)^2 \\\\ &\\implies 2dr - 2(r_1^2-r_2^2)-8rr_2-4r_2^2 = -2dr+2(r_1^2-r_2^2)-8rr_1-4r_1^2 \\\\ &\\implies 4dr = 8rr_2-8rr_1 \\\\ &\\implies d=2r_2-2r_1 \\end{align*} Therefore, $d=2(r_2-r_1) = 2(961-625)=672.",
"Let $O_{1}$, $O_{2}$, and $O$ be the centers of $\\omega_{1}$, $\\omega_{2}$, and $\\omega$ with $r_{1}$, $r_{2}$, and $r$ their radii, respectively. Then, the distance from $O$ to the radical axis $\\ell\\equiv\\overline{AB}$ of $\\omega_{1}, \\omega_{2}$ is equal to $\\frac{1}{2}r$. Let $x=O_{1}O_{2}$ and $O^{\\prime}$ the orthogonal projection of $O$ onto line $\\ell$. Define the function $f:\\mathbb{R}^{2}\\rightarrow\\mathbb{R}$ by \\[f(X)=\\text{Pow}_{\\omega_{1}}(X)-\\text{Pow}_{\\omega_{2}}(X).\\] Then \\begin{align*} f(O_{1})=-r_{1}^{2}-(x-r_{2})(x+r_{2})&=-x^{2}+r_{2}^{2}-r_{1}^{2}, \\\\ f(O_{2})=(x-r_{1})(x+r_{1})-(-r_{2}^{2})&=x^{2}+r_{2}^{2}-r_{1}^{2}, \\\\ f(O)=r(r+2r_{1})-r(r+2r_{2})&=2r(r_{1}-r_{2}), \\\\ f(O^{\\prime})&=0. \\end{align*} By linearity, \\[\\frac{f(O_{2})-f(O_{1})}{f(O)-f(O^{\\prime})}=\\frac{O_{1}O_{2}}{OO^{\\prime}}=\\frac{x}{\\tfrac{1}{2}r}=\\frac{2x}{r}.\\] Notice that $f(O_{2})-f(O_{1})=x^{2}-(-x^{2})=2x^{2}$ and $f(O)-f(O^{\\prime})=2r(r_{1}-r_{2})$, thus \\begin{align*}\\frac{2x^{2}}{2r(r_{1}-r_{2})}&=\\frac{2x}{r}\\end{align*} Dividing both sides by $\\frac{2x}{r}$ (which is obviously nonzero as $x$ is nonzero) gives us \\begin{align*}\\frac{x}{2(r_{1}-r_{2})}&=1\\end{align*} so $x=2(r_{1}-r_{2})$. Since $r_{1}=961$ and $r_{2}=625$, the answer is $x=2\\cdot(961-625)=672.",
"Let $O_{1}$, $O_{2}$, and $O$ be the centers of $\\omega_{1}$, $\\omega_{2}$, and $\\omega$ with $r_{1}$, $r_{2}$, and $r$ their radii, respectively. Then, the distance from $O$ to the radical axis $\\ell\\equiv\\overline{AB}$ of $\\omega_{1}, \\omega_{2}$ is equal to $\\frac{1}{2}r$. Let $x=O_{1}O_{2}$ and $O^{\\prime}$ the orthogonal projection of $O$ onto line $\\ell$. Define the function $f:\\mathbb{R}^{2}\\rightarrow\\mathbb{R}$ by \\[f(X)=\\text{Pow}_{\\omega_{1}}(X)-\\text{Pow}_{\\omega_{2}}(X).\\] Then \\begin{align*} f(O_{1})=-r_{1}^{2}-(x-r_{2})(x+r_{2})&=-x^{2}+r_{2}^{2}-r_{1}^{2}, \\\\ f(O_{2})=(x-r_{1})(x+r_{1})-(-r_{2}^{2})&=x^{2}+r_{2}^{2}-r_{1}^{2}, \\\\ f(O)=r(r+2r_{1})-r(r+2r_{2})&=2r(r_{1}-r_{2}), \\\\ f(O^{\\prime})&=0. \\end{align*} By linearity, \\[\\frac{f(O_{2})-f(O_{1})}{f(O)-f(O^{\\prime})}=\\frac{O_{1}O_{2}}{OO^{\\prime}}=\\frac{x}{\\tfrac{1}{2}r}=\\frac{2x}{r}.\\] Notice that $f(O_{2})-f(O_{1})=x^{2}-(-x^{2})=2x^{2}$ and $f(O)-f(O^{\\prime})=2r(r_{1}-r_{2})$, thus \\begin{align*}\\frac{2x^{2}}{2r(r_{1}-r_{2})}&=\\frac{2x}{r}\\end{align*} Dividing both sides by $\\frac{2x}{r}$ (which is obviously nonzero as $x$ is nonzero) gives us \\begin{align*}\\frac{x}{2(r_{1}-r_{2})}&=1\\end{align*} so $x=2(r_{1}-r_{2})$. Since $r_{1}=961$ and $r_{2}=625$, the answer is $x=2\\cdot(961-625)=672.",
"Denote by $O_1$, $O_2$, and $O$ the centers of $\\omega_1$, $\\omega_2$, and $\\omega$, respectively. Let $R_1 = 961$ and $R_2 = 625$ denote the radii of $\\omega_1$ and $\\omega_2$ respectively, $r$ be the radius of $\\omega$, and $\\ell$ the distance from $O$ to the line $AB$. We claim that\\[\\dfrac{\\ell}{r} = \\dfrac{R_2-R_1}{d},\\]where $d = O_1O_2$. This solves the problem, for then the $\\widehat{PQ} = 120^\\circ$ condition implies $\\tfrac{\\ell}r = \\cos 60^\\circ = \\tfrac{1}{2}$, and then we can solve to get $d = 672. Now recall that\\[d(O_2Y-O_1Y) = O_2Y^2 - O_1Y^2 = R_2^2 - R_1^2.\\]Furthermore, note that\\begin{align*}d(O_2X - O_1X) &= O_2X^2 - O_1X^2= O_2O^2 - O_1O^2 \\\\ &= (R_2 + r)^2 - (R_1+r)^2 = (R_2^2 - R_1^2) + 2r(R_2 - R_1).\\end{align*}Substituting the first equality into the second one and subtracting yields\\[2r(R_2 - R_1) = d(O_2X - O_1X) - d(O_2Y - O_1Y) = 2dXY,\\]which rearranges to the desired.",
"Suppose we label the points as shown below. [asy] defaultpen(fontsize(12)+0.6); size(300); pen p=fontsize(10)+royalblue+0.4; var r=1200; pair O1=origin, O2=(672,0), O=OP(CR(O1,961+r),CR(O2,625+r)); path c1=CR(O1,961), c2=CR(O2,625), c=CR(O,r); pair A=IP(CR(O1,961),CR(O2,625)), B=OP(CR(O1,961),CR(O2,625)), P=IP(L(A,B,0,0.2),c), Q=IP(L(A,B,0,200),c), F=IP(CR(O,625+r),O--O1), M=(F+O2)/2, D=IP(CR(O,r),O--O1), E=IP(CR(O,r),O--O2), X=extension(E,D,O,O+O1-O2), Y=extension(D,E,O1,O2); draw(c1^^c2); draw(c,blue+0.6); draw(O1--O2--O--cycle,black+0.6); draw(O--X^^Y--O2,black+0.6); draw(X--Y,heavygreen+0.6); draw((X+O)/2--O,MidArrow); draw(O2--Y-(300,0),MidArrow); dot(\"$A$\",A,dir(A-O2/2)); dot(\"$B$\",B,dir(B-O2/2)); dot(\"$O_2$\",O2,right+up); dot(\"$O_1$\",O1,left+up); dot(\"$O$\",O,dir(O-O2)); dot(\"$D$\",D,dir(170)); dot(\"$E$\",E,dir(E-O1)); dot(\"$X$\",X,dir(X-D)); dot(\"$Y$\",Y,dir(Y-D)); label(\"$R$\",O--E,right+up,p); label(\"$R$\",O--D,left+down,p); label(\"$2R$\",(X+O)/2-(150,0),down,p); label(\"$961$\",O1--D,2*(left+down),p); label(\"$625$\",O2--E,2*(right+up),p); MA(\"\",E,D,O1,100,fuchsia+linewidth(1)); MA(\"\",X,D,O,100,fuchsia+linewidth(1)); MA(\"\",Y,E,O2,100,orange+linewidth(1)); MA(\"\",D,E,O,100,orange+linewidth(1)); [/asy] By radical axis, the tangents to $\\omega$ at $D$ and $E$ intersect on $AB$. Thus $PDQE$ is harmonic, so the tangents to $\\omega$ at $P$ and $Q$ intersect at $X \\in DE$. Moreover, $OX \\parallel O_1O_2$ because both $OX$ and $O_1O_2$ are perpendicular to $AB$, and $OX = 2OP$ because $\\angle POQ = 120^{\\circ}$. Thus\\[O_1O_2 = O_1Y - O_2Y = 2 \\cdot 961 - 2\\cdot 625 = 672\\]by similar triangles. ~mathman3880",
"Suppose we label the points as shown below. [asy] defaultpen(fontsize(12)+0.6); size(300); pen p=fontsize(10)+royalblue+0.4; var r=1200; pair O1=origin, O2=(672,0), O=OP(CR(O1,961+r),CR(O2,625+r)); path c1=CR(O1,961), c2=CR(O2,625), c=CR(O,r); pair A=IP(CR(O1,961),CR(O2,625)), B=OP(CR(O1,961),CR(O2,625)), P=IP(L(A,B,0,0.2),c), Q=IP(L(A,B,0,200),c), F=IP(CR(O,625+r),O--O1), M=(F+O2)/2, D=IP(CR(O,r),O--O1), E=IP(CR(O,r),O--O2), X=extension(E,D,O,O+O1-O2), Y=extension(D,E,O1,O2); draw(c1^^c2); draw(c,blue+0.6); draw(O1--O2--O--cycle,black+0.6); draw(O--X^^Y--O2,black+0.6); draw(X--Y,heavygreen+0.6); draw((X+O)/2--O,MidArrow); draw(O2--Y-(300,0),MidArrow); dot(\"$A$\",A,dir(A-O2/2)); dot(\"$B$\",B,dir(B-O2/2)); dot(\"$O_2$\",O2,right+up); dot(\"$O_1$\",O1,left+up); dot(\"$O$\",O,dir(O-O2)); dot(\"$D$\",D,dir(170)); dot(\"$E$\",E,dir(E-O1)); dot(\"$X$\",X,dir(X-D)); dot(\"$Y$\",Y,dir(Y-D)); label(\"$R$\",O--E,right+up,p); label(\"$R$\",O--D,left+down,p); label(\"$2R$\",(X+O)/2-(150,0),down,p); label(\"$961$\",O1--D,2*(left+down),p); label(\"$625$\",O2--E,2*(right+up),p); MA(\"\",E,D,O1,100,fuchsia+linewidth(1)); MA(\"\",X,D,O,100,fuchsia+linewidth(1)); MA(\"\",Y,E,O2,100,orange+linewidth(1)); MA(\"\",D,E,O,100,orange+linewidth(1)); [/asy] By radical axis, the tangents to $\\omega$ at $D$ and $E$ intersect on $AB$. Thus $PDQE$ is harmonic, so the tangents to $\\omega$ at $P$ and $Q$ intersect at $X \\in DE$. Moreover, $OX \\parallel O_1O_2$ because both $OX$ and $O_1O_2$ are perpendicular to $AB$, and $OX = 2OP$ because $\\angle POQ = 120^{\\circ}$. Thus\\[O_1O_2 = O_1Y - O_2Y = 2 \\cdot 961 - 2\\cdot 625 = 672\\]by similar triangles. ~mathman3880",
"Like in other solutions, let $O$ be the center of $\\omega$ with $r$ its radius; also, let $O_{1}$ and $O_{2}$ be the centers of $\\omega_{1}$ and $\\omega_{2}$ with $R_{1}$ and $R_{2}$ their radii, respectively. Let line $OP$ intersect line $O_{1}O_{2}$ at $T$, and let $u=TO_{2}$, $v=TO_{1}$, $x=PT$, where the length $O_{1}O_{2}$ splits as $u+v$. Because the lines $PQ$ and $O_{1}O_{2}$ are perpendicular, lines $OT$ and $O_{1}O_{2}$ meet at a $60^{\\circ}$ angle. Applying the Law of Cosines to $\\triangle O_{2}PT$, $\\triangle O_{1}PT$, $\\triangle O_{2}OT$, and $\\triangle O_{1}OT$ gives \\begin{align*}\\triangle O_{2}PT&:O_{2}P^{2}=u^{2}+x^{2}-ux \\\\ \\triangle O_{1}PT&:O_{1}P^{2}=v^{2}+x^{2}+vx \\\\ \\triangle O_{2}OT&:(r+R_{2})^{2}=u^{2}+(r+x)^{2}-u(r+x) \\\\ \\triangle O_{1}OT&:(r+R_{1})^{2}=v^{2}+(r+x)^{2}+v(r+x)\\end{align*} Adding the first and fourth equations, then subtracting the second and third equations gives us \\[\\left(O_{2}P^{2}-O_{1}P^{2}\\right)+\\left(R_{1}^{2}-R_{2}^{2}\\right)+2r(R_{1}-R_{2})=r(u+v)\\] Since $P$ lies on the radical axis of $\\omega_{1}$ and $\\omega_{2}$, the power of point $P$ with respect to either circle is \\[O_{2}P^{2}-R_{2}^{2}=O_{1}P^{2}-R_{1}^{2}.\\] Hence $2r(R_{1}-R_{2})=r(u+v)$ which simplifies to \\[u+v=2(R_{1}-R_{2}).\\] The requested distance \\[O_{1}O_{2}=O_{1}T+O_{2}T=u+v\\] is therefore equal to $2\\cdot(961-625)=672.",
"Like in other solutions, let $O$ be the center of $\\omega$ with $r$ its radius; also, let $O_{1}$ and $O_{2}$ be the centers of $\\omega_{1}$ and $\\omega_{2}$ with $R_{1}$ and $R_{2}$ their radii, respectively. Let line $OP$ intersect line $O_{1}O_{2}$ at $T$, and let $u=TO_{2}$, $v=TO_{1}$, $x=PT$, where the length $O_{1}O_{2}$ splits as $u+v$. Because the lines $PQ$ and $O_{1}O_{2}$ are perpendicular, lines $OT$ and $O_{1}O_{2}$ meet at a $60^{\\circ}$ angle. Applying the Law of Cosines to $\\triangle O_{2}PT$, $\\triangle O_{1}PT$, $\\triangle O_{2}OT$, and $\\triangle O_{1}OT$ gives \\begin{align*}\\triangle O_{2}PT&:O_{2}P^{2}=u^{2}+x^{2}-ux \\\\ \\triangle O_{1}PT&:O_{1}P^{2}=v^{2}+x^{2}+vx \\\\ \\triangle O_{2}OT&:(r+R_{2})^{2}=u^{2}+(r+x)^{2}-u(r+x) \\\\ \\triangle O_{1}OT&:(r+R_{1})^{2}=v^{2}+(r+x)^{2}+v(r+x)\\end{align*} Adding the first and fourth equations, then subtracting the second and third equations gives us \\[\\left(O_{2}P^{2}-O_{1}P^{2}\\right)+\\left(R_{1}^{2}-R_{2}^{2}\\right)+2r(R_{1}-R_{2})=r(u+v)\\] Since $P$ lies on the radical axis of $\\omega_{1}$ and $\\omega_{2}$, the power of point $P$ with respect to either circle is \\[O_{2}P^{2}-R_{2}^{2}=O_{1}P^{2}-R_{1}^{2}.\\] Hence $2r(R_{1}-R_{2})=r(u+v)$ which simplifies to \\[u+v=2(R_{1}-R_{2}).\\] The requested distance \\[O_{1}O_{2}=O_{1}T+O_{2}T=u+v\\] is therefore equal to $2\\cdot(961-625)=672.",
"Let circle $\\omega$ tangent circles $\\omega_1$ and $\\omega_2,$ respectively at distinct points $C$ and $D$. Let $O, O_1, O_2 (r, r_1, r_2)$ be the centers (the radii) of $\\omega, \\omega_1$ and $\\omega_2,$ respectively. WLOG $r_1 < r_2.$ Let $F$ be the point of $OO_2$ such, that $OO_1 =OF.$ Let $M$ be the midpoint $FO_1, OE \\perp AB, CT$ be the radical axes of $\\omega_1$ and $\\omega, T \\in AB.$ Then $T$ is radical center of $\\omega, \\omega_1$ and $\\omega_2, TD = CT.$ In $\\triangle OFO_1 (OF = OO_1) OT$ is bisector of $\\angle O, OM$ is median $\\hspace{10mm} \\implies O,T,$ and $M$ are collinear. \\[\\angle OCT = \\angle ODT = \\angle OET = 90^\\circ \\implies\\] $OCTDE$ is cyclic (in $\\Omega), OT$ is diameter $\\Omega.$ $O_1O_2 \\perp AB, OM \\perp FO_1 \\implies \\angle FO_1O_2 = \\angle OTE$ $\\angle OTE = \\angle ODE$ as they intercept the same arc $\\overset{\\Large\\frown}{OE}$ in $\\Omega.$ \\[OE \\perp AB, O_1O_2 \\perp AB \\implies O_1 O_2 || OE \\implies\\] \\[\\angle OO_2O_1 = \\angle O_2 OE \\implies \\triangle ODE \\sim \\triangle O_2 O_1 F \\implies\\] \\[\\frac {OE}{OD} = \\frac {O_2F}{O_1O_2} \\implies \\cos \\frac {120^\\circ}{2} = \\frac{r_2 + r - r_1 -r} {O_1O_2}\\implies {O_1O_2}= 2|r_2 – r_1|.\\] Since $r_{1}=625$ and $r_{2}=961$, the answer is $2\\cdot|961-625|=672. vladimir.shelomovskii@gmail.com, vvsss",
"Let circle $\\omega$ tangent circles $\\omega_1$ and $\\omega_2,$ respectively at distinct points $C$ and $D$. Let $O, O_1, O_2 (r, r_1, r_2)$ be the centers (the radii) of $\\omega, \\omega_1$ and $\\omega_2,$ respectively. WLOG $r_1 < r_2.$ Let $F$ be the point of $OO_2$ such, that $OO_1 =OF.$ Let $M$ be the midpoint $FO_1, OE \\perp AB, CT$ be the radical axes of $\\omega_1$ and $\\omega, T \\in AB.$ Then $T$ is radical center of $\\omega, \\omega_1$ and $\\omega_2, TD = CT.$ In $\\triangle OFO_1 (OF = OO_1) OT$ is bisector of $\\angle O, OM$ is median $\\hspace{10mm} \\implies O,T,$ and $M$ are collinear. \\[\\angle OCT = \\angle ODT = \\angle OET = 90^\\circ \\implies\\] $OCTDE$ is cyclic (in $\\Omega), OT$ is diameter $\\Omega.$ $O_1O_2 \\perp AB, OM \\perp FO_1 \\implies \\angle FO_1O_2 = \\angle OTE$ $\\angle OTE = \\angle ODE$ as they intercept the same arc $\\overset{\\Large\\frown}{OE}$ in $\\Omega.$ \\[OE \\perp AB, O_1O_2 \\perp AB \\implies O_1 O_2 || OE \\implies\\] \\[\\angle OO_2O_1 = \\angle O_2 OE \\implies \\triangle ODE \\sim \\triangle O_2 O_1 F \\implies\\] \\[\\frac {OE}{OD} = \\frac {O_2F}{O_1O_2} \\implies \\cos \\frac {120^\\circ}{2} = \\frac{r_2 + r - r_1 -r} {O_1O_2}\\implies {O_1O_2}= 2|r_2 – r_1|.\\] Since $r_{1}=625$ and $r_{2}=961$, the answer is $2\\cdot|961-625|=672. vladimir.shelomovskii@gmail.com, vvsss",
"We are not given the radius of circle $w$, but based on the problem statement, that radius isn't important. We can set $w$ to have radius infinity (solution 8), but if you didn't observe that, you could also set the radius to be $2r$ so that the line containing the center of $w$, call it $W$, and $w_2$, call it $W_2$, is perpendicular to the line containing the center of $w_1$, call it $W_1$ and $w_2$. Let $AB = 2h$ and $W_1W_2 = x$. Also, let the projections of $W$ and $W_1$ onto line $AB$ be $X$ and $Y$, respectively. By Pythagorean Theorem on $\\triangle WW_1W_2$, we get \\[x^2+(625+2r)^2=(961+2r)^2 \\;(1)\\] Note that since $\\angle PWQ = 120$, $\\angle PWX = 60$. So, $WX = 2r/2 = r = W_1Y$. We now get two more equations from Pythag: \\[h^2+r^2 = 625^2 \\; (2)\\] \\[h^2+(x-r)^2 = 961^2 \\; (3)\\] From subtracting $(2)$ and $(3)$, $x^2-2rx=961^2-625^2 \\; (4)$. Rearranging $(1)$ yields $x^2-1344r = 961^2-625^2$. Plugging in our result from $(4)$, $x^2-1344r= x^2-2rx \\implies 1344r = 2rx \\implies x=672. ~sml1809",
"Let the circle $\\omega$ be infinitely big (a line). Then for it to be split into an arc of $120^{\\circ}$, $\\overline{PQ}$ must intersect at a $60^{\\circ}$ with line $\\omega$. Notice the 30-60-90 triangle in the image. $O_1R = 961 - 625$. Thus, the distance between the centers of $\\omega_1$ and $\\omega_2$ is $2(961-625)=672 picture by afly",
"Let the circle $\\omega$ be infinitely big (a line). Then for it to be split into an arc of $120^{\\circ}$, $\\overline{PQ}$ must intersect at a $60^{\\circ}$ with line $\\omega$. Notice the 30-60-90 triangle in the image. $O_1R = 961 - 625$. Thus, the distance between the centers of $\\omega_1$ and $\\omega_2$ is $2(961-625)=672 picture by afly"
] |
2021-I-14
| 2,021
| 14
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For any positive integer $a,$ $\sigma(a)$ denotes the sum of the positive integer divisors of $a$ . Let $n$ be the least positive integer such that $\sigma(a^n)-1$ is divisible by $2021$ for all positive integers $a$ . Find the sum of the prime factors in the prime factorization of $n$ .
| 125
|
I
|
[
"We first claim that $n$ must be divisible by $42$. Since $\\sigma(a^n)-1$ is divisible by $2021$ for all positive integers $a$, we can first consider the special case where $a$ is prime and $a \\neq 0,1 \\pmod{43}$. By Dirichlet's Theorem (Refer to the Remark section.), such $a$ always exists. Then $\\sigma(a^n)-1 = \\sum_{i=1}^n a^i = a\\left(\\frac{a^n - 1}{a-1}\\right)$. In order for this expression to be divisible by $2021=43\\cdot 47$, a necessary condition is $a^n - 1 \\equiv 0 \\pmod{43}$. By Fermat's Little Theorem, $a^{42} \\equiv 1 \\pmod{43}$. Moreover, if $a$ is a primitive root modulo $43$, then $\\text{ord}_{43}(a) = 42$, so $n$ must be divisible by $42$. By similar reasoning, $n$ must be divisible by $46$, by considering $a \\not\\equiv 0,1 \\pmod{47}$. We next claim that $n$ must be divisible by $43$. By Dirichlet, let $a$ be a prime that is congruent to $1 \\pmod{43}$. Then $\\sigma(a^n) \\equiv n+1 \\pmod{43}$, so since $\\sigma(a^n)-1$ is divisible by $43$, $n$ is a multiple of $43$. Alternatively, since $\\left(\\frac{a(a^n - 1^n)}{a-1}\\right)$ must be divisible by $43,$ by LTE, we have $v_{43}(a)+v_{43}{(a-1)}+v_{43}{(n)}-v_{43}{(a-1)} \\geq 1,$ which simplifies to $v_{43}(n) \\geq 1,$ which implies the desired result. Similarly, $n$ is a multiple of $47$. Lastly, we claim that if $n = \\text{lcm}(42, 46, 43, 47)$, then $\\sigma(a^n) - 1$ is divisible by $2021$ for all positive integers $a$. The claim is trivially true for $a=1$ so suppose $a>1$. Let $a = p_1^{e_1}\\ldots p_k^{e_k}$ be the prime factorization of $a$. Since $\\sigma(n)$ is multiplicative, we have \\[\\sigma(a^n) - 1 = \\prod_{i=1}^k \\sigma(p_i^{e_in}) - 1.\\] We can show that $\\sigma(p_i^{e_in}) \\equiv 1 \\pmod{2021}$ for all primes $p_i$ and integers $e_i \\ge 1$, so \\[\\sigma(p_i^{e_in}) = 1 + (p_i + p_i^2 + \\ldots + p_i^n) + (p_i^{n+1} + \\ldots + p_i^{2n}) + \\ldots + (p_i^{n(e_i-1)+1} + \\ldots + p_i^{e_in}),\\] where each expression in parentheses contains $n$ terms. It is easy to verify that if $p_i = 43$ or $p_i = 47$ then $\\sigma(p_i^{e_in}) \\equiv 1 \\pmod{2021}$ for this choice of $n$, so suppose $p_i \\not\\equiv 0 \\pmod{43}$ and $p_i \\not\\equiv 0 \\pmod{47}$. Each expression in parentheses equals $\\frac{p_i^n - 1}{p_i - 1}$ multiplied by some power of $p_i$. If $p_i \\not\\equiv 1 \\pmod{43}$, then FLT implies $p_i^n - 1 \\equiv 0 \\pmod{43}$, and if $p_i \\equiv 1 \\pmod{43}$, then $p_i + p_i^2 + \\ldots + p_i^n \\equiv 1 + 1 + \\ldots + 1 \\equiv 0 \\pmod{43}$ (since $n$ is also a multiple of $43$, by definition). Similarly, we can show $\\sigma(p_i^{e_in}) \\equiv 1 \\pmod{47}$, and a simple CRT argument shows $\\sigma(p_i^{e_in}) \\equiv 1 \\pmod{2021}$. Then $\\sigma(a^n) \\equiv 1^k \\equiv 1 \\pmod{2021}$. Then the prime factors of $n$ are $2,3,7,23,43,$ and $47,$ and the answer is $2+3+7+23+43+47 = 125. ~scrabbler94, Revised by wzs26843545602",
"$n$ only needs to satisfy $\\sigma(a^n)\\equiv 1 \\pmod{43}$ and $\\sigma(a^n)\\equiv 1 \\pmod{47}$ for all $a$. Let's work on the first requirement (mod 43) first. All $n$ works for $a=1$. If $a>1$, let $a$'s prime factorization be $a=p_1^{e_1}p_2^{e_2}\\cdots p_k^{e_k}$. The following three statements are the same: For all $a$, we have $\\sigma(a^n)=\\prod_{i=1}^k(1+p_i+\\cdots+p_i^{ne_i})\\equiv1\\pmod{43}$. For all $e>0$ and prime $p$, we have $1+p+\\cdots+p^{ne}\\equiv1\\pmod{43}$. For all prime $p$, we have $p+p^2+\\cdots+p^n \\equiv 0 \\pmod{43}$. We can show this by casework on $p$: For $p\\equiv0\\pmod{43}$, no matter what $n$ is, it is true that \\[p+p^2+\\cdots+p^n \\equiv 0 \\pmod{43}.\\] For all $p\\equiv1\\pmod{43}$, it is true that \\[p+p^2+\\cdots+p^n \\equiv n \\equiv 0 \\pmod{43}.\\] One can either use brute force or Dirichlet's Theorem to show such $p$ exists. Therefore, $43|n$. For all $p\\not\\equiv0,1\\pmod{43}$, it is true that \\[p+p^2+\\cdots+p^n \\equiv 0 \\pmod{43} \\Leftrightarrow p^n-1\\equiv0\\pmod{43}.\\] According to Fermat's Little Theorem, $42|n$ is sufficient. To show it's necessary, we just need to show $43$ has a prime primitive root. This can be done either by brute force or as follows. $43$ is prime and it must have a primitive root $t\\neq 1$ that's coprime to $43$. All numbers of the form $43k+t$ are also primitive roots of $43$. According to Dirichlet's Theorem there must be infinitely many primes among them. Similar arguments for modulo $47$ lead to $46|n$ and $47|n$. Therefore, we get $n=\\operatorname{lcm}[42,43,46,47]$. Following the last paragraph of Solution 1 gives the answer $125. ~Mryang6859 (Solution) ~MRENTHUSIASM (Reformatting)",
"$n$ only needs to satisfy $\\sigma(a^n)\\equiv 1 \\pmod{43}$ and $\\sigma(a^n)\\equiv 1 \\pmod{47}$ for all $a$. Let's work on the first requirement (mod 43) first. All $n$ works for $a=1$. If $a>1$, let $a$'s prime factorization be $a=p_1^{e_1}p_2^{e_2}\\cdots p_k^{e_k}$. The following three statements are the same: For all $a$, we have $\\sigma(a^n)=\\prod_{i=1}^k(1+p_i+\\cdots+p_i^{ne_i})\\equiv1\\pmod{43}$. For all $e>0$ and prime $p$, we have $1+p+\\cdots+p^{ne}\\equiv1\\pmod{43}$. For all prime $p$, we have $p+p^2+\\cdots+p^n \\equiv 0 \\pmod{43}$. We can show this by casework on $p$: For $p\\equiv0\\pmod{43}$, no matter what $n$ is, it is true that \\[p+p^2+\\cdots+p^n \\equiv 0 \\pmod{43}.\\] For all $p\\equiv1\\pmod{43}$, it is true that \\[p+p^2+\\cdots+p^n \\equiv n \\equiv 0 \\pmod{43}.\\] One can either use brute force or Dirichlet's Theorem to show such $p$ exists. Therefore, $43|n$. For all $p\\not\\equiv0,1\\pmod{43}$, it is true that \\[p+p^2+\\cdots+p^n \\equiv 0 \\pmod{43} \\Leftrightarrow p^n-1\\equiv0\\pmod{43}.\\] According to Fermat's Little Theorem, $42|n$ is sufficient. To show it's necessary, we just need to show $43$ has a prime primitive root. This can be done either by brute force or as follows. $43$ is prime and it must have a primitive root $t\\neq 1$ that's coprime to $43$. All numbers of the form $43k+t$ are also primitive roots of $43$. According to Dirichlet's Theorem there must be infinitely many primes among them. Similar arguments for modulo $47$ lead to $46|n$ and $47|n$. Therefore, we get $n=\\operatorname{lcm}[42,43,46,47]$. Following the last paragraph of Solution 1 gives the answer $125. ~Mryang6859 (Solution) ~MRENTHUSIASM (Reformatting)",
"We perform casework on $a:$ $a=1.$ For all positive integers $n,$ we have $\\sigma\\left(a^n\\right)-1=0.$ $a$ is a prime number. For all prime numbers $p,$ note that \\[\\sigma\\left(p^n\\right)-1=\\left(\\sum_{i=0}^{n}p^i\\right)-1=\\sum_{i=1}^{n}p^i=\\frac{p^{n+1}-p}{p-1}=\\frac{p\\left(p^n-1\\right)}{p-1}\\] by geometric series. Recall that $2021=43\\cdot47.$ Applying the Chinese Remainder Theorem, we get the following system of congruences: \\begin{align*} \\frac{p\\left(p^n-1\\right)}{p-1}&\\equiv0\\pmod{43}, &(1)\\\\ \\frac{p\\left(p^n-1\\right)}{p-1}&\\equiv0\\pmod{47}. &(2) \\end{align*} We perform casework on $(1):$ $p-1\\not\\equiv0\\pmod{43}.$ We need $p\\left(p^n-1\\right)\\equiv0\\pmod{43}.$ If $p=43,$ then this congruence is true for all positive integers $n.$ If $p\\neq43,$ then this congruence becomes $p^n-1\\equiv0\\pmod{43}.$ By Fermat's Little Theorem, we obtain $n\\equiv0\\pmod{42}.$ $p-1\\equiv0\\pmod{43}.$ We need $p^n-1$ to contain more factors of $43$ than $p-1$ does. More generally, for all positive integers $k,$ if $p-1\\equiv0 \\ \\left(\\operatorname{mod} \\ 43^k\\right),$ then $p^n-1\\equiv0 \\ \\left(\\operatorname{mod} \\ 43^{k+1}\\right).$ Let $p=43^km+1$ for some positive integer $m$ such that $m\\not\\equiv0\\pmod{43}.$ We substitute this into $p^n-1\\equiv0 \\ \\left(\\operatorname{mod} \\ 43^{k+1}\\right)$ and apply the Binomial Theorem: \\begin{align*} \\left(43^km+1\\right)^n-1&\\equiv0 \\ \\left(\\operatorname{mod} \\ 43^{k+1}\\right) \\\\ \\Biggl[\\binom{n}{0}\\left(43^km\\right)^0+\\binom{n}{1}\\left(43^km\\right)^1+\\phantom{ }\\underbrace{\\binom{n}{2}\\left(43^km\\right)^2+\\cdots+\\binom{n}{n}\\left(43^km\\right)^n}_{0 \\ \\left(\\operatorname{mod} \\ 43^{k+1}\\right)}\\phantom{ }\\Biggr]-1 &\\equiv0 \\ \\left(\\operatorname{mod} \\ 43^{k+1}\\right) \\\\ \\left[1+43^kmn\\right]-1&\\equiv0 \\ \\left(\\operatorname{mod} \\ 43^{k+1}\\right) \\\\ 43^kmn&\\equiv0 \\ \\left(\\operatorname{mod} \\ 43^{k+1}\\right), \\end{align*} from which $n\\equiv0\\pmod{43}.$ For $(1),$ we find that $n\\equiv0\\pmod{42}$ and $n\\equiv0\\pmod{43}.$ For $(2),$ we find that $n\\equiv0\\pmod{46}$ and $n\\equiv0\\pmod{47}$ by a similar argument. Together, we conclude that $n=\\operatorname{lcm}(42,43,46,47)$ is the least such positive integer $n$ for this case. $a$ is a composite number. Let $a=\\prod_{i=1}^{u}p_i^{e_i}$ be the prime factorization of $a.$ Note that $\\sigma(a)$ is a multiplicative function: \\[\\sigma(a)=\\sigma\\left(\\prod_{i=1}^{u}p_i^{e_i}\\right)=\\prod_{i=1}^{u}\\sigma\\left(p_i^{e_i}\\right)=\\prod_{i=1}^{u}\\left(\\sum_{j=0}^{e_i}p_i^j\\right),\\] as this product of geometric series spans all positive divisors of $a.$ At $n=\\operatorname{lcm}(42,43,46,47),$ it follows that \\begin{align*} \\sigma\\left(a^n\\right)-1&=\\left(\\prod_{i=1}^{u}\\sigma\\left(p_i^{e_in}\\right)\\right)-1 \\\\ &\\equiv\\left(\\prod_{i=1}^{u}1\\right)-1 &&\\pmod{2021} \\\\ &\\equiv0 &&\\pmod{2021}. \\end{align*} Finally, the least such positive integer $n$ for all cases is \\begin{align*} n&=\\operatorname{lcm}(42,43,46,47) \\\\ &=\\operatorname{lcm}(2\\cdot3\\cdot7,43,2\\cdot23,47) \\\\ &=2\\cdot3\\cdot7\\cdot23\\cdot43\\cdot47, \\end{align*} so the sum of its prime factors is $2+3+7+23+43+47=125 ~MRENTHUSIASM (inspired by Math Jams's 2021 AIME I Discussion)",
"We perform casework on $a:$ $a=1.$ For all positive integers $n,$ we have $\\sigma\\left(a^n\\right)-1=0.$ $a$ is a prime number. For all prime numbers $p,$ note that \\[\\sigma\\left(p^n\\right)-1=\\left(\\sum_{i=0}^{n}p^i\\right)-1=\\sum_{i=1}^{n}p^i=\\frac{p^{n+1}-p}{p-1}=\\frac{p\\left(p^n-1\\right)}{p-1}\\] by geometric series. Recall that $2021=43\\cdot47.$ Applying the Chinese Remainder Theorem, we get the following system of congruences: \\begin{align*} \\frac{p\\left(p^n-1\\right)}{p-1}&\\equiv0\\pmod{43}, &(1)\\\\ \\frac{p\\left(p^n-1\\right)}{p-1}&\\equiv0\\pmod{47}. &(2) \\end{align*} We perform casework on $(1):$ $p-1\\not\\equiv0\\pmod{43}.$ We need $p\\left(p^n-1\\right)\\equiv0\\pmod{43}.$ If $p=43,$ then this congruence is true for all positive integers $n.$ If $p\\neq43,$ then this congruence becomes $p^n-1\\equiv0\\pmod{43}.$ By Fermat's Little Theorem, we obtain $n\\equiv0\\pmod{42}.$ $p-1\\equiv0\\pmod{43}.$ We need $p^n-1$ to contain more factors of $43$ than $p-1$ does. More generally, for all positive integers $k,$ if $p-1\\equiv0 \\ \\left(\\operatorname{mod} \\ 43^k\\right),$ then $p^n-1\\equiv0 \\ \\left(\\operatorname{mod} \\ 43^{k+1}\\right).$ Let $p=43^km+1$ for some positive integer $m$ such that $m\\not\\equiv0\\pmod{43}.$ We substitute this into $p^n-1\\equiv0 \\ \\left(\\operatorname{mod} \\ 43^{k+1}\\right)$ and apply the Binomial Theorem: \\begin{align*} \\left(43^km+1\\right)^n-1&\\equiv0 \\ \\left(\\operatorname{mod} \\ 43^{k+1}\\right) \\\\ \\Biggl[\\binom{n}{0}\\left(43^km\\right)^0+\\binom{n}{1}\\left(43^km\\right)^1+\\phantom{ }\\underbrace{\\binom{n}{2}\\left(43^km\\right)^2+\\cdots+\\binom{n}{n}\\left(43^km\\right)^n}_{0 \\ \\left(\\operatorname{mod} \\ 43^{k+1}\\right)}\\phantom{ }\\Biggr]-1 &\\equiv0 \\ \\left(\\operatorname{mod} \\ 43^{k+1}\\right) \\\\ \\left[1+43^kmn\\right]-1&\\equiv0 \\ \\left(\\operatorname{mod} \\ 43^{k+1}\\right) \\\\ 43^kmn&\\equiv0 \\ \\left(\\operatorname{mod} \\ 43^{k+1}\\right), \\end{align*} from which $n\\equiv0\\pmod{43}.$ For $(1),$ we find that $n\\equiv0\\pmod{42}$ and $n\\equiv0\\pmod{43}.$ For $(2),$ we find that $n\\equiv0\\pmod{46}$ and $n\\equiv0\\pmod{47}$ by a similar argument. Together, we conclude that $n=\\operatorname{lcm}(42,43,46,47)$ is the least such positive integer $n$ for this case. $a$ is a composite number. Let $a=\\prod_{i=1}^{u}p_i^{e_i}$ be the prime factorization of $a.$ Note that $\\sigma(a)$ is a multiplicative function: \\[\\sigma(a)=\\sigma\\left(\\prod_{i=1}^{u}p_i^{e_i}\\right)=\\prod_{i=1}^{u}\\sigma\\left(p_i^{e_i}\\right)=\\prod_{i=1}^{u}\\left(\\sum_{j=0}^{e_i}p_i^j\\right),\\] as this product of geometric series spans all positive divisors of $a.$ At $n=\\operatorname{lcm}(42,43,46,47),$ it follows that \\begin{align*} \\sigma\\left(a^n\\right)-1&=\\left(\\prod_{i=1}^{u}\\sigma\\left(p_i^{e_in}\\right)\\right)-1 \\\\ &\\equiv\\left(\\prod_{i=1}^{u}1\\right)-1 &&\\pmod{2021} \\\\ &\\equiv0 &&\\pmod{2021}. \\end{align*} Finally, the least such positive integer $n$ for all cases is \\begin{align*} n&=\\operatorname{lcm}(42,43,46,47) \\\\ &=\\operatorname{lcm}(2\\cdot3\\cdot7,43,2\\cdot23,47) \\\\ &=2\\cdot3\\cdot7\\cdot23\\cdot43\\cdot47, \\end{align*} so the sum of its prime factors is $2+3+7+23+43+47=125 ~MRENTHUSIASM (inspired by Math Jams's 2021 AIME I Discussion)",
"Since the problem works for all positive integers $a$, let's plug in $a=2$ and see what we get. Since $\\sigma({2^n}) = 2^{n+1}-1,$ we have $2^{n+1} \\equiv 2 \\pmod{2021}.$ Simplifying using CRT and Fermat's Little Theorem, we get that $n \\equiv 0 \\pmod{42}$ and $n \\equiv 0 \\pmod{46}.$ Then, we can look at $a$ being a $1\\pmod{43}$ prime and a $1\\pmod{47}$ prime, just like in Solution 1, to find that $43$ and $47$ also divide $n$. There don't seem to be any other odd \"numbers\" to check, so we can hopefully assume that the answer is the sum of the prime factors of $\\text{lcm(42, 43, 46, 47)}.$ From here, follow solution 1 to get the final answer $125. ~PureSwag",
"Since the problem works for all positive integers $a$, let's plug in $a=2$ and see what we get. Since $\\sigma({2^n}) = 2^{n+1}-1,$ we have $2^{n+1} \\equiv 2 \\pmod{2021}.$ Simplifying using CRT and Fermat's Little Theorem, we get that $n \\equiv 0 \\pmod{42}$ and $n \\equiv 0 \\pmod{46}.$ Then, we can look at $a$ being a $1\\pmod{43}$ prime and a $1\\pmod{47}$ prime, just like in Solution 1, to find that $43$ and $47$ also divide $n$. There don't seem to be any other odd \"numbers\" to check, so we can hopefully assume that the answer is the sum of the prime factors of $\\text{lcm(42, 43, 46, 47)}.$ From here, follow solution 1 to get the final answer $125. ~PureSwag",
"Warning: This solution doesn't explain why $43*47\\mid n$. It says the problem implies that it works for all positive integers $a$, we basically know that If $p \\equiv 1 \\pmod q$, then from \"USEMO 2019 Problem 4\", if $p^n + p^{n-1} + \\cdots + 1 \\equiv 1 \\pmod{q}$, then \\[\\frac{p^{en+1}-1}{p-1} = p^{en} + p^{en-1} + \\cdots + 1 \\equiv 1 \\pmod{q}.\\] From here we can just let $\\sigma(2^n)$ or be a power of $2$ which we can do \\[\\sigma(2^n)=1+2+2^2+2^3+2^4+\\cdots+2^n=2^{n+1}-1,\\] which is a geometric series. We can plug in $a=2$ like in Solution 4 and use CRT. We have the prime factorization $2021 = 43 \\cdot 47$. We use CRT to find that \\begin{align*} 2^n &\\equiv 1 \\pmod{43}, \\\\ 2^n &\\equiv 1 \\pmod{47}. \\end{align*} We see that this is just FLT which is $a^{p-1} \\equiv 1 \\pmod p$ we see that all multiples of $42$ will work for first and $46$ for the second. We can figure out that it is just $\\text{lcm}(43-1,47-1)\\cdot43\\cdot47$ which when we add up we get that it's just the sum of the prime factors of $\\text{lcm}(42,43,46,47)$ which you can just look at Solution 1 to find out the sum of the prime factors and get the answer $125.",
"Warning: This solution doesn't explain why $43*47\\mid n$. It says the problem implies that it works for all positive integers $a$, we basically know that If $p \\equiv 1 \\pmod q$, then from \"USEMO 2019 Problem 4\", if $p^n + p^{n-1} + \\cdots + 1 \\equiv 1 \\pmod{q}$, then \\[\\frac{p^{en+1}-1}{p-1} = p^{en} + p^{en-1} + \\cdots + 1 \\equiv 1 \\pmod{q}.\\] From here we can just let $\\sigma(2^n)$ or be a power of $2$ which we can do \\[\\sigma(2^n)=1+2+2^2+2^3+2^4+\\cdots+2^n=2^{n+1}-1,\\] which is a geometric series. We can plug in $a=2$ like in Solution 4 and use CRT. We have the prime factorization $2021 = 43 \\cdot 47$. We use CRT to find that \\begin{align*} 2^n &\\equiv 1 \\pmod{43}, \\\\ 2^n &\\equiv 1 \\pmod{47}. \\end{align*} We see that this is just FLT which is $a^{p-1} \\equiv 1 \\pmod p$ we see that all multiples of $42$ will work for first and $46$ for the second. We can figure out that it is just $\\text{lcm}(43-1,47-1)\\cdot43\\cdot47$ which when we add up we get that it's just the sum of the prime factors of $\\text{lcm}(42,43,46,47)$ which you can just look at Solution 1 to find out the sum of the prime factors and get the answer $125."
] |
2021-I-15
| 2,021
| 15
|
Let $S$ be the set of positive integers $k$ such that the two parabolas \[y=x^2-k~~\text{and}~~x=2(y-20)^2-k\] intersect in four distinct points, and these four points lie on a circle with radius at most $21$ . Find the sum of the least element of $S$ and the greatest element of $S$ .
| 285
|
I
|
[
"Note that $y=x^2-k$ is an upward-opening parabola with the vertex at $(0,-k),$ and $x=2(y-20)^2-k$ is a rightward-opening parabola with the vertex at $(-k,20).$ We consider each condition separately: The two parabolas intersect at four distinct points. By a quick sketch, we have two subconditions: The point $(-k,20)$ is on or below the parabola $y=x^2-k.$ We need $20\\leq(-k)^2-k,$ from which $k\\geq5.$ Moreover, the point $(-k,20)$ is on the parabola $y=x^2-k$ when $k=5.$ We will prove that the two parabolas intersect at four distinct points at this value of $k:$ Substituting $y=x^2-5$ into $x=2(y-20)^2-5,$ we get $x=2\\left(\\left(x^2-5\\right)-20\\right)^2-5.$ Expanding and rearranging give \\[2x^4-100x^2-x+1245=0. \\hspace{20mm}(\\bigstar)\\] By either the graphs of the parabolas or the Rational Root Theorem, we conclude that $x=-5$ is a root of $(\\bigstar).$ So, we factor its left side: \\[(x+5)\\left(2x^3-10x^2-50x+249\\right)=0.\\] By either the graphs of the parabolas or Descartes' Rule of Signs, we conclude that $2x^3-10x^2-50x+249=0$ has two positive roots and one negative root such that $x\\neq-5.$ So, $(\\bigstar)$ has four distinct real roots, or the two parabolas intersect at four distinct points. For Subcondition A, we deduce that $k\\geq5.$ Remark for Subcondition A Recall that if $1\\leq k\\leq 4,$ then the point $(-k,20)$ is above the parabola $y=x^2-k.$ It follows that for $-k\\leq x\\leq0:$ The maximum value of $y$ for the parabola $y=x^2-k$ occurs at $x=-k,$ from which $y=k^2-k\\leq12.$ The minimum value of $y$ for the parabola $x=2(y-20)^2-k$ occurs at $x=0,$ from which $y=20-\\sqrt{\\frac k2}>18.$ Clearly, the parabola $x=2(y-20)^2-k$ and the left half of the parabola $y=x^2-k$ do not intersect. Therefore, the two parabolas do not intersect at four distinct points. The point $(0,-k)$ is on or below the parabola $x=2(y-20)^2-k.$ The lower half of the parabola $x=2(y-20)^2-k$ is $y=20-\\sqrt{\\frac{x+k}{2}}.$ We need $-k\\leq20-\\sqrt{\\frac k2},$ which holds for all values of $k.$ For Subcondition B, we deduce that $k$ can be any positive integer. For Condition 1, we obtain $\\boldsymbol{k\\geq5}$ by taking the intersection of Subconditions A and B. The four points of intersection lie on a circle with radius at most $21.$ For equations of circles, the coefficients of $x^2$ and $y^2$ must be the same. So, we add the equation $y=x^2-k$ to half the equation $x=2(y-20)^2-k:$ \\[y+\\frac12x=x^2+(y-20)^2-\\frac32k.\\] We expand, rearrange, and complete the squares: \\begin{align*} y+\\frac12x&=x^2+y^2-40y+400-\\frac32k \\\\ \\frac32k-400&=\\left(x^2-\\frac12x\\right)+\\left(y^2-41y\\right) \\\\ \\frac32k-400+\\frac{1}{16}+\\frac{1681}{4}&=\\left(x-\\frac14\\right)^2+\\left(y-\\frac{41}{2}\\right)^2. \\end{align*} We need $\\frac32k-400+\\frac{1}{16}+\\frac{1681}{4}\\leq21^2,$ from which $k\\leq\\left\\lfloor\\frac{6731}{24}\\right\\rfloor=280.$ For Condition 2, we obtain $\\boldsymbol{k\\leq280.}$ Taking the intersection of Conditions 1 and 2 produces $5\\leq k\\leq280.$ Therefore, the answer is $5+280=285 ~MRENTHUSIASM",
"Note that $y=x^2-k$ is an upward-opening parabola with the vertex at $(0,-k),$ and $x=2(y-20)^2-k$ is a rightward-opening parabola with the vertex at $(-k,20).$ We consider each condition separately: The two parabolas intersect at four distinct points. By a quick sketch, we have two subconditions: The point $(-k,20)$ is on or below the parabola $y=x^2-k.$ We need $20\\leq(-k)^2-k,$ from which $k\\geq5.$ Moreover, the point $(-k,20)$ is on the parabola $y=x^2-k$ when $k=5.$ We will prove that the two parabolas intersect at four distinct points at this value of $k:$ Substituting $y=x^2-5$ into $x=2(y-20)^2-5,$ we get $x=2\\left(\\left(x^2-5\\right)-20\\right)^2-5.$ Expanding and rearranging give \\[2x^4-100x^2-x+1245=0. \\hspace{20mm}(\\bigstar)\\] By either the graphs of the parabolas or the Rational Root Theorem, we conclude that $x=-5$ is a root of $(\\bigstar).$ So, we factor its left side: \\[(x+5)\\left(2x^3-10x^2-50x+249\\right)=0.\\] By either the graphs of the parabolas or Descartes' Rule of Signs, we conclude that $2x^3-10x^2-50x+249=0$ has two positive roots and one negative root such that $x\\neq-5.$ So, $(\\bigstar)$ has four distinct real roots, or the two parabolas intersect at four distinct points. For Subcondition A, we deduce that $k\\geq5.$ Remark for Subcondition A Recall that if $1\\leq k\\leq 4,$ then the point $(-k,20)$ is above the parabola $y=x^2-k.$ It follows that for $-k\\leq x\\leq0:$ The maximum value of $y$ for the parabola $y=x^2-k$ occurs at $x=-k,$ from which $y=k^2-k\\leq12.$ The minimum value of $y$ for the parabola $x=2(y-20)^2-k$ occurs at $x=0,$ from which $y=20-\\sqrt{\\frac k2}>18.$ Clearly, the parabola $x=2(y-20)^2-k$ and the left half of the parabola $y=x^2-k$ do not intersect. Therefore, the two parabolas do not intersect at four distinct points. The point $(0,-k)$ is on or below the parabola $x=2(y-20)^2-k.$ The lower half of the parabola $x=2(y-20)^2-k$ is $y=20-\\sqrt{\\frac{x+k}{2}}.$ We need $-k\\leq20-\\sqrt{\\frac k2},$ which holds for all values of $k.$ For Subcondition B, we deduce that $k$ can be any positive integer. For Condition 1, we obtain $\\boldsymbol{k\\geq5}$ by taking the intersection of Subconditions A and B. The four points of intersection lie on a circle with radius at most $21.$ For equations of circles, the coefficients of $x^2$ and $y^2$ must be the same. So, we add the equation $y=x^2-k$ to half the equation $x=2(y-20)^2-k:$ \\[y+\\frac12x=x^2+(y-20)^2-\\frac32k.\\] We expand, rearrange, and complete the squares: \\begin{align*} y+\\frac12x&=x^2+y^2-40y+400-\\frac32k \\\\ \\frac32k-400&=\\left(x^2-\\frac12x\\right)+\\left(y^2-41y\\right) \\\\ \\frac32k-400+\\frac{1}{16}+\\frac{1681}{4}&=\\left(x-\\frac14\\right)^2+\\left(y-\\frac{41}{2}\\right)^2. \\end{align*} We need $\\frac32k-400+\\frac{1}{16}+\\frac{1681}{4}\\leq21^2,$ from which $k\\leq\\left\\lfloor\\frac{6731}{24}\\right\\rfloor=280.$ For Condition 2, we obtain $\\boldsymbol{k\\leq280.}$ Taking the intersection of Conditions 1 and 2 produces $5\\leq k\\leq280.$ Therefore, the answer is $5+280=285 ~MRENTHUSIASM",
"Make the translation $y \\rightarrow y+20$ to obtain $20+y=x^2-k$ and $x=2y^2-k$. Multiply the first equation by $2$ and sum, we see that $2(x^2+y^2)=3k+40+2y+x$. Completing the square gives us $\\left(y- \\frac{1}{2}\\right)^2+\\left(x - \\frac{1}{4}\\right)^2 = \\frac{325+24k}{16}$; this explains why the two parabolas intersect at four points that lie on a circle*. For the upper bound, observe that $LHS \\leq 21^2=441 \\rightarrow 24k \\leq 6731$, so $k \\leq 280$. For the lower bound, we need to ensure there are $4$ intersections to begin with. (Here I'm using the un-translated coordinates.) Draw up a graph, and realize that two intersections are guaranteed, on the so called \"right branch\" of $y=x^2-k$. As we increase the value of $k$, two more intersections appear on the \"left branch\": $k=4$ does not work because the \"leftmost\" point of $x=2(y-20)^2-4$ is $(-4,20)$ which lies to the right of $\\left(-\\sqrt{24}, 20\\right)$, which is on the graph $y=x^2-4$. While technically speaking this doesn't prove that there are no intersections (why?), drawing the graph should convince you that this is the case. Clearly, $k<4$ does not work. $k=5$ does work because the two graphs intersect at $(-5,20)$, and by drawing the graph, you realize this is not a tangent point and there is in fact another intersection nearby, due to slope. Therefore, the answer is $5+280=285 intersection point(s), the statement that all these points lie on a circle is trivially true. ~Ross Gao",
"Make the translation $y \\rightarrow y+20$ to obtain $20+y=x^2-k$ and $x=2y^2-k$. Multiply the first equation by $2$ and sum, we see that $2(x^2+y^2)=3k+40+2y+x$. Completing the square gives us $\\left(y- \\frac{1}{2}\\right)^2+\\left(x - \\frac{1}{4}\\right)^2 = \\frac{325+24k}{16}$; this explains why the two parabolas intersect at four points that lie on a circle*. For the upper bound, observe that $LHS \\leq 21^2=441 \\rightarrow 24k \\leq 6731$, so $k \\leq 280$. For the lower bound, we need to ensure there are $4$ intersections to begin with. (Here I'm using the un-translated coordinates.) Draw up a graph, and realize that two intersections are guaranteed, on the so called \"right branch\" of $y=x^2-k$. As we increase the value of $k$, two more intersections appear on the \"left branch\": $k=4$ does not work because the \"leftmost\" point of $x=2(y-20)^2-4$ is $(-4,20)$ which lies to the right of $\\left(-\\sqrt{24}, 20\\right)$, which is on the graph $y=x^2-4$. While technically speaking this doesn't prove that there are no intersections (why?), drawing the graph should convince you that this is the case. Clearly, $k<4$ does not work. $k=5$ does work because the two graphs intersect at $(-5,20)$, and by drawing the graph, you realize this is not a tangent point and there is in fact another intersection nearby, due to slope. Therefore, the answer is $5+280=285 intersection point(s), the statement that all these points lie on a circle is trivially true. ~Ross Gao",
"Claim Let the axes of two parabolas be perpendicular, their focal parameters be $p_1$ and $p_2$ and the distances from the foci to the point of intersection of the axes be $x_2$ and $y_1$. Suppose that these parabolas intersect at four points. Then these points lie on the circle centered at point $(p_2, p_1)$ with radius $r = \\sqrt{2(p_1^2 + p_2^2 + p_1 y_1 + p_2 x_2)}.$ Proof Let's introduce a coordinate system with the center at the point of intersection of the axes. Let the first (red) parabola have axis $x = 0,$ focal parameter $p_1$ and focus at point $A(0, –y_1), y_1 > 0.$ Let second (blue) parabola have axis $y = 0,$ focal parameter $p_2$ and focus at point $B(–x_2,0), x_2 > 0.$ Let us denote the angle between the vector connecting the focus of the first parabola and its point and the positive direction of the ordinate axis $2\\theta,$ its length $\\rho_1(\\theta),$ the angle between the vector connecting the focus of the second parabola and its point and the positive direction of the abscissa axis $2\\phi,$ its length $\\rho_2(\\phi).$ Then \\[\\rho_1(\\theta) = \\frac{p_1}{1 - \\cos(2\\theta)}, \\rho_2(\\phi) = \\frac{p_2}{1 - \\cos(2\\phi)}.\\] Abscissa of the point of intersection is \\begin{align*} x =\\rho_1 \\sin(2\\theta) = p_1\\cot\\theta = \\rho_2 \\cos (2\\phi) - x_2 = \\frac{p_2}{2} (\\cot^2\\phi - 1)- x_2,\\end{align*} \\begin{align*} x^2 = p_1^2 \\cot ^2 \\theta , 2 p_1\\cot\\theta = p_2 \\cos^2 \\phi - p_2 - 2x_2 .\\end{align*} Ordinate of the point of intersection is \\begin{align*} y =\\rho_2 \\sin 2\\phi = p_2\\cot\\phi = \\rho_1 \\cos 2\\theta - y_1 = \\frac{p_1}{2} (\\cot^2\\theta - 1)- y_1,\\end{align*} \\begin{align*} y^2 = p_2^2 \\cot ^2 \\phi , 2 p_2\\cot\\phi = p_1 \\cos^2 \\theta - p_1 - 2y_1 .\\end{align*} The square of the distance from point of intersection to the point $(p_2, p_1)$ is \\begin{align*} r^2 = (x-p_2)^2 + (y-p_1)^2 = x^2 + y^2 - 2 p_1 y - 2 p_2 x + p_1^2 + p_2^2 .\\end{align*} After simple transformations, we get $r^2 = 2(p_1^2 + p_2^2 + p_1 y_1 + p_2 x_2).$ Hence, any intersection point has the same distance $r$ from the point $(p_2, p_1).$ Solution Parameters of the parabola $y = x^2 – k$ are $p_1 = \\frac{1}{2}, y_1 = 20 + k – \\frac{1}{2}.$ Parameters of the parabola $\\frac{x}{2} = (y – 20)^2 – \\frac{k}{2}$ are $p_2 = \\frac{1}{4}, x_2 = k – \\frac{1}{4} \\implies r^2 = 20 + \\frac{3k}{2}.$ If $r \\le 21, k \\le \\frac{842}{3},$ then integer $k \\le 280.$ The vertex of the second parabola is point $(– k,20)$ can be on the parabola $y = x^2 – k$ or below the point of the parabola with the same abscissa. So \\[20 \\ge (– k)^2 – k \\implies 5 \\le k \\le 280.\\] Therefore, the answer is $5+280=285. vladimir.shelomovskii@gmail.com, vvsss",
"Claim Let the axes of two parabolas be perpendicular, their focal parameters be $p_1$ and $p_2$ and the distances from the foci to the point of intersection of the axes be $x_2$ and $y_1$. Suppose that these parabolas intersect at four points. Then these points lie on the circle centered at point $(p_2, p_1)$ with radius $r = \\sqrt{2(p_1^2 + p_2^2 + p_1 y_1 + p_2 x_2)}.$ Proof Let's introduce a coordinate system with the center at the point of intersection of the axes. Let the first (red) parabola have axis $x = 0,$ focal parameter $p_1$ and focus at point $A(0, –y_1), y_1 > 0.$ Let second (blue) parabola have axis $y = 0,$ focal parameter $p_2$ and focus at point $B(–x_2,0), x_2 > 0.$ Let us denote the angle between the vector connecting the focus of the first parabola and its point and the positive direction of the ordinate axis $2\\theta,$ its length $\\rho_1(\\theta),$ the angle between the vector connecting the focus of the second parabola and its point and the positive direction of the abscissa axis $2\\phi,$ its length $\\rho_2(\\phi).$ Then \\[\\rho_1(\\theta) = \\frac{p_1}{1 - \\cos(2\\theta)}, \\rho_2(\\phi) = \\frac{p_2}{1 - \\cos(2\\phi)}.\\] Abscissa of the point of intersection is \\begin{align*} x =\\rho_1 \\sin(2\\theta) = p_1\\cot\\theta = \\rho_2 \\cos (2\\phi) - x_2 = \\frac{p_2}{2} (\\cot^2\\phi - 1)- x_2,\\end{align*} \\begin{align*} x^2 = p_1^2 \\cot ^2 \\theta , 2 p_1\\cot\\theta = p_2 \\cos^2 \\phi - p_2 - 2x_2 .\\end{align*} Ordinate of the point of intersection is \\begin{align*} y =\\rho_2 \\sin 2\\phi = p_2\\cot\\phi = \\rho_1 \\cos 2\\theta - y_1 = \\frac{p_1}{2} (\\cot^2\\theta - 1)- y_1,\\end{align*} \\begin{align*} y^2 = p_2^2 \\cot ^2 \\phi , 2 p_2\\cot\\phi = p_1 \\cos^2 \\theta - p_1 - 2y_1 .\\end{align*} The square of the distance from point of intersection to the point $(p_2, p_1)$ is \\begin{align*} r^2 = (x-p_2)^2 + (y-p_1)^2 = x^2 + y^2 - 2 p_1 y - 2 p_2 x + p_1^2 + p_2^2 .\\end{align*} After simple transformations, we get $r^2 = 2(p_1^2 + p_2^2 + p_1 y_1 + p_2 x_2).$ Hence, any intersection point has the same distance $r$ from the point $(p_2, p_1).$ Solution Parameters of the parabola $y = x^2 – k$ are $p_1 = \\frac{1}{2}, y_1 = 20 + k – \\frac{1}{2}.$ Parameters of the parabola $\\frac{x}{2} = (y – 20)^2 – \\frac{k}{2}$ are $p_2 = \\frac{1}{4}, x_2 = k – \\frac{1}{4} \\implies r^2 = 20 + \\frac{3k}{2}.$ If $r \\le 21, k \\le \\frac{842}{3},$ then integer $k \\le 280.$ The vertex of the second parabola is point $(– k,20)$ can be on the parabola $y = x^2 – k$ or below the point of the parabola with the same abscissa. So \\[20 \\ge (– k)^2 – k \\implies 5 \\le k \\le 280.\\] Therefore, the answer is $5+280=285. vladimir.shelomovskii@gmail.com, vvsss"
] |
2021-II-1
| 2,021
| 1
|
Find the arithmetic mean of all the three-digit palindromes. (Recall that a palindrome is a number that reads the same forward and backward, such as $777$ or $383$ .)
| 550
|
II
|
[
"Recall that the arithmetic mean of all the $n$ digit palindromes is just the average of the largest and smallest $n$ digit palindromes, and in this case the $2$ palindromes are $101$ and $999$ and $\\frac{101+999}{2}=550 which is the final answer. ~ math31415926535",
"For any palindrome $\\underline{ABA},$ note that $\\underline{ABA}$ is $100A + 10B + A = 101A + 10B.$ The average for $A$ is $5$ since $A$ can be any of $1, 2, 3, 4, 5, 6, 7, 8,$ or $9.$ The average for $B$ is $4.5$ since $B$ is either $0, 1, 2, 3, 4, 5, 6, 7, 8,$ or $9.$ Therefore, the answer is $505 + 45 = 550 - ARCTICTURN",
"For every three-digit palindrome $\\underline{ABA}$ with $A\\in\\{1,2,3,4,5,6,7,8,9\\}$ and $B\\in\\{0,1,2,3,4,5,6,7,8,9\\},$ note that $\\underline{(10-A)(9-B)(10-A)}$ must be another palindrome by symmetry. Therefore, we can pair each three-digit palindrome uniquely with another three-digit palindrome so that they sum to \\begin{align*} \\underline{ABA}+\\underline{(10-A)(9-B)(10-A)}&=\\left[100A+10B+A\\right]+\\left[100(10-A)+10(9-B)+(10-A)\\right] \\\\ &=\\left[100A+10B+A\\right]+\\left[1000-100A+90-10B+10-A\\right] \\\\ &=1000+90+10 \\\\ &=1100. \\end{align*} For instances: \\begin{align*} 171+929&=1100, \\\\ 262+838&=1100, \\\\ 303+797&=1100, \\\\ 414+686&=1100, \\\\ 545+555&=1100, \\end{align*} and so on. From this symmetry, the arithmetic mean of all the three-digit palindromes is $\\frac{1100}{2}=550 ~MRENTHUSIASM",
"For every three-digit palindrome $\\underline{ABA}$ with $A\\in\\{1,2,3,4,5,6,7,8,9\\}$ and $B\\in\\{0,1,2,3,4,5,6,7,8,9\\},$ note that $\\underline{(10-A)(9-B)(10-A)}$ must be another palindrome by symmetry. Therefore, we can pair each three-digit palindrome uniquely with another three-digit palindrome so that they sum to \\begin{align*} \\underline{ABA}+\\underline{(10-A)(9-B)(10-A)}&=\\left[100A+10B+A\\right]+\\left[100(10-A)+10(9-B)+(10-A)\\right] \\\\ &=\\left[100A+10B+A\\right]+\\left[1000-100A+90-10B+10-A\\right] \\\\ &=1000+90+10 \\\\ &=1100. \\end{align*} For instances: \\begin{align*} 171+929&=1100, \\\\ 262+838&=1100, \\\\ 303+797&=1100, \\\\ 414+686&=1100, \\\\ 545+555&=1100, \\end{align*} and so on. From this symmetry, the arithmetic mean of all the three-digit palindromes is $\\frac{1100}{2}=550 ~MRENTHUSIASM",
"We notice that a three-digit palindrome looks like this: $\\underline{aba}.$ And we know $a$ can be any digit from $1$ through $9,$ and $b$ can be any digit from $0$ through $9,$ so there are $9\\times{10}=90$ three-digit palindromes. We want to find the sum of these $90$ palindromes and divide it by $90$ to find the arithmetic mean. How can we do that? Instead of adding the numbers up, we can break each palindrome into two parts: $101a+10b.$ Thus, all of these $90$ palindromes can be broken into this form. Thus, the sum of these $90$ palindromes will be $101\\times{(1+2+...+9)}\\times{10}+10\\times{(0+1+2+...+9)}\\times{9},$ because each $a$ will be in $10$ different palindromes (since for each $a,$ there are $10$ choices for $b$). The same logic explains why we multiply by $9$ when computing the total sum of $b.$ We get a sum of $45\\times{1100},$ but don't compute this! Divide this by $90$ and you will get $550",
"We notice that a three-digit palindrome looks like this: $\\underline{aba}.$ And we know $a$ can be any digit from $1$ through $9,$ and $b$ can be any digit from $0$ through $9,$ so there are $9\\times{10}=90$ three-digit palindromes. We want to find the sum of these $90$ palindromes and divide it by $90$ to find the arithmetic mean. How can we do that? Instead of adding the numbers up, we can break each palindrome into two parts: $101a+10b.$ Thus, all of these $90$ palindromes can be broken into this form. Thus, the sum of these $90$ palindromes will be $101\\times{(1+2+...+9)}\\times{10}+10\\times{(0+1+2+...+9)}\\times{9},$ because each $a$ will be in $10$ different palindromes (since for each $a,$ there are $10$ choices for $b$). The same logic explains why we multiply by $9$ when computing the total sum of $b.$ We get a sum of $45\\times{1100},$ but don't compute this! Divide this by $90$ and you will get $550",
"The possible values of the first and last digits each are $1, 2, ..., 8, 9$ with a sum of $45$ so the average value is $5.$ The middle digit can be any digit from $0$ to $9$ with a sum of $45,$ so the average value is $4.5.$ The average of all three-digit palindromes is $5\\cdot 10^2+4.5\\cdot 10+5=550 ~MathIsFun286 ~MathFun1000 (Rephrasing with more clarity)",
"The possible values of the first and last digits each are $1, 2, ..., 8, 9$ with a sum of $45$ so the average value is $5.$ The middle digit can be any digit from $0$ to $9$ with a sum of $45,$ so the average value is $4.5.$ The average of all three-digit palindromes is $5\\cdot 10^2+4.5\\cdot 10+5=550 ~MathIsFun286 ~MathFun1000 (Rephrasing with more clarity)",
"Case 1 Consider palindromes of the form $5x5 = 505 + 10x.$ There are $10$ of them. The arithmetic mean of the first term is $505,$ and the second $\\frac {10 \\cdot(0 + 1 + ... + 9)}{10} = 45.$ The arithmetic mean of the sum is $505 + 45 = 550.$ Case 2 Consider palindromes of the form $yxy,$ where $y= {1,2,3,4,6,7,8,9}.$ Let $u = 10 – y, v = 9 – x.$ Then $uvu$ is a symmetric palindrome that can be constructed for each $yxy.$ The arithmetic mean of each such pair is $550.$ For example, $\\frac{737 + 363}{2} = 550.$ Thus, all palindromes are divided into groups of numbers with the arithmetic mean in each group equal to $550.$ The arithmetic mean of all numbers is also $550.$ vladimir.shelomovskii@gmail.com, vvsss",
"Case 1 Consider palindromes of the form $5x5 = 505 + 10x.$ There are $10$ of them. The arithmetic mean of the first term is $505,$ and the second $\\frac {10 \\cdot(0 + 1 + ... + 9)}{10} = 45.$ The arithmetic mean of the sum is $505 + 45 = 550.$ Case 2 Consider palindromes of the form $yxy,$ where $y= {1,2,3,4,6,7,8,9}.$ Let $u = 10 – y, v = 9 – x.$ Then $uvu$ is a symmetric palindrome that can be constructed for each $yxy.$ The arithmetic mean of each such pair is $550.$ For example, $\\frac{737 + 363}{2} = 550.$ Thus, all palindromes are divided into groups of numbers with the arithmetic mean in each group equal to $550.$ The arithmetic mean of all numbers is also $550.$ vladimir.shelomovskii@gmail.com, vvsss"
] |
2021-II-2
| 2,021
| 2
|
Equilateral triangle $ABC$ has side length $840$ . Point $D$ lies on the same side of line $BC$ as $A$ such that $\overline{BD} \perp \overline{BC}$ . The line $\ell$ through $D$ parallel to line $BC$ intersects sides $\overline{AB}$ and $\overline{AC}$ at points $E$ and $F$ , respectively. Point $G$ lies on $\ell$ such that $F$ is between $E$ and $G$ , $\triangle AFG$ is isosceles, and the ratio of the area of $\triangle AFG$ to the area of $\triangle BED$ is $8:9$ . Find $AF$ . [asy] pair A,B,C,D,E,F,G; B=origin; A=5*dir(60); C=(5,0); E=0.6*A+0.4*B; F=0.6*A+0.4*C; G=rotate(240,F)*A; D=extension(E,F,B,dir(90)); draw(D--G--A,grey); draw(B--0.5*A+rotate(60,B)*A*0.5,grey); draw(A--B--C--cycle,linewidth(1.5)); dot(A^^B^^C^^D^^E^^F^^G); label("$A$",A,dir(90)); label("$B$",B,dir(225)); label("$C$",C,dir(-45)); label("$D$",D,dir(180)); label("$E$",E,dir(-45)); label("$F$",F,dir(225)); label("$G$",G,dir(0)); label("$\ell$",midpoint(E--F),dir(90)); [/asy]
| 336
|
II
|
[
"By angle chasing, we conclude that $\\triangle AGF$ is a $30^\\circ\\text{-}30^\\circ\\text{-}120^\\circ$ triangle, and $\\triangle BED$ is a $30^\\circ\\text{-}60^\\circ\\text{-}90^\\circ$ triangle. Let $AF=x.$ It follows that $FG=x$ and $EB=FC=840-x.$ By the side-length ratios in $\\triangle BED,$ we have $DE=\\frac{840-x}{2}$ and $DB=\\frac{840-x}{2}\\cdot\\sqrt3.$ Let the brackets denote areas. We have \\[[AFG]=\\frac12\\cdot AF\\cdot FG\\cdot\\sin{\\angle AFG}=\\frac12\\cdot x\\cdot x\\cdot\\sin{120^\\circ}=\\frac12\\cdot x^2\\cdot\\frac{\\sqrt3}{2}\\] and \\[[BED]=\\frac12\\cdot DE\\cdot DB=\\frac12\\cdot\\frac{840-x}{2}\\cdot\\left(\\frac{840-x}{2}\\cdot\\sqrt3\\right).\\] We set up and solve an equation for $x:$ \\begin{align*} \\frac{[AFG]}{[BED]}&=\\frac89 \\\\ \\frac{\\frac12\\cdot x^2\\cdot\\frac{\\sqrt3}{2}}{\\frac12\\cdot\\frac{840-x}{2}\\cdot\\left(\\frac{840-x}{2}\\cdot\\sqrt3\\right)}&=\\frac89 \\\\ \\frac{2x^2}{(840-x)^2}&=\\frac89 \\\\ \\frac{x^2}{(840-x)^2}&=\\frac49. \\end{align*} Since $0<x<840,$ it is clear that $\\frac{x}{840-x}>0.$ Therefore, we take the positive square root for both sides: \\begin{align*} \\frac{x}{840-x}&=\\frac23 \\\\ 3x&=1680-2x \\\\ 5x&=1680 \\\\ x&=336. \\end{align*} ~MRENTHUSIASM",
"By angle chasing, we conclude that $\\triangle AGF$ is a $30^\\circ\\text{-}30^\\circ\\text{-}120^\\circ$ triangle, and $\\triangle BED$ is a $30^\\circ\\text{-}60^\\circ\\text{-}90^\\circ$ triangle. Let $AF=x.$ It follows that $FG=x$ and $EB=FC=840-x.$ By the side-length ratios in $\\triangle BED,$ we have $DE=\\frac{840-x}{2}$ and $DB=\\frac{840-x}{2}\\cdot\\sqrt3.$ Let the brackets denote areas. We have \\[[AFG]=\\frac12\\cdot AF\\cdot FG\\cdot\\sin{\\angle AFG}=\\frac12\\cdot x\\cdot x\\cdot\\sin{120^\\circ}=\\frac12\\cdot x^2\\cdot\\frac{\\sqrt3}{2}\\] and \\[[BED]=\\frac12\\cdot DE\\cdot DB=\\frac12\\cdot\\frac{840-x}{2}\\cdot\\left(\\frac{840-x}{2}\\cdot\\sqrt3\\right).\\] We set up and solve an equation for $x:$ \\begin{align*} \\frac{[AFG]}{[BED]}&=\\frac89 \\\\ \\frac{\\frac12\\cdot x^2\\cdot\\frac{\\sqrt3}{2}}{\\frac12\\cdot\\frac{840-x}{2}\\cdot\\left(\\frac{840-x}{2}\\cdot\\sqrt3\\right)}&=\\frac89 \\\\ \\frac{2x^2}{(840-x)^2}&=\\frac89 \\\\ \\frac{x^2}{(840-x)^2}&=\\frac49. \\end{align*} Since $0<x<840,$ it is clear that $\\frac{x}{840-x}>0.$ Therefore, we take the positive square root for both sides: \\begin{align*} \\frac{x}{840-x}&=\\frac23 \\\\ 3x&=1680-2x \\\\ 5x&=1680 \\\\ x&=336. \\end{align*} ~MRENTHUSIASM",
"We express the areas of $\\triangle BED$ and $\\triangle AFG$ in terms of $AF$ in order to solve for $AF.$ We let $x = AF.$ Because $\\triangle AFG$ is isosceles and $\\triangle AEF$ is equilateral, $AF = FG = EF = AE = x.$ Let the height of $\\triangle ABC$ be $h$ and the height of $\\triangle AEF$ be $h'.$ Then we have that $h = \\frac{\\sqrt{3}}{2}(840) = 420\\sqrt{3}$ and $h' = \\frac{\\sqrt{3}}{2}(EF) = \\frac{\\sqrt{3}}{2}x.$ Now we can find $DB$ and $BE$ in terms of $x.$ $DB = h - h' = 420\\sqrt{3} - \\frac{\\sqrt{3}}{2}x,$ $BE = AB - AE = 840 - x.$ Because we are given that $\\angle DBC = 90,$ $\\angle DBE = 30.$ This allows us to use the sin formula for triangle area: the area of $\\triangle BED$ is $\\frac{1}{2}(\\sin 30)\\left(420\\sqrt{3} - \\frac{\\sqrt{3}}{2}x\\right)(840-x).$ Similarly, because $\\angle AFG = 120,$ the area of $\\triangle AFG$ is $\\frac{1}{2}(\\sin 120)(x^2).$ Now we can make an equation: \\begin{align*} \\frac{\\triangle AFG}{\\triangle BED} &= \\frac{8}{9} \\\\ \\frac{\\frac{1}{2}(\\sin 120)(x^2)}{\\frac{1}{2}(\\sin 30)\\left(420\\sqrt{3} - \\frac{\\sqrt{3}}{2}x\\right)(840-x)} &= \\frac{8}{9} \\\\ \\frac{x^2}{\\left(420 - \\frac{x}{2}\\right)(840-x)} &= \\frac{8}{9}. \\end{align*} To make further calculations easier, we scale everything down by $420$ (while keeping the same variable names, so keep that in mind). \\begin{align*} \\frac{x^2}{\\left(1-\\frac{x}{2}\\right)(2-x)} &= \\frac{8}{9} \\\\ 8\\left(1-\\frac{x}{2}\\right)(2-x) &= 9x^2 \\\\ 16-16x + 4x^2 &= 9x^2 \\\\ 5x^2 + 16x -16 &= 0 \\\\ (5x-4)(x+4) &= 0. \\end{align*} Thus $x = \\frac{4}{5}.$ Because we scaled down everything by $420,$ the actual value of $AF$ is $\\frac{4}{5}(420) = 336 ~JimY",
"$\\angle AFE = \\angle AEF = \\angle EAF = 60^{0} \\Rightarrow \\angle AFG = 120^{0}$ So, If $\\Delta AFG$ is isosceles, it means that $AF = FG$. Let $AF = FG = AE = EF = x$ So, $[\\Delta AFG] = \\frac{1}{2} \\cdot x^{2} \\textup{sin} 120^{0} = \\frac{\\sqrt{3}}{4}x^{2}$ In $\\Delta BED$, $BE = 840 - x$, Hence $DE = \\frac{840 - x}{2}$ (because $\\textup{sin} 30^{0} = \\frac{1}{2}$) Therefore, $[\\Delta BED] = \\frac{1}{2} (840 - x) \\left (\\frac{840-x}{2} \\right) \\textup{sin} 60^{0}$ So, $[\\Delta BED] = \\frac{\\sqrt{3}}{4} (840 - x) \\left (\\frac{840-x}{2} \\right) = \\frac{\\sqrt{3}}{8} (840 - x)^{2}$ Now, as we know that the ratio of the areas of $\\Delta AFG$ and $\\Delta BED$ is $8:9$ Substituting the values, we get $\\frac{\\frac{\\sqrt{3}}{4}x^{2}}{\\frac{\\sqrt{3}}{8} (840 - x)^{2}} = \\frac{8}{9} \\Rightarrow \\left (\\frac{x}{840 - x} \\right)^{2} = \\frac{4}{9}$ Hence, $\\frac{x}{840 - x} = \\frac{2}{3}$. Solving this, we easily get $x = 336$ We have taken $AF = x$, Hence, $AF = 336 -Arnav Nigam",
"$\\angle AFE = \\angle AEF = \\angle EAF = 60^{0} \\Rightarrow \\angle AFG = 120^{0}$ So, If $\\Delta AFG$ is isosceles, it means that $AF = FG$. Let $AF = FG = AE = EF = x$ So, $[\\Delta AFG] = \\frac{1}{2} \\cdot x^{2} \\textup{sin} 120^{0} = \\frac{\\sqrt{3}}{4}x^{2}$ In $\\Delta BED$, $BE = 840 - x$, Hence $DE = \\frac{840 - x}{2}$ (because $\\textup{sin} 30^{0} = \\frac{1}{2}$) Therefore, $[\\Delta BED] = \\frac{1}{2} (840 - x) \\left (\\frac{840-x}{2} \\right) \\textup{sin} 60^{0}$ So, $[\\Delta BED] = \\frac{\\sqrt{3}}{4} (840 - x) \\left (\\frac{840-x}{2} \\right) = \\frac{\\sqrt{3}}{8} (840 - x)^{2}$ Now, as we know that the ratio of the areas of $\\Delta AFG$ and $\\Delta BED$ is $8:9$ Substituting the values, we get $\\frac{\\frac{\\sqrt{3}}{4}x^{2}}{\\frac{\\sqrt{3}}{8} (840 - x)^{2}} = \\frac{8}{9} \\Rightarrow \\left (\\frac{x}{840 - x} \\right)^{2} = \\frac{4}{9}$ Hence, $\\frac{x}{840 - x} = \\frac{2}{3}$. Solving this, we easily get $x = 336$ We have taken $AF = x$, Hence, $AF = 336 -Arnav Nigam",
"Since $\\triangle AFG$ is isosceles, $AF = FG$, and since $\\triangle AEF$ is equilateral, $AF = EF$. Thus, $EF = FG$, and since these triangles share an altitude, they must have the same area. Drop perpendiculars from $E$ and $F$ to line $BC$; call the meeting points $P$ and $Q$, respectively. $\\triangle BEP$ is clearly congruent to both $\\triangle BED$ and $\\triangle FQC$, and thus each of these new triangles has the same area as $\\triangle BED$. But we can \"slide\" $\\triangle BEP$ over to make it adjacent to $\\triangle FQC$, thus creating an equilateral triangle whose area has a ratio of $18:8$ when compared to $\\triangle AEF$ (based on our conclusion from the first paragraph). Since these triangles are both equilateral, they are similar, and since the area ratio $18:8$ reduces to $9:4$, the ratio of their sides must be $3:2$. So, because $FC$ and $AF$ represent sides of these triangles, and they add to $840$, $AF$ must equal two-fifths of $840$, or $336.",
"Since $\\triangle AFG$ is isosceles, $AF = FG$, and since $\\triangle AEF$ is equilateral, $AF = EF$. Thus, $EF = FG$, and since these triangles share an altitude, they must have the same area. Drop perpendiculars from $E$ and $F$ to line $BC$; call the meeting points $P$ and $Q$, respectively. $\\triangle BEP$ is clearly congruent to both $\\triangle BED$ and $\\triangle FQC$, and thus each of these new triangles has the same area as $\\triangle BED$. But we can \"slide\" $\\triangle BEP$ over to make it adjacent to $\\triangle FQC$, thus creating an equilateral triangle whose area has a ratio of $18:8$ when compared to $\\triangle AEF$ (based on our conclusion from the first paragraph). Since these triangles are both equilateral, they are similar, and since the area ratio $18:8$ reduces to $9:4$, the ratio of their sides must be $3:2$. So, because $FC$ and $AF$ represent sides of these triangles, and they add to $840$, $AF$ must equal two-fifths of $840$, or $336."
] |
2021-II-3
| 2,021
| 3
|
Find the number of permutations $x_1, x_2, x_3, x_4, x_5$ of numbers $1, 2, 3, 4, 5$ such that the sum of five products \[x_1x_2x_3 + x_2x_3x_4 + x_3x_4x_5 + x_4x_5x_1 + x_5x_1x_2\] is divisible by $3$ .
| 80
|
II
|
[
"Since $3$ is one of the numbers, a product with a $3$ in it is automatically divisible by $3,$ so WLOG $x_3=3,$ we will multiply by $5$ afterward since any of $x_1, x_2, \\ldots, x_5$ would be $3,$ after some cancelation we see that now all we need to find is the number of ways that $x_5x_1(x_4+x_2)$ is divisible by $3,$ since $x_5x_1$ is never divisible by $3,$ now we just need to find the number of ways $x_4+x_2$ is divisible by $3.$ Note that $x_2$ and $x_4$ can be $(1, 2), (2, 1), (1, 5), (5, 1), (2, 4), (4, 2), (4, 5),$ or $(5, 4).$ We have $2$ ways to designate $x_1$ and $x_5$ for a total of $8 \\cdot 2 = 16.$ So the desired answer is $16 \\cdot 5=080 ~math31415926535 ~MathFun1000 (Rephrasing for clarity)",
"The expression $x_1x_2x_3 + x_2x_3x_4 + x_3x_4x_5 + x_4x_5x_1 + x_5x_1x_2$ has cyclic symmetry. Without the loss of generality, let $x_1=3.$ It follows that $\\{x_2,x_3,x_4,x_5\\}=\\{1,2,4,5\\}.$ We have: $x_1x_2x_3 + x_2x_3x_4 + x_3x_4x_5 + x_4x_5x_1 + x_5x_1x_2\\equiv x_2x_3x_4 + x_3x_4x_5\\pmod{3}.$ $x_2,x_3,x_4,x_5$ are congruent to $1,2,1,2\\pmod{3}$ in some order. We construct the following table for the case $x_1=3,$ with all values in modulo $3:$ \\[\\begin{array}{c||c|c|c|c|c||c} & & & & & & \\\\ [-2.5ex] \\textbf{Row} & \\boldsymbol{x_2} & \\boldsymbol{x_3} & \\boldsymbol{x_4} & \\boldsymbol{x_5} & \\boldsymbol{x_2x_3x_4 + x_3x_4x_5} & \\textbf{Valid?} \\\\ [0.5ex] \\hline & & & & & & \\\\ [-2ex] 1 & 1 & 1 & 2 & 2 & 0 & \\checkmark \\\\ 2 & 1 & 2 & 1 & 2 & 0 & \\checkmark \\\\ 3 & 1 & 2 & 2 & 1 & 2 & \\\\ 4 & 2 & 1 & 1 & 2 & 1 & \\\\ 5 & 2 & 1 & 2 & 1 & 0 & \\checkmark \\\\ 6 & 2 & 2 & 1 & 1 & 0 & \\checkmark \\end{array}\\] For Row 1, $(x_2,x_3)$ can be either $(1,4)$ or $(4,1),$ and $(x_4,x_5)$ can be either $(2,5)$ or $(5,2).$ By the Multiplication Principle, Row 1 produces $2\\cdot2=4$ permutations. Similarly, Rows 2, 5, and 6 each produce $4$ permutations. Together, we get $4\\cdot4=16$ permutations for the case $x_1=3.$ By the cyclic symmetry, the cases $x_2=3, x_3=3, x_4=3,$ and $x_5=3$ all have the same count. Therefore, the total number of permutations $x_1, x_2, x_3, x_4, x_5$ is $16\\cdot5=080 ~MRENTHUSIASM",
"The expression $x_1x_2x_3 + x_2x_3x_4 + x_3x_4x_5 + x_4x_5x_1 + x_5x_1x_2$ has cyclic symmetry. Without the loss of generality, let $x_1=3.$ It follows that $\\{x_2,x_3,x_4,x_5\\}=\\{1,2,4,5\\}.$ We have: $x_1x_2x_3 + x_2x_3x_4 + x_3x_4x_5 + x_4x_5x_1 + x_5x_1x_2\\equiv x_2x_3x_4 + x_3x_4x_5\\pmod{3}.$ $x_2,x_3,x_4,x_5$ are congruent to $1,2,1,2\\pmod{3}$ in some order. We construct the following table for the case $x_1=3,$ with all values in modulo $3:$ \\[\\begin{array}{c||c|c|c|c|c||c} & & & & & & \\\\ [-2.5ex] \\textbf{Row} & \\boldsymbol{x_2} & \\boldsymbol{x_3} & \\boldsymbol{x_4} & \\boldsymbol{x_5} & \\boldsymbol{x_2x_3x_4 + x_3x_4x_5} & \\textbf{Valid?} \\\\ [0.5ex] \\hline & & & & & & \\\\ [-2ex] 1 & 1 & 1 & 2 & 2 & 0 & \\checkmark \\\\ 2 & 1 & 2 & 1 & 2 & 0 & \\checkmark \\\\ 3 & 1 & 2 & 2 & 1 & 2 & \\\\ 4 & 2 & 1 & 1 & 2 & 1 & \\\\ 5 & 2 & 1 & 2 & 1 & 0 & \\checkmark \\\\ 6 & 2 & 2 & 1 & 1 & 0 & \\checkmark \\end{array}\\] For Row 1, $(x_2,x_3)$ can be either $(1,4)$ or $(4,1),$ and $(x_4,x_5)$ can be either $(2,5)$ or $(5,2).$ By the Multiplication Principle, Row 1 produces $2\\cdot2=4$ permutations. Similarly, Rows 2, 5, and 6 each produce $4$ permutations. Together, we get $4\\cdot4=16$ permutations for the case $x_1=3.$ By the cyclic symmetry, the cases $x_2=3, x_3=3, x_4=3,$ and $x_5=3$ all have the same count. Therefore, the total number of permutations $x_1, x_2, x_3, x_4, x_5$ is $16\\cdot5=080 ~MRENTHUSIASM",
"WLOG, let $x_{3} = 3$ So, the terms $x_{1}x_{2}x_{3}, x_{2}x_{3}x_{4},x_{3}x_{4}x_{5}$ are divisible by $3$. We are left with $x_{4}x_{5}x_{1}$ and $x_{5}x_{1}x_{2}$. We need $x_{4}x_{5}x_{1} + x_{5}x_{1}x_{2} \\equiv 0 \\pmod{3}$. The only way is when They are $(+1,-1)$ or $(-1, +1) \\pmod{3}$. The numbers left with us are $1,2,4,5$ which are $+1,-1,+1,-1\\pmod{3}$ respectively. $+1$ (of $x_{4}x_{5}x_{1}$ or $x_{5}x_{1}x_{2}$) $= +1 \\cdot +1 \\cdot +1$ $\\;\\;\\; OR \\;\\;\\;+1$ (of $x_{4}x_{5}x_{1}$ or $x_{5}x_{1}x_{2}$) = $-1 \\cdot -1 \\cdot +1$. $-1$ (of $x_{4}x_{5}x_{1}$ or $x_{5}x_{1}x_{2}$) $= -1 \\cdot -1 \\cdot -1$ $\\;\\;\\; OR \\;\\;\\;-1$ (of $x_{4}x_{5}x_{1}$ or $x_{5}x_{1}x_{2}$) = $-1 \\cdot +1 \\cdot +1$ But, as we have just two $+1's$ and two $-1's$. Hence, We will have to take $+1 = +1 \\cdot -1 \\cdot -1$ and $-1 = -1 \\cdot +1 \\cdot +1$. Among these two, we have a $+1$ and $-1$ in common, i.e. $(x_{5}, x_{1}) = (+1, -1) or (-1, +1)$ (because $x_{1}$ and $x_{5}$. are common in $x_{4}x_{5}x_{1}$ and $x_{5}x_{1}x_{2}$). So, $(x_{5}, x_{1}) \\in {(1,2), (1,5), (4,2), (4,5), (2,1), (5,1), (2,4), (5,4)}$ i.e. $8$ values. For each value of $(x_{5}, x_{1})$ we get $2$ values for $(x_{2}, x_{4})$. Hence, in total, we have $8 \\times 2 = 16$ ways. But any of the $x_{i} 's$ can be $3$. So, $16 \\times 5 = 080. -Arnav Nigam",
"WLOG, let $x_{3} = 3$. Then: \\[x_{1}x_{2}x_{3} + x_{2}x_{3}x_{4} + x_{3}x_{4}x_{5} + x_{4}x_{5}x_{1} + x_{5}x_{1}x_{2} = 3 (x_1 x_2 + x_2 x_4 + x_4 x_5) + x_5 x_1 (x_2 + x_4).\\] The sum is divisible by $3$ if and only if $x_2 + x_4$ is divisible by $3$. The possible sums of $x_2 + x_4$ are $1 + 2, 1 + 4, 1 + 5, 2 + 4, 2 + 5, 4 + 5.$ Two of them are not multiples of $3$, but four of them are multiples. A total number of permutations is $5! = 120.$ $\\frac {2}{3}$ of this number, that is, $80,$ give sums that are multiples of $3.$ vladimir.shelomovskii@gmail.com, vvsss",
"WLOG, let $x_{3} = 3$. Then: \\[x_{1}x_{2}x_{3} + x_{2}x_{3}x_{4} + x_{3}x_{4}x_{5} + x_{4}x_{5}x_{1} + x_{5}x_{1}x_{2} = 3 (x_1 x_2 + x_2 x_4 + x_4 x_5) + x_5 x_1 (x_2 + x_4).\\] The sum is divisible by $3$ if and only if $x_2 + x_4$ is divisible by $3$. The possible sums of $x_2 + x_4$ are $1 + 2, 1 + 4, 1 + 5, 2 + 4, 2 + 5, 4 + 5.$ Two of them are not multiples of $3$, but four of them are multiples. A total number of permutations is $5! = 120.$ $\\frac {2}{3}$ of this number, that is, $80,$ give sums that are multiples of $3.$ vladimir.shelomovskii@gmail.com, vvsss",
"This is my first time doing a solution (feel free to edit it) We have \\[x_1x_2x_3 + x_2x_3x_4 + x_3x_4x_5 + x_4x_5x_1 + x_5x_1x_2.\\] We have $5$ numbers. Considering any $x$ as $3,$ we see that we are left with two terms that are not always divisible by $3,$ which means that already gives us 5 options. Let's now consider $x_1 =3:$We are left with \\[3x_2x_3 + x_2x_3x_4 + x_3x_4x_5 + 3x_4x_5 + 3x_5x_2.\\] The two terms left over are \\[x_2x_3x_4 + x_3x_4x_5 \\equiv 0 \\pmod{3}\\] since we already have used $3$ the remaining numbers are $1,2,4,5.$ We now factor \\begin{align*} (x_2 + x_5)(x_3x_4) &\\equiv 0 \\pmod{3} \\\\ (x_2 + x_5) &\\equiv 0 \\pmod{3} \\end{align*} since $1,2,4,5$ are all not factors of $3.$ Now using the number $1,2,4,5,$ we take two to get a number divisible by $3$ for $(x_2 + x_5):$ \\begin{align*} 1+5 &\\equiv 0 \\pmod{3}, \\\\ 4+2 &\\equiv 0 \\pmod{3}, \\\\ 4+5 &\\equiv 0 \\pmod{3}, \\\\ 1+2 &\\equiv 0 \\pmod{3}. \\end{align*} We have $4$ possibilities from above. Since we can also have $5+1$ or $2+4,$ there are $4\\cdot2=8$ possibilities in all. Now using \\[(x_2 + x_5)(x_3x_4) \\equiv 0 \\pmod{3},\\] we have $(x_3x_4),$ which results in $8$ more possibilities of $2$ times more. So, we get $2\\cdot2\\cdot4=16.$ Remember that $3$ can be any of $5$ different variables. So, we multiply by $5$ to get the answer $080",
"This is my first time doing a solution (feel free to edit it) We have \\[x_1x_2x_3 + x_2x_3x_4 + x_3x_4x_5 + x_4x_5x_1 + x_5x_1x_2.\\] We have $5$ numbers. Considering any $x$ as $3,$ we see that we are left with two terms that are not always divisible by $3,$ which means that already gives us 5 options. Let's now consider $x_1 =3:$We are left with \\[3x_2x_3 + x_2x_3x_4 + x_3x_4x_5 + 3x_4x_5 + 3x_5x_2.\\] The two terms left over are \\[x_2x_3x_4 + x_3x_4x_5 \\equiv 0 \\pmod{3}\\] since we already have used $3$ the remaining numbers are $1,2,4,5.$ We now factor \\begin{align*} (x_2 + x_5)(x_3x_4) &\\equiv 0 \\pmod{3} \\\\ (x_2 + x_5) &\\equiv 0 \\pmod{3} \\end{align*} since $1,2,4,5$ are all not factors of $3.$ Now using the number $1,2,4,5,$ we take two to get a number divisible by $3$ for $(x_2 + x_5):$ \\begin{align*} 1+5 &\\equiv 0 \\pmod{3}, \\\\ 4+2 &\\equiv 0 \\pmod{3}, \\\\ 4+5 &\\equiv 0 \\pmod{3}, \\\\ 1+2 &\\equiv 0 \\pmod{3}. \\end{align*} We have $4$ possibilities from above. Since we can also have $5+1$ or $2+4,$ there are $4\\cdot2=8$ possibilities in all. Now using \\[(x_2 + x_5)(x_3x_4) \\equiv 0 \\pmod{3},\\] we have $(x_3x_4),$ which results in $8$ more possibilities of $2$ times more. So, we get $2\\cdot2\\cdot4=16.$ Remember that $3$ can be any of $5$ different variables. So, we multiply by $5$ to get the answer $080"
] |
2021-II-4
| 2,021
| 4
|
There are real numbers $a, b, c,$ and $d$ such that $-20$ is a root of $x^3 + ax + b$ and $-21$ is a root of $x^3 + cx^2 + d.$ These two polynomials share a complex root $m + \sqrt{n} \cdot i,$ where $m$ and $n$ are positive integers and $i = \sqrt{-1}.$ Find $m+n.$
| 330
|
II
|
[
"By the Complex Conjugate Root Theorem, the imaginary roots for each of $x^3+ax+b$ and $x^3+cx^2+d$ are complex conjugates. Let $z=m+\\sqrt{n}\\cdot i$ and $\\overline{z}=m-\\sqrt{n}\\cdot i.$ It follows that the roots of $x^3+ax+b$ are $-20,z,\\overline{z},$ and the roots of $x^3+cx^2+d$ are $-21,z,\\overline{z}.$ We know that \\begin{align*} z+\\overline{z}&=2m, & (1) \\\\ z\\overline{z}&=m^2+n. & (2) \\end{align*} Applying Vieta's Formulas to $x^3+ax+b,$ we have $-20+z+\\overline{z}=0.$ Substituting $(1)$ into this equation, we get $m=10.$ Applying Vieta's Formulas to $x^3+cx^2+d,$ we have $-21z-21\\overline{z}+z\\overline{z}=0,$ or $-21(z+\\overline{z})+z\\overline{z}=0.$ Substituting $(1)$ and $(2)$ into this equation, we get $n=320.$ Finally, the answer is $m+n=330 ~MRENTHUSIASM",
"By the Complex Conjugate Root Theorem, the imaginary roots for each of $x^3+ax+b$ and $x^3+cx^2+d$ are complex conjugates. Let $z=m+\\sqrt{n}\\cdot i$ and $\\overline{z}=m-\\sqrt{n}\\cdot i.$ It follows that the roots of $x^3+ax+b$ are $-20,z,\\overline{z},$ and the roots of $x^3+cx^2+d$ are $-21,z,\\overline{z}.$ We know that \\begin{align*} z+\\overline{z}&=2m, & (1) \\\\ z\\overline{z}&=m^2+n. & (2) \\end{align*} Applying Vieta's Formulas to $x^3+ax+b,$ we have $-20+z+\\overline{z}=0.$ Substituting $(1)$ into this equation, we get $m=10.$ Applying Vieta's Formulas to $x^3+cx^2+d,$ we have $-21z-21\\overline{z}+z\\overline{z}=0,$ or $-21(z+\\overline{z})+z\\overline{z}=0.$ Substituting $(1)$ and $(2)$ into this equation, we get $n=320.$ Finally, the answer is $m+n=330 ~MRENTHUSIASM",
"$(-20)^{3} + (-20)a + b = 0$, hence $-20a + b = 8000$ Also, $(-21)^{3} + c(-21)^{2} + d = 0$, hence $441c + d = 9261$ $m + i \\sqrt{n}$ satisfies both $\\Rightarrow$ we can put it in both equations and equate to 0. In the first equation, we get $(m + i \\sqrt{n})^{3} + a(m + i \\sqrt{n}) + b = 0$ Simplifying this further, we get $(m^{3} - 3mn + am + b) + i(3m^{2} \\sqrt{n} - n\\sqrt{n} + a\\sqrt{n}) = 0$ Hence, $m^{3} - 3mn + am + b = 0$ and $3m^{2} \\sqrt{n} - n\\sqrt{n} + a\\sqrt{n} = 0 \\Rightarrow 3m^{2} - n + a = 0 \\rightarrow (1)$ In the second equation, we get $(m + i \\sqrt{n})^{3} + c(m + i \\sqrt{n})^{2} + d = 0$ Simplifying this further, we get $(m^{3} + m^{2}c - nc - 3mn + d) + i(3m^{2} \\sqrt{n} - n\\sqrt{n} + 2mc\\sqrt{n}) = 0$ Hence, $m^{3} + m^{2}c - nc - 3mn + d = 0$ and $3m^{2} \\sqrt{n} - n\\sqrt{n} + 2mc\\sqrt{n} = 0 \\Rightarrow 3m^{2} - n + 2mc = 0 \\rightarrow (2)$ Comparing (1) and (2), $a = 2mc$ and $am + b = m^{2}c - nc + d \\rightarrow (3)$ $b = 8000 + 20a \\Rightarrow b = 40mc + 8000$; $d = 9261 - 441c$ Substituting these in $(3)$ gives, $2m^{2}c + 8000 + 40mc = m^{2}c - nc + 9261 - 441c$ This simplifies to $m^{2}c + nc + 40mc + 441c = 1261 \\Rightarrow c(m^{2} + n + 40m + 441) = 1261$ Hence, $c|1261 \\Rightarrow c \\in {1,13,97,1261}$ Consider case of $c = 1$: $c = 1 \\Rightarrow d = 8820$ Also, $a = 2m, b = 8000 + 40m$ $am + b = m^{2} - n + 8820$ (because c = 1) Also, $m^{2} + n + 40m = 820 \\rightarrow (4)$ Also, Equation (2) gives $3m^{2} - n + 2m = 0 \\rightarrow (5)$ Solving (4) and (5) simultaneously gives $m = 10, n = 320$ [AIME can not have more than one answer, so we can stop here ... Not suitable for objective exam] Hence, $m + n = 10 + 320 = 330 -Arnav Nigam",
"$(-20)^{3} + (-20)a + b = 0$, hence $-20a + b = 8000$ Also, $(-21)^{3} + c(-21)^{2} + d = 0$, hence $441c + d = 9261$ $m + i \\sqrt{n}$ satisfies both $\\Rightarrow$ we can put it in both equations and equate to 0. In the first equation, we get $(m + i \\sqrt{n})^{3} + a(m + i \\sqrt{n}) + b = 0$ Simplifying this further, we get $(m^{3} - 3mn + am + b) + i(3m^{2} \\sqrt{n} - n\\sqrt{n} + a\\sqrt{n}) = 0$ Hence, $m^{3} - 3mn + am + b = 0$ and $3m^{2} \\sqrt{n} - n\\sqrt{n} + a\\sqrt{n} = 0 \\Rightarrow 3m^{2} - n + a = 0 \\rightarrow (1)$ In the second equation, we get $(m + i \\sqrt{n})^{3} + c(m + i \\sqrt{n})^{2} + d = 0$ Simplifying this further, we get $(m^{3} + m^{2}c - nc - 3mn + d) + i(3m^{2} \\sqrt{n} - n\\sqrt{n} + 2mc\\sqrt{n}) = 0$ Hence, $m^{3} + m^{2}c - nc - 3mn + d = 0$ and $3m^{2} \\sqrt{n} - n\\sqrt{n} + 2mc\\sqrt{n} = 0 \\Rightarrow 3m^{2} - n + 2mc = 0 \\rightarrow (2)$ Comparing (1) and (2), $a = 2mc$ and $am + b = m^{2}c - nc + d \\rightarrow (3)$ $b = 8000 + 20a \\Rightarrow b = 40mc + 8000$; $d = 9261 - 441c$ Substituting these in $(3)$ gives, $2m^{2}c + 8000 + 40mc = m^{2}c - nc + 9261 - 441c$ This simplifies to $m^{2}c + nc + 40mc + 441c = 1261 \\Rightarrow c(m^{2} + n + 40m + 441) = 1261$ Hence, $c|1261 \\Rightarrow c \\in {1,13,97,1261}$ Consider case of $c = 1$: $c = 1 \\Rightarrow d = 8820$ Also, $a = 2m, b = 8000 + 40m$ $am + b = m^{2} - n + 8820$ (because c = 1) Also, $m^{2} + n + 40m = 820 \\rightarrow (4)$ Also, Equation (2) gives $3m^{2} - n + 2m = 0 \\rightarrow (5)$ Solving (4) and (5) simultaneously gives $m = 10, n = 320$ [AIME can not have more than one answer, so we can stop here ... Not suitable for objective exam] Hence, $m + n = 10 + 320 = 330 -Arnav Nigam",
"We start off by applying Vieta's, and we find that $a=m^2+n-40m$, $b=20m^2+20n$, $c=21-2m$, and $d=21m^2+21n$. After that, we use the fact that $-20$ and $-21$ are roots of $x^3+ax+b$ and $x^3+cx^2+d$, respectively. Since substituting the roots back into the function returns zero, we have that $(-20)^3-20a+b=0$ and $(-21)^3+c\\cdot (-21)^2+d=0$. Setting these two equations equal to each other while also substituting the values of $a$, $b$, $c$, and $d$ above gives us $21m^2+21n-1682m+8000=0$. We then rearrange the equation into $21n = -21m^2+1682m-8000$. With this property, we know that $-21m^2+1682m-8000$ is divisible by $21$, so $1682m-8000=0 \\pmod{21}$. This results in $2m-20=0 \\pmod{21}$, which finally gives us $m=10 \\pmod{21}$. We can test the first obvious value of $m$, which is $10$, to find that this works, and we get $m=10$ and $n=320$. Therefore, the answer is $m + n = 10 + 320 = 330 ~Jske25 ~sidkris (formatting edits)",
"We start off by applying Vieta's, and we find that $a=m^2+n-40m$, $b=20m^2+20n$, $c=21-2m$, and $d=21m^2+21n$. After that, we use the fact that $-20$ and $-21$ are roots of $x^3+ax+b$ and $x^3+cx^2+d$, respectively. Since substituting the roots back into the function returns zero, we have that $(-20)^3-20a+b=0$ and $(-21)^3+c\\cdot (-21)^2+d=0$. Setting these two equations equal to each other while also substituting the values of $a$, $b$, $c$, and $d$ above gives us $21m^2+21n-1682m+8000=0$. We then rearrange the equation into $21n = -21m^2+1682m-8000$. With this property, we know that $-21m^2+1682m-8000$ is divisible by $21$, so $1682m-8000=0 \\pmod{21}$. This results in $2m-20=0 \\pmod{21}$, which finally gives us $m=10 \\pmod{21}$. We can test the first obvious value of $m$, which is $10$, to find that this works, and we get $m=10$ and $n=320$. Therefore, the answer is $m + n = 10 + 320 = 330 ~Jske25 ~sidkris (formatting edits)",
"We note that $x^3 + ax + b = (x+20)P(x)$ and $x^3 + cx^2 + d = (x+21)Q(x)$ for some polynomials $P(x)$ and $Q(x)$. Through synthetic division (ignoring the remainder as we can set $b$ and $d$ to constant values such that the remainder is zero), $P(x) = x^2 - 20x + (400+a)$, and $Q(x) = x^2 + (c-21)x + (441 - 21c)$. By the complex conjugate root theorem, we know that $P(x)$ and $Q(x)$ share the same roots, and they share the same leading coefficient, so $P(x) = Q(x)$. Therefore, $c-21 = -20$ and $441-21c = 400 + a$. Solving the system of equations, we get $a = 20$ and $c = 1$, so $P(x) = Q(x) = x^2 - 20x + 420$. Finally, by the quadratic formula, we have roots of $\\frac{20 \\pm \\sqrt{400 - 1680}}{2} = 10 \\pm \\sqrt{320}i$, so our final answer is $10 + 320 = 330 -faefeyfa",
"We note that $x^3 + ax + b = (x+20)P(x)$ and $x^3 + cx^2 + d = (x+21)Q(x)$ for some polynomials $P(x)$ and $Q(x)$. Through synthetic division (ignoring the remainder as we can set $b$ and $d$ to constant values such that the remainder is zero), $P(x) = x^2 - 20x + (400+a)$, and $Q(x) = x^2 + (c-21)x + (441 - 21c)$. By the complex conjugate root theorem, we know that $P(x)$ and $Q(x)$ share the same roots, and they share the same leading coefficient, so $P(x) = Q(x)$. Therefore, $c-21 = -20$ and $441-21c = 400 + a$. Solving the system of equations, we get $a = 20$ and $c = 1$, so $P(x) = Q(x) = x^2 - 20x + 420$. Finally, by the quadratic formula, we have roots of $\\frac{20 \\pm \\sqrt{400 - 1680}}{2} = 10 \\pm \\sqrt{320}i$, so our final answer is $10 + 320 = 330 -faefeyfa",
"We plug -20 into the equation obtaining $(-20)^3-20a+b$, likewise, plugging -21 into the second equation gets $(-21)^3+441c+d$. Both equations must have 3 solutions exactly, so the other two solutions must be $m + \\sqrt{n} \\cdot i$ and $m - \\sqrt{n} \\cdot i$. By Vieta's, the sum of the roots in the first equation is $0$, so $m$ must be $10$. Next, using Vieta's theorem on the second equation, you get $x1x2+x2x3+x1x3 = 0$. However, since we know that the sum of the roots with complex numbers are 20, we can factor out the terms with -21, so $-21*(20)+(m^2+n)=0$. Given that $m$ is $10$, then $n$ is equal to $320$. Therefore, the answer to the equation is $330",
"We plug -20 into the equation obtaining $(-20)^3-20a+b$, likewise, plugging -21 into the second equation gets $(-21)^3+441c+d$. Both equations must have 3 solutions exactly, so the other two solutions must be $m + \\sqrt{n} \\cdot i$ and $m - \\sqrt{n} \\cdot i$. By Vieta's, the sum of the roots in the first equation is $0$, so $m$ must be $10$. Next, using Vieta's theorem on the second equation, you get $x1x2+x2x3+x1x3 = 0$. However, since we know that the sum of the roots with complex numbers are 20, we can factor out the terms with -21, so $-21*(20)+(m^2+n)=0$. Given that $m$ is $10$, then $n$ is equal to $320$. Therefore, the answer to the equation is $330",
"Since $m+i\\sqrt{n}$ is a common root and all the coefficients are real, $m-i\\sqrt{n}$ must be a common root, too. Now that we know all three roots of both polynomials, we can match coefficients (or more specifically, the zero coefficients). First, however, the product of the two common roots is: \\begin{align*} &&&(x-m-i\\sqrt{n})(x-m+i\\sqrt{n})\\\\ &=&&x^2-x(m+i\\sqrt{n}+m-i\\sqrt{n})+(m+i\\sqrt{n})(m-i\\sqrt{n})\\\\ &=&&x^2-2xm+(m^2-i^2n)\\\\ &=&&x^2-2xm+m^2+n \\end{align*} Now, let's equate the two forms of both the polynomials: \\[x^3+ax+b=(x^2-2xm+m^2+n)(x+20)\\] \\[x^3+cx^2+d=(x^2-2xm+m^2+n)(x+21)\\] Now we can match the zero coefficients. \\[-2m+20=0\\to m=10\\text{ and}\\] \\[-42m+m^2+n=0\\to-420+100+n=0\\to n=320\\text{.}\\] Thus, $m+n=10+320=330.",
"Since $m+i\\sqrt{n}$ is a common root and all the coefficients are real, $m-i\\sqrt{n}$ must be a common root, too. Now that we know all three roots of both polynomials, we can match coefficients (or more specifically, the zero coefficients). First, however, the product of the two common roots is: \\begin{align*} &&&(x-m-i\\sqrt{n})(x-m+i\\sqrt{n})\\\\ &=&&x^2-x(m+i\\sqrt{n}+m-i\\sqrt{n})+(m+i\\sqrt{n})(m-i\\sqrt{n})\\\\ &=&&x^2-2xm+(m^2-i^2n)\\\\ &=&&x^2-2xm+m^2+n \\end{align*} Now, let's equate the two forms of both the polynomials: \\[x^3+ax+b=(x^2-2xm+m^2+n)(x+20)\\] \\[x^3+cx^2+d=(x^2-2xm+m^2+n)(x+21)\\] Now we can match the zero coefficients. \\[-2m+20=0\\to m=10\\text{ and}\\] \\[-42m+m^2+n=0\\to-420+100+n=0\\to n=320\\text{.}\\] Thus, $m+n=10+320=330."
] |
2021-II-5
| 2,021
| 5
|
For positive real numbers $s$ , let $\tau(s)$ denote the set of all obtuse triangles that have area $s$ and two sides with lengths $4$ and $10$ . The set of all $s$ for which $\tau(s)$ is nonempty, but all triangles in $\tau(s)$ are congruent, is an interval $[a,b)$ . Find $a^2+b^2$ .
| 736
|
II
|
[
"We start by defining a triangle. The two small sides MUST add to a larger sum than the long side. We are given $4$ and $10$ as the sides, so we know that the $3$rd side is between $6$ and $14$, exclusive. We also have to consider the word OBTUSE triangles. That means that the two small sides squared is less than the $3$rd side. So the triangles' sides are between $6$ and $\\sqrt{84}$ exclusive, and the larger bound is between $\\sqrt{116}$ and $14$, exclusive. The area of these triangles are from $0$ (straight line) to $2\\sqrt{84}$ on the first \"small bound\" and the larger bound is between $0$ and $20$. $0 < s < 2\\sqrt{84}$ is our first equation, and $0 < s < 20$ is our $2$nd equation. Therefore, the area is between $\\sqrt{336}$ and $\\sqrt{400}$, so our final answer is $736. ~ARCTICTURN",
"If $a,b,$ and $c$ are the side-lengths of an obtuse triangle with $a\\leq b\\leq c,$ then both of the following must be satisfied: Triangle Inequality Theorem: $a+b>c$ Pythagorean Inequality Theorem: $a^2+b^2<c^2$ For one such obtuse triangle, let $4,10,$ and $x$ be its side-lengths and $K$ be its area. We apply casework to its longest side: Case (1): The longest side has length $\\boldsymbol{10,}$ so $\\boldsymbol{0<x<10.}$ By the Triangle Inequality Theorem, we have $4+x>10,$ from which $x>6.$ By the Pythagorean Inequality Theorem, we have $4^2+x^2<10^2,$ from which $x<\\sqrt{84}.$ Taking the intersection produces $6<x<\\sqrt{84}$ for this case. At $x=6,$ the obtuse triangle degenerates into a straight line with area $K=0;$ at $x=\\sqrt{84},$ the obtuse triangle degenerates into a right triangle with area $K=\\frac12\\cdot4\\cdot\\sqrt{84}=2\\sqrt{84}.$ Together, we obtain $0<K<2\\sqrt{84},$ or $K\\in\\left(0,2\\sqrt{84}\\right).$ Case (2): The longest side has length $\\boldsymbol{x,}$ so $\\boldsymbol{x\\geq10.}$ By the Triangle Inequality Theorem, we have $4+10>x,$ from which $x<14.$ By the Pythagorean Inequality Theorem, we have $4^2+10^2<x^2,$ from which $x>\\sqrt{116}.$ Taking the intersection produces $\\sqrt{116}<x<14$ for this case. At $x=14,$ the obtuse triangle degenerates into a straight line with area $K=0;$ at $x=\\sqrt{116},$ the obtuse triangle degenerates into a right triangle with area $K=\\frac12\\cdot4\\cdot10=20.$ Together, we obtain $0<K<20,$ or $K\\in\\left(0,20\\right).$ Answer It is possible for noncongruent obtuse triangles to have the same area. Therefore, all such positive real numbers $s$ are in exactly one of $\\left(0,2\\sqrt{84}\\right)$ or $\\left(0,20\\right).$ By the exclusive disjunction, the set of all such $s$ is \\[[a,b)=\\left(0,2\\sqrt{84}\\right)\\oplus\\left(0,20\\right)=\\left[2\\sqrt{84},20\\right),\\] from which $a^2+b^2=736 ~MRENTHUSIASM",
"If $a,b,$ and $c$ are the side-lengths of an obtuse triangle with $a\\leq b\\leq c,$ then both of the following must be satisfied: Triangle Inequality Theorem: $a+b>c$ Pythagorean Inequality Theorem: $a^2+b^2<c^2$ For one such obtuse triangle, let $4,10,$ and $x$ be its side-lengths and $K$ be its area. We apply casework to its longest side: Case (1): The longest side has length $\\boldsymbol{10,}$ so $\\boldsymbol{0<x<10.}$ By the Triangle Inequality Theorem, we have $4+x>10,$ from which $x>6.$ By the Pythagorean Inequality Theorem, we have $4^2+x^2<10^2,$ from which $x<\\sqrt{84}.$ Taking the intersection produces $6<x<\\sqrt{84}$ for this case. At $x=6,$ the obtuse triangle degenerates into a straight line with area $K=0;$ at $x=\\sqrt{84},$ the obtuse triangle degenerates into a right triangle with area $K=\\frac12\\cdot4\\cdot\\sqrt{84}=2\\sqrt{84}.$ Together, we obtain $0<K<2\\sqrt{84},$ or $K\\in\\left(0,2\\sqrt{84}\\right).$ Case (2): The longest side has length $\\boldsymbol{x,}$ so $\\boldsymbol{x\\geq10.}$ By the Triangle Inequality Theorem, we have $4+10>x,$ from which $x<14.$ By the Pythagorean Inequality Theorem, we have $4^2+10^2<x^2,$ from which $x>\\sqrt{116}.$ Taking the intersection produces $\\sqrt{116}<x<14$ for this case. At $x=14,$ the obtuse triangle degenerates into a straight line with area $K=0;$ at $x=\\sqrt{116},$ the obtuse triangle degenerates into a right triangle with area $K=\\frac12\\cdot4\\cdot10=20.$ Together, we obtain $0<K<20,$ or $K\\in\\left(0,20\\right).$ Answer It is possible for noncongruent obtuse triangles to have the same area. Therefore, all such positive real numbers $s$ are in exactly one of $\\left(0,2\\sqrt{84}\\right)$ or $\\left(0,20\\right).$ By the exclusive disjunction, the set of all such $s$ is \\[[a,b)=\\left(0,2\\sqrt{84}\\right)\\oplus\\left(0,20\\right)=\\left[2\\sqrt{84},20\\right),\\] from which $a^2+b^2=736 ~MRENTHUSIASM",
"We have the diagram below. [asy] draw((0,0)--(1,2*sqrt(3))); draw((1,2*sqrt(3))--(10,0)); draw((10,0)--(0,0)); label(\"$A$\",(0,0),SW); label(\"$B$\",(1,2*sqrt(3)),N); label(\"$C$\",(10,0),SE); label(\"$\\theta$\",(0,0),NE); label(\"$\\alpha$\",(1,2*sqrt(3)),SSE); label(\"$4$\",(0,0)--(1,2*sqrt(3)),WNW); label(\"$10$\",(0,0)--(10,0),S); [/asy] We proceed by taking cases on the angles that can be obtuse, and finding the ranges for $s$ that they yield . If angle $\\theta$ is obtuse, then we have that $s \\in (0,20)$. This is because $s=20$ is attained at $\\theta = 90^{\\circ}$, and the area of the triangle is strictly decreasing as $\\theta$ increases beyond $90^{\\circ}$. This can be observed from \\[s=\\frac{1}{2}(4)(10)\\sin\\theta\\]by noting that $\\sin\\theta$ is decreasing in $\\theta \\in (90^{\\circ},180^{\\circ})$. Then, we note that if $\\alpha$ is obtuse, we have $s \\in (0,4\\sqrt{21})$. This is because we get $x=\\sqrt{10^2-4^2}=\\sqrt{84}=2\\sqrt{21}$ when $\\alpha=90^{\\circ}$, yileding $s=4\\sqrt{21}$. Then, $s$ is decreasing as $\\alpha$ increases by the same argument as before. $\\angle{ACB}$ cannot be obtuse since $AC>AB$. Now we have the intervals $s \\in (0,20)$ and $s \\in (0,4\\sqrt{21})$ for the cases where $\\theta$ and $\\alpha$ are obtuse, respectively. We are looking for the $s$ that are in exactly one of these intervals, and because $4\\sqrt{21}<20$, the desired range is \\[s\\in [4\\sqrt{21},20)\\]giving \\[a^2+b^2=736\\Box\\]",
"Note: Archimedes15 Solution which I added an answer here are two cases. Either the $4$ and $10$ are around an obtuse angle or the $4$ and $10$ are around an acute triangle. If they are around the obtuse angle, the area of that triangle is $<20$ as we have $\\frac{1}{2} \\cdot 40 \\cdot \\sin{\\alpha}$ and $\\sin$ is at most $1$. Note that for the other case, the side lengths around the obtuse angle must be $4$ and $x$ where we have $16+x^2 < 100 \\rightarrow x < 2\\sqrt{21}$. Using the same logic as the other case, the area is at most $4\\sqrt{21}$. Square and add $4\\sqrt{21}$ and $20$ to get the right answer \\[a^2+b^2= 736\\Box\\]",
"For $\\triangle ABC,$ we fix $AB=10$ and $BC=4.$ Without the loss of generality, we consider $C$ on only one side of $\\overline{AB}.$ As shown below, all locations for $C$ at which $\\triangle ABC$ is an obtuse triangle are indicated in red, excluding the endpoints. [asy] /* Made by MRENTHUSIASM */ size(300); pair A, B, O, P, Q, C1, C2, D; A = origin; B = (10,0); O = midpoint(A--B); P = B - (4,0); Q = B + (4,0); C1 = intersectionpoints(D--D+(100,0),Arc(B,Q,P))[1]; C2 = B + (0,4); D = intersectionpoint(Arc(O,B,A),Arc(B,Q,P)); draw(Arc(O,B,A)^^Arc(B,C2,D)^^A--Q); draw(Arc(B,Q,C2)^^Arc(B,D,P),red); dot(\"$A$\", A, 1.5*S, linewidth(4.5)); dot(\"$B$\", B, 1.5*S, linewidth(4.5)); dot(O, linewidth(4.5)); dot(P^^C2^^D^^Q, linewidth(0.8), UnFill); Label L1 = Label(\"$10$\", align=(0,0), position=MidPoint, filltype=Fill(3,0,white)); Label L2 = Label(\"$4$\", align=(0,0), position=MidPoint, filltype=Fill(3,0,white)); draw(A-(0,2)--B-(0,2), L=L1, arrow=Arrows(),bar=Bars(15)); draw(B-(0,2)--Q-(0,2), L=L2, arrow=Arrows(),bar=Bars(15)); label(\"$\\angle C$ obtuse\",(midpoint(Arc(B,D,P)).x,2),2.5*W,red); label(\"$\\angle B$ obtuse\",(midpoint(Arc(B,Q,C2)).x,2),5*E,red); [/asy] Note that: The region in which $\\angle B$ is obtuse is determined by construction. The region in which $\\angle C$ is obtuse is determined by the corollaries of the Inscribed Angle Theorem. For any fixed value of $s,$ the height from $C$ is fixed. We need obtuse $\\triangle ABC$ to be unique, so there can only be one possible location for $C.$ As shown below, all possible locations for $C$ are on minor arc $\\widehat{C_1C_2},$ including $C_1$ but excluding $C_2.$ [asy] /* Made by MRENTHUSIASM */ size(250); pair A, B, O, P, Q, C1, C2, D; A = origin; B = (10,0); O = midpoint(A--B); P = B - (4,0); Q = B + (4,0); C2 = B + (0,4); D = intersectionpoint(Arc(O,B,A),Arc(B,Q,P)); C1 = intersectionpoint(D--D+(100,0),Arc(B,Q,C2)); draw(Arc(O,B,A)^^Arc(B,C2,D)^^A--Q); draw(Arc(B,Q,C1)^^Arc(B,D,P),red); draw(Arc(B,C1,C2),green); draw((A.x,D.y)--(Q.x,D.y),dashed); dot(\"$A$\", A, 1.5*S, linewidth(4.5)); dot(\"$B$\", B, 1.5*S, linewidth(4.5)); dot(\"$D$\", D, 1.5*dir(75), linewidth(0.8), UnFill); dot(\"$C_2$\", C2, 1.5*N, linewidth(4.5)); dot(\"$C_1$\", C1, 1.5*dir(C1-B), linewidth(4.5)); dot(O, linewidth(4.5)); dot(P^^C2^^Q, linewidth(0.8), UnFill); dot(C1, green+linewidth(4.5)); Label L1 = Label(\"$10$\", align=(0,0), position=MidPoint, filltype=Fill(3,0,white)); Label L2 = Label(\"$4$\", align=(0,0), position=MidPoint, filltype=Fill(3,0,white)); draw(A-(0,2)--B-(0,2), L=L1, arrow=Arrows(),bar=Bars(15)); draw(B-(0,2)--Q-(0,2), L=L2, arrow=Arrows(),bar=Bars(15)); [/asy] Let the brackets denote areas: If $C=C_1,$ then $[ABC]$ will be minimized (attainable). By the same base and height and the Inscribed Angle Theorem, we have \\begin{align*} [ABC]&=[ABD] \\\\ &=\\frac12\\cdot BD\\cdot DA \\\\ &=\\frac12\\cdot BD\\cdot \\sqrt{AB^2-BD^2} \\\\ &=\\frac12\\cdot 4\\cdot \\sqrt{10^2-4^2} \\\\ &=2\\sqrt{84}. \\end{align*} If $C=C_2,$ then $[ABC]$ will be maximized (unattainable). For this right triangle, we have \\begin{align*} [ABC]&=\\frac12\\cdot AB\\cdot BC \\\\ &=\\frac12\\cdot 10\\cdot 4 \\\\ &=20. \\end{align*} Finally, the set of all such $s$ is $[a,b)=\\left[2\\sqrt{84},20\\right),$ from which $a^2+b^2=736 ~MRENTHUSIASM (credit given to Snowfan)",
"For $\\triangle ABC,$ we fix $AB=10$ and $BC=4.$ Without the loss of generality, we consider $C$ on only one side of $\\overline{AB}.$ As shown below, all locations for $C$ at which $\\triangle ABC$ is an obtuse triangle are indicated in red, excluding the endpoints. [asy] /* Made by MRENTHUSIASM */ size(300); pair A, B, O, P, Q, C1, C2, D; A = origin; B = (10,0); O = midpoint(A--B); P = B - (4,0); Q = B + (4,0); C1 = intersectionpoints(D--D+(100,0),Arc(B,Q,P))[1]; C2 = B + (0,4); D = intersectionpoint(Arc(O,B,A),Arc(B,Q,P)); draw(Arc(O,B,A)^^Arc(B,C2,D)^^A--Q); draw(Arc(B,Q,C2)^^Arc(B,D,P),red); dot(\"$A$\", A, 1.5*S, linewidth(4.5)); dot(\"$B$\", B, 1.5*S, linewidth(4.5)); dot(O, linewidth(4.5)); dot(P^^C2^^D^^Q, linewidth(0.8), UnFill); Label L1 = Label(\"$10$\", align=(0,0), position=MidPoint, filltype=Fill(3,0,white)); Label L2 = Label(\"$4$\", align=(0,0), position=MidPoint, filltype=Fill(3,0,white)); draw(A-(0,2)--B-(0,2), L=L1, arrow=Arrows(),bar=Bars(15)); draw(B-(0,2)--Q-(0,2), L=L2, arrow=Arrows(),bar=Bars(15)); label(\"$\\angle C$ obtuse\",(midpoint(Arc(B,D,P)).x,2),2.5*W,red); label(\"$\\angle B$ obtuse\",(midpoint(Arc(B,Q,C2)).x,2),5*E,red); [/asy] Note that: The region in which $\\angle B$ is obtuse is determined by construction. The region in which $\\angle C$ is obtuse is determined by the corollaries of the Inscribed Angle Theorem. For any fixed value of $s,$ the height from $C$ is fixed. We need obtuse $\\triangle ABC$ to be unique, so there can only be one possible location for $C.$ As shown below, all possible locations for $C$ are on minor arc $\\widehat{C_1C_2},$ including $C_1$ but excluding $C_2.$ [asy] /* Made by MRENTHUSIASM */ size(250); pair A, B, O, P, Q, C1, C2, D; A = origin; B = (10,0); O = midpoint(A--B); P = B - (4,0); Q = B + (4,0); C2 = B + (0,4); D = intersectionpoint(Arc(O,B,A),Arc(B,Q,P)); C1 = intersectionpoint(D--D+(100,0),Arc(B,Q,C2)); draw(Arc(O,B,A)^^Arc(B,C2,D)^^A--Q); draw(Arc(B,Q,C1)^^Arc(B,D,P),red); draw(Arc(B,C1,C2),green); draw((A.x,D.y)--(Q.x,D.y),dashed); dot(\"$A$\", A, 1.5*S, linewidth(4.5)); dot(\"$B$\", B, 1.5*S, linewidth(4.5)); dot(\"$D$\", D, 1.5*dir(75), linewidth(0.8), UnFill); dot(\"$C_2$\", C2, 1.5*N, linewidth(4.5)); dot(\"$C_1$\", C1, 1.5*dir(C1-B), linewidth(4.5)); dot(O, linewidth(4.5)); dot(P^^C2^^Q, linewidth(0.8), UnFill); dot(C1, green+linewidth(4.5)); Label L1 = Label(\"$10$\", align=(0,0), position=MidPoint, filltype=Fill(3,0,white)); Label L2 = Label(\"$4$\", align=(0,0), position=MidPoint, filltype=Fill(3,0,white)); draw(A-(0,2)--B-(0,2), L=L1, arrow=Arrows(),bar=Bars(15)); draw(B-(0,2)--Q-(0,2), L=L2, arrow=Arrows(),bar=Bars(15)); [/asy] Let the brackets denote areas: If $C=C_1,$ then $[ABC]$ will be minimized (attainable). By the same base and height and the Inscribed Angle Theorem, we have \\begin{align*} [ABC]&=[ABD] \\\\ &=\\frac12\\cdot BD\\cdot DA \\\\ &=\\frac12\\cdot BD\\cdot \\sqrt{AB^2-BD^2} \\\\ &=\\frac12\\cdot 4\\cdot \\sqrt{10^2-4^2} \\\\ &=2\\sqrt{84}. \\end{align*} If $C=C_2,$ then $[ABC]$ will be maximized (unattainable). For this right triangle, we have \\begin{align*} [ABC]&=\\frac12\\cdot AB\\cdot BC \\\\ &=\\frac12\\cdot 10\\cdot 4 \\\\ &=20. \\end{align*} Finally, the set of all such $s$ is $[a,b)=\\left[2\\sqrt{84},20\\right),$ from which $a^2+b^2=736 ~MRENTHUSIASM (credit given to Snowfan)",
"Let a triangle in $\\tau(s)$ be $ABC$, where $AB = 4$ and $BC = 10$. We will proceed with two cases: Case 1: $\\angle ABC$ is obtuse. If $\\angle ABC$ is obtuse, then, if we imagine $AB$ as the base of our triangle, the height can be anything in the range $(0,10)$; therefore, the area of the triangle will fall in the range of $(0, 20)$. Case 2: $\\angle BAC$ is obtuse. Then, if we imagine $AB$ as the base of our triangle, the height can be anything in the range $\\left(0, \\sqrt{10^{2} - 4^{2}}\\right)$. Therefore, the area of the triangle will fall in the range of $\\left(0, 2 \\sqrt{84}\\right)$. If $s < 2 \\sqrt{84}$, there will exist two types of triangles in $\\tau(s)$ - one type with $\\angle ABC$ obtuse; the other type with $\\angle BAC$ obtuse. If $s \\geq 2 \\sqrt{84}$, as we just found, $\\angle BAC$ cannot be obtuse, so therefore, there is only one type of triangle - the one in which $\\angle ABC$ is obtuse. Also, if $s > 20$, no triangle exists with lengths $4$ and $10$. Therefore, $s$ is in the range $\\left[ 2 \\sqrt{84}, 20\\right)$, so our answer is $\\left(2 \\sqrt{84}\\right)^{2} + 20^{2} = 736. Alternatively, refer to Solution 5 for the geometric interpretation. ~ihatemath123",
"Let's rephrase the condition. It is required to find such values of the area of an obtuse triangle with sides $4$ and $10,$ when there is exactly one such obtuse triangle. In the diagram, $AB = 4, AC = 10.$ The largest area of triangle with sides $4$ and $10$ is $20$ for a right triangle with legs $4$ and $10$ ($AC\\perp AB$). The diagram shows triangles with equal heights. The yellow triangle $ABC'$ has the longest side $BC',$ the blue triangle $ABC$ has the longest side $AC.$ If $BC\\perp AB,$ then$BC = \\sqrt {AC^2 – AB^2} = 2 \\sqrt{21}$ the area is equal to $4\\sqrt{21}.$ In the interval, the blue triangle $ABC$ is acute-angled, the yellow triangle $ABC'$ is obtuse-angled. Their heights and areas are equal. The condition is met. If the area is less than $4\\sqrt{21},$ both triangles are obtuse, not equal, so the condition is not met. Therefore, $s$ is in the range $\\left[ 4 \\sqrt{21}, 20\\right)$, so answer is $\\left(4 \\sqrt{21}\\right)^{2} + 20^{2} = 736 vladimir.shelomovskii@gmail.com, vvsss",
"If $4$ and $10$ are the shortest sides and $\\angle C$ is the included angle, then the area is \\[\\frac{4\\cdot10\\cdot\\sin\\angle C}{2} = 20\\sin\\angle C.\\] Because $0\\leq\\sin\\angle C\\leq1$, the maximum value of $20\\sin\\angle C$ is $20$, so $s\\leq20$. If $4$ is a shortest side and $10$ is the longest side, the length of the other short side is $4\\cos\\angle C+2\\sqrt{4\\cos^2 \\angle C+21}$ by law of cosines, and the area is $2\\left(4\\cos\\angle C+2\\sqrt{4\\cos^2\\angle C+21}\\right)\\sqrt{1-\\cos\\angle C}$. Because $-1\\le \\cos\\angle C\\le 0$, this is minimized if $\\cos\\angle C=0$, where $s=4\\sqrt{21}$. So, the answer is $20^2+\\left(4\\sqrt{21}\\right)^2=736. ~ryanbear"
] |
2021-II-6
| 2,021
| 6
|
For any finite set $S$ , let $|S|$ denote the number of elements in $S$ . Find the number of ordered pairs $(A,B)$ such that $A$ and $B$ are (not necessarily distinct) subsets of $\{1,2,3,4,5\}$ that satisfy \[|A| \cdot |B| = |A \cap B| \cdot |A \cup B|\]
| 454
|
II
|
[
"By PIE, $|A|+|B|-|A \\cap B| = |A \\cup B|$. Substituting into the equation and factoring, we get that $(|A| - |A \\cap B|)(|B| - |A \\cap B|) = 0$, so therefore $A \\subseteq B$ or $B \\subseteq A$. WLOG $A\\subseteq B$, then for each element there are $3$ possibilities, either it is in both $A$ and $B$, it is in $B$ but not $A$, or it is in neither $A$ nor $B$. This gives us $3^{5}$ possibilities, and we multiply by $2$ since it could have also been the other way around. Now we need to subtract the overlaps where $A=B$, and this case has $2^{5}=32$ ways that could happen. It is $32$ because each number could be in the subset or it could not be in the subset. So the final answer is $2\\cdot 3^5 - 2^5 = 454. ~math31415926535",
"We denote $\\Omega = \\left\\{ 1 , 2 , 3 , 4 , 5 \\right\\}$. We denote $X = A \\cap B$, $Y = A \\backslash \\left( A \\cap B \\right)$, $Z = B \\backslash \\left( A \\cap B \\right)$, $W = \\Omega \\backslash \\left( A \\cup B \\right)$. Therefore, $X \\cup Y \\cup Z \\cup W = \\Omega$ and the intersection of any two out of sets $X$, $Y$, $Z$, $W$ is an empty set. Therefore, $\\left( X , Y , Z , W \\right)$ is a partition of $\\Omega$. Following from our definition of $X$, $Y$, $Z$, we have $A \\cup B = X \\cup Y \\cup Z$. Therefore, the equation \\[|A| \\cdot |B| = |A \\cap B| \\cdot |A \\cup B|\\] can be equivalently written as \\[\\left( | X | + | Y | \\right) \\left( | X | + | Z | \\right) = | X | \\left( | X | + | Y | + | Z | \\right) .\\] This equality can be simplified as \\[| Y | \\cdot | Z | = 0 .\\] Therefore, we have the following three cases: (1) $|Y| = 0$ and $|Z| \\neq 0$, (2) $|Z| = 0$ and $|Y| \\neq 0$, (3) $|Y| = |Z| = 0$. Next, we analyze each of these cases, separately. Case 1: $|Y| = 0$ and $|Z| \\neq 0$. In this case, to count the number of solutions, we do the complementary counting. First, we count the number of solutions that satisfy $|Y| = 0$. Hence, each number in $\\Omega$ falls into exactly one out of these three sets: $X$, $Z$, $W$. Following from the rule of product, the number of solutions is $3^5$. Second, we count the number of solutions that satisfy $|Y| = 0$ and $|Z| = 0$. Hence, each number in $\\Omega$ falls into exactly one out of these two sets: $X$, $W$. Following from the rule of product, the number of solutions is $2^5$. Therefore, following from the complementary counting, the number of solutions in this case is equal to the number of solutions that satisfy $|Y| = 0$ minus the number of solutions that satisfy $|Y| = 0$ and $|Z| = 0$, i.e., $3^5 - 2^5$. Case 2: $|Z| = 0$ and $|Y| \\neq 0$. This case is symmetric to Case 1. Therefore, the number of solutions in this case is the same as the number of solutions in Case 1, i.e., $3^5 - 2^5$. Case 3: $|Y| = 0$ and $|Z| = 0$. Recall that this is one part of our analysis in Case 1. Hence, the number solutions in this case is $2^5$. By putting all cases together, following from the rule of sum, the total number of solutions is equal to \\begin{align*} \\left( 3^5 - 2^5 \\right) + \\left( 3^5 - 2^5 \\right) + 2^5 & = 2 \\cdot 3^5 - 2^5 \\\\ & = 454 . \\end{align*} ~ Steven Chen (www.professorchenedu.com)",
"By the Principle of Inclusion-Exclusion (abbreviated as PIE), we have $|A \\cup B|=|A|+|B|-|A \\cap B|,$ from which we rewrite the given equation as \\[|A| \\cdot |B| = |A \\cap B| \\cdot \\left(|A|+|B|-|A \\cap B|\\right).\\] Rearranging and applying Simon's Favorite Factoring Trick give \\begin{align*} |A| \\cdot |B| &= |A \\cap B|\\cdot|A| + |A \\cap B|\\cdot|B| - |A \\cap B|^2 \\\\ |A| \\cdot |B| - |A \\cap B|\\cdot|A| - |A \\cap B|\\cdot|B| &= - |A \\cap B|^2 \\\\ \\left(|A| - |A \\cap B|\\right)\\cdot\\left(|B| - |A \\cap B|\\right) &=0, \\end{align*} from which at least one of the following is true: $|A|=|A \\cap B|$ $|B|=|A \\cap B|$ Let $|A \\cap B|=k.$ For each value of $k\\in\\{0,1,2,3,4,5\\},$ we will use PIE to count the ordered pairs $(A,B):$ Suppose $|A|=k.$ There are $\\binom{5}{k}$ ways to choose the elements for $A.$ These $k$ elements must also appear in $B.$ Next, there are $2^{5-k}$ ways to add any number of the remaining $5-k$ elements to $B$ (Each element has $2$ options: in $B$ or not in $B.$). There are $\\binom{5}{k}2^{5-k}$ ordered pairs for $|A|=k.$ Similarly, there are $\\binom{5}{k}2^{5-k}$ ordered pairs for $|B|=k.$ To fix the overcount, we subtract the number of ordered pairs that are counted twice, in which $|A|=|B|=k.$ There are $\\binom{5}{k}$ such ordered pairs. Therefore, there are \\[2\\binom{5}{k}2^{5-k}-\\binom{5}{k}\\] ordered pairs for $|A \\cap B|=k.$ Two solutions follow from here: Solution 3.1 (Binomial Theorem) The answer is \\begin{align*} \\sum_{k=0}^{5}\\left[2\\binom{5}{k}2^{5-k}-\\binom{5}{k}\\right] &= 2\\sum_{k=0}^{5}\\binom{5}{k}2^{5-k}-\\sum_{k=0}^{5}\\binom{5}{k} \\\\ &=2(2+1)^5-(1+1)^5 \\\\ &=2(243)-32 \\\\ &=454. \\end{align*} ~MRENTHUSIASM Solution 3.2 (Bash) The answer is \\begin{align*} &\\hspace{5.125mm}\\sum_{k=0}^{5}\\left[2\\binom{5}{k}2^{5-k}-\\binom{5}{k}\\right] \\\\ &=\\left[2\\binom{5}{0}2^{5-0}-\\binom{5}{0}\\right] + \\left[2\\binom{5}{1}2^{5-1}-\\binom{5}{1}\\right] + \\left[2\\binom{5}{2}2^{5-2}-\\binom{5}{2}\\right] + \\left[2\\binom{5}{3}2^{5-3}-\\binom{5}{3}\\right] + \\left[2\\binom{5}{4}2^{5-4}-\\binom{5}{4}\\right] + \\left[2\\binom{5}{5}2^{5-5}-\\binom{5}{5}\\right] \\\\ &=\\left[2\\left(1\\right)2^5-1\\right] + \\left[2\\left(5\\right)2^4-5\\right] + \\left[2\\left(10\\right)2^3-10\\right] + \\left[2\\left(10\\right)2^2-10\\right] + \\left[2\\left(5\\right)2^1-5\\right] + \\left[2\\left(1\\right)2^0-1\\right] \\\\ &=63+155+150+70+15+1 \\\\ &=454. \\end{align*} ~MRENTHUSIASM",
"By the Principle of Inclusion-Exclusion (abbreviated as PIE), we have $|A \\cup B|=|A|+|B|-|A \\cap B|,$ from which we rewrite the given equation as \\[|A| \\cdot |B| = |A \\cap B| \\cdot \\left(|A|+|B|-|A \\cap B|\\right).\\] Rearranging and applying Simon's Favorite Factoring Trick give \\begin{align*} |A| \\cdot |B| &= |A \\cap B|\\cdot|A| + |A \\cap B|\\cdot|B| - |A \\cap B|^2 \\\\ |A| \\cdot |B| - |A \\cap B|\\cdot|A| - |A \\cap B|\\cdot|B| &= - |A \\cap B|^2 \\\\ \\left(|A| - |A \\cap B|\\right)\\cdot\\left(|B| - |A \\cap B|\\right) &=0, \\end{align*} from which at least one of the following is true: $|A|=|A \\cap B|$ $|B|=|A \\cap B|$ Let $|A \\cap B|=k.$ For each value of $k\\in\\{0,1,2,3,4,5\\},$ we will use PIE to count the ordered pairs $(A,B):$ Suppose $|A|=k.$ There are $\\binom{5}{k}$ ways to choose the elements for $A.$ These $k$ elements must also appear in $B.$ Next, there are $2^{5-k}$ ways to add any number of the remaining $5-k$ elements to $B$ (Each element has $2$ options: in $B$ or not in $B.$). There are $\\binom{5}{k}2^{5-k}$ ordered pairs for $|A|=k.$ Similarly, there are $\\binom{5}{k}2^{5-k}$ ordered pairs for $|B|=k.$ To fix the overcount, we subtract the number of ordered pairs that are counted twice, in which $|A|=|B|=k.$ There are $\\binom{5}{k}$ such ordered pairs. Therefore, there are \\[2\\binom{5}{k}2^{5-k}-\\binom{5}{k}\\] ordered pairs for $|A \\cap B|=k.$ Two solutions follow from here: Solution 3.1 (Binomial Theorem) The answer is \\begin{align*} \\sum_{k=0}^{5}\\left[2\\binom{5}{k}2^{5-k}-\\binom{5}{k}\\right] &= 2\\sum_{k=0}^{5}\\binom{5}{k}2^{5-k}-\\sum_{k=0}^{5}\\binom{5}{k} \\\\ &=2(2+1)^5-(1+1)^5 \\\\ &=2(243)-32 \\\\ &=454. \\end{align*} ~MRENTHUSIASM Solution 3.2 (Bash) The answer is \\begin{align*} &\\hspace{5.125mm}\\sum_{k=0}^{5}\\left[2\\binom{5}{k}2^{5-k}-\\binom{5}{k}\\right] \\\\ &=\\left[2\\binom{5}{0}2^{5-0}-\\binom{5}{0}\\right] + \\left[2\\binom{5}{1}2^{5-1}-\\binom{5}{1}\\right] + \\left[2\\binom{5}{2}2^{5-2}-\\binom{5}{2}\\right] + \\left[2\\binom{5}{3}2^{5-3}-\\binom{5}{3}\\right] + \\left[2\\binom{5}{4}2^{5-4}-\\binom{5}{4}\\right] + \\left[2\\binom{5}{5}2^{5-5}-\\binom{5}{5}\\right] \\\\ &=\\left[2\\left(1\\right)2^5-1\\right] + \\left[2\\left(5\\right)2^4-5\\right] + \\left[2\\left(10\\right)2^3-10\\right] + \\left[2\\left(10\\right)2^2-10\\right] + \\left[2\\left(5\\right)2^1-5\\right] + \\left[2\\left(1\\right)2^0-1\\right] \\\\ &=63+155+150+70+15+1 \\\\ &=454. \\end{align*} ~MRENTHUSIASM",
"The answer is \\begin{align*} \\sum_{k=0}^{5}\\left[2\\binom{5}{k}2^{5-k}-\\binom{5}{k}\\right] &= 2\\sum_{k=0}^{5}\\binom{5}{k}2^{5-k}-\\sum_{k=0}^{5}\\binom{5}{k} \\\\ &=2(2+1)^5-(1+1)^5 \\\\ &=2(243)-32 \\\\ &=454. \\end{align*} ~MRENTHUSIASM Solution 3.2 (Bash) The answer is \\begin{align*} &\\hspace{5.125mm}\\sum_{k=0}^{5}\\left[2\\binom{5}{k}2^{5-k}-\\binom{5}{k}\\right] \\\\ &=\\left[2\\binom{5}{0}2^{5-0}-\\binom{5}{0}\\right] + \\left[2\\binom{5}{1}2^{5-1}-\\binom{5}{1}\\right] + \\left[2\\binom{5}{2}2^{5-2}-\\binom{5}{2}\\right] + \\left[2\\binom{5}{3}2^{5-3}-\\binom{5}{3}\\right] + \\left[2\\binom{5}{4}2^{5-4}-\\binom{5}{4}\\right] + \\left[2\\binom{5}{5}2^{5-5}-\\binom{5}{5}\\right] \\\\ &=\\left[2\\left(1\\right)2^5-1\\right] + \\left[2\\left(5\\right)2^4-5\\right] + \\left[2\\left(10\\right)2^3-10\\right] + \\left[2\\left(10\\right)2^2-10\\right] + \\left[2\\left(5\\right)2^1-5\\right] + \\left[2\\left(1\\right)2^0-1\\right] \\\\ &=63+155+150+70+15+1 \\\\ &=454. \\end{align*} ~MRENTHUSIASM",
"The answer is \\begin{align*} \\sum_{k=0}^{5}\\left[2\\binom{5}{k}2^{5-k}-\\binom{5}{k}\\right] &= 2\\sum_{k=0}^{5}\\binom{5}{k}2^{5-k}-\\sum_{k=0}^{5}\\binom{5}{k} \\\\ &=2(2+1)^5-(1+1)^5 \\\\ &=2(243)-32 \\\\ &=454. \\end{align*} ~MRENTHUSIASM Solution 3.2 (Bash) The answer is \\begin{align*} &\\hspace{5.125mm}\\sum_{k=0}^{5}\\left[2\\binom{5}{k}2^{5-k}-\\binom{5}{k}\\right] \\\\ &=\\left[2\\binom{5}{0}2^{5-0}-\\binom{5}{0}\\right] + \\left[2\\binom{5}{1}2^{5-1}-\\binom{5}{1}\\right] + \\left[2\\binom{5}{2}2^{5-2}-\\binom{5}{2}\\right] + \\left[2\\binom{5}{3}2^{5-3}-\\binom{5}{3}\\right] + \\left[2\\binom{5}{4}2^{5-4}-\\binom{5}{4}\\right] + \\left[2\\binom{5}{5}2^{5-5}-\\binom{5}{5}\\right] \\\\ &=\\left[2\\left(1\\right)2^5-1\\right] + \\left[2\\left(5\\right)2^4-5\\right] + \\left[2\\left(10\\right)2^3-10\\right] + \\left[2\\left(10\\right)2^2-10\\right] + \\left[2\\left(5\\right)2^1-5\\right] + \\left[2\\left(1\\right)2^0-1\\right] \\\\ &=63+155+150+70+15+1 \\\\ &=454. \\end{align*} ~MRENTHUSIASM",
"The answer is \\begin{align*} &\\hspace{5.125mm}\\sum_{k=0}^{5}\\left[2\\binom{5}{k}2^{5-k}-\\binom{5}{k}\\right] \\\\ &=\\left[2\\binom{5}{0}2^{5-0}-\\binom{5}{0}\\right] + \\left[2\\binom{5}{1}2^{5-1}-\\binom{5}{1}\\right] + \\left[2\\binom{5}{2}2^{5-2}-\\binom{5}{2}\\right] + \\left[2\\binom{5}{3}2^{5-3}-\\binom{5}{3}\\right] + \\left[2\\binom{5}{4}2^{5-4}-\\binom{5}{4}\\right] + \\left[2\\binom{5}{5}2^{5-5}-\\binom{5}{5}\\right] \\\\ &=\\left[2\\left(1\\right)2^5-1\\right] + \\left[2\\left(5\\right)2^4-5\\right] + \\left[2\\left(10\\right)2^3-10\\right] + \\left[2\\left(10\\right)2^2-10\\right] + \\left[2\\left(5\\right)2^1-5\\right] + \\left[2\\left(1\\right)2^0-1\\right] \\\\ &=63+155+150+70+15+1 \\\\ &=454. \\end{align*} ~MRENTHUSIASM",
"The answer is \\begin{align*} &\\hspace{5.125mm}\\sum_{k=0}^{5}\\left[2\\binom{5}{k}2^{5-k}-\\binom{5}{k}\\right] \\\\ &=\\left[2\\binom{5}{0}2^{5-0}-\\binom{5}{0}\\right] + \\left[2\\binom{5}{1}2^{5-1}-\\binom{5}{1}\\right] + \\left[2\\binom{5}{2}2^{5-2}-\\binom{5}{2}\\right] + \\left[2\\binom{5}{3}2^{5-3}-\\binom{5}{3}\\right] + \\left[2\\binom{5}{4}2^{5-4}-\\binom{5}{4}\\right] + \\left[2\\binom{5}{5}2^{5-5}-\\binom{5}{5}\\right] \\\\ &=\\left[2\\left(1\\right)2^5-1\\right] + \\left[2\\left(5\\right)2^4-5\\right] + \\left[2\\left(10\\right)2^3-10\\right] + \\left[2\\left(10\\right)2^2-10\\right] + \\left[2\\left(5\\right)2^1-5\\right] + \\left[2\\left(1\\right)2^0-1\\right] \\\\ &=63+155+150+70+15+1 \\\\ &=454. \\end{align*} ~MRENTHUSIASM",
"Proceed with Solution 1 to get $(|A| - |A \\cap B|)(|B| - |A \\cap B|) = 0$. WLOG, assume $|A| = |A \\cap B|$. Thus, $A \\subseteq B$. Since $A \\subseteq B$, if $|B| = n$, there are $2^n$ possible sets $A$, and there are also ${5 \\choose n}$ ways of choosing such $B$. Therefore, the number of possible pairs of sets $(A, B)$ is \\[\\sum_{k=0}^{5} 2^n {5 \\choose n}\\] We can compute this manually since it's only from $k=0$ to $5$, and computing gives us $243$. We can double this result for $B \\subseteq A$, and we get $2(243) = 486$. However, we have double counted the cases where $A$ and $B$ are the same sets. There are $32$ possible such cases, so we subtract $32$ from $486$ to get $454. ~ adam_zheng",
"Proceed with Solution 1 to get $(|A| - |A \\cap B|)(|B| - |A \\cap B|) = 0$. WLOG, assume $|A| = |A \\cap B|$. Thus, $A \\subseteq B$. Since $A \\subseteq B$, if $|B| = n$, there are $2^n$ possible sets $A$, and there are also ${5 \\choose n}$ ways of choosing such $B$. Therefore, the number of possible pairs of sets $(A, B)$ is \\[\\sum_{k=0}^{5} 2^n {5 \\choose n}\\] We can compute this manually since it's only from $k=0$ to $5$, and computing gives us $243$. We can double this result for $B \\subseteq A$, and we get $2(243) = 486$. However, we have double counted the cases where $A$ and $B$ are the same sets. There are $32$ possible such cases, so we subtract $32$ from $486$ to get $454. ~ adam_zheng"
] |
2021-II-7
| 2,021
| 7
|
Let $a, b, c,$ and $d$ be real numbers that satisfy the system of equations \begin{align*} a + b &= -3, \\ ab + bc + ca &= -4, \\ abc + bcd + cda + dab &= 14, \\ abcd &= 30. \end{align*} There exist relatively prime positive integers $m$ and $n$ such that \[a^2 + b^2 + c^2 + d^2 = \frac{m}{n}.\] Find $m + n$ .
| 145
|
II
|
[
"From the fourth equation we get $d=\\frac{30}{abc}.$ Substitute this into the third equation and you get $abc + \\frac{30(ab + bc + ca)}{abc} = abc - \\frac{120}{abc} = 14$. Hence $(abc)^2 - 14(abc)-120 = 0$. Solving, we get $abc = -6$ or $abc = 20$. From the first and second equation, we get $ab + bc + ca = ab-3c = -4 \\Longrightarrow ab = 3c-4$. If $abc=-6$, substituting we get $c(3c-4)=-6$. If you try solving this you see that this does not have real solutions in $c$, so $abc$ must be $20$. So $d=\\frac{3}{2}$. Since $c(3c-4)=20$, $c=-2$ or $c=\\frac{10}{3}$. If $c=\\frac{10}{3}$, then the system $a+b=-3$ and $ab = 6$ does not give you real solutions. So $c=-2$. Since you already know $d=\\frac{3}{2}$ and $c=-2$, so you can solve for $a$ and $b$ pretty easily and see that $a^{2}+b^{2}+c^{2}+d^{2}=\\frac{141}{4}$. So the answer is $145. ~math31415926535 ~minor edit by Mathkiddie",
"Note that $ab + bc + ca = -4$ can be rewritten as $ab + c(a+b) = -4$. Hence, $ab = 3c - 4$. Rewriting $abc+bcd+cda+dab = 14$, we get $ab(c+d) + cd(a+b) = 14$. Substitute $ab = 3c - 4$ and solving, we get \\[3c^{2} - 4c - 4d - 14 = 0.\\] We refer to this as Equation 1. Note that $abcd = 30$ gives $(3c-4)cd = 30$. So, $3c^{2}d - 4cd = 30$, which implies $d(3c^{2} - 4c) = 30$ or \\[3c^{2} - 4c = \\frac{30}{d}.\\] We refer to this as Equation 2. Substituting Equation 2 into Equation 1 gives, $\\frac{30}{d} - 4d - 14 = 0$. Solving this quadratic yields that $d \\in \\left\\{-5, \\frac{3}{2}\\right\\}$. Now we just try these two cases: For $d = \\frac{3}{2}$ substituting in Equation 1 gives a quadratic in $c$ which has roots $c \\in \\left\\{\\frac{10}{3}, -2\\right\\}$. Again trying cases, by letting $c = -2$, we get $ab = 3c-4$, Hence $ab = -10$. We know that $a + b = -3$, Solving these we get $a = -5, b = 2$ or $a= 2, b = -5$ (doesn't matter due to symmetry in $a,b$). So, this case yields solutions $(a,b,c,d) = \\left(-5, 2 , -2, \\frac{3}{2}\\right)$. Similarly trying other three cases, we get no more solutions, Hence this is the solution for $(a,b,c,d)$. Finally, $a^{2} + b^{2} + c^{2} + d^{2} = 25 + 4 + 4 + \\frac{9}{4} = \\frac{141}{4} = \\frac{m}{n}$. Therefore, $m + n = 141 + 4 = 145. ~Arnav Nigam",
"For simplicity purposes, we number the given equations $(1),(2),(3),$ and $(4),$ in that order. Rearranging $(2)$ and solving for $c,$ we have \\begin{align*} ab+(a+b)c&=-4 \\\\ ab-3c&=-4 \\\\ c&=\\frac{ab+4}{3}. \\hspace{14mm} (5) \\end{align*} Substituting $(5)$ into $(4)$ and solving for $d,$ we get \\begin{align*} ab\\left(\\frac{ab+4}{3}\\right)d&=30 \\\\ d&=\\frac{90}{ab(ab+4)}. \\hspace{5mm} (6) \\end{align*} Substituting $(5)$ and $(6)$ into $(3)$ and simplifying, we rewrite the left side of $(3)$ in terms of $a$ and $b$ only: \\begin{align*} ab\\left[\\frac{ab+4}{3}\\right] + b\\left[\\frac{ab+4}{3}\\right]\\left[\\frac{90}{ab(ab+4)}\\right] + \\left[\\frac{ab+4}{3}\\right]\\left[\\frac{90}{ab(ab+4)}\\right]a + \\left[\\frac{90}{ab(ab+4)}\\right]ab &= 14 \\\\ ab\\left[\\frac{ab+4}{3}\\right] + \\underbrace{\\frac{30}{a} + \\frac{30}{b}}_{\\text{Group them.}} + \\frac{90}{ab+4} &= 14 \\\\ ab\\left[\\frac{ab+4}{3}\\right] + \\frac{30(a+b)}{ab} + \\frac{90}{ab+4} &= 14 \\\\ ab\\left[\\frac{ab+4}{3}\\right] + \\underbrace{\\frac{-90}{ab} + \\frac{90}{ab+4}}_{\\text{Group them.}} &= 14 \\\\ ab\\left[\\frac{ab+4}{3}\\right] - \\frac{360}{ab(ab+4)}&=14. \\end{align*} Let $t=ab(ab+4),$ from which \\[\\frac{t}{3}-\\frac{360}{t}=14.\\] Multiplying both sides by $3t,$ rearranging, and factoring give $(t+18)(t-60)=0.$ Substituting back and completing the squares produce \\begin{align*} \\left[ab(ab+4)+18\\right]\\left[ab(ab+4)-60\\right]&=0 \\\\ \\left[(ab)^2+4ab+18\\right]\\left[(ab)^2+4ab-60\\right]&=0 \\\\ \\underbrace{\\left[(ab+2)^2+14\\right]}_{ab+2=\\pm\\sqrt{14}i\\implies ab\\not\\in\\mathbb R}\\underbrace{\\left[(ab+2)^2-64\\right]}_{ab+2=\\pm8}&=0 \\\\ ab&=6,-10. \\end{align*} If $ab=6,$ then combining this with $(1),$ we know that $a$ and $b$ are the solutions of the quadratic $x^2+3x+6=0.$ Since the discriminant is negative, neither $a$ nor $b$ is a real number. If $ab=-10,$ then combining this with $(1),$ we know that $a$ and $b$ are the solutions of the quadratic $x^2+3x-10=0,$ or $(x+5)(x-2)=0,$ from which $\\{a,b\\}=\\{-5,2\\}.$ Substituting $ab=-10$ into $(5)$ and $(6),$ we obtain $c=-2$ and $d=\\frac32,$ respectively. Together, we have \\[a^2+b^2+c^2+d^2=\\frac{141}{4},\\] so the answer is $141+4=145 ~MRENTHUSIASM",
"Let the four equations from top to bottom be listed 1 through 4 respectively. We factor equation 3 like so: \\[abc+d(ab+bc+ca)=14\\] Then we plug in equation 2 to receive $abc-4d=14$. By equation 4 we get $abc=\\frac{30}{d}$. Plugging in, we get $\\frac{30}{d}-4d=14$. Multiply by $d$ on both sides to get the quadratic equation $4d^2+14d-30=0$. Solving using the quadratic equation, we receive $d=\\frac{3}{2},d=-5$. So, we have to test which one is correct. We repeat a similar process as we did above for equations 1 and 2. We factor equation 2 to get \\[ab+c(a+b)=-4\\] After plugging in equation 1, we get $ab-3c=-4$. Now we convert it into a quadratic to receive $3c^2-4c-abc=0$. The value of $abc$ will depend on $d$. So we obtain the discriminant $16+12abc$. Let d = -5. Then $abc = \\frac{30}{-5}$, so $abc=-6$, discriminant is $16-72$, which makes this a dead end. Thus $d=\\frac{3}{2}$ For $d=\\frac{3}{2}$, making $abc=20$. This means the discriminant is just $256$, so we obtain two values for $c$ as well. We get either $c=\\frac{10}{3}$ or $c=-2$. So, we must AGAIN test which one is correct. We know $ab=3c-4$, and $a+b=-3$, so we use these values for testing. Let $c=\\frac{10}{3}$. Then $ab=6$, so $a=\\frac{6}{b}$. We thus get $\\frac{6}{b}+b=-3$, which leads to the quadratic $b^2+3b+6$. The discriminant for this is $9-24$. That means this value of $c$ is wrong, so $c=-2$. Thus we get polynomial $b^2+3b-10$. The discriminant this time is $49$, so we get two values for $b$. Through simple inspection, you may see they are interchangeable, as if you take the value $b=2$, you get $a=-5$. If you take the value $b=-5$, you get $a=2$. So it doesn't matter. That means the sum of all their squares is \\[\\frac{9}{4}+4+4+25=\\frac{141}{4},\\] so the answer is $141+4=145 ~amcrunner",
"Let the four equations from top to bottom be listed 1 through 4 respectively. We factor equation 3 like so: \\[abc+d(ab+bc+ca)=14\\] Then we plug in equation 2 to receive $abc-4d=14$. By equation 4 we get $abc=\\frac{30}{d}$. Plugging in, we get $\\frac{30}{d}-4d=14$. Multiply by $d$ on both sides to get the quadratic equation $4d^2+14d-30=0$. Solving using the quadratic equation, we receive $d=\\frac{3}{2},d=-5$. So, we have to test which one is correct. We repeat a similar process as we did above for equations 1 and 2. We factor equation 2 to get \\[ab+c(a+b)=-4\\] After plugging in equation 1, we get $ab-3c=-4$. Now we convert it into a quadratic to receive $3c^2-4c-abc=0$. The value of $abc$ will depend on $d$. So we obtain the discriminant $16+12abc$. Let d = -5. Then $abc = \\frac{30}{-5}$, so $abc=-6$, discriminant is $16-72$, which makes this a dead end. Thus $d=\\frac{3}{2}$ For $d=\\frac{3}{2}$, making $abc=20$. This means the discriminant is just $256$, so we obtain two values for $c$ as well. We get either $c=\\frac{10}{3}$ or $c=-2$. So, we must AGAIN test which one is correct. We know $ab=3c-4$, and $a+b=-3$, so we use these values for testing. Let $c=\\frac{10}{3}$. Then $ab=6$, so $a=\\frac{6}{b}$. We thus get $\\frac{6}{b}+b=-3$, which leads to the quadratic $b^2+3b+6$. The discriminant for this is $9-24$. That means this value of $c$ is wrong, so $c=-2$. Thus we get polynomial $b^2+3b-10$. The discriminant this time is $49$, so we get two values for $b$. Through simple inspection, you may see they are interchangeable, as if you take the value $b=2$, you get $a=-5$. If you take the value $b=-5$, you get $a=2$. So it doesn't matter. That means the sum of all their squares is \\[\\frac{9}{4}+4+4+25=\\frac{141}{4},\\] so the answer is $141+4=145 ~amcrunner",
"Let the four equations from top to bottom be listed $(1)$ through $(4)$ respectively. Multiplying both sides of $(3)$ by $d$ and factoring some terms gives us $abcd + d^2(ab + ac + bc) = 14d$. Substituting using equations $(4)$ and $(2)$ gives us $30 -4 d^2 = 14d$, and solving gives us $d = -5$ or $d = \\frac{3}{2}$. Plugging this back into $(3)$ gives us $abc + d(ab + ac + bc) = abc + (-5)(-4) = abc + 20 = 14$, or using the other solution for $d$ gives us $abc - 6 = 14$. Solving both of these equations gives us $abc = -6$ when $d = -5$ and $abc = 20$ when $d = \\frac{3}{2}$. Multiplying both sides of $(2)$ by $c$ and factoring some terms gives us $abc + c^2 (a + b) = abc -3c^2 = -4c$. Testing $abc = -6$ will give us an imaginary solution for $c$, so therefore $abc = 20$ and $d = \\frac{3}{2}$. This gets us $20 - 3c^2 = -4c$. Solving for $c$ gives us $c = \\frac{3}{10}$ or $c = -2$. With a bit of testing, we can see that the correct value of $c$ is $c=-2$. Now we know $a+b = -3$ and $ab + bc + ca = ab + c(a+b) = ab + 6 = -4$, $ab = -10$, and it is obvious that $a = -5$ and $b = 2$ or the other way around, and therefore, $a^2 + b^2 + c^2 + d^2 = 25 + 4 + 4 + \\frac{9}{4} = \\frac{141}{4}$, giving us the answer $141 + 4 = 145. ~hihitherethere minor edit by ~Gustyrustypro"
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2021-II-8
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An ant makes a sequence of moves on a cube where a move consists of walking from one vertex to an adjacent vertex along an edge of the cube. Initially the ant is at a vertex of the bottom face of the cube and chooses one of the three adjacent vertices to move to as its first move. For all moves after the first move, the ant does not return to its previous vertex, but chooses to move to one of the other two adjacent vertices. All choices are selected at random so that each of the possible moves is equally likely. The probability that after exactly $8$ moves that ant is at a vertex of the top face on the cube is $\frac{m}{n}$ , where $m$ and $n$ are relatively prime positive integers. Find $m + n.$
| 49
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II
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[
"For all positive integers $k,$ let $N(k,\\mathrm{BB})$ be the number of ways to make a sequence of exactly $k$ moves, where the last move is from the bottom face to the bottom face. $N(k,\\mathrm{BT})$ be the number of ways to make a sequence of exactly $k$ moves, where the last move is from the bottom face to the top face. $N(k,\\mathrm{TB})$ be the number of ways to make a sequence of exactly $k$ moves, where the last move is from the top face to the bottom face. $N(k,\\mathrm{TT})$ be the number of ways to make a sequence of exactly $k$ moves, where the last move is from the top face to the top face. The base case occurs at $k=1,$ from which $\\left(N(1,\\mathrm{BB}),N(1,\\mathrm{BT}),N(1,\\mathrm{TB}),N(1,\\mathrm{TT})\\right)=(2,1,0,0).$ Suppose the ant makes exactly $k$ moves for some $k\\geq2.$ We perform casework on its last move: If its last move is from the bottom face to the bottom face, then its next move has $1$ way to move from the bottom face to the bottom face. $1$ way to move from the bottom face to the top face. If its last move is from the bottom face to the top face, then its next move has $2$ ways to move from the top face to the top face. If its last move is from the top face to the bottom face, then its next move has $2$ ways to move from the bottom face to the bottom face. If its last move is from the top face to the top face, then its next move has $1$ way to move from the top face to the bottom face. $1$ way to move from the top face to the top face. Alternatively, this recursion argument is illustrated below, where each dashed arrow indicates $1$ way, and each solid arrow indicates $2$ ways: [asy] /* Made by MRENTHUSIASM */ size(9cm); pair A, B, C, D, E, F, G, H, X, Y; A=(0,6); B=(0,4); C=(0,2); D=(0,0); E=(10,6); F=(10,4); G=(10,2); H=(10,0); X=(-1,8); Y=(11,8); label(\"BB\", A, (-2,0)); label(\"BT\", B, (-2,0)); label(\"TB\", C, (-2,0)); label(\"TT\", D, (-2,0)); label(\"BB\", E, (2,0)); label(\"BT\", F, (2,0)); label(\"TB\", G, (2,0)); label(\"TT\", H, (2,0)); label(\"\\textbf{The \\boldmath{$k$}th Move}\", shift(0.3,0)*X); label(\"\\textbf{The \\boldmath{$(k+1)$}th Move}\", shift(-0.3,-0.085)*Y); draw(A--E,0.8+black+dashed,EndArrow); draw(A--F,0.8+black+dashed,EndArrow); draw(B--H,0.8+black,EndArrow); draw(C--E,0.8+black,EndArrow); draw(D--G,0.8+black+dashed,EndArrow); draw(D--H,0.8+black+dashed,EndArrow); dot(A^^B^^C^^D^^E^^F^^G^^H, 5+black); [/asy] Therefore, we have the following relationships: \\begin{align*} N(1,\\mathrm{BB})&=2, \\\\ N(1,\\mathrm{BT})&=1, \\\\ N(1,\\mathrm{TB})&=0, \\\\ N(1,\\mathrm{TT})&=0, \\\\ N(k+1,\\mathrm{BB})&=N(k,\\mathrm{BB})+2\\cdot N(k,\\mathrm{TB}), \\\\ N(k+1,\\mathrm{BT})&=N(k,\\mathrm{BB}), \\\\ N(k+1,\\mathrm{TB})&=N(k,\\mathrm{TT}), \\\\ N(k+1,\\mathrm{TT})&=N(k,\\mathrm{TT})+2\\cdot N(k,\\mathrm{BT}). \\end{align*} Using these equations, we recursively fill out the table below: \\[\\begin{array}{c||c|c|c|c|c|c|c|c} \\hspace{7mm}&\\hspace{6.5mm}&\\hspace{6.5mm}&\\hspace{6.75mm}&\\hspace{6.75mm}&\\hspace{6.75mm}&\\hspace{6.75mm}&& \\\\ [-2.5ex] \\boldsymbol{k} & \\boldsymbol{1} & \\boldsymbol{2} & \\boldsymbol{3} & \\boldsymbol{4} & \\boldsymbol{5} & \\boldsymbol{6} & \\boldsymbol{7} & \\boldsymbol{8} \\\\ \\hline \\hline &&&&&&&& \\\\ [-2.25ex] \\boldsymbol{N(k,\\mathrm{BB})} &2&2&2&6&18&38&66&118 \\\\ \\hline &&&&&&&& \\\\ [-2.25ex] \\boldsymbol{N(k,\\mathrm{BT})} &1&2&2&2&6&18&38&66 \\\\ \\hline &&&&&&&& \\\\ [-2.25ex] \\boldsymbol{N(k,\\mathrm{TB})} &0&0&2&6&10&14&26&62 \\\\ \\hline &&&&&&&& \\\\ [-2.25ex] \\boldsymbol{N(k,\\mathrm{TT})} &0&2&6&10&14&26&62&138 \\\\ \\hline \\hline &&&&&&&& \\\\ [-2.25ex] \\textbf{Total}&\\boldsymbol{3}&\\boldsymbol{6}&\\boldsymbol{12}&\\boldsymbol{24}&\\boldsymbol{48}&\\boldsymbol{96}&\\boldsymbol{192}&\\boldsymbol{384} \\end{array}\\] By the Multiplication Principle, there are $3\\cdot2^{k-1}$ ways to make exactly $k$ moves. So, we must get \\[N(k,\\mathrm{BB})+N(k,\\mathrm{BT})+N(k,\\mathrm{TB})+N(k,\\mathrm{TT})=3\\cdot2^{k-1}\\] for all values of $k.$ Finally, the requested probability is \\[\\frac{N(8,\\mathrm{BT})+N(8,\\mathrm{TT})}{N(8,\\mathrm{BB})+N(8,\\mathrm{BT})+N(8,\\mathrm{TB})+N(8,\\mathrm{TT})}=\\frac{66+138}{118+66+62+138}=\\frac{204}{384}=\\frac{17}{32},\\] from which the answer is $17+32=049 ~Arcticturn ~MRENTHUSIASM",
"For all positive integers $k,$ let $N(k,\\mathrm{BB})$ be the number of ways to make a sequence of exactly $k$ moves, where the last move is from the bottom face to the bottom face. $N(k,\\mathrm{BT})$ be the number of ways to make a sequence of exactly $k$ moves, where the last move is from the bottom face to the top face. $N(k,\\mathrm{TB})$ be the number of ways to make a sequence of exactly $k$ moves, where the last move is from the top face to the bottom face. $N(k,\\mathrm{TT})$ be the number of ways to make a sequence of exactly $k$ moves, where the last move is from the top face to the top face. The base case occurs at $k=1,$ from which $\\left(N(1,\\mathrm{BB}),N(1,\\mathrm{BT}),N(1,\\mathrm{TB}),N(1,\\mathrm{TT})\\right)=(2,1,0,0).$ Suppose the ant makes exactly $k$ moves for some $k\\geq2.$ We perform casework on its last move: If its last move is from the bottom face to the bottom face, then its next move has $1$ way to move from the bottom face to the bottom face. $1$ way to move from the bottom face to the top face. If its last move is from the bottom face to the top face, then its next move has $2$ ways to move from the top face to the top face. If its last move is from the top face to the bottom face, then its next move has $2$ ways to move from the bottom face to the bottom face. If its last move is from the top face to the top face, then its next move has $1$ way to move from the top face to the bottom face. $1$ way to move from the top face to the top face. Alternatively, this recursion argument is illustrated below, where each dashed arrow indicates $1$ way, and each solid arrow indicates $2$ ways: [asy] /* Made by MRENTHUSIASM */ size(9cm); pair A, B, C, D, E, F, G, H, X, Y; A=(0,6); B=(0,4); C=(0,2); D=(0,0); E=(10,6); F=(10,4); G=(10,2); H=(10,0); X=(-1,8); Y=(11,8); label(\"BB\", A, (-2,0)); label(\"BT\", B, (-2,0)); label(\"TB\", C, (-2,0)); label(\"TT\", D, (-2,0)); label(\"BB\", E, (2,0)); label(\"BT\", F, (2,0)); label(\"TB\", G, (2,0)); label(\"TT\", H, (2,0)); label(\"\\textbf{The \\boldmath{$k$}th Move}\", shift(0.3,0)*X); label(\"\\textbf{The \\boldmath{$(k+1)$}th Move}\", shift(-0.3,-0.085)*Y); draw(A--E,0.8+black+dashed,EndArrow); draw(A--F,0.8+black+dashed,EndArrow); draw(B--H,0.8+black,EndArrow); draw(C--E,0.8+black,EndArrow); draw(D--G,0.8+black+dashed,EndArrow); draw(D--H,0.8+black+dashed,EndArrow); dot(A^^B^^C^^D^^E^^F^^G^^H, 5+black); [/asy] Therefore, we have the following relationships: \\begin{align*} N(1,\\mathrm{BB})&=2, \\\\ N(1,\\mathrm{BT})&=1, \\\\ N(1,\\mathrm{TB})&=0, \\\\ N(1,\\mathrm{TT})&=0, \\\\ N(k+1,\\mathrm{BB})&=N(k,\\mathrm{BB})+2\\cdot N(k,\\mathrm{TB}), \\\\ N(k+1,\\mathrm{BT})&=N(k,\\mathrm{BB}), \\\\ N(k+1,\\mathrm{TB})&=N(k,\\mathrm{TT}), \\\\ N(k+1,\\mathrm{TT})&=N(k,\\mathrm{TT})+2\\cdot N(k,\\mathrm{BT}). \\end{align*} Using these equations, we recursively fill out the table below: \\[\\begin{array}{c||c|c|c|c|c|c|c|c} \\hspace{7mm}&\\hspace{6.5mm}&\\hspace{6.5mm}&\\hspace{6.75mm}&\\hspace{6.75mm}&\\hspace{6.75mm}&\\hspace{6.75mm}&& \\\\ [-2.5ex] \\boldsymbol{k} & \\boldsymbol{1} & \\boldsymbol{2} & \\boldsymbol{3} & \\boldsymbol{4} & \\boldsymbol{5} & \\boldsymbol{6} & \\boldsymbol{7} & \\boldsymbol{8} \\\\ \\hline \\hline &&&&&&&& \\\\ [-2.25ex] \\boldsymbol{N(k,\\mathrm{BB})} &2&2&2&6&18&38&66&118 \\\\ \\hline &&&&&&&& \\\\ [-2.25ex] \\boldsymbol{N(k,\\mathrm{BT})} &1&2&2&2&6&18&38&66 \\\\ \\hline &&&&&&&& \\\\ [-2.25ex] \\boldsymbol{N(k,\\mathrm{TB})} &0&0&2&6&10&14&26&62 \\\\ \\hline &&&&&&&& \\\\ [-2.25ex] \\boldsymbol{N(k,\\mathrm{TT})} &0&2&6&10&14&26&62&138 \\\\ \\hline \\hline &&&&&&&& \\\\ [-2.25ex] \\textbf{Total}&\\boldsymbol{3}&\\boldsymbol{6}&\\boldsymbol{12}&\\boldsymbol{24}&\\boldsymbol{48}&\\boldsymbol{96}&\\boldsymbol{192}&\\boldsymbol{384} \\end{array}\\] By the Multiplication Principle, there are $3\\cdot2^{k-1}$ ways to make exactly $k$ moves. So, we must get \\[N(k,\\mathrm{BB})+N(k,\\mathrm{BT})+N(k,\\mathrm{TB})+N(k,\\mathrm{TT})=3\\cdot2^{k-1}\\] for all values of $k.$ Finally, the requested probability is \\[\\frac{N(8,\\mathrm{BT})+N(8,\\mathrm{TT})}{N(8,\\mathrm{BB})+N(8,\\mathrm{BT})+N(8,\\mathrm{TB})+N(8,\\mathrm{TT})}=\\frac{66+138}{118+66+62+138}=\\frac{204}{384}=\\frac{17}{32},\\] from which the answer is $17+32=049 ~Arcticturn ~MRENTHUSIASM",
"Let the state from bottom to top be $B2T,$ from top to top be $T2T,$ from top to bottom be $T2B,$ and from bottom to bottom be $B2B.$ We can draw the following State Transition Diagram with Markov Chain. The numbers on the transition arc are the transition probabilities. The probabilities of being in a state after $n$ steps and after $n-1$ steps has the following relationships: \\begin{align*} B2T(n) &= B2B(n-1) \\cdot \\frac12\\\\ T2T(n) &= B2T(n-1) + T2T(n-1) \\cdot \\frac12\\\\ T2B(n) &= T2T(n-1) \\cdot \\frac12\\\\ B2B(n) &= T2B(n-1) + B2B(n-1) \\cdot \\frac12 \\end{align*} Those probabilities are calculated by Dynamic Programming in the following table: \\[\\begin{array}{c|cccc} & & & & \\\\ [-2ex] n & B2T(n) & T2T(n) & T2B(n) & B2B(n) \\\\ [1ex] \\hline & & & & \\\\ [-1ex] 1 & \\frac13 & 0 & 0 & \\frac23\\\\ & & & & \\\\ 2 & \\frac23 \\cdot \\frac12 = \\frac13 & \\frac13 & 0 & \\frac23 \\cdot \\frac12 = \\frac13 \\\\ & & & & \\\\ 3 & \\frac13 \\cdot \\frac12 = \\frac16 & \\frac13 + \\frac13 \\cdot \\frac12 = \\frac12 & \\frac13 \\cdot \\frac12 = \\frac16 & \\frac13 \\cdot \\frac12 = \\frac16 \\\\ & & & & \\\\ 4 & \\frac16 \\cdot \\frac12 = \\frac{1}{12} & \\frac16 + \\frac12 \\cdot \\frac12 = \\frac{5}{12} & \\frac12 \\cdot \\frac12 = \\frac14 & \\frac16 + \\frac16 \\cdot \\frac12 = \\frac14 \\\\ & & & & \\\\ 5 & \\frac14 \\cdot \\frac12 = \\frac18 & \\frac{1}{12} + \\frac{5}{12} \\cdot \\frac{1}{2} = \\frac{7}{24} & \\frac{5}{12} \\cdot \\frac12 = \\frac{5}{24} & \\frac14 + \\frac14 \\cdot \\frac12 = \\frac38 \\\\ & & & & \\\\ 6 & \\frac38 \\cdot \\frac12 = \\frac{3}{16} & \\frac18 + \\frac{7}{24} \\cdot \\frac12 = \\frac{13}{48} & \\frac{7}{24} \\cdot \\frac12 = \\frac{7}{48} & \\frac{5}{24} + \\frac38 \\cdot \\frac12 = \\frac{19}{48} \\\\ & & & & \\\\ 7 & \\frac{19}{48} \\cdot \\frac12 = \\frac{19}{96}& \\frac{3}{16} + \\frac{13}{48} \\cdot \\frac{1}{2} = \\frac{31}{96}& \\frac{13}{48} \\cdot \\frac{1}{2} = \\frac{13}{96} & \\frac{7}{48} + \\frac{19}{48} \\cdot \\frac12 = \\frac{11}{32}\\\\ & & & & \\\\ 8 & \\frac{11}{32} \\cdot \\frac{1}{2} = \\frac{11}{64} & \\frac{19}{96} + \\frac{31}{96} \\cdot \\frac12 = \\frac{23}{64} & \\frac{31}{96} \\cdot \\frac12 = \\frac{31}{192} & \\frac{13}{96} + \\frac{11}{32} \\cdot \\frac12 = \\frac{59}{192} \\\\ [1ex] \\end{array}\\] Finally, the requested probability is $\\frac{11}{64} + \\frac{23}{64} = \\frac{17}{32},$ from which the answer is $17 + 32 = 049 ~isabelchen",
"Let the state from bottom to top be $B2T,$ from top to top be $T2T,$ from top to bottom be $T2B,$ and from bottom to bottom be $B2B.$ We can draw the following State Transition Diagram with Markov Chain. The numbers on the transition arc are the transition probabilities. The probabilities of being in a state after $n$ steps and after $n-1$ steps has the following relationships: \\begin{align*} B2T(n) &= B2B(n-1) \\cdot \\frac12\\\\ T2T(n) &= B2T(n-1) + T2T(n-1) \\cdot \\frac12\\\\ T2B(n) &= T2T(n-1) \\cdot \\frac12\\\\ B2B(n) &= T2B(n-1) + B2B(n-1) \\cdot \\frac12 \\end{align*} Those probabilities are calculated by Dynamic Programming in the following table: \\[\\begin{array}{c|cccc} & & & & \\\\ [-2ex] n & B2T(n) & T2T(n) & T2B(n) & B2B(n) \\\\ [1ex] \\hline & & & & \\\\ [-1ex] 1 & \\frac13 & 0 & 0 & \\frac23\\\\ & & & & \\\\ 2 & \\frac23 \\cdot \\frac12 = \\frac13 & \\frac13 & 0 & \\frac23 \\cdot \\frac12 = \\frac13 \\\\ & & & & \\\\ 3 & \\frac13 \\cdot \\frac12 = \\frac16 & \\frac13 + \\frac13 \\cdot \\frac12 = \\frac12 & \\frac13 \\cdot \\frac12 = \\frac16 & \\frac13 \\cdot \\frac12 = \\frac16 \\\\ & & & & \\\\ 4 & \\frac16 \\cdot \\frac12 = \\frac{1}{12} & \\frac16 + \\frac12 \\cdot \\frac12 = \\frac{5}{12} & \\frac12 \\cdot \\frac12 = \\frac14 & \\frac16 + \\frac16 \\cdot \\frac12 = \\frac14 \\\\ & & & & \\\\ 5 & \\frac14 \\cdot \\frac12 = \\frac18 & \\frac{1}{12} + \\frac{5}{12} \\cdot \\frac{1}{2} = \\frac{7}{24} & \\frac{5}{12} \\cdot \\frac12 = \\frac{5}{24} & \\frac14 + \\frac14 \\cdot \\frac12 = \\frac38 \\\\ & & & & \\\\ 6 & \\frac38 \\cdot \\frac12 = \\frac{3}{16} & \\frac18 + \\frac{7}{24} \\cdot \\frac12 = \\frac{13}{48} & \\frac{7}{24} \\cdot \\frac12 = \\frac{7}{48} & \\frac{5}{24} + \\frac38 \\cdot \\frac12 = \\frac{19}{48} \\\\ & & & & \\\\ 7 & \\frac{19}{48} \\cdot \\frac12 = \\frac{19}{96}& \\frac{3}{16} + \\frac{13}{48} \\cdot \\frac{1}{2} = \\frac{31}{96}& \\frac{13}{48} \\cdot \\frac{1}{2} = \\frac{13}{96} & \\frac{7}{48} + \\frac{19}{48} \\cdot \\frac12 = \\frac{11}{32}\\\\ & & & & \\\\ 8 & \\frac{11}{32} \\cdot \\frac{1}{2} = \\frac{11}{64} & \\frac{19}{96} + \\frac{31}{96} \\cdot \\frac12 = \\frac{23}{64} & \\frac{31}{96} \\cdot \\frac12 = \\frac{31}{192} & \\frac{13}{96} + \\frac{11}{32} \\cdot \\frac12 = \\frac{59}{192} \\\\ [1ex] \\end{array}\\] Finally, the requested probability is $\\frac{11}{64} + \\frac{23}{64} = \\frac{17}{32},$ from which the answer is $17 + 32 = 049 ~isabelchen",
"Note that we don't care which exact vertex the ant is located at, just which level (either top face or bottom face). Consider the ant to be on any of the two levels and having moved at least one move. Define $p_i$ to be the probability that after $i$ moves, the ant ends up on the level it started on. On the first move, the ant can stay on the bottom level with $\\frac{2}{3}$ chance and $7$ moves left. Or, it can move to the top level with $\\frac{1}{3}$ chance and $6$ moves left (it has to spend another on the top as it can not return immediately). So the requested probability is $P = \\frac{2}{3}(1 - p_7) + \\frac{1}{3}p_6$. Consider when the ant has $i$ moves left (and it's not the ant's first move). It can either stay on its current level with $\\frac{1}{2}$ chance and $i - 1$ moves left, or travel to the opposite level with $\\frac{1}{2}$ chance, then move to another vertex on the opposite level, to have $i - 2$ moves left. Thus we obtain the recurrence \\[p_i = \\frac{1}{2}p_{i - 1} + \\frac{1}{2}(1 - p_{i - 2})\\] Computing $p_i$ with the starting conditions $p_0 = 1$ and $p_1 = \\frac{1}{2}$, we obtain $p_6 = \\frac{33}{64}$ and $p_7 = \\frac{59}{128}$. Hence $P = \\frac{2}{3}(1 - p_7) + \\frac{1}{3}p_6= \\frac{17}{32}$ as desired; $049. ~polarity",
"Note that we don't care which exact vertex the ant is located at, just which level (either top face or bottom face). Consider the ant to be on any of the two levels and having moved at least one move. Define $p_i$ to be the probability that after $i$ moves, the ant ends up on the level it started on. On the first move, the ant can stay on the bottom level with $\\frac{2}{3}$ chance and $7$ moves left. Or, it can move to the top level with $\\frac{1}{3}$ chance and $6$ moves left (it has to spend another on the top as it can not return immediately). So the requested probability is $P = \\frac{2}{3}(1 - p_7) + \\frac{1}{3}p_6$. Consider when the ant has $i$ moves left (and it's not the ant's first move). It can either stay on its current level with $\\frac{1}{2}$ chance and $i - 1$ moves left, or travel to the opposite level with $\\frac{1}{2}$ chance, then move to another vertex on the opposite level, to have $i - 2$ moves left. Thus we obtain the recurrence \\[p_i = \\frac{1}{2}p_{i - 1} + \\frac{1}{2}(1 - p_{i - 2})\\] Computing $p_i$ with the starting conditions $p_0 = 1$ and $p_1 = \\frac{1}{2}$, we obtain $p_6 = \\frac{33}{64}$ and $p_7 = \\frac{59}{128}$. Hence $P = \\frac{2}{3}(1 - p_7) + \\frac{1}{3}p_6= \\frac{17}{32}$ as desired; $049. ~polarity",
"On each move, we can either stay on the level we previously were (stay on the bottom/top) or switch levels (go from top to bottom and vise versa). Since we start on the bottom, ending on the top means that we will have to switch an odd number of times; since we cannot switch twice in a row, over an eight-move period we can either make one or three switches. Furthermore, once we switch to a level we can choose one of two directions of traveling on that level: clockwise or counterclockwise (since we can't go back to our previous move, our first move on the level after switching determines our direction). Case 1: one switch. Our one switch can either happen at the start/end of our moves, or in the middle. There are $4 + 24 = 28$ ways to do this, outlined below. Subcase 1: switch happens at ends. If our first move is a switch, then there are two ways to determine the direction we travel along the top layer. Multiply by $2$ to count for symmetry (last move is a switch) so this case yields $2^2 = 4$ possibilities. Subcase 2: switch happens in the middle. There are six places for the switch to happen; the switch breaks the sequences of moves into two chains, with each having $2$ ways to choose their direction of travel. This case yields $6 \\cdot 2^2 = 24$ possibilities. Case 2: three switches. Either two, one, or none of our switches occur at the start/end of our moves. There are $16 + 96 + 64 = 176$ ways to do this, outlined below. (Keep in mind we can't have two switches in a row.) Subcase 1: start and end with a switch. Since our third switch can't be in moves $2$ or $7$, there are four ways to place our switch, breaking our sequence into two chains. This case yields $4 \\cdot 2^2 = 16$ possibilities. Subcase 2: one of our switches is at the start/end. WLOG our first move is a switch; moves $2$ and $8$ cannot be switches. We can choose $2$ from any of the remaining $5$ moves to be switches, but we have to subtract the $4$ illegal cases where the two switches are in a row (3-4, 4-5, 5-6, 6-7). These three switches break our sequence into three chains; accounting for symmetry this case yields $2\\left(\\binom{5}{2} - 4\\right) \\cdot 2^3 = 96$ possibilities. Subcase 3: all our switches are in the middle. We choose $3$ from any of the $6$ middle moves to be our switches, but have to subtract the cases where at least two of them are in a row. If at least two switches are in a row, there are five places for the group of $2$ and four places for the third switch; however this overcounts the case where all three are in a row, which has $4$ possibilities. These three switches break our sequence into four chains, so this case yields $\\left(\\binom{6}{3} - 5 \\cdot 4 + 4\\right) \\cdot 2^4 = 64$ possibilities. Our probability is then $\\frac{176 + 28}{3 \\cdot 2^7} = \\frac{17}{32}$, so the answer is $17+32=049.",
"On each move, we can either stay on the level we previously were (stay on the bottom/top) or switch levels (go from top to bottom and vise versa). Since we start on the bottom, ending on the top means that we will have to switch an odd number of times; since we cannot switch twice in a row, over an eight-move period we can either make one or three switches. Furthermore, once we switch to a level we can choose one of two directions of traveling on that level: clockwise or counterclockwise (since we can't go back to our previous move, our first move on the level after switching determines our direction). Case 1: one switch. Our one switch can either happen at the start/end of our moves, or in the middle. There are $4 + 24 = 28$ ways to do this, outlined below. Subcase 1: switch happens at ends. If our first move is a switch, then there are two ways to determine the direction we travel along the top layer. Multiply by $2$ to count for symmetry (last move is a switch) so this case yields $2^2 = 4$ possibilities. Subcase 2: switch happens in the middle. There are six places for the switch to happen; the switch breaks the sequences of moves into two chains, with each having $2$ ways to choose their direction of travel. This case yields $6 \\cdot 2^2 = 24$ possibilities. Case 2: three switches. Either two, one, or none of our switches occur at the start/end of our moves. There are $16 + 96 + 64 = 176$ ways to do this, outlined below. (Keep in mind we can't have two switches in a row.) Subcase 1: start and end with a switch. Since our third switch can't be in moves $2$ or $7$, there are four ways to place our switch, breaking our sequence into two chains. This case yields $4 \\cdot 2^2 = 16$ possibilities. Subcase 2: one of our switches is at the start/end. WLOG our first move is a switch; moves $2$ and $8$ cannot be switches. We can choose $2$ from any of the remaining $5$ moves to be switches, but we have to subtract the $4$ illegal cases where the two switches are in a row (3-4, 4-5, 5-6, 6-7). These three switches break our sequence into three chains; accounting for symmetry this case yields $2\\left(\\binom{5}{2} - 4\\right) \\cdot 2^3 = 96$ possibilities. Subcase 3: all our switches are in the middle. We choose $3$ from any of the $6$ middle moves to be our switches, but have to subtract the cases where at least two of them are in a row. If at least two switches are in a row, there are five places for the group of $2$ and four places for the third switch; however this overcounts the case where all three are in a row, which has $4$ possibilities. These three switches break our sequence into four chains, so this case yields $\\left(\\binom{6}{3} - 5 \\cdot 4 + 4\\right) \\cdot 2^4 = 64$ possibilities. Our probability is then $\\frac{176 + 28}{3 \\cdot 2^7} = \\frac{17}{32}$, so the answer is $17+32=049.",
"Let $p(n)$ be the probability that we are on the top when you get to the $n$th move, and $p(n-1)$ and $p(n-2)$ be the probability that you are on the top when you get to the $(n-1)$th move and $(n-2)$th move respectively. Now you can do some recursion, splitting up into cases: Case 1: You are not on the top for the $(n-1)$th move or the $(n-2)$th move. In this case, it is a $1-p(n-1)$ chance that you were not on the top for the $(n-1)$th move, and a $1-p(n-2)$ that you are not on the top for the $(n-2)$th move. This leads you to a $\\frac{1}{2}\\cdot(1-p(n-1))\\cdot(1-p(n-2))$ chance for the first case. (The $\\frac12$ is there because of the fact that you can go up and you can also stay on the bottom side, as you cannot return.) Case 2: You are on the top for the $(n-1)$th move, but not on the top for the $(n-2)$th move. This leads you to a $p(n-1)\\cdot(1-p(n-2))$ probability (no extra components because you cannot return). Case 3: You are on top for both the $(n-1)$th move and the $(n-2)$th move. This leads to a probability of $\\frac{1}{2}\\cdot p(n-1)\\cdot p(n-2)$ (adding the extra $\\frac12$ because you can either stay on the top or go down. As the 4th case requires you to go down then up, but you cannot retrace, there is no $4$th case. These cases, added up, lead you to \\[p(n)=\\frac{1}{2}\\cdot(1-p(n-1))(1-p(n-2))+p(n-1)(1-p(n-2))+\\frac{1}{2}\\cdot p(n-1)p(n-2).\\] This can be further simplified down by expanding and combining like terms to \\[p(n)=\\frac{1+p(n-1)-p(n-2)}{2}.\\] Then we must find $p(1)$ and $p(2)$. p(1) is trivially $\\frac{1}{3}$. You can find $p(2)$ using basic probability techniques that is left as an exercise to the reader to get $\\frac{2}{3}$. In the end, you plug in to get \\[p(1)=\\frac{1}{3}, P(2)=\\frac{2}{3}, P(3)=\\frac{2}{3}, P(4)=\\frac{1}{2}, P(5)=\\frac{5}{12}, P(6)=\\frac{11}{24}, P(7)=\\frac{25}{48}, P(8)=\\frac{17}{32}. ~dragoon",
"Let $p(n)$ be the probability that we are on the top when you get to the $n$th move, and $p(n-1)$ and $p(n-2)$ be the probability that you are on the top when you get to the $(n-1)$th move and $(n-2)$th move respectively. Now you can do some recursion, splitting up into cases: Case 1: You are not on the top for the $(n-1)$th move or the $(n-2)$th move. In this case, it is a $1-p(n-1)$ chance that you were not on the top for the $(n-1)$th move, and a $1-p(n-2)$ that you are not on the top for the $(n-2)$th move. This leads you to a $\\frac{1}{2}\\cdot(1-p(n-1))\\cdot(1-p(n-2))$ chance for the first case. (The $\\frac12$ is there because of the fact that you can go up and you can also stay on the bottom side, as you cannot return.) Case 2: You are on the top for the $(n-1)$th move, but not on the top for the $(n-2)$th move. This leads you to a $p(n-1)\\cdot(1-p(n-2))$ probability (no extra components because you cannot return). Case 3: You are on top for both the $(n-1)$th move and the $(n-2)$th move. This leads to a probability of $\\frac{1}{2}\\cdot p(n-1)\\cdot p(n-2)$ (adding the extra $\\frac12$ because you can either stay on the top or go down. As the 4th case requires you to go down then up, but you cannot retrace, there is no $4$th case. These cases, added up, lead you to \\[p(n)=\\frac{1}{2}\\cdot(1-p(n-1))(1-p(n-2))+p(n-1)(1-p(n-2))+\\frac{1}{2}\\cdot p(n-1)p(n-2).\\] This can be further simplified down by expanding and combining like terms to \\[p(n)=\\frac{1+p(n-1)-p(n-2)}{2}.\\] Then we must find $p(1)$ and $p(2)$. p(1) is trivially $\\frac{1}{3}$. You can find $p(2)$ using basic probability techniques that is left as an exercise to the reader to get $\\frac{2}{3}$. In the end, you plug in to get \\[p(1)=\\frac{1}{3}, P(2)=\\frac{2}{3}, P(3)=\\frac{2}{3}, P(4)=\\frac{1}{2}, P(5)=\\frac{5}{12}, P(6)=\\frac{11}{24}, P(7)=\\frac{25}{48}, P(8)=\\frac{17}{32}. ~dragoon"
] |
2021-II-9
| 2,021
| 9
|
Find the number of ordered pairs $(m, n)$ such that $m$ and $n$ are positive integers in the set $\{1, 2, ..., 30\}$ and the greatest common divisor of $2^m + 1$ and $2^n - 1$ is not $1$ .
| 295
|
II
|
[
"This solution refers to the Remarks section. By the Euclidean Algorithm, we have \\[\\gcd\\left(2^m+1,2^m-1\\right)=\\gcd\\left(2,2^m-1\\right)=1.\\] We are given that $\\gcd\\left(2^m+1,2^n-1\\right)>1.$ Multiplying both sides by $\\gcd\\left(2^m-1,2^n-1\\right)$ gives \\begin{align*} \\gcd\\left(2^m+1,2^n-1\\right)\\cdot\\gcd\\left(2^m-1,2^n-1\\right)&>\\gcd\\left(2^m-1,2^n-1\\right) \\\\ \\gcd\\left(\\left(2^m+1\\right)\\left(2^m-1\\right),2^n-1\\right)&>\\gcd\\left(2^m-1,2^n-1\\right) \\hspace{12mm} &&\\text{by }\\textbf{Claim 1} \\\\ \\gcd\\left(2^{2m}-1,2^n-1\\right)&>\\gcd\\left(2^m-1,2^n-1\\right) \\\\ 2^{\\gcd(2m,n)}-1&>2^{\\gcd(m,n)}-1 &&\\text{by }\\textbf{Claim 2} \\\\ \\gcd(2m,n)&>\\gcd(m,n), \\end{align*} which implies that $n$ must have more factors of $2$ than $m$ does. We construct the following table for the first $30$ positive integers: \\[\\begin{array}{c|c|c} && \\\\ [-2.5ex] \\boldsymbol{\\#}\\textbf{ of Factors of }\\boldsymbol{2} & \\textbf{Numbers} & \\textbf{Count} \\\\ \\hline && \\\\ [-2.25ex] 0 & 1,3,5,7,9,11,13,15,17,19,21,23,25,27,29 & 15 \\\\ && \\\\ [-2.25ex] 1 & 2,6,10,14,18,22,26,30 & 8 \\\\ && \\\\ [-2.25ex] 2 & 4,12,20,28 & 4 \\\\ && \\\\ [-2.25ex] 3 & 8,24 & 2 \\\\ && \\\\ [-2.25ex] 4 & 16 & 1 \\\\ \\end{array}\\] To count the ordered pairs $(m,n),$ we perform casework on the number of factors of $2$ that $m$ has: If $m$ has $0$ factors of $2,$ then $m$ has $15$ options and $n$ has $8+4+2+1=15$ options. So, this case has $15\\cdot15=225$ ordered pairs. If $m$ has $1$ factor of $2,$ then $m$ has $8$ options and $n$ has $4+2+1=7$ options. So, this case has $8\\cdot7=56$ ordered pairs. If $m$ has $2$ factors of $2,$ then $m$ has $4$ options and $n$ has $2+1=3$ options. So, this case has $4\\cdot3=12$ ordered pairs. If $m$ has $3$ factors of $2,$ then $m$ has $2$ options and $n$ has $1$ option. So, this case has $2\\cdot1=2$ ordered pairs. Together, the answer is $225+56+12+2=295 ~Lcz ~MRENTHUSIASM",
"Consider any ordered pair $(m,n)$ such that $\\gcd(2^m+1, 2^n-1) > 1$. There must exist some odd number $p\\ne 1$ such that $2^m \\equiv -1 \\pmod{p}$ and $2^n \\equiv 1 \\pmod{p}$. Let $d$ be the order of $2$ modulo $p$. Note that $2^{2m} \\equiv 1 \\pmod{p}$. From this, we can say that $2m$ and $n$ are both multiples of $d$, but $m$ is not. Thus, we have $v_2(n) \\ge v_2(d)$ and $v_2(m) + 1 = v_2(d)$. Substituting the latter equation into the inequality before gives $v_2(n) \\ge v_2(m)+1$. Since $v_2(n)$ and $v_2(m)$ are integers, this implies $v_2(n)>v_2(m)$. The rest of the solution now proceeds as in Solution 1. ~Sedro"
] |
2021-II-10
| 2,021
| 10
|
Two spheres with radii $36$ and one sphere with radius $13$ are each externally tangent to the other two spheres and to two different planes $\mathcal{P}$ and $\mathcal{Q}$ . The intersection of planes $\mathcal{P}$ and $\mathcal{Q}$ is the line $\ell$ . The distance from line $\ell$ to the point where the sphere with radius $13$ is tangent to plane $\mathcal{P}$ is $\tfrac{m}{n}$ , where $m$ and $n$ are relatively prime positive integers. Find $m + n$ .
| 335
|
II
|
[
"This solution refers to the Diagram section. As shown below, let $O_1,O_2,O_3$ be the centers of the spheres (where sphere $O_3$ has radius $13$) and $T_1,T_2,T_3$ be their respective points of tangency to plane $\\mathcal{P}.$ Let $\\mathcal{R}$ be the plane that is determined by $O_1,O_2,$ and $O_3.$ Suppose $A$ is the foot of the perpendicular from $O_3$ to line $\\ell,$ so $\\overleftrightarrow{O_3A}$ is the perpendicular bisector of $\\overline{O_1O_2}.$ We wish to find $T_3A.$ [asy] /* Made by MRENTHUSIASM */ size(300); import graph3; import solids; currentprojection=orthographic((1,1/2,0)); triple O1, O2, O3, T1, T2, T3, A, L1, L2; O1 = (0,-36,0); O2 = (0,36,0); O3 = (0,0,-sqrt(1105)); T1 = (864*sqrt(1105)/1105,-36,-828*sqrt(1105)/1105); T2 = (864*sqrt(1105)/1105,36,-828*sqrt(1105)/1105); T3 = (24*sqrt(1105)/85,0,-108*sqrt(1105)/85); A = (0,0,-36*sqrt(1105)/23); L1 = shift(0,-80,0)*A; L2 = shift(0,80,0)*A; draw(surface(L1--L2--(-T2.x,L2.y,T2.z)--(-T1.x,L1.y,T1.z)--cycle),pink); draw(surface(L1--L2--(L2.x,L2.y,40)--(L1.x,L1.y,40)--cycle),gray); draw(shift(O1)*rotate(90,O1,O2)*scale3(36)*unithemisphere,yellow,light=White); draw(shift(O2)*rotate(90,O1,O2)*scale3(36)*unithemisphere,yellow,light=White); draw(shift(O3)*rotate(90,O1,O2)*scale3(13)*unithemisphere,red,light=White); draw(surface(L1--L2--(T2.x,L2.y,T2.z)--(T1.x,L1.y,T1.z)--cycle),palegreen); draw(surface(L1--L2--(-T2.x,L2.y,L2.z-abs(T2.z))--(-T1.x,L1.y,L2.z-abs(T1.z))--cycle),palegreen); draw(surface(L1--L2--(L2.x,L2.y,L2.z-abs(T1.z))--(L1.x,L1.y,L1.z-abs(T2.z))--cycle),gray); draw(surface(L1--L2--(T2.x,L2.y,L2.z-abs(T1.z))--(T1.x,L1.y,L1.z-abs(T2.z))--cycle),pink); draw(O1--O2--O3--cycle^^O3--A,dashed); draw(T1--T2--T3--cycle^^T3--A,heavygreen); draw(O1--T1^^O2--T2^^O3--T3,mediumblue+dashed); draw(L1--L2,L=Label(\"$\\ell$\",position=EndPoint,align=3*E),red); label(\"$\\mathcal{P}$\",midpoint(L1--(T1.x,L1.y,T1.z)),(0,-3,0),heavygreen); label(\"$\\mathcal{Q}$\",midpoint(L1--(T1.x,L1.y,L1.z-abs(T2.z))),(0,-3,0),heavymagenta); label(\"$\\mathcal{R}$\",O1,(0,-24,0)); dot(\"$O_1$\",O1,(0,-1,1),linewidth(4.5)); dot(\"$O_2$\",O2,(0,1,1),linewidth(4.5)); dot(\"$O_3$\",O3,(0,-1.5,0),linewidth(4.5)); dot(\"$T_1$\",T1,(0,-1,-1),heavygreen+linewidth(4.5)); dot(\"$T_2$\",T2,(0,1,-1),heavygreen+linewidth(4.5)); dot(\"$T_3$\",T3,(0,-1,-1),heavygreen+linewidth(4.5)); dot(\"$A$\",A,(0,0,-2),red+linewidth(4.5)); [/asy] Note that: In $\\triangle O_1O_2O_3,$ we get $O_1O_2=72$ and $O_1O_3=O_2O_3=49.$ Both $\\triangle O_1O_2O_3$ and $\\overline{O_3A}$ lie in plane $\\mathcal{R}.$ Both $\\triangle T_1T_2T_3$ and $\\overline{T_3A}$ lie in plane $\\mathcal{P}.$ By symmetry, since planes $\\mathcal{P}$ and $\\mathcal{Q}$ are reflections of each other about plane $\\mathcal{R},$ the three planes are concurrent to line $\\ell.$ Since $\\overline{O_1T_1}\\perp\\mathcal{P}$ and $\\overline{O_3T_3}\\perp\\mathcal{P},$ it follows that $\\overline{O_1T_1}\\parallel\\overline{O_3T_3},$ from which $O_1,O_3,T_1,$ and $T_3$ are coplanar. Now, we focus on cross-sections $O_1O_3T_3T_1$ and $\\mathcal{R}:$ In the three-dimensional space, the intersection of a line and a plane must be exactly one of the empty set, a point, or a line. Clearly, cross-section $O_1O_3T_3T_1$ intersects line $\\ell$ at exactly one point. Furthermore, as the intersection of planes $\\mathcal{R}$ and $\\mathcal{P}$ is line $\\ell,$ we conclude that $\\overrightarrow{O_1O_3}$ and $\\overrightarrow{T_1T_3}$ must intersect line $\\ell$ at the same point. Let $B$ be the point of concurrency of $\\overrightarrow{O_1O_3},\\overrightarrow{T_1T_3},$ and line $\\ell.$ In cross-section $\\mathcal{R},$ let $C$ be the foot of the perpendicular from $O_1$ to line $\\ell,$ and $D$ be the foot of the perpendicular from $O_3$ to $\\overline{O_1C}.$ We have the following diagram: [asy] size(300); import graph3; import solids; currentprojection=orthographic((1,1/2,0)); triple O1, O3, T1, T3, A, B, C, D; O1 = (0,-36,0); O3 = (0,0,-sqrt(1105)); T1 = (864*sqrt(1105)/1105,-36,-828*sqrt(1105)/1105); T3 = (24*sqrt(1105)/85,0,-108*sqrt(1105)/85); A = (0,0,-36*sqrt(1105)/23); B = intersectionpoint(O1--O1+100*(O3-O1),T1--T1+100*(T3-T1)); C = (0,-36,-36*sqrt(1105)/23); D = (0,-36,-sqrt(1105)); draw(C--O1--O3--A^^D--O3--B,dashed); draw(T1--T3--A^^T3--B,heavygreen); draw(O1--T1^^O3--T3,mediumblue+dashed); draw(shift(0,-80,0)*A--shift(0,80,0)*A,L=Label(\"$\\ell$\",position=EndPoint,align=3*E),red); dot(\"$O_1$\",O1,(0,-1,1),linewidth(4.5)); dot(\"$O_3$\",O3,(0,1,1),linewidth(4.5)); dot(\"$T_1$\",T1,(0,-1,-1),heavygreen+linewidth(4.5)); dot(\"$T_3$\",T3,(0,-1,-1),heavygreen+linewidth(4.5)); dot(\"$A$\",A,(0,0,-2),red+linewidth(4.5)); dot(\"$B$\",B,(0,0,-2),red+linewidth(4.5)); dot(\"$C$\",C,(0,0,-2),red+linewidth(4.5)); dot(\"$D$\",D,(0,-2,0),linewidth(4.5)); [/asy] In cross-section $O_1O_3T_3T_1,$ since $\\overline{O_1T_1}\\parallel\\overline{O_3T_3}$ as discussed, we obtain $\\triangle O_1T_1B\\sim\\triangle O_3T_3B$ by AA, with the ratio of similitude $\\frac{O_1T_1}{O_3T_3}=\\frac{36}{13}.$ Therefore, we get $\\frac{O_1B}{O_3B}=\\frac{49+O_3B}{O_3B}=\\frac{36}{13},$ or $O_3B=\\frac{637}{23}.$ In cross-section $\\mathcal{R},$ note that $O_1O_3=49$ and $DO_3=\\frac{O_1O_2}{2}=36.$ Applying the Pythagorean Theorem to right $\\triangle O_1DO_3,$ we have $O_1D=\\sqrt{1105}.$ Moreover, since $\\ell\\perp\\overline{O_1C}$ and $\\overline{DO_3}\\perp\\overline{O_1C},$ we obtain $\\ell\\parallel\\overline{DO_3}$ so that $\\triangle O_1CB\\sim\\triangle O_1DO_3$ by AA, with the ratio of similitude $\\frac{O_1B}{O_1O_3}=\\frac{49+\\frac{637}{23}}{49}.$ Therefore, we get $\\frac{O_1C}{O_1D}=\\frac{\\sqrt{1105}+DC}{\\sqrt{1105}}=\\frac{49+\\frac{637}{23}}{49},$ or $DC=\\frac{13\\sqrt{1105}}{23}.$ Finally, note that $\\overline{O_3T_3}\\perp\\overline{T_3A}$ and $O_3T_3=13.$ Since quadrilateral $DCAO_3$ is a rectangle, we have $O_3A=DC=\\frac{13\\sqrt{1105}}{23}.$ Applying the Pythagorean Theorem to right $\\triangle O_3T_3A$ gives $T_3A=\\frac{312}{23},$ from which the answer is $312+23=335 ~MRENTHUSIASM",
"This solution refers to the Diagram section. As shown below, let $O_1,O_2,O_3$ be the centers of the spheres (where sphere $O_3$ has radius $13$) and $T_1,T_2,T_3$ be their respective points of tangency to plane $\\mathcal{P}.$ Let $\\mathcal{R}$ be the plane that is determined by $O_1,O_2,$ and $O_3.$ Suppose $A$ is the foot of the perpendicular from $O_3$ to line $\\ell,$ so $\\overleftrightarrow{O_3A}$ is the perpendicular bisector of $\\overline{O_1O_2}.$ We wish to find $T_3A.$ [asy] /* Made by MRENTHUSIASM */ size(300); import graph3; import solids; currentprojection=orthographic((1,1/2,0)); triple O1, O2, O3, T1, T2, T3, A, L1, L2; O1 = (0,-36,0); O2 = (0,36,0); O3 = (0,0,-sqrt(1105)); T1 = (864*sqrt(1105)/1105,-36,-828*sqrt(1105)/1105); T2 = (864*sqrt(1105)/1105,36,-828*sqrt(1105)/1105); T3 = (24*sqrt(1105)/85,0,-108*sqrt(1105)/85); A = (0,0,-36*sqrt(1105)/23); L1 = shift(0,-80,0)*A; L2 = shift(0,80,0)*A; draw(surface(L1--L2--(-T2.x,L2.y,T2.z)--(-T1.x,L1.y,T1.z)--cycle),pink); draw(surface(L1--L2--(L2.x,L2.y,40)--(L1.x,L1.y,40)--cycle),gray); draw(shift(O1)*rotate(90,O1,O2)*scale3(36)*unithemisphere,yellow,light=White); draw(shift(O2)*rotate(90,O1,O2)*scale3(36)*unithemisphere,yellow,light=White); draw(shift(O3)*rotate(90,O1,O2)*scale3(13)*unithemisphere,red,light=White); draw(surface(L1--L2--(T2.x,L2.y,T2.z)--(T1.x,L1.y,T1.z)--cycle),palegreen); draw(surface(L1--L2--(-T2.x,L2.y,L2.z-abs(T2.z))--(-T1.x,L1.y,L2.z-abs(T1.z))--cycle),palegreen); draw(surface(L1--L2--(L2.x,L2.y,L2.z-abs(T1.z))--(L1.x,L1.y,L1.z-abs(T2.z))--cycle),gray); draw(surface(L1--L2--(T2.x,L2.y,L2.z-abs(T1.z))--(T1.x,L1.y,L1.z-abs(T2.z))--cycle),pink); draw(O1--O2--O3--cycle^^O3--A,dashed); draw(T1--T2--T3--cycle^^T3--A,heavygreen); draw(O1--T1^^O2--T2^^O3--T3,mediumblue+dashed); draw(L1--L2,L=Label(\"$\\ell$\",position=EndPoint,align=3*E),red); label(\"$\\mathcal{P}$\",midpoint(L1--(T1.x,L1.y,T1.z)),(0,-3,0),heavygreen); label(\"$\\mathcal{Q}$\",midpoint(L1--(T1.x,L1.y,L1.z-abs(T2.z))),(0,-3,0),heavymagenta); label(\"$\\mathcal{R}$\",O1,(0,-24,0)); dot(\"$O_1$\",O1,(0,-1,1),linewidth(4.5)); dot(\"$O_2$\",O2,(0,1,1),linewidth(4.5)); dot(\"$O_3$\",O3,(0,-1.5,0),linewidth(4.5)); dot(\"$T_1$\",T1,(0,-1,-1),heavygreen+linewidth(4.5)); dot(\"$T_2$\",T2,(0,1,-1),heavygreen+linewidth(4.5)); dot(\"$T_3$\",T3,(0,-1,-1),heavygreen+linewidth(4.5)); dot(\"$A$\",A,(0,0,-2),red+linewidth(4.5)); [/asy] Note that: In $\\triangle O_1O_2O_3,$ we get $O_1O_2=72$ and $O_1O_3=O_2O_3=49.$ Both $\\triangle O_1O_2O_3$ and $\\overline{O_3A}$ lie in plane $\\mathcal{R}.$ Both $\\triangle T_1T_2T_3$ and $\\overline{T_3A}$ lie in plane $\\mathcal{P}.$ By symmetry, since planes $\\mathcal{P}$ and $\\mathcal{Q}$ are reflections of each other about plane $\\mathcal{R},$ the three planes are concurrent to line $\\ell.$ Since $\\overline{O_1T_1}\\perp\\mathcal{P}$ and $\\overline{O_3T_3}\\perp\\mathcal{P},$ it follows that $\\overline{O_1T_1}\\parallel\\overline{O_3T_3},$ from which $O_1,O_3,T_1,$ and $T_3$ are coplanar. Now, we focus on cross-sections $O_1O_3T_3T_1$ and $\\mathcal{R}:$ In the three-dimensional space, the intersection of a line and a plane must be exactly one of the empty set, a point, or a line. Clearly, cross-section $O_1O_3T_3T_1$ intersects line $\\ell$ at exactly one point. Furthermore, as the intersection of planes $\\mathcal{R}$ and $\\mathcal{P}$ is line $\\ell,$ we conclude that $\\overrightarrow{O_1O_3}$ and $\\overrightarrow{T_1T_3}$ must intersect line $\\ell$ at the same point. Let $B$ be the point of concurrency of $\\overrightarrow{O_1O_3},\\overrightarrow{T_1T_3},$ and line $\\ell.$ In cross-section $\\mathcal{R},$ let $C$ be the foot of the perpendicular from $O_1$ to line $\\ell,$ and $D$ be the foot of the perpendicular from $O_3$ to $\\overline{O_1C}.$ We have the following diagram: [asy] size(300); import graph3; import solids; currentprojection=orthographic((1,1/2,0)); triple O1, O3, T1, T3, A, B, C, D; O1 = (0,-36,0); O3 = (0,0,-sqrt(1105)); T1 = (864*sqrt(1105)/1105,-36,-828*sqrt(1105)/1105); T3 = (24*sqrt(1105)/85,0,-108*sqrt(1105)/85); A = (0,0,-36*sqrt(1105)/23); B = intersectionpoint(O1--O1+100*(O3-O1),T1--T1+100*(T3-T1)); C = (0,-36,-36*sqrt(1105)/23); D = (0,-36,-sqrt(1105)); draw(C--O1--O3--A^^D--O3--B,dashed); draw(T1--T3--A^^T3--B,heavygreen); draw(O1--T1^^O3--T3,mediumblue+dashed); draw(shift(0,-80,0)*A--shift(0,80,0)*A,L=Label(\"$\\ell$\",position=EndPoint,align=3*E),red); dot(\"$O_1$\",O1,(0,-1,1),linewidth(4.5)); dot(\"$O_3$\",O3,(0,1,1),linewidth(4.5)); dot(\"$T_1$\",T1,(0,-1,-1),heavygreen+linewidth(4.5)); dot(\"$T_3$\",T3,(0,-1,-1),heavygreen+linewidth(4.5)); dot(\"$A$\",A,(0,0,-2),red+linewidth(4.5)); dot(\"$B$\",B,(0,0,-2),red+linewidth(4.5)); dot(\"$C$\",C,(0,0,-2),red+linewidth(4.5)); dot(\"$D$\",D,(0,-2,0),linewidth(4.5)); [/asy] In cross-section $O_1O_3T_3T_1,$ since $\\overline{O_1T_1}\\parallel\\overline{O_3T_3}$ as discussed, we obtain $\\triangle O_1T_1B\\sim\\triangle O_3T_3B$ by AA, with the ratio of similitude $\\frac{O_1T_1}{O_3T_3}=\\frac{36}{13}.$ Therefore, we get $\\frac{O_1B}{O_3B}=\\frac{49+O_3B}{O_3B}=\\frac{36}{13},$ or $O_3B=\\frac{637}{23}.$ In cross-section $\\mathcal{R},$ note that $O_1O_3=49$ and $DO_3=\\frac{O_1O_2}{2}=36.$ Applying the Pythagorean Theorem to right $\\triangle O_1DO_3,$ we have $O_1D=\\sqrt{1105}.$ Moreover, since $\\ell\\perp\\overline{O_1C}$ and $\\overline{DO_3}\\perp\\overline{O_1C},$ we obtain $\\ell\\parallel\\overline{DO_3}$ so that $\\triangle O_1CB\\sim\\triangle O_1DO_3$ by AA, with the ratio of similitude $\\frac{O_1B}{O_1O_3}=\\frac{49+\\frac{637}{23}}{49}.$ Therefore, we get $\\frac{O_1C}{O_1D}=\\frac{\\sqrt{1105}+DC}{\\sqrt{1105}}=\\frac{49+\\frac{637}{23}}{49},$ or $DC=\\frac{13\\sqrt{1105}}{23}.$ Finally, note that $\\overline{O_3T_3}\\perp\\overline{T_3A}$ and $O_3T_3=13.$ Since quadrilateral $DCAO_3$ is a rectangle, we have $O_3A=DC=\\frac{13\\sqrt{1105}}{23}.$ Applying the Pythagorean Theorem to right $\\triangle O_3T_3A$ gives $T_3A=\\frac{312}{23},$ from which the answer is $312+23=335 ~MRENTHUSIASM",
"The centers of the three spheres form a $49$-$49$-$72$ triangle. Consider the points at which the plane is tangent to the two bigger spheres; the line segment connecting these two points should be parallel to the $72$ side of this triangle. Take its midpoint $M$, which is $36$ away from the midpoint $A$ of the $72$ side, and connect these two midpoints. Now consider the point at which the plane is tangent to the small sphere, and connect $M$ with the small sphere's tangent point $B$. Extend $\\overline{MB}$ through $B$ until it hits the ray from $A$ through the center of the small sphere (convince yourself that these two intersect). Call this intersection $D$, the center of the small sphere $C$, we want to find $BD$. By Pythagoras, $AC=\\sqrt{49^2-36^2}=\\sqrt{1105}$, and we know that $MA=36$ and $BC=13$. We know that $\\overline{MA}$ and $\\overline{BC}$ must be parallel, using ratios we realize that $CD=\\frac{13}{23}\\sqrt{1105}$. Apply the Pythagorean theorem to $\\triangle BCD$, $BD=\\frac{312}{23}$, so $312 + 23 = 335. ~Ross Gao",
"The centers of the three spheres form a $49$-$49$-$72$ triangle. Consider the points at which the plane is tangent to the two bigger spheres; the line segment connecting these two points should be parallel to the $72$ side of this triangle. Take its midpoint $M$, which is $36$ away from the midpoint $A$ of the $72$ side, and connect these two midpoints. Now consider the point at which the plane is tangent to the small sphere, and connect $M$ with the small sphere's tangent point $B$. Extend $\\overline{MB}$ through $B$ until it hits the ray from $A$ through the center of the small sphere (convince yourself that these two intersect). Call this intersection $D$, the center of the small sphere $C$, we want to find $BD$. By Pythagoras, $AC=\\sqrt{49^2-36^2}=\\sqrt{1105}$, and we know that $MA=36$ and $BC=13$. We know that $\\overline{MA}$ and $\\overline{BC}$ must be parallel, using ratios we realize that $CD=\\frac{13}{23}\\sqrt{1105}$. Apply the Pythagorean theorem to $\\triangle BCD$, $BD=\\frac{312}{23}$, so $312 + 23 = 335. ~Ross Gao",
"This solution refers to the Diagram section. [asy] /* Made by MRENTHUSIASM */ size(300); import graph3; import solids; currentprojection=orthographic((10,-3,-40)); triple O1, O2, O3, T1, T2, T3, A, L1, L2, M; O1 = (0,-36,0); O2 = (0,36,0); O3 = (0,0,-sqrt(1105)); T1 = (864*sqrt(1105)/1105,-36,-828*sqrt(1105)/1105); T2 = (864*sqrt(1105)/1105,36,-828*sqrt(1105)/1105); T3 = (24*sqrt(1105)/85,0,-108*sqrt(1105)/85); A = (0,0,-36*sqrt(1105)/23); L1 = shift(0,-80,0)*A; L2 = shift(0,80,0)*A; M = midpoint(T1--T2); draw(shift(O1)*rotate(-90,O1,O2)*scale3(36)*unithemisphere,yellow,light=White); draw(shift(O2)*rotate(-90,O1,O2)*scale3(36)*unithemisphere,yellow,light=White); draw(shift(O3)*rotate(-90,O1,O2)*scale3(13)*unithemisphere,red,light=White); draw(surface(L1--L2--(L2.x,L2.y,40)--(L1.x,L1.y,40)--cycle),gray); draw(shift(O1)*rotate(90,O1,O2)*scale3(36)*unithemisphere,yellow,light=White); draw(shift(O2)*rotate(90,O1,O2)*scale3(36)*unithemisphere,yellow,light=White); draw(shift(O3)*rotate(90,O1,O2)*scale3(13)*unithemisphere,red,light=White); draw(surface(T2--T1--T3--A--cycle),cyan); draw(surface(L1--L2--(L2.x,L2.y,L2.z-abs(T1.z))--(L1.x,L1.y,L1.z-abs(T2.z))--cycle),gray); draw(T1--T2--T3--cycle^^M--A--T2,blue); dot(\"$O_1$\",O1,(0,-1,1),linewidth(4.5)); dot(\"$O_2$\",O2,(0,1,1),linewidth(4.5)); dot(\"$O$\",O3,(0.5,-1,0),linewidth(4.5)); dot(\"$T_1$\",T1,(0,-1,-1),blue+linewidth(4.5)); dot(\"$T_2$\",T2,(0,1,-1),blue+linewidth(4.5)); dot(\"$T$\",T3,(1,1,2),blue+linewidth(4.5)); dot(\"$M$\",M,(0,0,5),blue+linewidth(4.5)); dot(\"$A$\",A,(-0.5,-1.5,0),red+linewidth(4.5)); [/asy] The isosceles triangle of centers $O_1 O_2 O$ ($O$ is the center of sphere of radii $13$) has sides $O_1 O = O_2 O = 36 + 13 = 49,$ and $O_1 O_2 = 36 + 36 = 72.$ Let $N$ be the midpoint $O_1 O_2$. The isosceles triangle of points of tangency $T_1 T_2 T$ has sides $T_1 T = T_2 T = 2 \\sqrt{13 \\cdot 36} = 12 \\sqrt{13}$ and $T_1 T_2 = 72.$ Let $M$ be the midpoint $T_1 T_2.$ The height $TM$ is $\\sqrt {12^2 \\cdot 13 - 36^2} = 12 \\sqrt {13-9} = 24.$ The tangents of the half-angle between the planes is $\\frac {TO}{AT} = \\frac {MN - TO}{TM},$ so $\\frac {13}{AT} = \\frac {36 - 13}{24},$ \\[AT = \\frac{24\\cdot 13}{23} = \\frac {312}{23} \\implies 312 + 23 = 335.\\] vladimir.shelomovskii@gmail.com, vvsss",
"This solution refers to the Diagram section. [asy] /* Made by MRENTHUSIASM */ size(300); import graph3; import solids; currentprojection=orthographic((10,-3,-40)); triple O1, O2, O3, T1, T2, T3, A, L1, L2, M; O1 = (0,-36,0); O2 = (0,36,0); O3 = (0,0,-sqrt(1105)); T1 = (864*sqrt(1105)/1105,-36,-828*sqrt(1105)/1105); T2 = (864*sqrt(1105)/1105,36,-828*sqrt(1105)/1105); T3 = (24*sqrt(1105)/85,0,-108*sqrt(1105)/85); A = (0,0,-36*sqrt(1105)/23); L1 = shift(0,-80,0)*A; L2 = shift(0,80,0)*A; M = midpoint(T1--T2); draw(shift(O1)*rotate(-90,O1,O2)*scale3(36)*unithemisphere,yellow,light=White); draw(shift(O2)*rotate(-90,O1,O2)*scale3(36)*unithemisphere,yellow,light=White); draw(shift(O3)*rotate(-90,O1,O2)*scale3(13)*unithemisphere,red,light=White); draw(surface(L1--L2--(L2.x,L2.y,40)--(L1.x,L1.y,40)--cycle),gray); draw(shift(O1)*rotate(90,O1,O2)*scale3(36)*unithemisphere,yellow,light=White); draw(shift(O2)*rotate(90,O1,O2)*scale3(36)*unithemisphere,yellow,light=White); draw(shift(O3)*rotate(90,O1,O2)*scale3(13)*unithemisphere,red,light=White); draw(surface(T2--T1--T3--A--cycle),cyan); draw(surface(L1--L2--(L2.x,L2.y,L2.z-abs(T1.z))--(L1.x,L1.y,L1.z-abs(T2.z))--cycle),gray); draw(T1--T2--T3--cycle^^M--A--T2,blue); dot(\"$O_1$\",O1,(0,-1,1),linewidth(4.5)); dot(\"$O_2$\",O2,(0,1,1),linewidth(4.5)); dot(\"$O$\",O3,(0.5,-1,0),linewidth(4.5)); dot(\"$T_1$\",T1,(0,-1,-1),blue+linewidth(4.5)); dot(\"$T_2$\",T2,(0,1,-1),blue+linewidth(4.5)); dot(\"$T$\",T3,(1,1,2),blue+linewidth(4.5)); dot(\"$M$\",M,(0,0,5),blue+linewidth(4.5)); dot(\"$A$\",A,(-0.5,-1.5,0),red+linewidth(4.5)); [/asy] The isosceles triangle of centers $O_1 O_2 O$ ($O$ is the center of sphere of radii $13$) has sides $O_1 O = O_2 O = 36 + 13 = 49,$ and $O_1 O_2 = 36 + 36 = 72.$ Let $N$ be the midpoint $O_1 O_2$. The isosceles triangle of points of tangency $T_1 T_2 T$ has sides $T_1 T = T_2 T = 2 \\sqrt{13 \\cdot 36} = 12 \\sqrt{13}$ and $T_1 T_2 = 72.$ Let $M$ be the midpoint $T_1 T_2.$ The height $TM$ is $\\sqrt {12^2 \\cdot 13 - 36^2} = 12 \\sqrt {13-9} = 24.$ The tangents of the half-angle between the planes is $\\frac {TO}{AT} = \\frac {MN - TO}{TM},$ so $\\frac {13}{AT} = \\frac {36 - 13}{24},$ \\[AT = \\frac{24\\cdot 13}{23} = \\frac {312}{23} \\implies 312 + 23 = 335.\\] vladimir.shelomovskii@gmail.com, vvsss"
] |
2021-II-11
| 2,021
| 11
|
A teacher was leading a class of four perfectly logical students. The teacher chose a set $S$ of four integers and gave a different number in $S$ to each student. Then the teacher announced to the class that the numbers in $S$ were four consecutive two-digit positive integers, that some number in $S$ was divisible by $6$ , and a different number in $S$ was divisible by $7$ . The teacher then asked if any of the students could deduce what $S$ is, but in unison, all of the students replied no. However, upon hearing that all four students replied no, each student was able to determine the elements of $S$ . Find the sum of all possible values of the greatest element of $S$ .
| 258
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II
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[
"Note that $\\operatorname{lcm}(6,7)=42.$ It is clear that $42\\not\\in S$ and $84\\not\\in S,$ otherwise the three other elements in $S$ are divisible by neither $6$ nor $7.$ In the table below, the multiples of $6$ are colored in yellow, and the multiples of $7$ are colored in green. By the least common multiple, we obtain cycles: If $n$ is a possible maximum value of $S,$ then $n+42$ must be another possible maximum value of $S,$ and vice versa. By observations, we circle all possible maximum values of $S.$ [asy] /* Made by MRENTHUSIASM */ size(20cm); fill((5,0)--(6,0)--(6,2)--(5,2)--cycle,yellow); fill((11,0)--(12,0)--(12,3)--(11,3)--cycle,yellow); fill((17,1)--(18,1)--(18,3)--(17,3)--cycle,yellow); fill((23,1)--(24,1)--(24,3)--(23,3)--cycle,yellow); fill((29,1)--(30,1)--(30,3)--(29,3)--cycle,yellow); fill((35,1)--(36,1)--(36,3)--(35,3)--cycle,yellow); fill((6,0)--(7,0)--(7,2)--(6,2)--cycle,green); fill((13,0)--(14,0)--(14,3)--(13,3)--cycle,green); fill((20,1)--(21,1)--(21,3)--(20,3)--cycle,green); fill((27,1)--(28,1)--(28,3)--(27,3)--cycle,green); fill((34,1)--(35,1)--(35,3)--(34,3)--cycle,green); fill((42,3)--(41,3)--(41,2)--cycle,yellow); fill((42,2)--(41,2)--(41,1)--cycle,yellow); fill((42,3)--(42,2)--(41,2)--cycle,green); fill((42,2)--(42,1)--(41,1)--cycle,green); for (real i=9.5; i<=41.5; ++i) { label(\"$\"+string(i+0.5)+\"$\",(i,2.5),fontsize(9pt)); } for (real i=0.5; i<=41.5; ++i) { label(\"$\"+string(i+42.5)+\"$\",(i,1.5),fontsize(9pt)); } for (real i=0.5; i<=14.5; ++i) { label(\"$\"+string(i+84.5)+\"$\",(i,0.5),fontsize(9pt)); } draw(circle((6.5,1.5),0.45)); draw(circle((6.5,0.5),0.45)); draw(circle((7.5,1.5),0.45)); draw(circle((7.5,0.5),0.45)); draw(circle((8.5,1.5),0.45)); draw(circle((8.5,0.5),0.45)); draw(circle((13.5,2.5),0.45)); draw(circle((13.5,1.5),0.45)); draw(circle((13.5,0.5),0.45)); draw(circle((14.5,2.5),0.45)); draw(circle((14.5,1.5),0.45)); draw(circle((14.5,0.5),0.45)); draw(circle((20.5,2.5),0.45)); draw(circle((20.5,1.5),0.45)); draw(circle((23.5,2.5),0.45)); draw(circle((23.5,1.5),0.45)); draw(circle((29.5,2.5),0.45)); draw(circle((29.5,1.5),0.45)); draw(circle((30.5,2.5),0.45)); draw(circle((30.5,1.5),0.45)); draw(circle((35.5,2.5),0.45)); draw(circle((35.5,1.5),0.45)); draw(circle((36.5,2.5),0.45)); draw(circle((36.5,1.5),0.45)); draw(circle((37.5,2.5),0.45)); draw(circle((37.5,1.5),0.45)); draw((9,3)--(42,3)); draw((0,2)--(42,2)); draw((0,1)--(42,1)); draw((0,0)--(15,0)); for (real i=0; i<9; ++i) { draw((i,2)--(i,0)); } for (real i=9; i<16; ++i) { draw((i,3)--(i,0)); } for (real i=16; i<=42; ++i) { draw((i,3)--(i,1)); } [/asy] From the second row of the table above, we perform casework on the possible maximum value of $S:$ \\[\\begin{array}{c||c|c|l} & & & \\\\ [-2.5ex] \\textbf{Max Value} & \\boldsymbol{S} & \\textbf{Valid?} & \\hspace{16.25mm}\\textbf{Reasoning/Conclusion} \\\\ [0.5ex] \\hline & & & \\\\ [-2ex] 49 & \\{46,47,48,49\\} & & \\text{The student who gets } 46 \\text{ will reply yes.} \\\\ 50 & \\{47,48,49,50\\} & \\checkmark & \\text{Another possibility is } S=\\{89,90,91,92\\}. \\\\ 51 & \\{48,49,50,51\\} & & \\text{The student who gets } 51 \\text{ will reply yes.} \\\\ 56 & \\{53,54,55,56\\} & & \\text{The student who gets } 53 \\text{ will reply yes.} \\\\ 57 & \\{54,55,56,57\\} & & \\text{The student who gets } 57 \\text{ will reply yes.} \\\\ 63 & \\{60,61,62,63\\} & & \\text{The students who get } 60,61,62 \\text{ will reply yes.} \\\\ 66 & \\{63,64,65,66\\} & & \\text{The students who get } 64,65,66 \\text{ will reply yes.} \\\\ 72 & \\{69,70,71,72\\} & & \\text{The student who gets } 69 \\text{ will reply yes.} \\\\ 73 & \\{70,71,72,73\\} & & \\text{The student who gets } 73 \\text{ will reply yes.} \\\\ 78 & \\{75,76,77,78\\} & & \\text{The student who gets } 75 \\text{ will reply yes.} \\\\ 79 & \\{76,77,78,79\\} & \\checkmark & \\text{Another possibility is } S=\\{34,35,36,37\\}. \\\\ 80 & \\{77,78,79,80\\} & & \\text{The student who gets } 80 \\text{ will reply yes.} \\end{array}\\] Finally, all possibilities for $S$ are $\\{34,35,36,37\\}, \\{47,48,49,50\\}, \\{76,77,78,79\\},$ and $\\{89,90,91,92\\},$ from which the answer is $37+50+79+92=258 has only one possibility. ~MRENTHUSIASM",
"We know right away that $42\\not\\in S$ and $84\\not\\in S$ as stated in Solution 1. To get a feel for the problem, let’s write out some possible values of $S$ based on the teacher’s remarks. The first multiple of 7 that is two-digit is 14. The closest multiple of six from 14 is 12, and therefore there are two possible sets of four consecutive integers containing 12 and 14; $\\{11,12,13,14\\}$ and $\\{12,13,14,15\\}$. Here we get our first crucial idea; that if the multiples of 6 and 7 differ by two, there will be 2 possible sets of $S$ without any student input. Similarly, if they differ by 3, there will be only 1 possible set, and if they differ by 1, 3 possible sets. Now we read the student input. Each student says they can’t figure out what $S$ is just based on the teacher’s information, which means each student has to have a number that would be in 2 or 3 of the possible sets (This is based off of the first line of student input). However, now that each student knows that all of them have numbers that fit into more than one possible set, this means that S cannot have two possible sets because otherwise, when shifting from one set to the other, one of the end numbers would not be in the shifted set, but we know each number has to fall in two or more possible sets. For example, take $\\{11,12,13,14\\}$ and $\\{12,13,14,15\\}$. The numbers at the end, 11 and 15, only fall in one set, but each number must fall in at least two sets. This means that there must be three possible sets of S, in which case the actual S would be the middle S. Take for example $\\{33,34,35,36\\}$, $\\{34,35,36,37\\}$, and $\\{35,36,37,38\\}$. 37 and 34 fall in two sets while 35 and 36 fall in all three sets, so the condition is met. Now, this means that the multiple of 6 and 7 must differ by 1. Since 42 means the difference is 0, when you add/subtract 6 and 7, you will obtain the desired difference of 1, and similarly adding/subtracting 6 or 7 from 84 will also obtain the difference of 1. Thus there are four possible sets of $S$; $\\{34,35,36,37\\}$, $\\{47,48,49,50\\}$, $\\{76,77,78,79\\}$ and $\\{89,90,91,92.\\}$. Therefore the sum of the greatest elements of the possible sets $S$ is $37+50+79+92=258 ~KingRavi",
"In a solution that satisfies these constraints, the multiple of 6 must be adjacent to multiple of 7. The other two numbers must be on either side. WLOG assume the set is $\\{a,6j,7k,b\\}$. The student with numbers $a$, $6j$, and $7k$ can think the set is $\\{a-1, a,6j,7k\\}$ or $\\{a,6j,7k,b\\}$, and the students with number $6j$, $7k$, and $b$ can think the set is $\\{a,6j,7k,b\\}$ or $\\{6j,7k,b, b+1\\}$. Therefore, none of the students know the set for sure. Playing around with the arrangement of the multiple of 6 and multiple of 7 shows that this is the only configuration viewed as ambiguous to all the students. (Therefore when they hear nobody else knows either, they can find out it is this configuration) Considering $S$ as $\\{a,6j,7k,b\\}$,b is 2 mod 6 and 1 mod 7, so $b$ is 8 mod 42. (since it is all 2-digit, the values are either 50 or 92). Similarly, considering $S$ as $\\{a,7j,6k,b\\}$, $b$ is 1 mod 6 and 2 mod 7, so $b$ is 37 mod 42. The values that satisfy that are 37 and 79. The total sum of all these values is therefore $50+92+37+79=258. This solution will even work when the bounds (in this question, 2-digit so <100) are much larger and it is impractical to perform casework. ~balav123",
"Consider the tuple $(a, a+1, a+2, a+3)$ as a possible $S$. If one of the values in $S$ is $3$ or $4 \\pmod{7}$, observe the student will be able to deduce $S$ with no additional information. This is because, if a value is $b = 3 \\pmod{7}$ and $S$ contains a $0 \\pmod{7}$, then the values of $S$ must be $(b-3, b-2, b-1, b)$. Similarly, if we are given a $b \\equiv 4 \\pmod{7}$ and we know that $0 \\pmod{7}$ is in $S$, $S$ must be $(b, b+1, b+2, b+3).$ Hence, the only possibility for $a$ is $5, 6 \\pmod{7}.$ In either case, we are guaranteed there is a $6, 0, 1 \\pmod{7}$ value in $7$. The difference comes down to if there is a $5 \\pmod{7}$ value or a $1 \\pmod{7}$ value. The person receiving such value will be able to determine all of $S$ but the $6, 0, 1 \\pmod{7}$ people will not be able to differentiate the two cases ... yet. Now consider which value among the consecutive integers is $c \\equiv 3 \\pmod{6}$, if any. The person will know that $S$ is either $(c-3, c-2, c-1, c)$ or $(c, c+1, c+2, c+3)$ to have a $0 \\pmod{6}$ value in $S$. Neither the $6, 1 \\pmod{7}$ person can be $3 \\pmod{7}$, else they can decipher what $S$ is right off the bat by considering which set has $0 \\pmod{7}.$ This translates to the possible $5 \\pmod{7}$ or $1 \\pmod{7}$ person cannot be $0 \\pmod{6}$. We are given that $0 \\pmod{7}$ and $0 \\pmod{6}$ cannot be the same person. Hence we conclude one of the $6 \\pmod{7}$ or $1 \\pmod{7}$ must be the $0 \\pmod{6}$ person. Let the $6 \\pmod{7}$ person be $0 \\pmod{6}.$ Then--hypothetical--$c \\equiv 2 \\pmod{7}$ person is $3 \\pmod{6}.$ After the first round, the $6, 0, 1 \\pmod{7}$ people realize that $2 \\pmod{7}$ is not in $S$ else they would have deduced $S$ by noting $S$ was either $(c-3, c-2, c-1, c)$ or $(c, c+1, c+2, c+3)$ to have a $0 \\pmod{6}$ and choosing the former based on where $0 \\pmod{7}$ is. Hence they figure out $S$ by knowing $5 \\pmod{7}.$ So $a \\equiv 5 \\pmod{7}$ and $a \\equiv 5 \\pmod{6}$ (from $6 \\pmod{7}$ person being $0 \\pmod{6}$). Similarly, if $1 \\pmod{7}$ person is $0 \\pmod{6}$ we find that $a \\equiv 6 \\pmod{7}$ (so a $2 \\pmod{7}$ is in $S$) and $a \\equiv 4 \\pmod{6}.$ By CRT, the possibilities are $a \\equiv 5, 34 \\pmod{42}.$ The sum of the greatest values of $S$ are the sum of $a + 3$ and so we get $(34 + 3) + (47 + 3) + (76 + 3) + (89 + 3) = 258 ~Aaryabhatta1",
"Consider the tuple $(a, a+1, a+2, a+3)$ as a possible $S$. If one of the values in $S$ is $3$ or $4 \\pmod{7}$, observe the student will be able to deduce $S$ with no additional information. This is because, if a value is $b = 3 \\pmod{7}$ and $S$ contains a $0 \\pmod{7}$, then the values of $S$ must be $(b-3, b-2, b-1, b)$. Similarly, if we are given a $b \\equiv 4 \\pmod{7}$ and we know that $0 \\pmod{7}$ is in $S$, $S$ must be $(b, b+1, b+2, b+3).$ Hence, the only possibility for $a$ is $5, 6 \\pmod{7}.$ In either case, we are guaranteed there is a $6, 0, 1 \\pmod{7}$ value in $7$. The difference comes down to if there is a $5 \\pmod{7}$ value or a $1 \\pmod{7}$ value. The person receiving such value will be able to determine all of $S$ but the $6, 0, 1 \\pmod{7}$ people will not be able to differentiate the two cases ... yet. Now consider which value among the consecutive integers is $c \\equiv 3 \\pmod{6}$, if any. The person will know that $S$ is either $(c-3, c-2, c-1, c)$ or $(c, c+1, c+2, c+3)$ to have a $0 \\pmod{6}$ value in $S$. Neither the $6, 1 \\pmod{7}$ person can be $3 \\pmod{7}$, else they can decipher what $S$ is right off the bat by considering which set has $0 \\pmod{7}.$ This translates to the possible $5 \\pmod{7}$ or $1 \\pmod{7}$ person cannot be $0 \\pmod{6}$. We are given that $0 \\pmod{7}$ and $0 \\pmod{6}$ cannot be the same person. Hence we conclude one of the $6 \\pmod{7}$ or $1 \\pmod{7}$ must be the $0 \\pmod{6}$ person. Let the $6 \\pmod{7}$ person be $0 \\pmod{6}.$ Then--hypothetical--$c \\equiv 2 \\pmod{7}$ person is $3 \\pmod{6}.$ After the first round, the $6, 0, 1 \\pmod{7}$ people realize that $2 \\pmod{7}$ is not in $S$ else they would have deduced $S$ by noting $S$ was either $(c-3, c-2, c-1, c)$ or $(c, c+1, c+2, c+3)$ to have a $0 \\pmod{6}$ and choosing the former based on where $0 \\pmod{7}$ is. Hence they figure out $S$ by knowing $5 \\pmod{7}.$ So $a \\equiv 5 \\pmod{7}$ and $a \\equiv 5 \\pmod{6}$ (from $6 \\pmod{7}$ person being $0 \\pmod{6}$). Similarly, if $1 \\pmod{7}$ person is $0 \\pmod{6}$ we find that $a \\equiv 6 \\pmod{7}$ (so a $2 \\pmod{7}$ is in $S$) and $a \\equiv 4 \\pmod{6}.$ By CRT, the possibilities are $a \\equiv 5, 34 \\pmod{42}.$ The sum of the greatest values of $S$ are the sum of $a + 3$ and so we get $(34 + 3) + (47 + 3) + (76 + 3) + (89 + 3) = 258 ~Aaryabhatta1",
"Start by writing down every two-digit multiple of $6$ and $7$. First, $42$ and $84$ will not be in $S$ because we need two distinct numbers where one is a multiple of $6$ and ANOTHER is a multiple of $7$. Furthermore, it is clear at this point that if the multiples of $6$ and $7$ are exactly $3$ apart, then it will be obvious what $S$ is; it is just the smaller multiple to the larger multiple. Therefore, we cross out numbers $18, 21, 24, 60, 63, 66$. That leaves us with pairs of multiples of $6$ and $7$ that are exactly $1$ or $2$ apart. In this case, if the multiples of $6$ and $7$ are two apart, such as $12$ and $14$, then the non-multiples will be able to deduce what the set is; this leaves that the $6$ and $7$ must be one apart. Specifically, these are pairs $(35, 36), (48, 49), (77, 78),$ and $(90, 91)$. In each of these pairs, respectively, the largest number will be one greater than the larger multiple, namely $37, 50, 79, 92$. Therefore, our answer is $37 + 50 + 79 + 92 = 258. ~ xHypotenuse",
"Start by writing down every two-digit multiple of $6$ and $7$. First, $42$ and $84$ will not be in $S$ because we need two distinct numbers where one is a multiple of $6$ and ANOTHER is a multiple of $7$. Furthermore, it is clear at this point that if the multiples of $6$ and $7$ are exactly $3$ apart, then it will be obvious what $S$ is; it is just the smaller multiple to the larger multiple. Therefore, we cross out numbers $18, 21, 24, 60, 63, 66$. That leaves us with pairs of multiples of $6$ and $7$ that are exactly $1$ or $2$ apart. In this case, if the multiples of $6$ and $7$ are two apart, such as $12$ and $14$, then the non-multiples will be able to deduce what the set is; this leaves that the $6$ and $7$ must be one apart. Specifically, these are pairs $(35, 36), (48, 49), (77, 78),$ and $(90, 91)$. In each of these pairs, respectively, the largest number will be one greater than the larger multiple, namely $37, 50, 79, 92$. Therefore, our answer is $37 + 50 + 79 + 92 = 258. ~ xHypotenuse"
] |
2021-II-12
| 2,021
| 12
|
A convex quadrilateral has area $30$ and side lengths $5, 6, 9,$ and $7,$ in that order. Denote by $\theta$ the measure of the acute angle formed by the diagonals of the quadrilateral. Then $\tan \theta$ can be written in the form $\tfrac{m}{n}$ , where $m$ and $n$ are relatively prime positive integers. Find $m + n$ .
| 47
|
II
|
[
"Since we are asked to find $\\tan \\theta$, we can find $\\sin \\theta$ and $\\cos \\theta$ separately and use their values to get $\\tan \\theta$. We can start by drawing a diagram. Let the vertices of the quadrilateral be $A$, $B$, $C$, and $D$. Let $AB = 5$, $BC = 6$, $CD = 9$, and $DA = 7$. Let $AX = a$, $BX = b$, $CX = c$, and $DX = d$. We know that $\\theta$ is the acute angle formed between the intersection of the diagonals $AC$ and $BD$. [asy] unitsize(4cm); pair A,B,C,D,X; A = (0,0); B = (1,0); C = (1.25,-1); D = (-0.75,-0.75); draw(A--B--C--D--cycle,black+1bp); X = intersectionpoint(A--C,B--D); draw(A--C); draw(B--D); label(\"$A$\",A,NW); label(\"$B$\",B,NE); label(\"$C$\",C,SE); label(\"$D$\",D,SW); dot(X); label(\"$X$\",X,S); label(\"$5$\",(A+B)/2,N); label(\"$6$\",(B+C)/2,E); label(\"$9$\",(C+D)/2,S); label(\"$7$\",(D+A)/2,W); label(\"$\\theta$\",X,2.5E); label(\"$a$\",(A+X)/2,NE); label(\"$b$\",(B+X)/2,NW); label(\"$c$\",(C+X)/2,SW); label(\"$d$\",(D+X)/2,SE); [/asy] We are given that the area of quadrilateral $ABCD$ is $30$. We can express this area using the areas of triangles $AXB$, $BXC$, $CXD$, and $DXA$. Since we want to find $\\sin \\theta$ and $\\cos \\theta$, we can represent these areas using $\\sin \\theta$ as follows: \\begin{align*} 30 &=[ABCD] \\\\ &=[AXB] + [BXC] + [CXD] + [DXA] \\\\ &=\\frac{1}{2} ab \\sin (\\angle AXB) + \\frac{1}{2} bc \\sin (\\angle BXC) + \\frac{1}{2} cd \\sin (\\angle CXD) + \\frac{1}{2} da \\sin (\\angle AXD) \\\\ &=\\frac{1}{2} ab \\sin (180^\\circ - \\theta) + \\frac{1}{2} bc \\sin (\\theta) + \\frac{1}{2} cd \\sin (180^\\circ - \\theta) + \\frac{1}{2} da \\sin (\\theta). \\end{align*} We know that $\\sin (180^\\circ - \\theta) = \\sin \\theta$. Therefore it follows that: \\begin{align*} 30 &=\\frac{1}{2} ab \\sin (180^\\circ - \\theta) + \\frac{1}{2} bc \\sin (\\theta) + \\frac{1}{2} cd \\sin (180^\\circ - \\theta) + \\frac{1}{2} da \\sin (\\theta) \\\\ &=\\frac{1}{2} ab \\sin (\\theta) + \\frac{1}{2} bc \\sin (\\theta) + \\frac{1}{2} cd \\sin (\\theta) + \\frac{1}{2} da \\sin (\\theta) \\\\ &=\\frac{1}{2}\\sin\\theta (ab + bc + cd + da). \\end{align*} From here we see that $\\sin \\theta = \\frac{60}{ab + bc + cd + da}$. Now we need to find $\\cos \\theta$. Using the Law of Cosines on each of the four smaller triangles, we get following equations: \\begin{align*} 5^2 &= a^2 + b^2 - 2ab\\cos(180^\\circ-\\theta), \\\\ 6^2 &= b^2 + c^2 - 2bc\\cos \\theta, \\\\ 9^2 &= c^2 + d^2 - 2cd\\cos(180^\\circ-\\theta), \\\\ 7^2 &= d^2 + a^2 - 2da\\cos \\theta. \\end{align*} We know that $\\cos (180^\\circ - \\theta) = -\\cos \\theta$ for all $\\theta$. We can substitute this value into our equations to get: \\begin{align*} 5^2 &= a^2 + b^2 + 2ab\\cos \\theta, &&(1) \\\\ 6^2 &= b^2 + c^2 - 2bc\\cos \\theta, &&(2) \\\\ 9^2 &= c^2 + d^2 + 2cd\\cos \\theta, &&(3) \\\\ 7^2 &= d^2 + a^2 - 2da\\cos \\theta. &&(4) \\end{align*} If we subtract $(2)+(4)$ from $(1)+(3)$, the squared terms cancel, leaving us with: \\begin{align*} 5^2 + 9^2 - 6^2 - 7^2 &= 2ab \\cos \\theta + 2bc \\cos \\theta + 2cd \\cos \\theta + 2da \\cos \\theta \\\\ 21 &= 2\\cos \\theta (ab + bc + cd + da). \\end{align*} From here we see that $\\cos \\theta = \\frac{21/2}{ab + bc + cd + da}$. Since we have figured out $\\sin \\theta$ and $\\cos \\theta$, we can calculate $\\tan \\theta$: \\[\\tan \\theta = \\frac{\\sin \\theta}{\\cos \\theta} = \\frac{\\frac{60}{ab + bc + cd + da}}{\\frac{21/2}{ab + bc + cd + da}} = \\frac{60}{21/2} = \\frac{120}{21} = \\frac{40}{7}.\\] Therefore our answer is $40 + 7 = 047. ~ Steven Chen (www.professorchenedu.com) ~ my_aops_lessons",
"Since we are asked to find $\\tan \\theta$, we can find $\\sin \\theta$ and $\\cos \\theta$ separately and use their values to get $\\tan \\theta$. We can start by drawing a diagram. Let the vertices of the quadrilateral be $A$, $B$, $C$, and $D$. Let $AB = 5$, $BC = 6$, $CD = 9$, and $DA = 7$. Let $AX = a$, $BX = b$, $CX = c$, and $DX = d$. We know that $\\theta$ is the acute angle formed between the intersection of the diagonals $AC$ and $BD$. [asy] unitsize(4cm); pair A,B,C,D,X; A = (0,0); B = (1,0); C = (1.25,-1); D = (-0.75,-0.75); draw(A--B--C--D--cycle,black+1bp); X = intersectionpoint(A--C,B--D); draw(A--C); draw(B--D); label(\"$A$\",A,NW); label(\"$B$\",B,NE); label(\"$C$\",C,SE); label(\"$D$\",D,SW); dot(X); label(\"$X$\",X,S); label(\"$5$\",(A+B)/2,N); label(\"$6$\",(B+C)/2,E); label(\"$9$\",(C+D)/2,S); label(\"$7$\",(D+A)/2,W); label(\"$\\theta$\",X,2.5E); label(\"$a$\",(A+X)/2,NE); label(\"$b$\",(B+X)/2,NW); label(\"$c$\",(C+X)/2,SW); label(\"$d$\",(D+X)/2,SE); [/asy] We are given that the area of quadrilateral $ABCD$ is $30$. We can express this area using the areas of triangles $AXB$, $BXC$, $CXD$, and $DXA$. Since we want to find $\\sin \\theta$ and $\\cos \\theta$, we can represent these areas using $\\sin \\theta$ as follows: \\begin{align*} 30 &=[ABCD] \\\\ &=[AXB] + [BXC] + [CXD] + [DXA] \\\\ &=\\frac{1}{2} ab \\sin (\\angle AXB) + \\frac{1}{2} bc \\sin (\\angle BXC) + \\frac{1}{2} cd \\sin (\\angle CXD) + \\frac{1}{2} da \\sin (\\angle AXD) \\\\ &=\\frac{1}{2} ab \\sin (180^\\circ - \\theta) + \\frac{1}{2} bc \\sin (\\theta) + \\frac{1}{2} cd \\sin (180^\\circ - \\theta) + \\frac{1}{2} da \\sin (\\theta). \\end{align*} We know that $\\sin (180^\\circ - \\theta) = \\sin \\theta$. Therefore it follows that: \\begin{align*} 30 &=\\frac{1}{2} ab \\sin (180^\\circ - \\theta) + \\frac{1}{2} bc \\sin (\\theta) + \\frac{1}{2} cd \\sin (180^\\circ - \\theta) + \\frac{1}{2} da \\sin (\\theta) \\\\ &=\\frac{1}{2} ab \\sin (\\theta) + \\frac{1}{2} bc \\sin (\\theta) + \\frac{1}{2} cd \\sin (\\theta) + \\frac{1}{2} da \\sin (\\theta) \\\\ &=\\frac{1}{2}\\sin\\theta (ab + bc + cd + da). \\end{align*} From here we see that $\\sin \\theta = \\frac{60}{ab + bc + cd + da}$. Now we need to find $\\cos \\theta$. Using the Law of Cosines on each of the four smaller triangles, we get following equations: \\begin{align*} 5^2 &= a^2 + b^2 - 2ab\\cos(180^\\circ-\\theta), \\\\ 6^2 &= b^2 + c^2 - 2bc\\cos \\theta, \\\\ 9^2 &= c^2 + d^2 - 2cd\\cos(180^\\circ-\\theta), \\\\ 7^2 &= d^2 + a^2 - 2da\\cos \\theta. \\end{align*} We know that $\\cos (180^\\circ - \\theta) = -\\cos \\theta$ for all $\\theta$. We can substitute this value into our equations to get: \\begin{align*} 5^2 &= a^2 + b^2 + 2ab\\cos \\theta, &&(1) \\\\ 6^2 &= b^2 + c^2 - 2bc\\cos \\theta, &&(2) \\\\ 9^2 &= c^2 + d^2 + 2cd\\cos \\theta, &&(3) \\\\ 7^2 &= d^2 + a^2 - 2da\\cos \\theta. &&(4) \\end{align*} If we subtract $(2)+(4)$ from $(1)+(3)$, the squared terms cancel, leaving us with: \\begin{align*} 5^2 + 9^2 - 6^2 - 7^2 &= 2ab \\cos \\theta + 2bc \\cos \\theta + 2cd \\cos \\theta + 2da \\cos \\theta \\\\ 21 &= 2\\cos \\theta (ab + bc + cd + da). \\end{align*} From here we see that $\\cos \\theta = \\frac{21/2}{ab + bc + cd + da}$. Since we have figured out $\\sin \\theta$ and $\\cos \\theta$, we can calculate $\\tan \\theta$: \\[\\tan \\theta = \\frac{\\sin \\theta}{\\cos \\theta} = \\frac{\\frac{60}{ab + bc + cd + da}}{\\frac{21/2}{ab + bc + cd + da}} = \\frac{60}{21/2} = \\frac{120}{21} = \\frac{40}{7}.\\] Therefore our answer is $40 + 7 = 047. ~ Steven Chen (www.professorchenedu.com) ~ my_aops_lessons",
"In convex quadrilateral $ABCD,$ let $AB=5,BC=6,CD=9,$ and $DA=7.$ Let $A'$ and $C'$ be the feet of the perpendiculars from $A$ and $C,$ respectively, to $\\overline{BD}.$ We obtain the following diagram: [asy] /* Made by MRENTHUSIASM */ size(500); pair A, B, C, D, P, A1, C1; B = origin; D = (3*sqrt(32498*(29400*sqrt(47)+312523))/32498,0); A = intersectionpoints(Circle(B,5),Circle(D,7))[0]; C = intersectionpoints(Circle(B,6),Circle(D,9))[1]; P = intersectionpoint(A--C,B--D); A1 = foot(A,B,D); C1 = foot(C,B,D); markscalefactor=3/160; draw(rightanglemark(A,A1,D),red); draw(rightanglemark(C,C1,B),red); dot(\"$A$\",A,1.5*dir(aCos(7/sqrt(1649)))); dot(\"$B$\",B,1.5*W); dot(\"$C$\",C,1.5*dir(180+aCos(7/sqrt(1649)))); dot(\"$D$\",D,1.5*E); dot(\"$E$\",P,dir(180-(180-aCos(7/sqrt(1649)))/2)); dot(\"$A'$\",A1,dir(-75)); dot(\"$C'$\",C1,N); label(\"$\\theta$\",P,dir(180+aCos(7/sqrt(1649))/2),red); draw(A--A1^^C--C1,dashed); draw(A--B--C--D--cycle^^A--C^^B--D); [/asy] Let $BC'=p,C'E=q,EA'=r,A'D=s,AA'=h_1,$ and $CC'=h_2.$ We apply the Pythagorean Theorem to right triangles $\\triangle ABA',\\triangle BCC',\\triangle CDC',$ and $\\triangle DAA',$ respectively: \\[\\begin{array}{ccccccccccccccccc} (p+q+r)^2&+&h_1^2&=&5^2, &&&&&&&&&&&&\\hspace{36mm}(1) \\\\ [1ex] p^2&+&h_2^2&=&6^2, &&&&&&&&&&&&\\hspace{36mm}(2) \\\\ [1ex] (q+r+s)^2&+&h_2^2&=&9^2, &&&&&&&&&&&&\\hspace{36mm}(3) \\\\ [1ex] s^2&+&h_1^2&=&7^2. &&&&&&&&&&&&\\hspace{36mm}(4) \\end{array}\\] Let the brackets denote areas. We get \\begin{align*} [ABD]+[CBD]&=[ABCD] \\\\ \\frac12(p+q+r+s)h_1+\\frac12(p+q+r+s)h_2&=30 \\\\ \\frac12(p+q+r+s)(h_1+h_2)&=30 \\\\ (p+q+r+s)(h_1+h_2)&=60. \\hspace{49.25mm}(5) \\end{align*} We subtract $(2)+(4)$ from $(1)+(3):$ \\begin{align*} (p+q+r)^2+(q+r+s)^2-p^2-s^2&=21 \\\\ \\left[(p+q+r)^2-s^2\\right]+\\left[(q+r+s)^2-p^2\\right]&=21 \\\\ (p+q+r+s)(p+q+r-s)+(p+q+r+s)(-p+q+r+s)&=21 \\\\ (p+q+r+s)(2q+2r)&=21 \\\\ 2(p+q+r+s)(q+r)&=21 \\\\ (p+q+r+s)(q+r)&=\\frac{21}{2}. \\hspace{9.5mm}(6) \\end{align*} From right triangles $\\triangle AEA'$ and $\\triangle CEC',$ we have $\\tan\\theta=\\frac{h_1}{r}=\\frac{h_2}{q}.$ It follows that \\begin{alignat*}{8} \\tan\\theta&=\\frac{h_1}{r}\\qquad&\\implies\\qquad h_1&=r\\tan\\theta, \\hspace{64mm}&(1\\star)\\\\ \\tan\\theta&=\\frac{h_2}{q}\\qquad&\\implies\\qquad h_2&=q\\tan\\theta. &(2\\star) \\end{alignat*} Finally, we divide $(5)$ by $(6):$ \\begin{align*} \\frac{h_1+h_2}{q+r}&=\\frac{40}{7} \\\\ \\frac{r\\tan\\theta+q\\tan\\theta}{q+r}&=\\frac{40}{7} \\hspace{15mm} &&\\text{by }(1\\star)\\text{ and }(2\\star)\\\\ \\frac{(r+q)\\tan\\theta}{q+r}&=\\frac{40}{7} \\\\ \\tan\\theta&=\\frac{40}{7}, \\end{align*} from which the answer is $40+7=047 ~MRENTHUSIASM",
"In convex quadrilateral $ABCD,$ let $AB=5,BC=6,CD=9,$ and $DA=7.$ Let $A'$ and $C'$ be the feet of the perpendiculars from $A$ and $C,$ respectively, to $\\overline{BD}.$ We obtain the following diagram: [asy] /* Made by MRENTHUSIASM */ size(500); pair A, B, C, D, P, A1, C1; B = origin; D = (3*sqrt(32498*(29400*sqrt(47)+312523))/32498,0); A = intersectionpoints(Circle(B,5),Circle(D,7))[0]; C = intersectionpoints(Circle(B,6),Circle(D,9))[1]; P = intersectionpoint(A--C,B--D); A1 = foot(A,B,D); C1 = foot(C,B,D); markscalefactor=3/160; draw(rightanglemark(A,A1,D),red); draw(rightanglemark(C,C1,B),red); dot(\"$A$\",A,1.5*dir(aCos(7/sqrt(1649)))); dot(\"$B$\",B,1.5*W); dot(\"$C$\",C,1.5*dir(180+aCos(7/sqrt(1649)))); dot(\"$D$\",D,1.5*E); dot(\"$E$\",P,dir(180-(180-aCos(7/sqrt(1649)))/2)); dot(\"$A'$\",A1,dir(-75)); dot(\"$C'$\",C1,N); label(\"$\\theta$\",P,dir(180+aCos(7/sqrt(1649))/2),red); draw(A--A1^^C--C1,dashed); draw(A--B--C--D--cycle^^A--C^^B--D); [/asy] Let $BC'=p,C'E=q,EA'=r,A'D=s,AA'=h_1,$ and $CC'=h_2.$ We apply the Pythagorean Theorem to right triangles $\\triangle ABA',\\triangle BCC',\\triangle CDC',$ and $\\triangle DAA',$ respectively: \\[\\begin{array}{ccccccccccccccccc} (p+q+r)^2&+&h_1^2&=&5^2, &&&&&&&&&&&&\\hspace{36mm}(1) \\\\ [1ex] p^2&+&h_2^2&=&6^2, &&&&&&&&&&&&\\hspace{36mm}(2) \\\\ [1ex] (q+r+s)^2&+&h_2^2&=&9^2, &&&&&&&&&&&&\\hspace{36mm}(3) \\\\ [1ex] s^2&+&h_1^2&=&7^2. &&&&&&&&&&&&\\hspace{36mm}(4) \\end{array}\\] Let the brackets denote areas. We get \\begin{align*} [ABD]+[CBD]&=[ABCD] \\\\ \\frac12(p+q+r+s)h_1+\\frac12(p+q+r+s)h_2&=30 \\\\ \\frac12(p+q+r+s)(h_1+h_2)&=30 \\\\ (p+q+r+s)(h_1+h_2)&=60. \\hspace{49.25mm}(5) \\end{align*} We subtract $(2)+(4)$ from $(1)+(3):$ \\begin{align*} (p+q+r)^2+(q+r+s)^2-p^2-s^2&=21 \\\\ \\left[(p+q+r)^2-s^2\\right]+\\left[(q+r+s)^2-p^2\\right]&=21 \\\\ (p+q+r+s)(p+q+r-s)+(p+q+r+s)(-p+q+r+s)&=21 \\\\ (p+q+r+s)(2q+2r)&=21 \\\\ 2(p+q+r+s)(q+r)&=21 \\\\ (p+q+r+s)(q+r)&=\\frac{21}{2}. \\hspace{9.5mm}(6) \\end{align*} From right triangles $\\triangle AEA'$ and $\\triangle CEC',$ we have $\\tan\\theta=\\frac{h_1}{r}=\\frac{h_2}{q}.$ It follows that \\begin{alignat*}{8} \\tan\\theta&=\\frac{h_1}{r}\\qquad&\\implies\\qquad h_1&=r\\tan\\theta, \\hspace{64mm}&(1\\star)\\\\ \\tan\\theta&=\\frac{h_2}{q}\\qquad&\\implies\\qquad h_2&=q\\tan\\theta. &(2\\star) \\end{alignat*} Finally, we divide $(5)$ by $(6):$ \\begin{align*} \\frac{h_1+h_2}{q+r}&=\\frac{40}{7} \\\\ \\frac{r\\tan\\theta+q\\tan\\theta}{q+r}&=\\frac{40}{7} \\hspace{15mm} &&\\text{by }(1\\star)\\text{ and }(2\\star)\\\\ \\frac{(r+q)\\tan\\theta}{q+r}&=\\frac{40}{7} \\\\ \\tan\\theta&=\\frac{40}{7}, \\end{align*} from which the answer is $40+7=047 ~MRENTHUSIASM",
"Bretschneider's Formula [asy] size(200); import olympiad; defaultpen(linewidth(0.8)+fontsize(10)); pair A,B,C,D; A=origin; B=(1.25,0); D=dir(65); C=D+0.85*dir(90)*(A-D); draw(A--B--C--D--cycle); draw(A--C^^B--D, gray+0.5); dot(\"$A$\", A, SW); dot(\"$B$\", B, SE); dot(\"$C$\", C, NE); dot(\"$D$\",D,NW); label(\"$a$\", A--B, S); label(\"$b$\", B--C, E); label(\"$c$\", D--C, N); label(\"$d$\",D--A,W); label(\"$u$\",D--B,2*dir(170)); label(\"$v$\",A--C,S); [/asy] Given quadrilateral $ABCD$, let, $a, b, c, d$, be the sides, $s$ the semiperimeter, and $u, v$, the diagonals. Then the area, $K$, is given by \\[K = \\tfrac 14 \\sqrt {4u^2v^2-(b^2+d^2-a^2-c^2)^2}\\] Solution By Bretschneider's Formula, \\[30=\\tfrac{1}{4}\\sqrt{4u^2v^2-(b^2+d^2-a^2-c^2)^2}=\\tfrac{1}{4}\\sqrt{4u^2v^2-441}.\\] Thus, $uv=3\\sqrt{1649}$. Also, \\[[ABCD]=\\tfrac 12 \\cdot uv\\sin{\\theta};\\] solving for $\\sin{\\theta}$ yields $\\sin{\\theta}=\\tfrac{40}{\\sqrt{1649}}$. Since $\\theta$ is acute, $\\cos{\\theta}$ is positive, from which $\\cos{\\theta}=\\tfrac{7}{\\sqrt{1649}}$. Solving for $\\tan{\\theta}$ yields \\[\\tan{\\theta}=\\frac{\\sin{\\theta}}{\\cos{\\theta}}=\\frac{40}{7},\\] for a final answer of $047. ~ Leo.Euler",
"Bretschneider's Formula [asy] size(200); import olympiad; defaultpen(linewidth(0.8)+fontsize(10)); pair A,B,C,D; A=origin; B=(1.25,0); D=dir(65); C=D+0.85*dir(90)*(A-D); draw(A--B--C--D--cycle); draw(A--C^^B--D, gray+0.5); dot(\"$A$\", A, SW); dot(\"$B$\", B, SE); dot(\"$C$\", C, NE); dot(\"$D$\",D,NW); label(\"$a$\", A--B, S); label(\"$b$\", B--C, E); label(\"$c$\", D--C, N); label(\"$d$\",D--A,W); label(\"$u$\",D--B,2*dir(170)); label(\"$v$\",A--C,S); [/asy] Given quadrilateral $ABCD$, let, $a, b, c, d$, be the sides, $s$ the semiperimeter, and $u, v$, the diagonals. Then the area, $K$, is given by \\[K = \\tfrac 14 \\sqrt {4u^2v^2-(b^2+d^2-a^2-c^2)^2}\\] Solution By Bretschneider's Formula, \\[30=\\tfrac{1}{4}\\sqrt{4u^2v^2-(b^2+d^2-a^2-c^2)^2}=\\tfrac{1}{4}\\sqrt{4u^2v^2-441}.\\] Thus, $uv=3\\sqrt{1649}$. Also, \\[[ABCD]=\\tfrac 12 \\cdot uv\\sin{\\theta};\\] solving for $\\sin{\\theta}$ yields $\\sin{\\theta}=\\tfrac{40}{\\sqrt{1649}}$. Since $\\theta$ is acute, $\\cos{\\theta}$ is positive, from which $\\cos{\\theta}=\\tfrac{7}{\\sqrt{1649}}$. Solving for $\\tan{\\theta}$ yields \\[\\tan{\\theta}=\\frac{\\sin{\\theta}}{\\cos{\\theta}}=\\frac{40}{7},\\] for a final answer of $047. ~ Leo.Euler",
"By Bretschneider's Formula, \\[30=\\tfrac{1}{4}\\sqrt{4u^2v^2-(b^2+d^2-a^2-c^2)^2}=\\tfrac{1}{4}\\sqrt{4u^2v^2-441}.\\] Thus, $uv=3\\sqrt{1649}$. Also, \\[[ABCD]=\\tfrac 12 \\cdot uv\\sin{\\theta};\\] solving for $\\sin{\\theta}$ yields $\\sin{\\theta}=\\tfrac{40}{\\sqrt{1649}}$. Since $\\theta$ is acute, $\\cos{\\theta}$ is positive, from which $\\cos{\\theta}=\\tfrac{7}{\\sqrt{1649}}$. Solving for $\\tan{\\theta}$ yields \\[\\tan{\\theta}=\\frac{\\sin{\\theta}}{\\cos{\\theta}}=\\frac{40}{7},\\] for a final answer of $047. ~ Leo.Euler",
"Claim Given an inscribed quadrilateral $ABCD$ with sides $AB = a, BC = b, CD = c,$ and $DA = d.$ Prove that the $\\angle \\theta < 90^\\circ$ between the diagonals is given by \\begin{align*}2(ac + bd) \\cos \\theta = {|d^2 – c^2 + b^2 – a^2|}.\\end{align*} Proof Let the point $B'$ be symmetric to $B$ with respect to the perpendicular bisector $AC.$ Then the quadrilateral $AB'CD$ is an inscribed one, $AB' = b, B'C = a.$ \\[2 \\angle AEB = \\overset{\\Large\\frown} {AB} + \\overset{\\Large\\frown} {CD}.\\] \\begin{align*} 2\\angle B'AD = \\overset{\\Large\\frown} {B'C} + \\overset{\\Large\\frown} {CD} = \\overset{\\Large\\frown} {AB} + \\overset{\\Large\\frown} {CD} \\implies \\angle AEB = \\angle B'AD.\\end{align*} We apply the Law of Cosines to $\\triangle AB'D$ and $\\triangle CB'D$: \\begin{align*} B'D^2 = AD^2 + AB'^2 – 2 AD \\cdot AB' \\cos \\theta, \\end{align*} \\begin{align*} B'D^2 = CD^2 + CB'^2 + 2 CD \\cdot CB' \\cos \\theta,\\end{align*} \\begin{align*} d^2 + b^2 – 2 bd \\cos \\theta = c^2 + a^2 + 2ac \\cos \\theta,\\end{align*} \\begin{align*} 2(ac + bd) \\cos \\theta = |d^2 – c^2 + b^2 – a^2|.\\end{align*} vladimir.shelomovskii@gmail.com, vvsss",
"Claim Given an inscribed quadrilateral $ABCD$ with sides $AB = a, BC = b, CD = c,$ and $DA = d.$ Prove that the $\\angle \\theta < 90^\\circ$ between the diagonals is given by \\begin{align*}2(ac + bd) \\cos \\theta = {|d^2 – c^2 + b^2 – a^2|}.\\end{align*} Proof Let the point $B'$ be symmetric to $B$ with respect to the perpendicular bisector $AC.$ Then the quadrilateral $AB'CD$ is an inscribed one, $AB' = b, B'C = a.$ \\[2 \\angle AEB = \\overset{\\Large\\frown} {AB} + \\overset{\\Large\\frown} {CD}.\\] \\begin{align*} 2\\angle B'AD = \\overset{\\Large\\frown} {B'C} + \\overset{\\Large\\frown} {CD} = \\overset{\\Large\\frown} {AB} + \\overset{\\Large\\frown} {CD} \\implies \\angle AEB = \\angle B'AD.\\end{align*} We apply the Law of Cosines to $\\triangle AB'D$ and $\\triangle CB'D$: \\begin{align*} B'D^2 = AD^2 + AB'^2 – 2 AD \\cdot AB' \\cos \\theta, \\end{align*} \\begin{align*} B'D^2 = CD^2 + CB'^2 + 2 CD \\cdot CB' \\cos \\theta,\\end{align*} \\begin{align*} d^2 + b^2 – 2 bd \\cos \\theta = c^2 + a^2 + 2ac \\cos \\theta,\\end{align*} \\begin{align*} 2(ac + bd) \\cos \\theta = |d^2 – c^2 + b^2 – a^2|.\\end{align*} vladimir.shelomovskii@gmail.com, vvsss"
] |
2021-II-13
| 2,021
| 13
|
Find the least positive integer $n$ for which $2^n + 5^n - n$ is a multiple of $1000$ .
| 797
|
II
|
[
"Recall that $1000$ divides this expression if $8$ and $125$ both divide it. It should be fairly obvious that $n \\geq 3$; so we may break up the initial condition into two sub-conditions. (1) $5^n \\equiv n \\pmod{8}$. Notice that the square of any odd integer is $1$ modulo $8$ (proof by plugging in $1^2,3^2,5^2,7^2$ into modulo $8$), so the LHS of this expression goes $5,1,5,1,\\ldots$, while the RHS goes $1,2,3,4,5,6,7,8,1,\\ldots$. The cycle length of the LHS is $2$, RHS is $8$, so the cycle length of the solution is $\\operatorname{lcm}(2,8)=8$. Indeed, the $n$ that solve this congruence are exactly those such that $n \\equiv 5 \\pmod{8}$. (2) $2^n \\equiv n \\pmod{125}$. This is extremely computationally intensive if we try to calculate all $2^1,2^2,\\ldots,2^{100} \\pmod{125}$, so we take a divide-and-conquer approach instead. In order for this expression to be true, $2^n \\equiv n \\pmod{5}$ is necessary; it shouldn't take too long for us to go through the $20$ possible LHS-RHS combinations, considering that $n$ must be odd. We only need to test $10$ values of $n$ and obtain $n \\equiv 3 \\pmod{20}$ or $n \\equiv 17 \\pmod{20}$. With this in mind we consider $2^n \\equiv n \\pmod{25}$. By the Generalized Fermat's Little Theorem, $2^{20} \\equiv 1 \\pmod{25}$, but we already have $n$ modulo $20$. Our calculation is greatly simplified. The LHS's cycle length is $20$ and the RHS's cycle length is $25$, from which their least common multiple is $100$. In this step we need to test all the numbers between $1$ to $100$ that $n \\equiv 3 \\pmod{20}$ or $n \\equiv 17 \\pmod{20}$. In the case that $n \\equiv 3 \\pmod{20}$, the RHS goes $3,23,43,63,83$, and we need $2^n \\equiv n \\equiv 2^3 \\pmod{25}$; clearly $n \\equiv 83 \\pmod{100}$. In the case that $n \\equiv 17 \\pmod{20}$, by a similar argument, $n \\equiv 97 \\pmod{100}$. In the final step, we need to calculate $2^{97}$ and $2^{83}$ modulo $125$: Note that $2^{97}\\equiv2^{-3}$; because $8\\cdot47=376\\equiv1\\pmod{125},$ we get $2^{97} \\equiv 47\\pmod{125}$. Note that $2^{83}$ is $2^{-17}=2^{-16}\\cdot2^{-1}$. We have \\begin{align*} 2^{-1}&\\equiv63, \\\\ 2^{-2}&\\equiv63^2=3969\\equiv-31, \\\\ 2^{-4}&\\equiv(-31)^2=961\\equiv-39, \\\\ 2^{-8}&\\equiv1521\\equiv21, \\\\ 2^{-16}&\\equiv441, \\\\ 2^{-17}&\\equiv63\\cdot441\\equiv7\\cdot(-31)=-217\\equiv33. \\end{align*} This time, LHS cycle is $100$, RHS cycle is $125$, so we need to figure out $n$ modulo $500$. It should be $n \\equiv 283,297 \\pmod{500}$. Put everything together. By the second subcondition, the only candidates less than $1000$ are $283,297,783,797$. Apply the first subcondition, $n=797 is the desired answer. ~Ross Gao (Solution) ~MRENTHUSIASM (Minor Reformatting)",
"We have that $2^n + 5^n \\equiv n\\pmod{1000}$, or $2^n + 5^n \\equiv n \\pmod{8}$ and $2^n + 5^n \\equiv n \\pmod{125}$ by CRT. It is easy to check $n < 3$ don't work, so we have that $n \\geq 3$. Then, $2^n \\equiv 0 \\pmod{8}$ and $5^n \\equiv 0 \\pmod{125}$, so we just have $5^n \\equiv n \\pmod{8}$ and $2^n \\equiv n \\pmod{125}$. Let us consider both of these congruences separately. First, we look at $5^n \\equiv n \\pmod{8}$. By Euler's Totient Theorem (ETT), we have $5^4 \\equiv 1 \\pmod{8}$, so $5^5 \\equiv 5 \\pmod{8}$. On the RHS of the congruence, the possible values of $n$ are all nonnegative integers less than $8$ and on the RHS the only possible values are $5$ and $1$. However, for $5^n$ to be $1 \\pmod{8}$ we must have $n \\equiv 0 \\pmod{4}$, a contradiction. So, the only possible values of $n$ are when $n \\equiv 5 \\pmod{8} \\implies n = 8k+5$. Now we look at $2^n \\equiv n \\pmod{125}$. Plugging in $n = 8k+5$, we get $2^{8k+5} \\equiv 8k+5 \\pmod{125} \\implies 2^{8k} \\cdot 32 \\equiv 8k+5 \\pmod{125}$. Note, for $2^n \\equiv n\\pmod{125}$ to be satisfied, we must have $2^n \\equiv n \\pmod{5}$ and $2^n \\equiv n\\pmod{25}$. Since $2^{8k} \\equiv 1\\pmod{5}$ as $8k \\equiv 0\\pmod{4}$, we have $2 \\equiv -2k \\pmod{5} \\implies k = 5m-1$. Then, $n = 8(5m-1) + 5 = 40m-3$. Now, we get $2^{40m-3} \\equiv 40m-3 \\pmod{125}$. Using the fact that $2^n \\equiv n\\pmod{25}$, we get $2^{-3} \\equiv 15m-3 \\pmod{25}$. The inverse of $2$ modulo $25$ is obviously $13$, so $2^{-3} \\equiv 13^3 \\equiv 22 \\pmod{25}$, so $15m \\equiv 0 \\pmod{25} \\implies m = 5s$. Plugging in $m = 5s$, we get $n = 200s - 3$. Now, we are finally ready to plug $n$ into the congruence modulo $125$. Plugging in, we get $2^{200s-3} \\equiv 200s - 3 \\pmod{125}$. By ETT, we get $2^{100} \\equiv 1 \\pmod{125}$, so $2^{200s- 3} \\equiv 2^{-3} \\equiv 47 \\pmod{125}$. Then, $200s \\equiv 50 \\pmod{125} \\implies s \\equiv 4 \\pmod{5} \\implies s = 5y+4$. Plugging this in, we get $n = 200(5y+4) - 3 = 1000y+797$, implying the smallest value of $n$ is simply $797. ~rocketsri",
"We wish to find the least positive integer $n$ for which $2^n+5^n-n\\equiv0\\pmod{1000}.$ Rearranging gives \\[2^n+5^n\\equiv n\\pmod{1000}.\\] Applying the Chinese Remainder Theorem, we get the following system of congruences: \\begin{align*} 2^n+5^n &\\equiv n \\pmod{8}, \\\\ 2^n+5^n &\\equiv n \\pmod{125}. \\end{align*} It is clear that $n\\geq3,$ from which we simplify to \\begin{align*} 5^n &\\equiv n \\pmod{8}, \\hspace{15mm} &(1) \\\\ 2^n &\\equiv n \\pmod{125}. &(2) \\end{align*} We solve each congruence separately: For $(1),$ quick inspections produce that $5^1,5^2,5^3,5^4,\\ldots$ are congruent to $5,1,5,1,\\ldots$ modulo $8,$ respectively. More generally, $5^n \\equiv 5 \\pmod{8}$ if $n$ is odd, and $5^n \\equiv 1 \\pmod{8}$ if $n$ is even. As $5^n$ is always odd (so is $n$), we must have $n\\equiv5\\pmod{8}.$ That is, $\\boldsymbol{n=8r+5}$ for some nonnegative integer $\\boldsymbol{r.}$ For $(2),$ we substitute the result from $(1)$ and simplify: \\begin{align*} 2^{8r+5}&\\equiv8r+5\\pmod{125} \\\\ \\left(2^8\\right)^r\\cdot2^5&\\equiv8r+5\\pmod{125} \\\\ 256^r\\cdot32&\\equiv8r+5\\pmod{125} \\\\ 6^r\\cdot32&\\equiv8r+5\\pmod{125}. \\end{align*} Note that $5^3=125$ and $6=5+1,$ so we apply the Binomial Theorem to the left side: \\begin{align*} (5+1)^r\\cdot32&\\equiv8r+5\\pmod{125} \\\\ \\Biggl[\\binom{r}{0}5^0+\\binom{r}{1}5^1+\\binom{r}{2}5^2+\\phantom{ }\\underbrace{\\binom{r}{3}5^3+\\cdots+\\binom{r}{r}5^r}_{0\\pmod{125}}\\phantom{ }\\Biggr]\\cdot32&\\equiv8r+5\\pmod{125} \\\\ \\left[1+5r+\\frac{25r(r-1)}{2}\\right]\\cdot32&\\equiv8r+5\\pmod{125} \\\\ 32+160r+400r(r-1)&\\equiv8r+5\\pmod{125} \\\\ 32+35r+25r(r-1)&\\equiv8r+5\\pmod{125} \\\\ 25r^2+2r+27&\\equiv0\\phantom{r+5}\\pmod{125}. \\hspace{15mm} (*) \\end{align*} Since $125\\equiv0\\pmod{5},$ it follows that \\begin{align*} 25r^2+2r+27&\\equiv0\\pmod{5} \\\\ 2r+2&\\equiv0\\pmod{5} \\\\ r&\\equiv4\\pmod{5}. \\end{align*} That is, $\\boldsymbol{r=5s+4}$ for some nonnegative integer $\\boldsymbol{s.}$ Substituting this back into $(*),$ we get \\begin{align*} 25(5s+4)^2+2(5s+4)+27&\\equiv0\\pmod{125} \\\\ 625s^2+1010s+435&\\equiv0\\pmod{125} \\\\ 10s+60&\\equiv0\\pmod{125} \\\\ 10(s+6)&\\equiv0\\pmod{125}. \\end{align*} As $10(s+6)$ is a multiple of $125,$ it has at least three factors of $5.$ Since $10$ contributes one factor, $s+6$ contributes at least two factors, or $s+6$ must be a multiple of $25.$ Therefore, the least such nonnegative integer $s$ is $19.$ Finally, combining the two results from above (bolded) generates the least such positive integer $n$ at $s=19:$ \\begin{align*} n&=8r+5 \\\\ &=8(5s+4)+5 \\\\ &=40s+37 \\\\ &=797. \\end{align*} ~MRENTHUSIASM (inspired by Math Jams's 2021 AIME II Discussion)",
"We wish to find the least positive integer $n$ for which $2^n+5^n-n\\equiv0\\pmod{1000}.$ Rearranging gives \\[2^n+5^n\\equiv n\\pmod{1000}.\\] Applying the Chinese Remainder Theorem, we get the following system of congruences: \\begin{align*} 2^n+5^n &\\equiv n \\pmod{8}, \\\\ 2^n+5^n &\\equiv n \\pmod{125}. \\end{align*} It is clear that $n\\geq3,$ from which we simplify to \\begin{align*} 5^n &\\equiv n \\pmod{8}, \\hspace{15mm} &(1) \\\\ 2^n &\\equiv n \\pmod{125}. &(2) \\end{align*} We solve each congruence separately: For $(1),$ quick inspections produce that $5^1,5^2,5^3,5^4,\\ldots$ are congruent to $5,1,5,1,\\ldots$ modulo $8,$ respectively. More generally, $5^n \\equiv 5 \\pmod{8}$ if $n$ is odd, and $5^n \\equiv 1 \\pmod{8}$ if $n$ is even. As $5^n$ is always odd (so is $n$), we must have $n\\equiv5\\pmod{8}.$ That is, $\\boldsymbol{n=8r+5}$ for some nonnegative integer $\\boldsymbol{r.}$ For $(2),$ we substitute the result from $(1)$ and simplify: \\begin{align*} 2^{8r+5}&\\equiv8r+5\\pmod{125} \\\\ \\left(2^8\\right)^r\\cdot2^5&\\equiv8r+5\\pmod{125} \\\\ 256^r\\cdot32&\\equiv8r+5\\pmod{125} \\\\ 6^r\\cdot32&\\equiv8r+5\\pmod{125}. \\end{align*} Note that $5^3=125$ and $6=5+1,$ so we apply the Binomial Theorem to the left side: \\begin{align*} (5+1)^r\\cdot32&\\equiv8r+5\\pmod{125} \\\\ \\Biggl[\\binom{r}{0}5^0+\\binom{r}{1}5^1+\\binom{r}{2}5^2+\\phantom{ }\\underbrace{\\binom{r}{3}5^3+\\cdots+\\binom{r}{r}5^r}_{0\\pmod{125}}\\phantom{ }\\Biggr]\\cdot32&\\equiv8r+5\\pmod{125} \\\\ \\left[1+5r+\\frac{25r(r-1)}{2}\\right]\\cdot32&\\equiv8r+5\\pmod{125} \\\\ 32+160r+400r(r-1)&\\equiv8r+5\\pmod{125} \\\\ 32+35r+25r(r-1)&\\equiv8r+5\\pmod{125} \\\\ 25r^2+2r+27&\\equiv0\\phantom{r+5}\\pmod{125}. \\hspace{15mm} (*) \\end{align*} Since $125\\equiv0\\pmod{5},$ it follows that \\begin{align*} 25r^2+2r+27&\\equiv0\\pmod{5} \\\\ 2r+2&\\equiv0\\pmod{5} \\\\ r&\\equiv4\\pmod{5}. \\end{align*} That is, $\\boldsymbol{r=5s+4}$ for some nonnegative integer $\\boldsymbol{s.}$ Substituting this back into $(*),$ we get \\begin{align*} 25(5s+4)^2+2(5s+4)+27&\\equiv0\\pmod{125} \\\\ 625s^2+1010s+435&\\equiv0\\pmod{125} \\\\ 10s+60&\\equiv0\\pmod{125} \\\\ 10(s+6)&\\equiv0\\pmod{125}. \\end{align*} As $10(s+6)$ is a multiple of $125,$ it has at least three factors of $5.$ Since $10$ contributes one factor, $s+6$ contributes at least two factors, or $s+6$ must be a multiple of $25.$ Therefore, the least such nonnegative integer $s$ is $19.$ Finally, combining the two results from above (bolded) generates the least such positive integer $n$ at $s=19:$ \\begin{align*} n&=8r+5 \\\\ &=8(5s+4)+5 \\\\ &=40s+37 \\\\ &=797. \\end{align*} ~MRENTHUSIASM (inspired by Math Jams's 2021 AIME II Discussion)",
"\\[5^n \\equiv n \\pmod{8}\\] Note that $5^n \\equiv 5,1,5,1,...$ and $n \\equiv 0,..,7$ so $n$ is periodic every $[2,8]=8$. Easy to check that only $n\\equiv 5 \\pmod{8}$ satisfy. Let $n=8a+5$. Note that by binomial theorem, \\[2^{8a+5}=32\\cdot 2^{8a} \\equiv 7(1+15)^{2a} \\equiv 7(1+30a)\\pmod{25}\\] So we have $7(1+30a) \\equiv 8a+5\\pmod{25} \\implies 202a \\equiv 23 \\implies 2a \\equiv 23+25 \\implies a \\equiv 24 \\pmod{25}$. Combining $a\\equiv 24\\pmod{25}$ with $n \\equiv 5 \\pmod{8}$ gives that $n$ is in the form of $197+200k$, $n=197,397,597,797$. Note that since $\\phi(125)=100$ \\[2^{197+200k} \\equiv 2^{197} \\equiv 2^{-3} \\equiv \\underbrace{47}_{\\text{Extended Euclidean Algorithm}}\\pmod{125}\\] Easy to check that only $797 ~Afo",
"\\[5^n \\equiv n \\pmod{8}\\] Note that $5^n \\equiv 5,1,5,1,...$ and $n \\equiv 0,..,7$ so $n$ is periodic every $[2,8]=8$. Easy to check that only $n\\equiv 5 \\pmod{8}$ satisfy. Let $n=8a+5$. Note that by binomial theorem, \\[2^{8a+5}=32\\cdot 2^{8a} \\equiv 7(1+15)^{2a} \\equiv 7(1+30a)\\pmod{25}\\] So we have $7(1+30a) \\equiv 8a+5\\pmod{25} \\implies 202a \\equiv 23 \\implies 2a \\equiv 23+25 \\implies a \\equiv 24 \\pmod{25}$. Combining $a\\equiv 24\\pmod{25}$ with $n \\equiv 5 \\pmod{8}$ gives that $n$ is in the form of $197+200k$, $n=197,397,597,797$. Note that since $\\phi(125)=100$ \\[2^{197+200k} \\equiv 2^{197} \\equiv 2^{-3} \\equiv \\underbrace{47}_{\\text{Extended Euclidean Algorithm}}\\pmod{125}\\] Easy to check that only $797 ~Afo",
"1. The desired $n$ has the form $n = m + 20l + 100p,$ where $m, l, p$ are integers and $m < 20.$ Really: \\[(2^{m+ 20} – (m + 20)) – (2^m – m) = (2^{m+ 20} – 2^m) – 20.\\] The first term is a multiple of $100$(Claim). We denote step an increase in $m$ by 20. At each step, the divisibility by $10$ is preserved, the number of tens is reduced by $2$. \\[(2^{m+ 100} – (m + 100)) – (2^m – m) = (2^{m+ 100} – 2^m) – 100.\\] We denote STEP an increase in $m$ by $100.$ At each STEP, the first term is a multiple of $1000,$ which means that at each STEP the divisibility by $100$ is preserved, the number of hundreds decreases by $1.$ 2. If the expression $2^n + 5^n – n$ is a multiple of $1000,$ the number $n$ is odd ($2^n$ is even, $5^n$ is odd), which means that $5^n$ ends in $125.$ Therefore, the number $2^n – n$ ends in $875.$ 3. $2^3 – 3 = 8 – 3 = 5.$ The tens digit is even $(0),$ it cannot be transformed into $7$ in several steps, since at each step this digit changes by $2.$ $17 = 20 – 3,$ so $2^{17} + 2^3$ is a multiple of $10,$ $(2^{17} – 17) + (2^3 – 3) = 2^{17} + 2^3 – 20$ is a multiple of $10.$ \\[2^{17} \\pmod {100} = ((2^{10} \\pmod {100}) \\cdot (2^7 \\pmod {100})) \\pmod {100} = (24 \\cdot 28) \\pmod {100} = 72.\\] \\[(2^{17} – 17 ) \\pmod {100} = 55.\\] The tens digit $5$ is odd, subtracting $2$ at each step in $4$ steps of $20$ will turn it into $7.$ So \\[(2^{97} – 97 ) \\pmod {100} = 75.\\] \\[2^{97} \\pmod {1000} = ((2^{49} \\pmod {1000}) \\cdot 2^7) \\pmod {1000} = 672.\\] \\[(2^{97} – 97 ) \\pmod {1000} = 575.\\] We transform $5$ into $8$ in $7$ STEPS, so \\[(2^{797} – 797 ) \\pmod {1000} = 875.\\] \\[(5^{797} + 2^{797} – 797 ) \\pmod {1000} = 0.\\] Note, that for $n = 797 + 1000k, k$ is an integer, the expression $2^n + 5^n – n$ is a multiple of $1000.$ Claim The numbers $2^{n+2\\cdot 5^k}+2^n=2^n(2^{2\\cdot 5^k}+1)$ and $2^{n+4\\cdot 5^k}-2^n=2^n(2^{4\\cdot 5^k}-1)$ are a multiple of $10^{k+1}$ for any $n > k$, where $n,k$ are an integer. Proof First, if $m \\pmod{10} \\equiv 4,$ then $(m^4 – m^3 + m^2 – m + 1) \\pmod{10} \\equiv 6 – 4 + 6 – 4 + 1 = 5.$ $2^{4n+2}\\pmod{10} \\equiv 4.$ So, if the number $2^{4n+2} + 1$ is a multiple of $5^k$, then $2^{20n+10} + 1$ is a multiple of $5^{k +1}.$ Really, denote $m = 2^{4n+2},$ then, $2^{20n+10} + 1 = m^5 + 1 = (m + 1)\\cdot (m^4 – m^3 + m^2 – m + 1).$ The first factor is a multiple of $5^k$, the second factor is a multiple of $5.$ Their product is a multiple of $5^{k +1}.$ For odd $n,$ using Newton's binomial for $4^n$, we get $2^{2n} + 1$ is a multiple of $5.$ \\begin{align*} 2^{2n} + 1 = 4^n + 1 = (5 - 1)^n + 1 = 5^n -n\\cdot 5^{n-1} +...+5n\\end{align*} If $n = 5^k,$ then $n > k,$ each term is a multiple of $5n$. Therefore, the number $4^{5^k}+1$ is a multiple of $5^{k+1}$ and $2^{k+1} (2^{2\\cdot 5^k}+1)$ is a multiple of $10^{k+1}$. The difference \\[2^{n+4\\cdot 5^k}-2^n=2^n(2^{4\\cdot 5^k}-1) = 2^n(2^{2\\cdot 5^k}-1)(2^{2\\cdot 5^k}+1)\\] is a multiple of $10^{k+1}.$ vladimir.shelomovskii@gmail.com, vvsss",
"1. The desired $n$ has the form $n = m + 20l + 100p,$ where $m, l, p$ are integers and $m < 20.$ Really: \\[(2^{m+ 20} – (m + 20)) – (2^m – m) = (2^{m+ 20} – 2^m) – 20.\\] The first term is a multiple of $100$(Claim). We denote step an increase in $m$ by 20. At each step, the divisibility by $10$ is preserved, the number of tens is reduced by $2$. \\[(2^{m+ 100} – (m + 100)) – (2^m – m) = (2^{m+ 100} – 2^m) – 100.\\] We denote STEP an increase in $m$ by $100.$ At each STEP, the first term is a multiple of $1000,$ which means that at each STEP the divisibility by $100$ is preserved, the number of hundreds decreases by $1.$ 2. If the expression $2^n + 5^n – n$ is a multiple of $1000,$ the number $n$ is odd ($2^n$ is even, $5^n$ is odd), which means that $5^n$ ends in $125.$ Therefore, the number $2^n – n$ ends in $875.$ 3. $2^3 – 3 = 8 – 3 = 5.$ The tens digit is even $(0),$ it cannot be transformed into $7$ in several steps, since at each step this digit changes by $2.$ $17 = 20 – 3,$ so $2^{17} + 2^3$ is a multiple of $10,$ $(2^{17} – 17) + (2^3 – 3) = 2^{17} + 2^3 – 20$ is a multiple of $10.$ \\[2^{17} \\pmod {100} = ((2^{10} \\pmod {100}) \\cdot (2^7 \\pmod {100})) \\pmod {100} = (24 \\cdot 28) \\pmod {100} = 72.\\] \\[(2^{17} – 17 ) \\pmod {100} = 55.\\] The tens digit $5$ is odd, subtracting $2$ at each step in $4$ steps of $20$ will turn it into $7.$ So \\[(2^{97} – 97 ) \\pmod {100} = 75.\\] \\[2^{97} \\pmod {1000} = ((2^{49} \\pmod {1000}) \\cdot 2^7) \\pmod {1000} = 672.\\] \\[(2^{97} – 97 ) \\pmod {1000} = 575.\\] We transform $5$ into $8$ in $7$ STEPS, so \\[(2^{797} – 797 ) \\pmod {1000} = 875.\\] \\[(5^{797} + 2^{797} – 797 ) \\pmod {1000} = 0.\\] Note, that for $n = 797 + 1000k, k$ is an integer, the expression $2^n + 5^n – n$ is a multiple of $1000.$ Claim The numbers $2^{n+2\\cdot 5^k}+2^n=2^n(2^{2\\cdot 5^k}+1)$ and $2^{n+4\\cdot 5^k}-2^n=2^n(2^{4\\cdot 5^k}-1)$ are a multiple of $10^{k+1}$ for any $n > k$, where $n,k$ are an integer. Proof First, if $m \\pmod{10} \\equiv 4,$ then $(m^4 – m^3 + m^2 – m + 1) \\pmod{10} \\equiv 6 – 4 + 6 – 4 + 1 = 5.$ $2^{4n+2}\\pmod{10} \\equiv 4.$ So, if the number $2^{4n+2} + 1$ is a multiple of $5^k$, then $2^{20n+10} + 1$ is a multiple of $5^{k +1}.$ Really, denote $m = 2^{4n+2},$ then, $2^{20n+10} + 1 = m^5 + 1 = (m + 1)\\cdot (m^4 – m^3 + m^2 – m + 1).$ The first factor is a multiple of $5^k$, the second factor is a multiple of $5.$ Their product is a multiple of $5^{k +1}.$ For odd $n,$ using Newton's binomial for $4^n$, we get $2^{2n} + 1$ is a multiple of $5.$ \\begin{align*} 2^{2n} + 1 = 4^n + 1 = (5 - 1)^n + 1 = 5^n -n\\cdot 5^{n-1} +...+5n\\end{align*} If $n = 5^k,$ then $n > k,$ each term is a multiple of $5n$. Therefore, the number $4^{5^k}+1$ is a multiple of $5^{k+1}$ and $2^{k+1} (2^{2\\cdot 5^k}+1)$ is a multiple of $10^{k+1}$. The difference \\[2^{n+4\\cdot 5^k}-2^n=2^n(2^{4\\cdot 5^k}-1) = 2^n(2^{2\\cdot 5^k}-1)(2^{2\\cdot 5^k}+1)\\] is a multiple of $10^{k+1}.$ vladimir.shelomovskii@gmail.com, vvsss",
"Problem require us to find minimum $n$ where $2^n + 5^n - n$ is a multiple of $1000$. Since we already know that multiplication does not affect mod,we can simply just focus on enumerating and hope to find some pattern. The pattern of $5^n$ is pretty simple, it starts with 5,25, and then repeat the pattern 125,625, in other words, $5^n\\equiv 125 \\mod{1000}$ when n is odd, and $5^n\\equiv 625 \\mod{1000}$ when n is even if we ignore $n=1$ and $n=2$. The pattern of $2^n$, however, is quite long. In fact, it happens when $n=103$, where $2^{103} \\equiv 008 \\mod{1000}$. More specifically, the pattern starts with $n=3$, which means for every 100 n, $2^n \\mod 1000$ will repeat in a loop: $008{,}016{,}032{,}064{,}128{,}256{,}512{,}024{,}048{,}096{,}$ $192{,}384{,}768{,}536{,}072{,}144{,}288{,}576{,}152{,}304{,}$ $608{,}216{,}432{,}864{,}728{,}456{,}912{,}824{,}648{,}296{,}$ $592{,}184{,}368{,}736{,}472{,}944{,}888{,}776{,}552{,}104{,}$ $208{,}416{,}832{,}664{,}328{,}656{,}312{,}624{,}248{,}496{,}$ $992{,}984{,}968{,}936{,}872{,}744{,}488{,}976{,}952{,}904{,}$ $808{,}616{,}232{,}464{,}928{,}856{,}712{,}424{,}848{,}696{,}$ $392{,}784{,}568{,}136{,}272{,}544{,}088{,}176{,}352{,}704{,}$ $408{,}816{,}632{,}264{,}528{,}056{,}112{,}224{,}448{,}896{,}$ $792{,}584{,}168{,}336{,}672{,}344{,}688{,}376{,}752{,}504$ Moreover, we might found that for every 20 consequtive integer, $2^n \\mod 100$ will repeat in a loop. Thus, by adding $5^n$ to first twenty element, which is $n=3$ to $n=22$, we can get a pattern of last two digit of $2^n + 5^n \\mod{1000}$, which is $33{,}41{,}57{,}89{,}53{,}$ $81{,}37{,}49{,}73{,}21{,}$ $17{,}09{,}93{,}61{,}97{,}$ $69{,}13{,}01{,}77{,}29$ By subtracting n, we can also get following sequence: $30{,}37{,}52{,}83{,}46{,}$ $73{,}28{,}39{,}62{,}09{,}$ $04{,}95{,}78{,}45{,}80{,}$ $51{,}94{,}81{,}56{,}07$ Noticing that only when $n=3$ and $n=17$, $2^n+5^n-n \\equiv 0 \\mod{10}$, so $n=3+20k$ and $17+20k$ are our possible candidates. However, since the length of our pattern is 20, we know that $n=3+20k$ is definetely not answer, which only left us $n=17+20k$. Now, we calculate the last three digit of $2^n+5^n-n$ at $n=17, n=37, n=57, n=77, n=97$, which is $180{,}560{,}940{,}320{,}700$. Since the pattern will repeat for every 100 consequtive integer, the only thing that could possibly change last three digit is -n, noticing that only when $n=97$, $2^n+5^n-n \\equiv 0 \\mod 100$. So the answer only could be $n=97+100k_2$, by subtracting 700, which means adding 700 to n, we can get our final answer $n=797. ~henry_in_out",
"Problem require us to find minimum $n$ where $2^n + 5^n - n$ is a multiple of $1000$. Since we already know that multiplication does not affect mod,we can simply just focus on enumerating and hope to find some pattern. The pattern of $5^n$ is pretty simple, it starts with 5,25, and then repeat the pattern 125,625, in other words, $5^n\\equiv 125 \\mod{1000}$ when n is odd, and $5^n\\equiv 625 \\mod{1000}$ when n is even if we ignore $n=1$ and $n=2$. The pattern of $2^n$, however, is quite long. In fact, it happens when $n=103$, where $2^{103} \\equiv 008 \\mod{1000}$. More specifically, the pattern starts with $n=3$, which means for every 100 n, $2^n \\mod 1000$ will repeat in a loop: $008{,}016{,}032{,}064{,}128{,}256{,}512{,}024{,}048{,}096{,}$ $192{,}384{,}768{,}536{,}072{,}144{,}288{,}576{,}152{,}304{,}$ $608{,}216{,}432{,}864{,}728{,}456{,}912{,}824{,}648{,}296{,}$ $592{,}184{,}368{,}736{,}472{,}944{,}888{,}776{,}552{,}104{,}$ $208{,}416{,}832{,}664{,}328{,}656{,}312{,}624{,}248{,}496{,}$ $992{,}984{,}968{,}936{,}872{,}744{,}488{,}976{,}952{,}904{,}$ $808{,}616{,}232{,}464{,}928{,}856{,}712{,}424{,}848{,}696{,}$ $392{,}784{,}568{,}136{,}272{,}544{,}088{,}176{,}352{,}704{,}$ $408{,}816{,}632{,}264{,}528{,}056{,}112{,}224{,}448{,}896{,}$ $792{,}584{,}168{,}336{,}672{,}344{,}688{,}376{,}752{,}504$ Moreover, we might found that for every 20 consequtive integer, $2^n \\mod 100$ will repeat in a loop. Thus, by adding $5^n$ to first twenty element, which is $n=3$ to $n=22$, we can get a pattern of last two digit of $2^n + 5^n \\mod{1000}$, which is $33{,}41{,}57{,}89{,}53{,}$ $81{,}37{,}49{,}73{,}21{,}$ $17{,}09{,}93{,}61{,}97{,}$ $69{,}13{,}01{,}77{,}29$ By subtracting n, we can also get following sequence: $30{,}37{,}52{,}83{,}46{,}$ $73{,}28{,}39{,}62{,}09{,}$ $04{,}95{,}78{,}45{,}80{,}$ $51{,}94{,}81{,}56{,}07$ Noticing that only when $n=3$ and $n=17$, $2^n+5^n-n \\equiv 0 \\mod{10}$, so $n=3+20k$ and $17+20k$ are our possible candidates. However, since the length of our pattern is 20, we know that $n=3+20k$ is definetely not answer, which only left us $n=17+20k$. Now, we calculate the last three digit of $2^n+5^n-n$ at $n=17, n=37, n=57, n=77, n=97$, which is $180{,}560{,}940{,}320{,}700$. Since the pattern will repeat for every 100 consequtive integer, the only thing that could possibly change last three digit is -n, noticing that only when $n=97$, $2^n+5^n-n \\equiv 0 \\mod 100$. So the answer only could be $n=97+100k_2$, by subtracting 700, which means adding 700 to n, we can get our final answer $n=797. ~henry_in_out"
] |
2021-II-14
| 2,021
| 14
|
Let $\Delta ABC$ be an acute triangle with circumcenter $O$ and centroid $G$ . Let $X$ be the intersection of the line tangent to the circumcircle of $\Delta ABC$ at $A$ and the line perpendicular to $GO$ at $G$ . Let $Y$ be the intersection of lines $XG$ and $BC$ . Given that the measures of $\angle ABC, \angle BCA,$ and $\angle XOY$ are in the ratio $13 : 2 : 17,$ the degree measure of $\angle BAC$ can be written as $\frac{m}{n},$ where $m$ and $n$ are relatively prime positive integers. Find $m+n$ .
| 592
|
II
|
[
"In this solution, all angle measures are in degrees. Let $M$ be the midpoint of $\\overline{BC}$ so that $\\overline{OM}\\perp\\overline{BC}$ and $A,G,M$ are collinear. Let $\\angle ABC=13k,\\angle BCA=2k$ and $\\angle XOY=17k.$ Note that: Since $\\angle OGX = \\angle OAX = 90,$ quadrilateral $OGAX$ is cyclic by the Converse of the Inscribed Angle Theorem.It follows that $\\angle OAG = \\angle OXG,$ as they share the same intercepted arc $\\widehat{OG}.$ Since $\\angle OGY = \\angle OMY = 90,$ quadrilateral $OGYM$ is cyclic by the supplementary opposite angles.It follows that $\\angle OMG = \\angle OYG,$ as they share the same intercepted arc $\\widehat{OG}.$ Together, we conclude that $\\triangle OAM \\sim \\triangle OXY$ by AA, so $\\angle AOM = \\angle XOY = 17k.$ Next, we express $\\angle BAC$ in terms of $k.$ By angle addition, we have \\begin{align*} \\angle AOM &= \\angle AOB + \\angle BOM \\\\ &= 2\\angle BCA + \\frac12\\angle BOC \\hspace{10mm} &&\\text{by Inscribed Angle Theorem and Perpendicular Bisector Property} \\\\ &= 2\\angle BCA + \\angle BAC. &&\\text{by Inscribed Angle Theorem} \\end{align*} Substituting back gives $17k=2(2k)+\\angle BAC,$ from which $\\angle BAC=13k.$ For the sum of the interior angles of $\\triangle ABC,$ we get \\begin{align*} \\angle ABC + \\angle BCA + \\angle BAC &= 180 \\\\ 13k+2k+13k&=180 \\\\ 28k&=180 \\\\ k&=\\frac{45}{7}. \\end{align*} Finally, we obtain $\\angle BAC=13k=\\frac{585}{7},$ from which the answer is $585+7=592 ~Constance-variance ~MRENTHUSIASM",
"Let $M$ be the midpoint of $BC$. Because $\\angle{OAX}=\\angle{OGX}=\\angle{OGY}=\\angle{OMY}=90^o$, $AXOG$ and $OMYG$ are cyclic, so $O$ is the center of the spiral similarity sending $AM$ to $XY$, and $\\angle{XOY}=\\angle{AOM}$. Because $\\angle{AOM}=2\\angle{BCA}+\\angle{BAC}$, it's easy to get $\\frac{585}{7} \\implies 592 from here. ~Lcz",
"Firstly, let $M$ be the midpoint of $BC$. Then, $\\angle OMB = 90^o$. Now, note that since $\\angle OGX = \\angle XAO = 90^o$, quadrilateral $AGOX$ is cyclic. Also, because $\\angle OMY + \\angle OGY = 180^o$, $OMYG$ is also cyclic. Now, we define some variables: let $\\alpha$ be the constant such that $\\angle ABC = 13\\alpha, \\angle ACB = 2\\alpha,$ and $\\angle XOY = 17\\alpha$. Also, let $\\beta = \\angle OMG = \\angle OYG$ and $\\theta = \\angle OXG = \\angle OAG$ (due to the fact that $AGOX$ and $OMYG$ are cyclic). Then, \\[\\angle XOY = 180 - \\beta - \\theta = 17\\alpha \\implies \\beta + \\theta = 180 - 17\\alpha.\\] Now, because $AX$ is tangent to the circumcircle at $A$, $\\angle XAC = \\angle CBA = 13\\alpha$, and $\\angle CAO = \\angle OAX - \\angle CAX = 90 - 13\\alpha$. Finally, notice that $\\angle AMB = \\angle OMB - \\angle OMG = 90 - \\beta$. Then, \\[\\angle BAM = 180 - \\angle ABC - \\angle AMB = 180 - 13\\alpha - (90 - \\beta) = 90 + \\beta - 13\\alpha.\\] Thus, \\[\\angle BAC = \\angle BAM + \\angle MAO + \\angle OAC = 90 + \\beta - 13\\alpha + \\theta + 90 - 13\\alpha = 180 - 26\\alpha + (\\beta + \\theta),\\] and \\[180 = \\angle BAC + 13\\alpha + 2\\alpha = 180 - 11\\alpha + \\beta + \\theta \\implies \\beta + \\theta = 11\\alpha.\\] However, from before, $\\beta+\\theta = 180 - 17 \\alpha$, so $11 \\alpha = 180 - 17 \\alpha \\implies 180 = 28 \\alpha \\implies \\alpha = \\frac{180}{28}$. To finish the problem, we simply compute \\[\\angle BAC = 180 - 15 \\alpha = 180 \\cdot \\left(1 - \\frac{15}{28}\\right) = 180 \\cdot \\frac{13}{28} = \\frac{585}{7},\\] so our final answer is $585+7=592. ~advanture",
"Firstly, let $M$ be the midpoint of $BC$. Then, $\\angle OMB = 90^o$. Now, note that since $\\angle OGX = \\angle XAO = 90^o$, quadrilateral $AGOX$ is cyclic. Also, because $\\angle OMY + \\angle OGY = 180^o$, $OMYG$ is also cyclic. Now, we define some variables: let $\\alpha$ be the constant such that $\\angle ABC = 13\\alpha, \\angle ACB = 2\\alpha,$ and $\\angle XOY = 17\\alpha$. Also, let $\\beta = \\angle OMG = \\angle OYG$ and $\\theta = \\angle OXG = \\angle OAG$ (due to the fact that $AGOX$ and $OMYG$ are cyclic). Then, \\[\\angle XOY = 180 - \\beta - \\theta = 17\\alpha \\implies \\beta + \\theta = 180 - 17\\alpha.\\] Now, because $AX$ is tangent to the circumcircle at $A$, $\\angle XAC = \\angle CBA = 13\\alpha$, and $\\angle CAO = \\angle OAX - \\angle CAX = 90 - 13\\alpha$. Finally, notice that $\\angle AMB = \\angle OMB - \\angle OMG = 90 - \\beta$. Then, \\[\\angle BAM = 180 - \\angle ABC - \\angle AMB = 180 - 13\\alpha - (90 - \\beta) = 90 + \\beta - 13\\alpha.\\] Thus, \\[\\angle BAC = \\angle BAM + \\angle MAO + \\angle OAC = 90 + \\beta - 13\\alpha + \\theta + 90 - 13\\alpha = 180 - 26\\alpha + (\\beta + \\theta),\\] and \\[180 = \\angle BAC + 13\\alpha + 2\\alpha = 180 - 11\\alpha + \\beta + \\theta \\implies \\beta + \\theta = 11\\alpha.\\] However, from before, $\\beta+\\theta = 180 - 17 \\alpha$, so $11 \\alpha = 180 - 17 \\alpha \\implies 180 = 28 \\alpha \\implies \\alpha = \\frac{180}{28}$. To finish the problem, we simply compute \\[\\angle BAC = 180 - 15 \\alpha = 180 \\cdot \\left(1 - \\frac{15}{28}\\right) = 180 \\cdot \\frac{13}{28} = \\frac{585}{7},\\] so our final answer is $585+7=592. ~advanture",
"[asy] /* Made by MRENTHUSIASM */ size(375); pair A, B, C, O, G, X, Y; A = origin; B = (1,0); C = extension(A,A+10*dir(585/7),B,B+10*dir(180-585/7)); O = circumcenter(A,B,C); G = centroid(A,B,C); Y = intersectionpoint(G--G+(100,0),B--C); X = intersectionpoint(G--G-(100,0),A--scale(100)*rotate(90)*dir(O-A)); pair O1=circumcenter(O,G,A); real r1=length(O1-O); markscalefactor=3/160; filldraw(O--X--Y--cycle, rgb(255,255,0)); draw(rightanglemark(O,G,X),red); draw(A--O--B,fuchsia+0.4); draw(Arc(O1,r1,-40,50),royalblue+0.5); draw(circumcircle(O,G,Y), heavygreen+0.5); dot(\"$A$\",A,1.5*dir(180+585/7),linewidth(4)); dot(\"$B$\",B,1.5*dir(-585/7),linewidth(4)); dot(\"$C$\",C,1.5N,linewidth(4)); dot(\"$O$\",O,1.5N,linewidth(4)); dot(\"$G$\",G,1.5S,linewidth(4)); dot(\"$Y$\",Y,1.5E,linewidth(4)); dot(\"$X$\",X,1.5W,linewidth(4)); draw(A--B--C--cycle^^X--O--Y--cycle^^A--X^^O--G^^circumcircle(A,B,C)); [/asy] $\\angle OAX = \\angle OGX = 90^\\circ \\implies$ quadrilateral $XAGO$ is cyclic $\\implies$ $\\angle GXO = \\angle GAO,$ as they share the same intersept $\\overset{\\Large\\frown} {GO}.$ $\\angle OGY = \\angle OMY = 90^\\circ \\implies$ quadrilateral $OGYM$ is cyclic $\\implies$ $\\angle GYO = \\angle OMG,$ as they share the same intercept $\\overset{\\Large\\frown} {GO}.$ In triangles $\\triangle XOY$ and $\\triangle AOM,$ two pairs of angles are equal, which means that the third angles $\\angle XOY = \\angle AOM$ are also equal. $\\angle ABC : \\angle BCA : \\angle AOM = 13 : 2 : 17,$ so $\\angle AOM = \\angle ABC + 2 \\angle BCA.$ According to the Claim, $\\triangle ABC$ is isosceles, \\[\\angle ABC : \\angle BCA : \\angle BAC = 13 : 2 : 13.\\] \\[\\angle BAC = \\frac{13} {13 + 2 + 13} \\cdot 180^\\circ = \\frac {585^\\circ}{7} \\implies 585 + 7 = 592 vladimir.shelomovskii@gmail.com, vvsss",
"[asy] /* Made by MRENTHUSIASM */ size(375); pair A, B, C, O, G, X, Y; A = origin; B = (1,0); C = extension(A,A+10*dir(585/7),B,B+10*dir(180-585/7)); O = circumcenter(A,B,C); G = centroid(A,B,C); Y = intersectionpoint(G--G+(100,0),B--C); X = intersectionpoint(G--G-(100,0),A--scale(100)*rotate(90)*dir(O-A)); pair O1=circumcenter(O,G,A); real r1=length(O1-O); markscalefactor=3/160; filldraw(O--X--Y--cycle, rgb(255,255,0)); draw(rightanglemark(O,G,X),red); draw(A--O--B,fuchsia+0.4); draw(Arc(O1,r1,-40,50),royalblue+0.5); draw(circumcircle(O,G,Y), heavygreen+0.5); dot(\"$A$\",A,1.5*dir(180+585/7),linewidth(4)); dot(\"$B$\",B,1.5*dir(-585/7),linewidth(4)); dot(\"$C$\",C,1.5N,linewidth(4)); dot(\"$O$\",O,1.5N,linewidth(4)); dot(\"$G$\",G,1.5S,linewidth(4)); dot(\"$Y$\",Y,1.5E,linewidth(4)); dot(\"$X$\",X,1.5W,linewidth(4)); draw(A--B--C--cycle^^X--O--Y--cycle^^A--X^^O--G^^circumcircle(A,B,C)); [/asy] $\\angle OAX = \\angle OGX = 90^\\circ \\implies$ quadrilateral $XAGO$ is cyclic $\\implies$ $\\angle GXO = \\angle GAO,$ as they share the same intersept $\\overset{\\Large\\frown} {GO}.$ $\\angle OGY = \\angle OMY = 90^\\circ \\implies$ quadrilateral $OGYM$ is cyclic $\\implies$ $\\angle GYO = \\angle OMG,$ as they share the same intercept $\\overset{\\Large\\frown} {GO}.$ In triangles $\\triangle XOY$ and $\\triangle AOM,$ two pairs of angles are equal, which means that the third angles $\\angle XOY = \\angle AOM$ are also equal. $\\angle ABC : \\angle BCA : \\angle AOM = 13 : 2 : 17,$ so $\\angle AOM = \\angle ABC + 2 \\angle BCA.$ According to the Claim, $\\triangle ABC$ is isosceles, \\[\\angle ABC : \\angle BCA : \\angle BAC = 13 : 2 : 13.\\] \\[\\angle BAC = \\frac{13} {13 + 2 + 13} \\cdot 180^\\circ = \\frac {585^\\circ}{7} \\implies 585 + 7 = 592 vladimir.shelomovskii@gmail.com, vvsss",
"Extend $XA$ and meet line $CB$ at $P$. Extend $AG$ to meet $BC$ at $F$. Since $AF$ is the median from $A$ to $BC$, $A,G,F$ are collinear. Furthermore, $OF$ is perpendicular to $BC$ Draw the circumcircle of $\\triangle{XPY}$, as $OA\\bot XP, OG\\bot XY, OF\\bot PY$, $A,G,F$ are collinear, $O$ lies on $(XYP)$ as $AGF$ is the Simson line of $O$ with respect to $\\triangle{XPY}$. Thus, $\\angle{P}=180-17x, \\angle{PAB}=\\angle{C}=2x, 180-15x=13x, x=\\frac{45}{7}$, the answer is $180-15\\cdot \\frac{45}{7}=\\frac{585}{7}$ which is $592. ~bluesoul"
] |
2021-II-15
| 2,021
| 15
|
Let $f(n)$ and $g(n)$ be functions satisfying \[f(n) = \begin{cases} \sqrt{n} & \text{ if } \sqrt{n} \text{ is an integer}\\ 1 + f(n+1) & \text{ otherwise} \end{cases}\] and \[g(n) = \begin{cases}\sqrt{n} & \text{ if } \sqrt{n} \text{ is an integer}\\ 2 + g(n+2) & \text{ otherwise} \end{cases}\] for positive integers $n$ . Find the least positive integer $n$ such that $\tfrac{f(n)}{g(n)} = \tfrac{4}{7}$ .
| 258
|
II
|
[
"Consider what happens when we try to calculate $f(n)$ where n is not a square. If $k^2<n<(k+1)^2$ for (positive) integer k, recursively calculating the value of the function gives us $f(n)=(k+1)^2-n+f((k+1)^2)=k^2+3k+2-n$. Note that this formula also returns the correct value when $n=(k+1)^2$, but not when $n=k^2$. Thus $f(n)=k^2+3k+2-n$ for $k^2<n \\leq (k+1)^2$. If $2 \\mid (k+1)^2-n$, $g(n)$ returns the same value as $f(n)$. This is because the recursion once again stops at $(k+1)^2$. We seek a case in which $f(n)<g(n)$, so obviously this is not what we want. We want $(k+1)^2,n$ to have a different parity, or $n, k$ have the same parity. When this is the case, $g(n)$ instead returns $(k+2)^2-n+g((k+2)^2)=k^2+5k+6-n$. Write $7f(n)=4g(n)$, which simplifies to $3k^2+k-10=3n$. Notice that we want the LHS expression to be divisible by 3; as a result, $k \\equiv 1 \\pmod{3}$. We also want n to be strictly greater than $k^2$, so $k-10>0, k>10$. The LHS expression is always even (since $3k^2+k-10$ factors to $k(3k+1)-10$, and one of $k$ and $3k+1$ will be even), so to ensure that $k$ and $n$ share the same parity, $k$ should be even. Then the least $k$ that satisfies these requirements is $k=16$, giving $n=258$. Indeed - if we check our answer, it works. Therefore, the answer is $258. -Ross Gao",
"We consider $f(n)$ and $g(n)$ separately: $\\boldsymbol{f(n)}$ We restrict $n$ in which $k^2<n\\leq(k+1)^2$ for some positive integer $k,$ or \\[n=(k+1)^2-p\\hspace{40mm}(1)\\] for some integer $p$ such that $0\\leq p<2k+1.$ By recursion, we get \\begin{align*} f\\left((k+1)^2\\right)&=k+1, \\\\ f\\left((k+1)^2-1\\right)&=k+2, \\\\ f\\left((k+1)^2-2\\right)&=k+3, \\\\ & \\ \\vdots \\\\ f\\bigl(\\phantom{ }\\underbrace{(k+1)^2-p}_{n}\\phantom{ }\\bigr)&=k+p+1. \\hspace{19mm}(2) \\\\ \\end{align*} $\\boldsymbol{g(n)}$ If $n$ and $(k+1)^2$ have the same parity, then we get $g(n)=f(n)$ by a similar process from $g\\left((k+1)^2\\right)=k+1.$ This contradicts the precondition $\\frac{f(n)}{g(n)} = \\frac{4}{7}.$ Therefore, $n$ and $(k+1)^2$ must have different parities, from which $n$ and $(k+2)^2$ must have the same parity. It follows that $k^2<n<(k+2)^2,$ or \\[n=(k+2)^2-2q\\hspace{38.25mm}(3)\\] for some integer $q$ such that $0<q<2k+2.$ By recursion, we get \\begin{align*} g\\left((k+2)^2\\right)&=k+2, \\\\ g\\left((k+2)^2-2\\right)&=k+4, \\\\ g\\left((k+2)^2-4\\right)&=k+6, \\\\ & \\ \\vdots \\\\ g\\bigl(\\phantom{ }\\underbrace{(k+2)^2-2q}_{n}\\phantom{ }\\bigr)&=k+2q+2. \\hspace{15.5mm}(4) \\\\ \\end{align*} Answer By $(2)$ and $(4),$ we have \\[\\frac{f(n)}{g(n)}=\\frac{k+p+1}{k+2q+2}=\\frac{4}{7}. \\hspace{27mm}(5)\\] From $(1)$ and $(3),$ equating the expressions for $n$ gives $(k+1)^2-p=(k+2)^2-2q.$ Solving for $k$ produces \\[k=\\frac{2q-p-3}{2}. \\hspace{41.25mm}(6)\\] We substitute $(6)$ into $(5),$ then simplify, cross-multiply, and rearrange: \\begin{align*} \\frac{\\tfrac{2q-p-3}{2}+p+1}{\\tfrac{2q-p-3}{2}+2q+2}&=\\frac{4}{7} \\\\ \\frac{p+2q-1}{-p+6q+1}&=\\frac{4}{7} \\\\ 7p+14q-7&=-4p+24q+4 \\\\ 11p-11&=10q \\\\ 11(p-1)&=10q. \\hspace{29mm}(7) \\end{align*} Since $\\gcd(11,10)=1,$ we know that $p-1$ must be divisible by $10,$ and $q$ must be divisible by $11.$ Recall that the restrictions on $(1)$ and $(2)$ are $0\\leq p<2k+1$ and $0<q<2k+2,$ respectively. Substituting $(6)$ into either inequality gives $p+1<q.$ Combining all these results produces \\[0<p+1<q<2k+2. \\hspace{28mm}(8)\\] To minimize $n$ in either $(1)$ or $(3),$ we minimize $k,$ so we minimize $p$ and $q$ in $(8).$ From $(6)$ and $(7),$ we construct the following table: \\[\\begin{array}{c|c|c|c} & & & \\\\ [-2.5ex] \\boldsymbol{p} & \\boldsymbol{q} & \\boldsymbol{k} & \\textbf{Satisfies }\\boldsymbol{(8)?} \\\\ [0.5ex] \\hline & & & \\\\ [-2ex] 11 & 11 & 4 & \\\\ 21 & 22 & 10 & \\\\ 31 & 33 & 16 & \\checkmark \\\\ \\geq41 & \\geq44 & \\geq22 & \\checkmark \\\\ \\end{array}\\] Finally, we have $(p,q,k)=(31,33,16).$ Substituting this result into either $(1)$ or $(3)$ generates $n=258 Remark We can verify that \\[\\frac{f(258)}{g(258)}=\\frac{1\\cdot31+f(258+1\\cdot31)}{2\\cdot33+g(258+2\\cdot33)}=\\frac{31+\\overbrace{f(289)}^{17}}{66+\\underbrace{g(324)}_{18}}=\\frac{48}{84}=\\frac47.\\] ~MRENTHUSIASM",
"We consider $f(n)$ and $g(n)$ separately: $\\boldsymbol{f(n)}$ We restrict $n$ in which $k^2<n\\leq(k+1)^2$ for some positive integer $k,$ or \\[n=(k+1)^2-p\\hspace{40mm}(1)\\] for some integer $p$ such that $0\\leq p<2k+1.$ By recursion, we get \\begin{align*} f\\left((k+1)^2\\right)&=k+1, \\\\ f\\left((k+1)^2-1\\right)&=k+2, \\\\ f\\left((k+1)^2-2\\right)&=k+3, \\\\ & \\ \\vdots \\\\ f\\bigl(\\phantom{ }\\underbrace{(k+1)^2-p}_{n}\\phantom{ }\\bigr)&=k+p+1. \\hspace{19mm}(2) \\\\ \\end{align*} $\\boldsymbol{g(n)}$ If $n$ and $(k+1)^2$ have the same parity, then we get $g(n)=f(n)$ by a similar process from $g\\left((k+1)^2\\right)=k+1.$ This contradicts the precondition $\\frac{f(n)}{g(n)} = \\frac{4}{7}.$ Therefore, $n$ and $(k+1)^2$ must have different parities, from which $n$ and $(k+2)^2$ must have the same parity. It follows that $k^2<n<(k+2)^2,$ or \\[n=(k+2)^2-2q\\hspace{38.25mm}(3)\\] for some integer $q$ such that $0<q<2k+2.$ By recursion, we get \\begin{align*} g\\left((k+2)^2\\right)&=k+2, \\\\ g\\left((k+2)^2-2\\right)&=k+4, \\\\ g\\left((k+2)^2-4\\right)&=k+6, \\\\ & \\ \\vdots \\\\ g\\bigl(\\phantom{ }\\underbrace{(k+2)^2-2q}_{n}\\phantom{ }\\bigr)&=k+2q+2. \\hspace{15.5mm}(4) \\\\ \\end{align*} Answer By $(2)$ and $(4),$ we have \\[\\frac{f(n)}{g(n)}=\\frac{k+p+1}{k+2q+2}=\\frac{4}{7}. \\hspace{27mm}(5)\\] From $(1)$ and $(3),$ equating the expressions for $n$ gives $(k+1)^2-p=(k+2)^2-2q.$ Solving for $k$ produces \\[k=\\frac{2q-p-3}{2}. \\hspace{41.25mm}(6)\\] We substitute $(6)$ into $(5),$ then simplify, cross-multiply, and rearrange: \\begin{align*} \\frac{\\tfrac{2q-p-3}{2}+p+1}{\\tfrac{2q-p-3}{2}+2q+2}&=\\frac{4}{7} \\\\ \\frac{p+2q-1}{-p+6q+1}&=\\frac{4}{7} \\\\ 7p+14q-7&=-4p+24q+4 \\\\ 11p-11&=10q \\\\ 11(p-1)&=10q. \\hspace{29mm}(7) \\end{align*} Since $\\gcd(11,10)=1,$ we know that $p-1$ must be divisible by $10,$ and $q$ must be divisible by $11.$ Recall that the restrictions on $(1)$ and $(2)$ are $0\\leq p<2k+1$ and $0<q<2k+2,$ respectively. Substituting $(6)$ into either inequality gives $p+1<q.$ Combining all these results produces \\[0<p+1<q<2k+2. \\hspace{28mm}(8)\\] To minimize $n$ in either $(1)$ or $(3),$ we minimize $k,$ so we minimize $p$ and $q$ in $(8).$ From $(6)$ and $(7),$ we construct the following table: \\[\\begin{array}{c|c|c|c} & & & \\\\ [-2.5ex] \\boldsymbol{p} & \\boldsymbol{q} & \\boldsymbol{k} & \\textbf{Satisfies }\\boldsymbol{(8)?} \\\\ [0.5ex] \\hline & & & \\\\ [-2ex] 11 & 11 & 4 & \\\\ 21 & 22 & 10 & \\\\ 31 & 33 & 16 & \\checkmark \\\\ \\geq41 & \\geq44 & \\geq22 & \\checkmark \\\\ \\end{array}\\] Finally, we have $(p,q,k)=(31,33,16).$ Substituting this result into either $(1)$ or $(3)$ generates $n=258 Remark We can verify that \\[\\frac{f(258)}{g(258)}=\\frac{1\\cdot31+f(258+1\\cdot31)}{2\\cdot33+g(258+2\\cdot33)}=\\frac{31+\\overbrace{f(289)}^{17}}{66+\\underbrace{g(324)}_{18}}=\\frac{48}{84}=\\frac47.\\] ~MRENTHUSIASM",
"Since $n$ isn't a perfect square, let $n=m^2+k$ with $0<k<2m+1$. If $k$ is odd, then $f(n)=g(n)$. If $k$ is even, then \\begin{align*} f(n)&=(m+1)^2-(m^2+k)+(m+1)=3m+2-k, \\\\ g(n)&=(m+2)^2-(m^2+k)+(m+2)=5m+6-k, \\end{align*} from which \\begin{align*} 7(3m+2-k)&=4(5m+6-k) \\\\ m&=3k+10. \\end{align*} Since $k$ is even, $m$ is even. Since $k\\neq 0$, the smallest $k$ is $2$ which produces the smallest $n$: \\[k=2 \\implies m=16 \\implies n=16^2+2=258.\\] ~Afo",
"To begin, note that if $n$ is a perfect square, $f(n)=g(n)$, so $f(n)/g(n)=1$, so we must look at values of $n$ that are not perfect squares (what a surprise). First, let the distance between $n$ and the first perfect square greater than or equal to it be $k$, making the values of $f(n+k)$ and $g(n+k)$ integers. Using this notation, we see that $f(n)=k+f(n+k)$, giving us a formula for the numerator of our ratio. However, since the function of $g(n)$ does not add one to the previous inputs in the function until a perfect square is achieved, but adds values of two, we can not achieve the value of $\\sqrt{n+k}$ in $g(n)$ unless $k$ is an even number. However, this is impossible, since if $k$ was an even number, $f(n)=g(n)$, giving a ratio of one. Thus, $k$ must be an odd number. Thus, since $k$ must be an odd number, regardless of whether $n$ is even or odd, to get an integral value in $g(n)$, we must get to the next perfect square after $n+k$. To make matters easier, let $z^2=n+k$. Thus, in $g(n)$, we want to achieve $(z+1)^2$. Expanding $(z+1)^2$ and substituting in the fact that $z=\\sqrt{n+k}$ yields: \\[(z+1)^2=z^2+2z+1=n+k+2\\sqrt{n+k}+1\\] Thus, we must add the quantity $k+2z+1$ to $n$ to achieve a integral value in the function $g(n)$. Thus. \\[g(n)=(k+2z+1)+\\sqrt{n+k+2\\sqrt{n+k}+1}\\] However, note that the quantity within the square root is just $(z+1)^2$, and so: \\[g(n)=k+3z+2\\] Thus, \\[\\frac{f(n)}{g(n)}=\\frac{k+z}{k+3z+2}\\] Since we want this quantity to equal $\\frac{4}{7}$, we can set the above equation equal to this number and collect all the variables to one side to achieve \\[3k-5z=8\\] Substituting back in that $z=\\sqrt{n+k}$, and then separating variables and squaring yields that \\[9k^2-73k+64=25n\\] Now, if we treat $n$ as a constant, we can use the quadratic formula in respect to $k$ to get an equation for $k$ in terms of $n$ (without all the squares). Doing so yields \\[\\frac{73\\pm\\sqrt{3025+900n}}{18}=k\\] Now, since $n$ and $k$ are integers, we want the quantity within the square root to be a perfect square. Note that $55^2=3025$. Thus, assume that the quantity within the root is equal to the perfect square, $m^2$. Thus, after using a difference of squares, we have \\[(m-55)(m+55)=900n\\] Since we want $n$ to be an integer, we know that the $LHS$ should be divisible by five, so, let's assume that we should have $m$ divisible by five. If so, the quantity $18k-73$ must be divisible by five, meaning that $k$ leaves a remainder of one when divided by 5 (if the reader knows LaTeX well enough to write this as a modulo argument, please go ahead and do so!). Thus, we see that to achieve integers $n$ and $k$ that could potentially satisfy the problem statement, we must try the values of $k$ congruent to one modulo five. However, if we recall a statement made earlier in the solution, we see that we can skip all even values of $k$ produced by this modulo argument. Also, note that $k=1,6$ won't work, as they are too small, and will give an erroneous value for $n$. After trying $k=11,21,31$, we see that $k=31$ will give a value of $m=485$, which yields $n=258, will yield the desired ratio, and we're done. Side Note: If any part of this solution is not rigorous, or too vague, please label it in the margin with \"needs proof\". If you can prove it, please add a lemma to the solution doing so :) -Azeem H. (mathislife52)",
"To begin, note that if $n$ is a perfect square, $f(n)=g(n)$, so $f(n)/g(n)=1$, so we must look at values of $n$ that are not perfect squares (what a surprise). First, let the distance between $n$ and the first perfect square greater than or equal to it be $k$, making the values of $f(n+k)$ and $g(n+k)$ integers. Using this notation, we see that $f(n)=k+f(n+k)$, giving us a formula for the numerator of our ratio. However, since the function of $g(n)$ does not add one to the previous inputs in the function until a perfect square is achieved, but adds values of two, we can not achieve the value of $\\sqrt{n+k}$ in $g(n)$ unless $k$ is an even number. However, this is impossible, since if $k$ was an even number, $f(n)=g(n)$, giving a ratio of one. Thus, $k$ must be an odd number. Thus, since $k$ must be an odd number, regardless of whether $n$ is even or odd, to get an integral value in $g(n)$, we must get to the next perfect square after $n+k$. To make matters easier, let $z^2=n+k$. Thus, in $g(n)$, we want to achieve $(z+1)^2$. Expanding $(z+1)^2$ and substituting in the fact that $z=\\sqrt{n+k}$ yields: \\[(z+1)^2=z^2+2z+1=n+k+2\\sqrt{n+k}+1\\] Thus, we must add the quantity $k+2z+1$ to $n$ to achieve a integral value in the function $g(n)$. Thus. \\[g(n)=(k+2z+1)+\\sqrt{n+k+2\\sqrt{n+k}+1}\\] However, note that the quantity within the square root is just $(z+1)^2$, and so: \\[g(n)=k+3z+2\\] Thus, \\[\\frac{f(n)}{g(n)}=\\frac{k+z}{k+3z+2}\\] Since we want this quantity to equal $\\frac{4}{7}$, we can set the above equation equal to this number and collect all the variables to one side to achieve \\[3k-5z=8\\] Substituting back in that $z=\\sqrt{n+k}$, and then separating variables and squaring yields that \\[9k^2-73k+64=25n\\] Now, if we treat $n$ as a constant, we can use the quadratic formula in respect to $k$ to get an equation for $k$ in terms of $n$ (without all the squares). Doing so yields \\[\\frac{73\\pm\\sqrt{3025+900n}}{18}=k\\] Now, since $n$ and $k$ are integers, we want the quantity within the square root to be a perfect square. Note that $55^2=3025$. Thus, assume that the quantity within the root is equal to the perfect square, $m^2$. Thus, after using a difference of squares, we have \\[(m-55)(m+55)=900n\\] Since we want $n$ to be an integer, we know that the $LHS$ should be divisible by five, so, let's assume that we should have $m$ divisible by five. If so, the quantity $18k-73$ must be divisible by five, meaning that $k$ leaves a remainder of one when divided by 5 (if the reader knows LaTeX well enough to write this as a modulo argument, please go ahead and do so!). Thus, we see that to achieve integers $n$ and $k$ that could potentially satisfy the problem statement, we must try the values of $k$ congruent to one modulo five. However, if we recall a statement made earlier in the solution, we see that we can skip all even values of $k$ produced by this modulo argument. Also, note that $k=1,6$ won't work, as they are too small, and will give an erroneous value for $n$. After trying $k=11,21,31$, we see that $k=31$ will give a value of $m=485$, which yields $n=258, will yield the desired ratio, and we're done. Side Note: If any part of this solution is not rigorous, or too vague, please label it in the margin with \"needs proof\". If you can prove it, please add a lemma to the solution doing so :) -Azeem H. (mathislife52)",
"First of all, if $n$ is a perfect square, $f(n)=g(n)=\\sqrt{n}$ and their quotient is $1.$ So, for the rest of this solution, assume $n$ is not a perfect square. Let $a^2$ be the smallest perfect square greater than $n$ and let $b^2$ be the smallest perfect square greater than $n$ with the same parity as $n,$ and note that either $b=a$ or $b=a+1.$ Notice that $(a-1)^2 < n < a^2.$ With a bit of inspection, it becomes clear that $f(n) = a+(a^2-n)$ and $g(n) = b+(b^2-n).$ If $a$ and $n$ have the same parity, we get $a=b$ so $f(n) = g(n)$ and their quotient is $1.$ So, for the rest of this solution, we let $a$ and $n$ have opposite parity. We have two cases to consider. Case 1: $n$ is odd and $a$ is even Here, we get $a=2k$ for some positive integer $k.$ Then, $b = 2k+1.$ We let $n = a^2-(2m+1)$ for some positive integer $m$ so $f(n) = 2k+2m+1$ and $g(n) = 2k+1+2m+1+4k+1 = 6k+2m+3.$ We set $\\frac{2k+2m+1}{6k+2m+3}=\\frac{4}{7},$ cross multiply, and rearrange to get $6m-10k=5.$ Since $k$ and $m$ are integers, the LHS will always be even and the RHS will always be odd, and thus, this case yields no solutions. Case 2: $n$ is even and $a$ is odd Here, we get $a=2k+1$ for some positive ineger $k.$ Then, $b=2k+2.$ We let $n = a^2-(2m+1)$ for some positive integer $m$ so $f(n)=2k+1+2m+1=2k+2m+2$ and $g(n)=2k+2+2m+1+4k+3 = 6k+2m+6.$ We set $\\frac{2k+2m+2}{6k+2m+6} = \\frac{4}{7},$ cross multiply, and rearrange to get $5k=3m-5,$ or $k=\\frac{3}{5}m-1.$ Since $k$ and $m$ are integers, $m$ must be a multiple of $5.$ Some possible solutions for $(k,m)$ with the least $k$ and $m$ are $(2,5), (5,10), (8,15),$ and $(11,20).$ We wish to minimize $k$ since $a=2k+1.$ One thing to keep in mind is the initial assumption $(a-1)^2 < n < a^2.$ The pair $(2,5)$ gives $a=2(2)+1=5$ and $n=5^2-(2(5)+1)=14.$ But $4^2<14<5^2$ is clearly false, so we discard this case. The pair $(5,10)$ gives $a=2(5)+1=11$ and $n=11^2-(2(10)+1)=100,$ which is a perfect square and therefore can be discarded. The pair $(8,15)$ gives $a=2(8)+1=17$ and $n=17^2-(2(15)+1)=258,$ which is between $16^2$ and $17^2$ so it is our smallest solution. So, $258 is the correct answer. ~mc21s",
"First of all, if $n$ is a perfect square, $f(n)=g(n)=\\sqrt{n}$ and their quotient is $1.$ So, for the rest of this solution, assume $n$ is not a perfect square. Let $a^2$ be the smallest perfect square greater than $n$ and let $b^2$ be the smallest perfect square greater than $n$ with the same parity as $n,$ and note that either $b=a$ or $b=a+1.$ Notice that $(a-1)^2 < n < a^2.$ With a bit of inspection, it becomes clear that $f(n) = a+(a^2-n)$ and $g(n) = b+(b^2-n).$ If $a$ and $n$ have the same parity, we get $a=b$ so $f(n) = g(n)$ and their quotient is $1.$ So, for the rest of this solution, we let $a$ and $n$ have opposite parity. We have two cases to consider. Case 1: $n$ is odd and $a$ is even Here, we get $a=2k$ for some positive integer $k.$ Then, $b = 2k+1.$ We let $n = a^2-(2m+1)$ for some positive integer $m$ so $f(n) = 2k+2m+1$ and $g(n) = 2k+1+2m+1+4k+1 = 6k+2m+3.$ We set $\\frac{2k+2m+1}{6k+2m+3}=\\frac{4}{7},$ cross multiply, and rearrange to get $6m-10k=5.$ Since $k$ and $m$ are integers, the LHS will always be even and the RHS will always be odd, and thus, this case yields no solutions. Case 2: $n$ is even and $a$ is odd Here, we get $a=2k+1$ for some positive ineger $k.$ Then, $b=2k+2.$ We let $n = a^2-(2m+1)$ for some positive integer $m$ so $f(n)=2k+1+2m+1=2k+2m+2$ and $g(n)=2k+2+2m+1+4k+3 = 6k+2m+6.$ We set $\\frac{2k+2m+2}{6k+2m+6} = \\frac{4}{7},$ cross multiply, and rearrange to get $5k=3m-5,$ or $k=\\frac{3}{5}m-1.$ Since $k$ and $m$ are integers, $m$ must be a multiple of $5.$ Some possible solutions for $(k,m)$ with the least $k$ and $m$ are $(2,5), (5,10), (8,15),$ and $(11,20).$ We wish to minimize $k$ since $a=2k+1.$ One thing to keep in mind is the initial assumption $(a-1)^2 < n < a^2.$ The pair $(2,5)$ gives $a=2(2)+1=5$ and $n=5^2-(2(5)+1)=14.$ But $4^2<14<5^2$ is clearly false, so we discard this case. The pair $(5,10)$ gives $a=2(5)+1=11$ and $n=11^2-(2(10)+1)=100,$ which is a perfect square and therefore can be discarded. The pair $(8,15)$ gives $a=2(8)+1=17$ and $n=17^2-(2(15)+1)=258,$ which is between $16^2$ and $17^2$ so it is our smallest solution. So, $258 is the correct answer. ~mc21s",
"Say the answer is in the form $n^{2}-x$, then $x$ must be odd or else $f(x) = g(x)$. Say $y = n^{2}-x$. $f(y) = x+n$, $g(y) = 3n+2+x$. Because $f(y)/g(y)$ = $4$*(an integer)/$7$*(an integer), $f(y)$ is $4$*(an integer) so $n$ must be odd or else $f(y)$ would be odd. Solving for $x$ in terms of $n$ gives integer $x = (5/3)n+8/3$ which means $n$ is $2$ mod $3$, because $n$ is also odd, $n$ is $5$ mod $6$. $x$ must be less than $2n-1$ or else the minimum square above $y$ would be $(n-1)^{2}$. We set an inequality $(5/3)n+8/3<2n-1 => 5n+8<6n-3 => n>11$. Since $n$ is $5$ mod $6$, $n = 17$ and $x = 31$ giving $17^{2}-31$ = $258 added by Bread10",
"Say the answer is in the form $n^{2}-x$, then $x$ must be odd or else $f(x) = g(x)$. Say $y = n^{2}-x$. $f(y) = x+n$, $g(y) = 3n+2+x$. Because $f(y)/g(y)$ = $4$*(an integer)/$7$*(an integer), $f(y)$ is $4$*(an integer) so $n$ must be odd or else $f(y)$ would be odd. Solving for $x$ in terms of $n$ gives integer $x = (5/3)n+8/3$ which means $n$ is $2$ mod $3$, because $n$ is also odd, $n$ is $5$ mod $6$. $x$ must be less than $2n-1$ or else the minimum square above $y$ would be $(n-1)^{2}$. We set an inequality $(5/3)n+8/3<2n-1 => 5n+8<6n-3 => n>11$. Since $n$ is $5$ mod $6$, $n = 17$ and $x = 31$ giving $17^{2}-31$ = $258 added by Bread10",
"Define a nonnegative integer $k$ such that $g(n)-f(n)$ = $k$. Since $\\frac{f(n)}{g(n)} = \\frac{4}{7},$ $k$ is a positive integer. Now, suppose 3 consecutive integers $a-1$, $a$, and $a+1$, and $(a-1)^2 < n < a^2$. When $g(n) > f(n)$, $g(n)$ must have a different parity than $a$, so that $f(n)$'s recursive sequence ends on $a^2$, while $g(n)$ continues to $(a+1)^2$. If this condition is satisfied, we can figure out the value of $k$ based on $a$. According to the definitions of $f(n)$ and $g(n)$, $f(n) = a+a^2-n,$ and $g(n) = (a+1)+(a+1)^2-n,$ which gives $k = 2a+2$. And because of $f(n) + k = g(n)$, $\\frac{k}{g(n)} = \\frac{3}{7},$ so cross multiplying gives $3g(n) = 14a+14$. This means that $14a+14$ is divisible by 3, and thus $a \\equiv 2 \\pmod{3}$. The final thing left is to find the smallest $a$ such that the corresponding value of $g(n)$ exist. Simple guess and check should give that the smallest value of $a$ is $a = 17$, which yields an answer of $n = 258. ~ Marchk26"
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2022-I-1
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Quadratic polynomials $P(x)$ and $Q(x)$ have leading coefficients $2$ and $-2,$ respectively. The graphs of both polynomials pass through the two points $(16,54)$ and $(20,53).$ Find $P(0) + Q(0).$
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I
|
[
"Let $R(x)=P(x)+Q(x).$ Since the $x^2$-terms of $P(x)$ and $Q(x)$ cancel, we conclude that $R(x)$ is a linear polynomial. Note that \\begin{alignat*}{8} R(16) &= P(16)+Q(16) &&= 54+54 &&= 108, \\\\ R(20) &= P(20)+Q(20) &&= 53+53 &&= 106, \\end{alignat*} so the slope of $R(x)$ is $\\frac{106-108}{20-16}=-\\frac12.$ It follows that the equation of $R(x)$ is \\[R(x)=-\\frac12x+c\\] for some constant $c,$ and we wish to find $R(0)=c.$ We substitute $x=20$ into this equation to get $106=-\\frac12\\cdot20+c,$ from which $c=116 ~MRENTHUSIASM",
"Let $R(x)=P(x)+Q(x).$ Since the $x^2$-terms of $P(x)$ and $Q(x)$ cancel, we conclude that $R(x)$ is a linear polynomial. Note that \\begin{alignat*}{8} R(16) &= P(16)+Q(16) &&= 54+54 &&= 108, \\\\ R(20) &= P(20)+Q(20) &&= 53+53 &&= 106, \\end{alignat*} so the slope of $R(x)$ is $\\frac{106-108}{20-16}=-\\frac12.$ It follows that the equation of $R(x)$ is \\[R(x)=-\\frac12x+c\\] for some constant $c,$ and we wish to find $R(0)=c.$ We substitute $x=20$ into this equation to get $106=-\\frac12\\cdot20+c,$ from which $c=116 ~MRENTHUSIASM",
"Let \\begin{alignat*}{8} P(x) &= &2x^2 + ax + b, \\\\ Q(x) &= &\\hspace{1mm}-2x^2 + cx + d, \\end{alignat*} for some constants $a,b,c$ and $d.$ We are given that \\begin{alignat*}{8} P(16) &= &512 + 16a + b &= 54, \\hspace{20mm}&&(1) \\\\ Q(16) &= &\\hspace{1mm}-512 + 16c + d &= 54, &&(2) \\\\ P(20) &= &800 + 20a + b &= 53, &&(3) \\\\ Q(20) &= &\\hspace{1mm}-800 + 20c + d &= 53, &&(4) \\end{alignat*} and we wish to find \\[P(0)+Q(0)=b+d.\\] We need to cancel $a$ and $c.$ Since $\\operatorname{lcm}(16,20)=80,$ we subtract $4\\cdot[(3)+(4)]$ from $5\\cdot[(1)+(2)]$ to get \\[b+d=5\\cdot(54+54)-4\\cdot(53+53)=116.\\] ~MRENTHUSIASM",
"Let \\begin{alignat*}{8} P(x) &= &2x^2 + ax + b, \\\\ Q(x) &= &\\hspace{1mm}-2x^2 + cx + d, \\end{alignat*} for some constants $a,b,c$ and $d.$ We are given that \\begin{alignat*}{8} P(16) &= &512 + 16a + b &= 54, \\hspace{20mm}&&(1) \\\\ Q(16) &= &\\hspace{1mm}-512 + 16c + d &= 54, &&(2) \\\\ P(20) &= &800 + 20a + b &= 53, &&(3) \\\\ Q(20) &= &\\hspace{1mm}-800 + 20c + d &= 53, &&(4) \\end{alignat*} and we wish to find \\[P(0)+Q(0)=b+d.\\] We need to cancel $a$ and $c.$ Since $\\operatorname{lcm}(16,20)=80,$ we subtract $4\\cdot[(3)+(4)]$ from $5\\cdot[(1)+(2)]$ to get \\[b+d=5\\cdot(54+54)-4\\cdot(53+53)=116.\\] ~MRENTHUSIASM",
"Like Solution 2, we can begin by setting $P$ and $Q$ to the quadratic above, giving us \\begin{alignat*}{8} P(16) &= &512 + 16a + b &= 54, \\hspace{20mm}&&(1) \\\\ Q(16) &= &\\hspace{1mm}-512 + 16c + d &= 54, &&(2) \\\\ P(20) &= &800 + 20a + b &= 53, &&(3) \\\\ Q(20) &= &\\hspace{1mm}-800 + 20c + d &= 53, &&(4) \\end{alignat*} We can first add $(1)$ and $(2)$ to obtain $16(a-c) + (b+d) = 108.$ Next, we can add $(3)$ and $(4)$ to obtain $20(a-c) + (b+d) = 106.$ By subtracting these two equations, we find that $4(a-c) = -2,$ so substituting this into equation $[(1) + (2)],$ we know that $4 \\cdot (-2) + (b+d) = 108,$ so therefore $b+d = 116 ~jessiewang28",
"Like Solution 2, we can begin by setting $P$ and $Q$ to the quadratic above, giving us \\begin{alignat*}{8} P(16) &= &512 + 16a + b &= 54, \\hspace{20mm}&&(1) \\\\ Q(16) &= &\\hspace{1mm}-512 + 16c + d &= 54, &&(2) \\\\ P(20) &= &800 + 20a + b &= 53, &&(3) \\\\ Q(20) &= &\\hspace{1mm}-800 + 20c + d &= 53, &&(4) \\end{alignat*} We can first add $(1)$ and $(2)$ to obtain $16(a-c) + (b+d) = 108.$ Next, we can add $(3)$ and $(4)$ to obtain $20(a-c) + (b+d) = 106.$ By subtracting these two equations, we find that $4(a-c) = -2,$ so substituting this into equation $[(1) + (2)],$ we know that $4 \\cdot (-2) + (b+d) = 108,$ so therefore $b+d = 116 ~jessiewang28",
"Let \\begin{alignat*}{8} P(x) &= &2x^2 + ax + b, \\\\ Q(x) &= &\\hspace{1mm}-2x^2 + cx + d, \\end{alignat*} By substituting $(16, 54)$ and $(20, 53)$ into these equations, we can get: \\begin{align*} 2(16)^2 + 16a + b &= 54, \\\\ 2(20)^2 + 20a + b &= 53. \\end{align*} Hence, $a = -72.25$ and $b = 698.$ Similarly, \\begin{align*} -2(16)^2 + 16c + d &= 54, \\\\ -2(20)^2 + 20c + d &= 53. \\end{align*} Hence, $c = 71.75$ and $d = -582.$ Notice that $b = P(0)$ and $d = Q(0).$ Therefore \\[P(0) + Q(0) = 698 + (-582) = 116.\\] ~Littlemouse",
"Let \\begin{alignat*}{8} P(x) &= &2x^2 + ax + b, \\\\ Q(x) &= &\\hspace{1mm}-2x^2 + cx + d, \\end{alignat*} By substituting $(16, 54)$ and $(20, 53)$ into these equations, we can get: \\begin{align*} 2(16)^2 + 16a + b &= 54, \\\\ 2(20)^2 + 20a + b &= 53. \\end{align*} Hence, $a = -72.25$ and $b = 698.$ Similarly, \\begin{align*} -2(16)^2 + 16c + d &= 54, \\\\ -2(20)^2 + 20c + d &= 53. \\end{align*} Hence, $c = 71.75$ and $d = -582.$ Notice that $b = P(0)$ and $d = Q(0).$ Therefore \\[P(0) + Q(0) = 698 + (-582) = 116.\\] ~Littlemouse",
"Add the equations of the polynomials $y=2x^2+ax+b$ and $y=-2x^2+cx+d$ to get $2y=(a+c)x+(b+d)$. This equation must also pass through the two points $(16,54)$ and $(20,53)$. Let $m=a+c$ and $n=b+d$. We then have two equations: \\begin{align*} 108&=16m+n, \\\\ 106&=20m+n. \\end{align*} We are trying to solve for $n=P(0)$. Using elimination: \\begin{align*} 540&=80m+5n, \\\\ 424&=80m+4n. \\end{align*} Subtracting both equations, we find that $n=116. ~eevee9406"
] |
2022-I-2
| 2,022
| 2
|
Find the three-digit positive integer $\underline{a}\,\underline{b}\,\underline{c}$ whose representation in base nine is $\underline{b}\,\underline{c}\,\underline{a}_{\,\text{nine}},$ where $a,$ $b,$ and $c$ are (not necessarily distinct) digits.
| 227
|
I
|
[
"We are given that \\[100a + 10b + c = 81b + 9c + a,\\] which rearranges to \\[99a = 71b + 8c.\\] Taking both sides modulo $71,$ we have \\begin{align*} 28a &\\equiv 8c \\pmod{71} \\\\ 7a &\\equiv 2c \\pmod{71}. \\end{align*} The only solution occurs at $(a,c)=(2,7),$ from which $b=2.$ Therefore, the requested three-digit positive integer is $\\underline{a}\\,\\underline{b}\\,\\underline{c}=227 ~MRENTHUSIASM",
"As shown in Solution 1, we get $99a = 71b+8c$. Note that $99$ and $71$ are large numbers comparatively to $8$, so we hypothesize that $a$ and $b$ are equal and $8c$ fills the gap between them. The difference between $99$ and $71$ is $28$, which is a multiple of $4$. So, if we multiply this by $2$, it will be a multiple of $8$ and thus the gap can be filled. Therefore, the only solution is $(a,b,c)=(2,2,7)$, and the answer is $\\underline{a}\\,\\underline{b}\\,\\underline{c}=227 is very reasonable. ~Technodoggo",
"A little bit more motivation: taking mod $8$ on both sides, we find that $7a\\equiv7b\\pmod8\\implies a\\equiv b\\pmod8$, so either $a=b$ or they differ by a multiple of $8$ (okok technically $a=b$ is a subcase of the other one but shh for the sake of clarity). $99$ and $71$ differ by $28$, so a difference of merely $8$ in $a$ and $b$ accounts for a huge difference on the scale of $28\\cdot8=224$ (which is obviously unfillable by the mere $8c$), so $a=b$ is very reasonable. ~Technodoggo",
"As shown in Solution 1, we get $99a = 71b+8c.$ We list a few multiples of $99$ out: \\[99,198,297,396.\\] Of course, $99$ can't be made of just $8$'s. If we use one $71$, we get a remainder of $28$, which can't be made of $8$'s either. So $99$ doesn't work. $198$ can't be made up of just $8$'s. If we use one $71$, we get a remainder of $127$, which can't be made of $8$'s. If we use two $71$'s, we get a remainder of $56$, which can be made of $8$'s. Therefore we get $99\\cdot2=71\\cdot2+8\\cdot7$ so $a=2,b=2,$ and $c=7$. Plugging this back into the original problem shows that this answer is indeed correct. Therefore, $\\underline{a}\\,\\underline{b}\\,\\underline{c}=227 ~Technodoggo",
"As shown in Solution 1, we get $99a = 71b+8c$. We can see that $99$ is $28$ larger than $71$, and we have an $8c$. We can clearly see that $56$ is a multiple of $8$, and any larger than $56$ would result in $c$ being larger than $9$. Therefore, our only solution is $a = 2, b = 2, c = 7$. Our answer is $\\underline{a}\\,\\underline{b}\\,\\underline{c}=227. ~Arcticturn",
"As shown in Solution 1, we get $99a = 71b+8c,$ which rearranges to \\[99(a – b) = 8c – 28 b = 4(2c – 7b) \\le 4(2\\cdot 9 - 0 ) = 72.\\] So $a=b, 2c = 7b \\implies c=7, b=2,a=2.$ vladimir.shelomovskii@gmail.com, vvsss",
"As shown in Solution 1, we have that $99a = 71b + 8c$. Note that by the divisibility rule for $9$, we have $a+b+c \\equiv a \\pmod{9}$. Since $b$ and $c$ are base-$9$ digits, we can say that $b+c = 0$ or $b+c=9$. The former possibility can be easily eliminated, and thus $b+c=9$. Next, we write the equation from Solution 1 as $99a = 63b + 8(b+c)$, and dividing this by $9$ gives $11a = 7b+8$. Taking both sides modulo $7$, we have $4a \\equiv 1 \\pmod{7}$. Multiplying both sides by $2$ gives $a\\equiv 2 \\pmod{7}$, which implies $a=2$. From here, we can find that $b=2$ and $c=7$, giving an answer of $227. ~Sedro"
] |
2022-I-3
| 2,022
| 3
|
In isosceles trapezoid $ABCD,$ parallel bases $\overline{AB}$ and $\overline{CD}$ have lengths $500$ and $650,$ respectively, and $AD=BC=333.$ The angle bisectors of $\angle A$ and $\angle D$ meet at $P,$ and the angle bisectors of $\angle B$ and $\angle C$ meet at $Q.$ Find $PQ.$
| 242
|
I
|
[
"We have the following diagram: [asy] /* Made by MRENTHUSIASM , modified by Cytronical */ size(300); pair A, B, C, D, A1, B1, C1, D1, P, Q, X, Y, Z, W; A = (-250,6*sqrt(731)); B = (250,6*sqrt(731)); C = (325,-6*sqrt(731)); D = (-325,-6*sqrt(731)); A1 = bisectorpoint(B,A,D); B1 = bisectorpoint(A,B,C); C1 = bisectorpoint(B,C,D); D1 = bisectorpoint(A,D,C); P = intersectionpoint(A--300*(A1-A)+A,D--300*(D1-D)+D); Q = intersectionpoint(B--300*(B1-B)+B,C--300*(C1-C)+C); X = intersectionpoint(B--5*(Q-B)+B,C--D); Y = (0,6*sqrt(731)); Z = intersectionpoint(A--4*(P-A)+A,B--4*(Q-B)+B); W = intersectionpoint(A--5*(P-A)+A,C--D); draw(anglemark(P,A,B,1000),red); draw(anglemark(D,A,P,1000),red); draw(anglemark(A,B,Q,1000),red); draw(anglemark(Q,B,C,1000),red); draw(anglemark(P,D,A,1000),red); draw(anglemark(C,D,P,1000),red); draw(anglemark(Q,C,D,1000),red); draw(anglemark(B,C,Q,1000),red); add(pathticks(anglemark(P,A,B,1000), n = 1, r = 0.15, s = 750, red)); add(pathticks(anglemark(D,A,P,1000), n = 1, r = 0.15, s = 750, red)); add(pathticks(anglemark(A,B,Q,1000), n = 1, r = 0.15, s = 750, red)); add(pathticks(anglemark(Q,B,C,1000), n = 1, r = 0.15, s = 750, red)); add(pathticks(anglemark(P,D,A,1000), n = 2, r = 0.12, spacing = 150, s = 750, red)); add(pathticks(anglemark(C,D,P,1000), n = 2, r = 0.12, spacing = 150, s = 750, red)); add(pathticks(anglemark(Q,C,D,1000), n = 2, r = 0.12, spacing = 150, s = 750, red)); add(pathticks(anglemark(B,C,Q,1000), n = 2, r = 0.12, spacing = 150, s = 750, red)); dot(\"$A$\",A,1.5*dir(A),linewidth(4)); dot(\"$B$\",B,1.5*dir(B),linewidth(4)); dot(\"$C$\",C,1.5*dir(C),linewidth(4)); dot(\"$D$\",D,1.5*dir(D),linewidth(4)); dot(\"$P$\",P,1.5*(-1,0),linewidth(4)); dot(\"$Q$\",Q,1.5*E,linewidth(4)); dot(\"$X$\",X,1.5*dir(-105),linewidth(4)); dot(\"$Y$\",Y,1.5*N,linewidth(4)); dot(\"$Z$\",Z,4.5*dir(75),linewidth(4)); dot(\"$W$\",W,1.5*dir(-75),linewidth(4)); draw(A--B--C--D--cycle^^A--P--D^^B--Q--C^^P--Q); draw(P--W^^Q--X^^Y--Z,dashed); [/asy] Let $X$ and $W$ be the points where $AP$ and $BQ$ extend to meet $CD$, and $YZ$ be the height of $\\triangle AZB$. As proven in Solution 2, triangles $APD$ and $DPW$ are congruent right triangles. Therefore, $AD = DW = 333$. We can apply this logic to triangles $BCQ$ and $XCQ$ as well, giving us $BC = CX = 333$. Since $CD = 650$, $XW = DW + CX - CD = 16$. Additionally, we can see that $\\triangle XZW$ is similar to $\\triangle PQZ$ and $\\triangle AZB$. We know that $\\frac{XW}{AB} = \\frac{16}{500}$. So, we can say that the height of the triangle $AZB$ is $500u$ while the height of the triangle $XZW$ is $16u$. After that, we can figure out the distance from $Y$ to $PQ: \\frac{500+16}{2} = 258u$ and the height of triangle $PZQ: 500-258 = 242u$. Finally, since the ratio between the height of $PZQ$ to the height of $AZB$ is $242:500$ and $AB$ is $500$, $PQ = 242 ~Cytronical",
"Extend line $PQ$ to meet $AD$ at $P'$ and $BC$ at $Q'$. The diagram looks like this: [asy] /* Made by MRENTHUSIASM */ size(300); pair A, B, C, D, A1, B1, C1, D1, P, Q, P1, Q1; A = (-250,6*sqrt(731)); B = (250,6*sqrt(731)); C = (325,-6*sqrt(731)); D = (-325,-6*sqrt(731)); A1 = bisectorpoint(B,A,D); B1 = bisectorpoint(A,B,C); C1 = bisectorpoint(B,C,D); D1 = bisectorpoint(A,D,C); P = intersectionpoint(A--300*(A1-A)+A,D--300*(D1-D)+D); Q = intersectionpoint(B--300*(B1-B)+B,C--300*(C1-C)+C); P1 = intersectionpoint(A--D,P--(-300)*(Q-P)+P); Q1 = intersectionpoint(B--C,Q--300*(Q-P)+Q); draw(anglemark(P,A,B,1000),red); draw(anglemark(D,A,P,1000),red); draw(anglemark(A,B,Q,1000),red); draw(anglemark(Q,B,C,1000),red); draw(anglemark(P,D,A,1000),red); draw(anglemark(C,D,P,1000),red); draw(anglemark(Q,C,D,1000),red); draw(anglemark(B,C,Q,1000),red); add(pathticks(anglemark(P,A,B,1000), n = 1, r = 0.15, s = 750, red)); add(pathticks(anglemark(D,A,P,1000), n = 1, r = 0.15, s = 750, red)); add(pathticks(anglemark(A,B,Q,1000), n = 1, r = 0.15, s = 750, red)); add(pathticks(anglemark(Q,B,C,1000), n = 1, r = 0.15, s = 750, red)); add(pathticks(anglemark(P,D,A,1000), n = 2, r = 0.12, spacing = 150, s = 750, red)); add(pathticks(anglemark(C,D,P,1000), n = 2, r = 0.12, spacing = 150, s = 750, red)); add(pathticks(anglemark(Q,C,D,1000), n = 2, r = 0.12, spacing = 150, s = 750, red)); add(pathticks(anglemark(B,C,Q,1000), n = 2, r = 0.12, spacing = 150, s = 750, red)); dot(\"$A$\",A,1.5*dir(A),linewidth(4)); dot(\"$B$\",B,1.5*dir(B),linewidth(4)); dot(\"$C$\",C,1.5*dir(C),linewidth(4)); dot(\"$D$\",D,1.5*dir(D),linewidth(4)); dot(\"$P$\",P,1.5*NE,linewidth(4)); dot(\"$Q$\",Q,1.5*NW,linewidth(4)); dot(\"$P'$\",P1,1.5*W,linewidth(4)); dot(\"$Q'$\",Q1,1.5*E,linewidth(4)); draw(A--B--C--D--cycle^^A--P--D^^B--Q--C^^P--Q); draw(P--P1^^Q--Q1,dashed); [/asy] Because the trapezoid is isosceles, by symmetry $PQ$ is parallel to $AB$ and $CD$. Therefore, $\\angle PAB \\cong \\angle APP'$ by interior angles and $\\angle PAB \\cong \\angle PAD$ by the problem statement. Thus, $\\triangle P'AP$ is isosceles with $P'P = P'A$. By symmetry, $P'DP$ is also isosceles, and thus $P'A = \\frac{AD}{2}$. Similarly, the same thing is happening on the right side of the trapezoid, and thus $P'Q'$ is the midline of the trapezoid. Then, $PQ = P'Q' - (P'P + Q'Q)$. Since $P'P = P'A = \\frac{AD}{2}, Q'Q = Q'B = \\frac{BC}{2}$ and $AD = BC = 333$, we have $P'P + Q'Q = \\frac{333}{2} + \\frac{333}{2} = 333$. The length of the midline of a trapezoid is the average of their bases, so $P'Q' = \\frac{500+650}{2} = 575$. Finally, $PQ = 575 - 333 = 242. ~KingRavi",
"We have the following diagram: [asy] /* Made by MRENTHUSIASM */ size(300); pair A, B, C, D, A1, B1, C1, D1, P, Q, X, Y, Z, W; A = (-250,6*sqrt(731)); B = (250,6*sqrt(731)); C = (325,-6*sqrt(731)); D = (-325,-6*sqrt(731)); A1 = bisectorpoint(B,A,D); B1 = bisectorpoint(A,B,C); C1 = bisectorpoint(B,C,D); D1 = bisectorpoint(A,D,C); P = intersectionpoint(A--300*(A1-A)+A,D--300*(D1-D)+D); Q = intersectionpoint(B--300*(B1-B)+B,C--300*(C1-C)+C); X = intersectionpoint(B--5*(Q-B)+B,C--D); Y = intersectionpoint(C--5*(Q-C)+C,A--B); Z = intersectionpoint(D--5*(P-D)+D,A--B); W = intersectionpoint(A--5*(P-A)+A,C--D); draw(anglemark(P,A,B,1000),red); draw(anglemark(D,A,P,1000),red); draw(anglemark(A,B,Q,1000),red); draw(anglemark(Q,B,C,1000),red); draw(anglemark(P,D,A,1000),red); draw(anglemark(C,D,P,1000),red); draw(anglemark(Q,C,D,1000),red); draw(anglemark(B,C,Q,1000),red); add(pathticks(anglemark(P,A,B,1000), n = 1, r = 0.15, s = 750, red)); add(pathticks(anglemark(D,A,P,1000), n = 1, r = 0.15, s = 750, red)); add(pathticks(anglemark(A,B,Q,1000), n = 1, r = 0.15, s = 750, red)); add(pathticks(anglemark(Q,B,C,1000), n = 1, r = 0.15, s = 750, red)); add(pathticks(anglemark(P,D,A,1000), n = 2, r = 0.12, spacing = 150, s = 750, red)); add(pathticks(anglemark(C,D,P,1000), n = 2, r = 0.12, spacing = 150, s = 750, red)); add(pathticks(anglemark(Q,C,D,1000), n = 2, r = 0.12, spacing = 150, s = 750, red)); add(pathticks(anglemark(B,C,Q,1000), n = 2, r = 0.12, spacing = 150, s = 750, red)); dot(\"$A$\",A,1.5*dir(A),linewidth(4)); dot(\"$B$\",B,1.5*dir(B),linewidth(4)); dot(\"$C$\",C,1.5*dir(C),linewidth(4)); dot(\"$D$\",D,1.5*dir(D),linewidth(4)); dot(\"$P$\",P,1.5*(-1,0),linewidth(4)); dot(\"$Q$\",Q,1.5*E,linewidth(4)); dot(\"$X$\",X,1.5*dir(-105),linewidth(4)); dot(\"$Y$\",Y,1.5*N,linewidth(4)); dot(\"$Z$\",Z,1.5*N,linewidth(4)); dot(\"$W$\",W,1.5*dir(-75),linewidth(4)); draw(A--B--C--D--cycle^^A--P--D^^B--Q--C^^P--Q); draw(P--Z^^P--W^^Q--X^^Q--Y,dashed); [/asy] Extend lines $AP$ and $BQ$ to meet line $DC$ at points $W$ and $X$, respectively, and extend lines $DP$ and $CQ$ to meet $AB$ at points $Z$ and $Y$, respectively. Claim: quadrilaterals $AZWD$ and $BYXC$ are rhombuses. Proof: Since $\\angle DAB + \\angle ADC = 180^{\\circ}$, $\\angle ADP + \\angle PAD = 90^{\\circ}$. Therefore, triangles $APD$, $APZ$, $DPW$ and $PZW$ are all right triangles. By SAA congruence, the first three triangles are congruent; by SAS congruence, $\\triangle PZW$ is congruent to the other three. Therefore, $AD = DW = WZ = AZ$, so $AZWD$ is a rhombus. By symmetry, $BYXC$ is also a rhombus. Extend line $PQ$ to meet $\\overline{AD}$ and $\\overline{BC}$ at $R$ and $S$, respectively. Because of rhombus properties, $RP = QS = \\frac{333}{2}$. Also, by rhombus properties, $R$ and $S$ are the midpoints of segments $AD$ and $BC$, respectively; therefore, by trapezoid properties, $RS = \\frac{AB + CD}{2} = 575$. Finally, $PQ = RS - RP - QS = 242. ~ihatemath123",
"[asy] /* Made by MRENTHUSIASM */ size(300); pair A, B, C, D, A1, B1, C1, D1, P, Q, X, Y, Z, W; A = (-250,6*sqrt(731)); B = (250,6*sqrt(731)); C = (325,-6*sqrt(731)); D = (-325,-6*sqrt(731)); A1 = bisectorpoint(B,A,D); B1 = bisectorpoint(A,B,C); C1 = bisectorpoint(B,C,D); D1 = bisectorpoint(A,D,C); P = intersectionpoint(A--300*(A1-A)+A,D--300*(D1-D)+D); Q = intersectionpoint(B--300*(B1-B)+B,C--300*(C1-C)+C); draw(anglemark(P,A,B,1000),red); draw(anglemark(D,A,P,1000),red); draw(anglemark(A,B,Q,1000),red); draw(anglemark(Q,B,C,1000),red); draw(anglemark(P,D,A,1000),red); draw(anglemark(C,D,P,1000),red); draw(anglemark(Q,C,D,1000),red); draw(anglemark(B,C,Q,1000),red); add(pathticks(anglemark(P,A,B,1000), n = 1, r = 0.15, s = 750, red)); add(pathticks(anglemark(D,A,P,1000), n = 1, r = 0.15, s = 750, red)); add(pathticks(anglemark(A,B,Q,1000), n = 1, r = 0.15, s = 750, red)); add(pathticks(anglemark(Q,B,C,1000), n = 1, r = 0.15, s = 750, red)); add(pathticks(anglemark(P,D,A,1000), n = 2, r = 0.12, spacing = 150, s = 750, red)); add(pathticks(anglemark(C,D,P,1000), n = 2, r = 0.12, spacing = 150, s = 750, red)); add(pathticks(anglemark(Q,C,D,1000), n = 2, r = 0.12, spacing = 150, s = 750, red)); add(pathticks(anglemark(B,C,Q,1000), n = 2, r = 0.12, spacing = 150, s = 750, red)); dot(\"$A$\",A,1.5*dir(A),linewidth(4)); dot(\"$B$\",B,1.5*dir(B),linewidth(4)); dot(\"$C$\",C,1.5*dir(C),linewidth(4)); dot(\"$D$\",D,1.5*dir(D),linewidth(4)); dot(\"$P$\",P,1.5*NE,linewidth(4)); dot(\"$Q$\",Q,1.5*NW,linewidth(4)); draw(A--B--C--D--cycle^^A--P--D^^B--Q--C^^P--Q); X = (-121,6*sqrt(731)); Y = (121,6*sqrt(731)); Z = (-121,-6*sqrt(731)); W = (121,-6*sqrt(731)); draw(X--Z^^Y--W,dashed); draw(rightanglemark(A,X,Z,500),red); draw(rightanglemark(B,Y,W,500),red); draw(rightanglemark(C,W,Y,500),red); draw(rightanglemark(D,Z,X,500),red); dot(\"$X$\",X,1.5*N,linewidth(4)); dot(\"$Y$\",Y,1.5*N,linewidth(4)); dot(\"$Z$\",Z,1.5*S,linewidth(4)); dot(\"$W$\",W,1.5*S,linewidth(4)); [/asy] Let $X$ and $Y$ be the feet of the altitudes from $P$ and $Q$, respectively, to $AB$, and let $Z$ and $W$ be the feet of the altitudes from $P$ and $Q$, respectively, to $CD$. Side $AB$ is parallel to side $CD$, so $XYWZ$ is a rectangle with width $PQ$. Furthermore, because $CD - AB = 650-500 = 150$ and trapezoid $ABCD$ is isosceles, $WC - YB = ZD - XA = 75$. Also because $ABCD$ is isosceles, $\\angle ABC + \\angle BCD$ is half the total sum of angles in $ABCD$, or $180^{\\circ}$. Since $BQ$ and $CQ$ bisect $\\angle ABC$ and $\\angle BCD$, respectively, we have $\\angle QBC + \\angle QCB = 90^{\\circ}$, so $\\angle BQC = 90^{\\circ}$. Letting $BQ = 333k$, applying Pythagoras to $\\triangle BQC$ yields $QC = 333\\sqrt{1-k^2}$. We then proceed using similar triangles: $\\angle BYQ = \\angle BQC = 90^{\\circ}$ and $\\angle YBQ = \\angle QBC$, so by AA similarity $YB = 333k^2$. Likewise, $\\angle CWQ = \\angle BQC = 90^{\\circ}$ and $\\angle WCQ = \\angle QCB$, so by AA similarity $WC = 333(1 - k^2)$. Thus $WC + YB = 333$. Adding our two equations for $WC$ and $YB$ gives $2WC = 75 + 333 = 408$. Therefore, the answer is $PQ = ZW = CD - 2WC = 650 - 408 = 242. ~Orange_Quail_9",
"This will be my first solution on AoPS. My apologies in advance for any errors. Angle bisectors can be thought of as the locus of all points equidistant from the lines whose angle they bisect. It can thus be seen that $P$ is equidistant from $AB, AD,$ and $CD$ and $Q$ is equidistant from $AB, BC,$ and $CD.$ If we let the feet of the altitudes from $P$ to $AB, AD,$ and $CD$ be called $E, F,$ and $G$ respectively, we can say that $PE = PF = PG.$ Analogously, we let the feet of the altitudes from $Q$ to $AB, BC,$ and $CD$ be $H, I,$ and $J$ respectively. Thus, $QH = QI = QJ.$ Because $ABCD$ is an isosceles trapezoid, we can say that all of the altitudes are equal to each other. By SA as well as SS congruence for right triangles, we find that triangles $AEP, AFP, BHQ,$ and $BIQ$ are congruent. Similarly, $DFP, DGP, CJQ,$ and $CIQ$ by the same reasoning. Additionally, $EH = GJ = PQ$ since $EHQP$ and $GJQP$ are congruent rectangles. If we then let $x = AE = AF = BH = BI,$ let $y = CI = CJ = DG = DF,$ and let $z = EH = GJ = PQ,$ we can create the following system of equations with the given side length information: \\begin{align*} 2x + z &= 500, \\\\ 2y + z &= 650, \\\\ x + y &= 333. \\end{align*} Adding the first two equations, subtracting by twice the third, and dividing by $2$ yields $z = PQ = 242 ~regular",
"Extend line $PQ$ to meet $AD$ at $P'$ and $BC$ at $Q'$. The diagram looks like this: [asy] /* Made by MRENTHUSIASM */ size(300); pair A, B, C, D, A1, B1, C1, D1, P, Q, P1, Q1; A = (-250,6*sqrt(731)); B = (250,6*sqrt(731)); C = (325,-6*sqrt(731)); D = (-325,-6*sqrt(731)); A1 = bisectorpoint(B,A,D); B1 = bisectorpoint(A,B,C); C1 = bisectorpoint(B,C,D); D1 = bisectorpoint(A,D,C); P = intersectionpoint(A--300*(A1-A)+A,D--300*(D1-D)+D); Q = intersectionpoint(B--300*(B1-B)+B,C--300*(C1-C)+C); P1 = intersectionpoint(A--D,P--(-300)*(Q-P)+P); Q1 = intersectionpoint(B--C,Q--300*(Q-P)+Q); draw(anglemark(P,A,B,1000),red); draw(anglemark(D,A,P,1000),red); draw(anglemark(A,B,Q,1000),red); draw(anglemark(Q,B,C,1000),red); draw(anglemark(P,D,A,1000),red); draw(anglemark(C,D,P,1000),red); draw(anglemark(Q,C,D,1000),red); draw(anglemark(B,C,Q,1000),red); add(pathticks(anglemark(P,A,B,1000), n = 1, r = 0.15, s = 750, red)); add(pathticks(anglemark(D,A,P,1000), n = 1, r = 0.15, s = 750, red)); add(pathticks(anglemark(A,B,Q,1000), n = 1, r = 0.15, s = 750, red)); add(pathticks(anglemark(Q,B,C,1000), n = 1, r = 0.15, s = 750, red)); add(pathticks(anglemark(P,D,A,1000), n = 2, r = 0.12, spacing = 150, s = 750, red)); add(pathticks(anglemark(C,D,P,1000), n = 2, r = 0.12, spacing = 150, s = 750, red)); add(pathticks(anglemark(Q,C,D,1000), n = 2, r = 0.12, spacing = 150, s = 750, red)); add(pathticks(anglemark(B,C,Q,1000), n = 2, r = 0.12, spacing = 150, s = 750, red)); dot(\"$A$\",A,1.5*dir(A),linewidth(4)); dot(\"$B$\",B,1.5*dir(B),linewidth(4)); dot(\"$C$\",C,1.5*dir(C),linewidth(4)); dot(\"$D$\",D,1.5*dir(D),linewidth(4)); dot(\"$P$\",P,1.5*NE,linewidth(4)); dot(\"$Q$\",Q,1.5*NW,linewidth(4)); dot(\"$P'$\",P1,1.5*W,linewidth(4)); dot(\"$Q'$\",Q1,1.5*E,linewidth(4)); draw(A--B--C--D--cycle^^A--P--D^^B--Q--C^^P--Q); draw(P--P1^^Q--Q1,dashed); [/asy] Since $\\angle A + \\angle D=\\angle B + \\angle C = 180^{\\circ}$, it follows that $\\angle P'AP+\\angle P'DP = \\angle Q'BQ + \\angle Q'CQ = 90^{\\circ}$. Thus, $\\angle APD = \\angle BQC = 90^{\\circ}$, implying that $\\triangle APD$ and $\\triangle BQC$ are right triangles. Since $P'P$ and $Q'Q$ are medians, $P'P+Q'Q=\\frac{333\\times2}{2}=333$. Since $P'Q'=\\frac{500+650}{2}=575$, we have $PQ+P'P+Q'Q=575$, or $PQ=575-333=242. ~sigma",
"Let $PQ = x$. Note that since $AP$ bisects $\\angle{A}$ and $DP$ bisects $\\angle{D}$, we have \\[\\angle{APD} = 180^{\\circ}-\\tfrac12 \\angle{A}-\\tfrac12 \\angle{D}=90^{\\circ}.\\] Let $\\angle{ADP}=\\theta$. We have that $\\angle{ADC} = 2\\theta.$ Now, drop an altitude from $A$ to $CD$ at $E$. Notice that $DE=\\tfrac{650-500}{2}=75$. By the definition of cosine, we have \\[\\cos{2\\theta}=1-2\\cos^2{\\theta}=\\tfrac{75}{333}=\\tfrac{25}{111} \\implies \\cos{\\theta}=\\tfrac{2\\sqrt{1887}}{111}.\\] Notice, however, that we can also apply this to $\\triangle{APD}$; we have \\[\\cos{\\theta}=\\tfrac{DP}{333} \\implies DP=6\\sqrt{1887}.\\] By the Pythagorean Theorem, we get \\[AP=\\sqrt{333^2-(6\\sqrt{1887})^2}=3\\sqrt{4773}.\\] Then, drop an altitude from $P$ to $AB$ at $F$; if $AF=y$, then $PQ=x=500-2y$. Because $AP$ is an angle bisector, we see that $\\angle{BAP}=\\angle{DAP}=90^{\\circ}-\\theta$. Again, by the definition of cosine, we have \\[\\cos{(90^{\\circ}-\\theta)}=\\sin{\\theta}=\\tfrac{\\sqrt{4773}}{111}=\\tfrac{y}{3\\sqrt{4773}} \\implies y=129.\\] Finally, $PQ=500-2y=242. ~pqr.",
"Let $PQ = x$. Note that since $AP$ bisects $\\angle{A}$ and $DP$ bisects $\\angle{D}$, we have \\[\\angle{APD} = 180^{\\circ}-\\tfrac12 \\angle{A}-\\tfrac12 \\angle{D}=90^{\\circ}.\\] Let $\\angle{ADP}=\\theta$. We have that $\\angle{ADC} = 2\\theta.$ Now, drop an altitude from $A$ to $CD$ at $E$. Notice that $DE=\\tfrac{650-500}{2}=75$. By the definition of cosine, we have \\[\\cos{2\\theta}=1-2\\cos^2{\\theta}=\\tfrac{75}{333}=\\tfrac{25}{111} \\implies \\cos{\\theta}=\\tfrac{2\\sqrt{1887}}{111}.\\] Notice, however, that we can also apply this to $\\triangle{APD}$; we have \\[\\cos{\\theta}=\\tfrac{DP}{333} \\implies DP=6\\sqrt{1887}.\\] By the Pythagorean Theorem, we get \\[AP=\\sqrt{333^2-(6\\sqrt{1887})^2}=3\\sqrt{4773}.\\] Then, drop an altitude from $P$ to $AB$ at $F$; if $AF=y$, then $PQ=x=500-2y$. Because $AP$ is an angle bisector, we see that $\\angle{BAP}=\\angle{DAP}=90^{\\circ}-\\theta$. Again, by the definition of cosine, we have \\[\\cos{(90^{\\circ}-\\theta)}=\\sin{\\theta}=\\tfrac{\\sqrt{4773}}{111}=\\tfrac{y}{3\\sqrt{4773}} \\implies y=129.\\] Finally, $PQ=500-2y=242. ~pqr.",
"[asy] /* Made by MRENTHUSIASM */ size(300); pair A, B, C, D, A1, B1, C1, D1, P, Q, X, Y, Z, W; A = (-250,6*sqrt(731)); B = (250,6*sqrt(731)); C = (325,-6*sqrt(731)); D = (-325,-6*sqrt(731)); A1 = bisectorpoint(B,A,D); B1 = bisectorpoint(A,B,C); C1 = bisectorpoint(B,C,D); D1 = bisectorpoint(A,D,C); P = intersectionpoint(A--300*(A1-A)+A,D--300*(D1-D)+D); Q = intersectionpoint(B--300*(B1-B)+B,C--300*(C1-C)+C); draw(anglemark(P,A,B,1000),red); draw(anglemark(D,A,P,1000),red); draw(anglemark(A,B,Q,1000),red); draw(anglemark(Q,B,C,1000),red); draw(anglemark(P,D,A,1000),red); draw(anglemark(C,D,P,1000),red); draw(anglemark(Q,C,D,1000),red); draw(anglemark(B,C,Q,1000),red); add(pathticks(anglemark(P,A,B,1000), n = 1, r = 0.15, s = 750, red)); add(pathticks(anglemark(D,A,P,1000), n = 1, r = 0.15, s = 750, red)); add(pathticks(anglemark(A,B,Q,1000), n = 1, r = 0.15, s = 750, red)); add(pathticks(anglemark(Q,B,C,1000), n = 1, r = 0.15, s = 750, red)); add(pathticks(anglemark(P,D,A,1000), n = 2, r = 0.12, spacing = 150, s = 750, red)); add(pathticks(anglemark(C,D,P,1000), n = 2, r = 0.12, spacing = 150, s = 750, red)); add(pathticks(anglemark(Q,C,D,1000), n = 2, r = 0.12, spacing = 150, s = 750, red)); add(pathticks(anglemark(B,C,Q,1000), n = 2, r = 0.12, spacing = 150, s = 750, red)); dot(\"$A$\",A,1.5*dir(A),linewidth(4)); dot(\"$B$\",B,1.5*dir(B),linewidth(4)); dot(\"$C$\",C,1.5*dir(C),linewidth(4)); dot(\"$D$\",D,1.5*dir(D),linewidth(4)); dot(\"$P$\",P,1.5*NE,linewidth(4)); dot(\"$Q$\",Q,1.5*NW,linewidth(4)); draw(A--B--C--D--cycle^^A--P--D^^B--Q--C^^P--Q); X = (-121,6*sqrt(731)); Y = (121,6*sqrt(731)); Z = (-121,-6*sqrt(731)); W = (121,-6*sqrt(731)); draw(X--Z^^Y--W,dashed); draw(rightanglemark(A,X,Z,500),red); draw(rightanglemark(B,Y,W,500),red); draw(rightanglemark(C,W,Y,500),red); draw(rightanglemark(D,Z,X,500),red); dot(\"$X$\",X,1.5*N,linewidth(4)); dot(\"$Y$\",Y,1.5*N,linewidth(4)); dot(\"$Z$\",Z,1.5*S,linewidth(4)); dot(\"$W$\",W,1.5*S,linewidth(4)); [/asy] As in solution 4, $\\angle APD = 90^{\\circ}$. Set $k = AX$ and $x = DP$. We know that $DZ = AX + \\frac{DC-AB}{2}$, so $DZ = k + \\frac{650-500}{2} = k + 75$. $\\triangle DPZ \\sim \\triangle APD$ by AA, so we have $\\frac{PD}{AD} = \\frac{ZD}{PD}$, resulting in \\[\\frac{x}{333} = \\frac{k+75}{x} \\text{ (1)}\\] $\\triangle APX \\sim \\triangle ADP$ by AA, so we have $\\frac{AP}{AD} = \\frac{AX}{AP}$, resulting in \\[\\frac{\\sqrt{333^2-x^2}}{333} = \\frac{k}{\\sqrt{333^2-k^2}} \\text{ (2)}\\] From $\\text{(1)}$, we have $x^2 = 333k + 333(75) = 333k + 24975$. From $\\text{(2)}$, we have $333^2 - x^2 = 333k$, or $x^2 = 333^2 - 333k$. Thus, $333k + 24975 = 333^2 - 333k$. Solving for $k$ yields $k = 129$. By symmetry, $YB = AX = 129$. Thus, $PQ = XY = AB - 2AX = 500 - 2(129) = 242. ~ adam_zheng",
"[asy] /* Made by MRENTHUSIASM */ size(300); pair A, B, C, D, A1, B1, C1, D1, P, Q, X, Y, Z, W; A = (-250,6*sqrt(731)); B = (250,6*sqrt(731)); C = (325,-6*sqrt(731)); D = (-325,-6*sqrt(731)); A1 = bisectorpoint(B,A,D); B1 = bisectorpoint(A,B,C); C1 = bisectorpoint(B,C,D); D1 = bisectorpoint(A,D,C); P = intersectionpoint(A--300*(A1-A)+A,D--300*(D1-D)+D); Q = intersectionpoint(B--300*(B1-B)+B,C--300*(C1-C)+C); draw(anglemark(P,A,B,1000),red); draw(anglemark(D,A,P,1000),red); draw(anglemark(A,B,Q,1000),red); draw(anglemark(Q,B,C,1000),red); draw(anglemark(P,D,A,1000),red); draw(anglemark(C,D,P,1000),red); draw(anglemark(Q,C,D,1000),red); draw(anglemark(B,C,Q,1000),red); add(pathticks(anglemark(P,A,B,1000), n = 1, r = 0.15, s = 750, red)); add(pathticks(anglemark(D,A,P,1000), n = 1, r = 0.15, s = 750, red)); add(pathticks(anglemark(A,B,Q,1000), n = 1, r = 0.15, s = 750, red)); add(pathticks(anglemark(Q,B,C,1000), n = 1, r = 0.15, s = 750, red)); add(pathticks(anglemark(P,D,A,1000), n = 2, r = 0.12, spacing = 150, s = 750, red)); add(pathticks(anglemark(C,D,P,1000), n = 2, r = 0.12, spacing = 150, s = 750, red)); add(pathticks(anglemark(Q,C,D,1000), n = 2, r = 0.12, spacing = 150, s = 750, red)); add(pathticks(anglemark(B,C,Q,1000), n = 2, r = 0.12, spacing = 150, s = 750, red)); dot(\"$A$\",A,1.5*dir(A),linewidth(4)); dot(\"$B$\",B,1.5*dir(B),linewidth(4)); dot(\"$C$\",C,1.5*dir(C),linewidth(4)); dot(\"$D$\",D,1.5*dir(D),linewidth(4)); dot(\"$P$\",P,1.5*NE,linewidth(4)); dot(\"$Q$\",Q,1.5*NW,linewidth(4)); draw(A--B--C--D--cycle^^A--P--D^^B--Q--C^^P--Q); X = (-121,6*sqrt(731)); Y = (121,6*sqrt(731)); Z = (-121,-6*sqrt(731)); W = (121,-6*sqrt(731)); draw(X--Z^^Y--W,dashed); draw(rightanglemark(A,X,Z,500),red); draw(rightanglemark(B,Y,W,500),red); draw(rightanglemark(C,W,Y,500),red); draw(rightanglemark(D,Z,X,500),red); dot(\"$X$\",X,1.5*N,linewidth(4)); dot(\"$Y$\",Y,1.5*N,linewidth(4)); dot(\"$Z$\",Z,1.5*S,linewidth(4)); dot(\"$W$\",W,1.5*S,linewidth(4)); [/asy] As in solution 4, $\\angle APD = 90^{\\circ}$. Set $k = AX$ and $x = DP$. We know that $DZ = AX + \\frac{DC-AB}{2}$, so $DZ = k + \\frac{650-500}{2} = k + 75$. $\\triangle DPZ \\sim \\triangle APD$ by AA, so we have $\\frac{PD}{AD} = \\frac{ZD}{PD}$, resulting in \\[\\frac{x}{333} = \\frac{k+75}{x} \\text{ (1)}\\] $\\triangle APX \\sim \\triangle ADP$ by AA, so we have $\\frac{AP}{AD} = \\frac{AX}{AP}$, resulting in \\[\\frac{\\sqrt{333^2-x^2}}{333} = \\frac{k}{\\sqrt{333^2-k^2}} \\text{ (2)}\\] From $\\text{(1)}$, we have $x^2 = 333k + 333(75) = 333k + 24975$. From $\\text{(2)}$, we have $333^2 - x^2 = 333k$, or $x^2 = 333^2 - 333k$. Thus, $333k + 24975 = 333^2 - 333k$. Solving for $k$ yields $k = 129$. By symmetry, $YB = AX = 129$. Thus, $PQ = XY = AB - 2AX = 500 - 2(129) = 242. ~ adam_zheng"
] |
2022-I-4
| 2,022
| 4
|
Let $w = \dfrac{\sqrt{3} + i}{2}$ and $z = \dfrac{-1 + i\sqrt{3}}{2},$ where $i = \sqrt{-1}.$ Find the number of ordered pairs $(r,s)$ of positive integers not exceeding $100$ that satisfy the equation $i \cdot w^r = z^s.$
| 834
|
I
|
[
"We rewrite $w$ and $z$ in polar form: \\begin{align*} w &= e^{i\\cdot\\frac{\\pi}{6}}, \\\\ z &= e^{i\\cdot\\frac{2\\pi}{3}}. \\end{align*} The equation $i \\cdot w^r = z^s$ becomes \\begin{align*} e^{i\\cdot\\frac{\\pi}{2}} \\cdot \\left(e^{i\\cdot\\frac{\\pi}{6}}\\right)^r &= \\left(e^{i\\cdot\\frac{2\\pi}{3}}\\right)^s \\\\ e^{i\\left(\\frac{\\pi}{2}+\\frac{\\pi}{6}r\\right)} &= e^{i\\left(\\frac{2\\pi}{3}s\\right)} \\\\ \\frac{\\pi}{2}+\\frac{\\pi}{6}r &= \\frac{2\\pi}{3}s+2\\pi k \\\\ 3+r &= 4s+12k \\\\ 3+r &= 4(s+3k). \\end{align*} for some integer $k.$ Since $4\\leq 3+r\\leq 103$ and $4\\mid 3+r,$ we conclude that \\begin{align*} 3+r &\\in \\{4,8,12,\\ldots,100\\}, \\\\ s+3k &\\in \\{1,2,3,\\ldots,25\\}. \\end{align*} Note that the values for $s+3k$ and the values for $r$ have one-to-one correspondence. We apply casework to the values for $s+3k:$ $s+3k\\equiv0\\pmod{3}$ There are $8$ values for $s+3k,$ so there are $8$ values for $r.$ It follows that $s\\equiv0\\pmod{3},$ so there are $33$ values for $s.$ There are $8\\cdot33=264$ ordered pairs $(r,s)$ in this case. $s+3k\\equiv1\\pmod{3}$ There are $9$ values for $s+3k,$ so there are $9$ values for $r.$ It follows that $s\\equiv1\\pmod{3},$ so there are $34$ values for $s.$ There are $9\\cdot34=306$ ordered pairs $(r,s)$ in this case. $s+3k\\equiv2\\pmod{3}$ There are $8$ values for $s+3k,$ so there are $8$ values for $r.$ It follows that $s\\equiv2\\pmod{3},$ so there are $33$ values for $s.$ There are $8\\cdot33=264$ ordered pairs $(r,s)$ in this case. Together, the answer is $264+306+264=834 ~MRENTHUSIASM",
"First we recognize that $w = \\operatorname{cis}(30^{\\circ})$ and $z = \\operatorname{cis}(120^{\\circ})$ because the cosine and sine sums of those angles give the values of $w$ and $z$, respectively. By De Moivre's theorem, $\\operatorname{cis}(\\theta)^n = \\operatorname{cis}(n\\theta)$. When you multiply by $i$, we can think of that as rotating the complex number $90^{\\circ}$ counterclockwise in the complex plane. Therefore, by the equation we know that $30r + 90$ and $120s$ land on the same angle. This means that \\[30r + 90 \\equiv 120s \\pmod{360},\\] which we can simplify to \\[r+3 \\equiv 4s \\pmod{12}.\\] Notice that this means that $r$ cycles by $12$ for every value of $s$. This is because once $r$ hits $12$, we get an angle of $360^{\\circ}$ and the angle laps onto itself again. By a similar reasoning, $s$ laps itself every $3$ times, which is much easier to count. By listing the possible values out, we get the pairs $(r,s)$: \\[\\begin{array}{cccccccc} (1,1) & (5,2) & (9,3) & (13,1) & (17,2) & (21,3) & \\ldots & (97,1) \\\\ (1,4) & (5,5) & (9,6) & (13,4) & (17,5) & (21,6) & \\ldots & (97,4) \\\\ (1,7) & (5,8) & (9,9) & (13,7) & (17,8) & (21,9) & \\ldots & (97,7) \\\\ [-1ex] \\vdots & \\vdots & \\vdots & \\vdots & \\vdots & \\vdots & \\vdots & \\vdots \\\\ (1,100) & (5,98) & (9,99) & (13,100) & (17,98) & (21,99) & \\ldots & (97,100) \\end{array}\\] We have $25$ columns in total: $34$ values for the first column, $33$ for the second, $33$ for the third, and then $34$ for the fourth, $33$ for the fifth, $33$ for the sixth, etc. Therefore, this cycle repeats every $3$ columns and our total sum is $(34+33+33) \\cdot 8 + 34 = 100 \\cdot 8 + 34 = 834. ~KingRavi"
] |
2022-I-5
| 2,022
| 5
|
A straight river that is $264$ meters wide flows from west to east at a rate of $14$ meters per minute. Melanie and Sherry sit on the south bank of the river with Melanie a distance of $D$ meters downstream from Sherry. Relative to the water, Melanie swims at $80$ meters per minute, and Sherry swims at $60$ meters per minute. At the same time, Melanie and Sherry begin swimming in straight lines to a point on the north bank of the river that is equidistant from their starting positions. The two women arrive at this point simultaneously. Find $D.$
| 550
|
I
|
[
"Define $m$ as the number of minutes they swim for. Let their meeting point be $A$. Melanie is swimming against the current, so she must aim upstream from point $A$, to compensate for this; in particular, since she is swimming for $m$ minutes, the current will push her $14m$ meters downstream in that time, so she must aim for a point $B$ that is $14m$ meters upstream from point $A$. Similarly, Sherry is swimming downstream for $m$ minutes, so she must also aim at point $B$ to compensate for the flow of the current. If Melanie and Sherry were to both aim at point $B$ in a currentless river with the same dimensions, they would still both meet at that point simultaneously. Since there is no current in this scenario, the distances that Melanie and Sherry travel, respectively, are $80m$ and $60m$ meters. We can draw out this new scenario, with the dimensions that we have: [asy] unitsize(0.02cm); draw((0,0)--(0,264)--(550,264)--(550,0)--cycle); pair B = (198,264); dot(B^^(0,0)^^(550,0),linewidth(5)); draw((0,0)--B,dashed); draw((550,0)--B,dashed); label(\"$60m$\", (0,0)--B, E); label(\"$80m$\", (550,0)--B, W); label(\"$264$\", (0,0)--(0,264), W); label(\"$\\frac{D}{2} - 14m$\", (0,264)--B, N); label(\"$\\frac{D}{2} + 14m$\", B--(550,264), N); label(\"$D$\", (0,0)--(550,0), S); label(\"$B$\", B, N); label(\"Downstream\", (350,325), E); label(\"Upstream\", (200,325), W); draw((225,325)--(325,325), Arrows); [/asy] (While it is indeed true that the triangle above with side lengths $60m$, $80m$ and $D$ is a right triangle, we do not know this yet, so we cannot assume this based on the diagram.) By the Pythagorean Theorem, we have \\begin{align*} 264^{2} + \\left( \\frac{D}{2} - 14m \\right) ^{2} &= 3600m^{2} \\\\ 264^{2} + \\left( \\frac{D}{2} + 14m \\right) ^{2} &= 6400m^{2}. \\end{align*} Subtracting the first equation from the second gives us $28Dm = 2800m^{2}$, so $D = 100m$. Substituting this into our first equation, we have that \\begin{align*}264^{2} + 36^{2} m^{2} &= 60^{2}m^{2} \\\\ 264^{2} &= 96 \\cdot 24 \\cdot m^{2} \\\\ 11^{2} &= 4 \\cdot m^{2} \\\\ m &= \\frac{11}{2}. \\end{align*} So $D = 100m = 550. ~ihatemath123",
"Define $m$ as the number of minutes they swim for. Let their meeting point be $A$. Melanie is swimming against the current, so she must aim upstream from point $A$, to compensate for this; in particular, since she is swimming for $m$ minutes, the current will push her $14m$ meters downstream in that time, so she must aim for a point $B$ that is $14m$ meters upstream from point $A$. Similarly, Sherry is swimming downstream for $m$ minutes, so she must also aim at point $B$ to compensate for the flow of the current. If Melanie and Sherry were to both aim at point $B$ in a currentless river with the same dimensions, they would still both meet at that point simultaneously. Since there is no current in this scenario, the distances that Melanie and Sherry travel, respectively, are $80m$ and $60m$ meters. We can draw out this new scenario, with the dimensions that we have: [asy] unitsize(0.02cm); draw((0,0)--(0,264)--(550,264)--(550,0)--cycle); pair B = (198,264); dot(B^^(0,0)^^(550,0),linewidth(5)); draw((0,0)--B,dashed); draw((550,0)--B,dashed); label(\"$60m$\", (0,0)--B, E); label(\"$80m$\", (550,0)--B, W); label(\"$264$\", (0,0)--(0,264), W); label(\"$\\frac{D}{2} - 14m$\", (0,264)--B, N); label(\"$\\frac{D}{2} + 14m$\", B--(550,264), N); label(\"$D$\", (0,0)--(550,0), S); label(\"$B$\", B, N); label(\"Downstream\", (350,325), E); label(\"Upstream\", (200,325), W); draw((225,325)--(325,325), Arrows); [/asy] (While it is indeed true that the triangle above with side lengths $60m$, $80m$ and $D$ is a right triangle, we do not know this yet, so we cannot assume this based on the diagram.) By the Pythagorean Theorem, we have \\begin{align*} 264^{2} + \\left( \\frac{D}{2} - 14m \\right) ^{2} &= 3600m^{2} \\\\ 264^{2} + \\left( \\frac{D}{2} + 14m \\right) ^{2} &= 6400m^{2}. \\end{align*} Subtracting the first equation from the second gives us $28Dm = 2800m^{2}$, so $D = 100m$. Substituting this into our first equation, we have that \\begin{align*}264^{2} + 36^{2} m^{2} &= 60^{2}m^{2} \\\\ 264^{2} &= 96 \\cdot 24 \\cdot m^{2} \\\\ 11^{2} &= 4 \\cdot m^{2} \\\\ m &= \\frac{11}{2}. \\end{align*} So $D = 100m = 550. ~ihatemath123",
"Claim Median $AM$ and altitude $AH$ are drawn in triangle $ABC$. $AB = c, AC = b < c, BC = a$ are known. Let's denote $MH = x$. Prove that \\begin{align*}2ax = c^{2} - b^{2}\\end{align*} Proof \\[BH + CH = a,\\] \\begin{align*} BH^{2} - CH^{2} = c^{2} - b^{2}\\implies BH - CH &= \\frac{c^{2} - b^{2}} {a},\\end{align*} \\[BH = \\frac{c^{2} - b^{2}}{2a} + \\frac{a}{2},\\] \\begin{align*}MH = BH - BM &= \\frac{c^{2} - b^{2}} {2a}.\\end{align*} Solution In the coordinate system associated with water, the movement is described by the scheme in the form of a triangle, the side on which Melanie floats is $80t$, where t is the time of Melanie's movement, the side along which Sherry floats is $60t$. The meeting point floated away at a distance of $14t$ from the midpoint between the starting points of Melanie and Sherry. In the notation of the Claim, \\begin{align*} c = 80t, b = 60t, x = 14t \\implies a = \\frac{(80t)^2-(60t)^2}{2 \\cdot 14t}=\\frac{20^2}{4}\\cdot \\frac{16-9}{7}t = 100t.\\end{align*} Hence, \\begin{align*} AH = \\sqrt{BC^2-BH^2}= \\sqrt{(80t)^2-(50t+14t)^2}=16t \\cdot \\sqrt{5^2-4^2}= 48t = 264 \\implies t = 5.5.\\end{align*} \\[D = a = 100t = 550\\] vladimir.shelomovskii@gmail.com, vvsss",
"Claim Median $AM$ and altitude $AH$ are drawn in triangle $ABC$. $AB = c, AC = b < c, BC = a$ are known. Let's denote $MH = x$. Prove that \\begin{align*}2ax = c^{2} - b^{2}\\end{align*} Proof \\[BH + CH = a,\\] \\begin{align*} BH^{2} - CH^{2} = c^{2} - b^{2}\\implies BH - CH &= \\frac{c^{2} - b^{2}} {a},\\end{align*} \\[BH = \\frac{c^{2} - b^{2}}{2a} + \\frac{a}{2},\\] \\begin{align*}MH = BH - BM &= \\frac{c^{2} - b^{2}} {2a}.\\end{align*} Solution In the coordinate system associated with water, the movement is described by the scheme in the form of a triangle, the side on which Melanie floats is $80t$, where t is the time of Melanie's movement, the side along which Sherry floats is $60t$. The meeting point floated away at a distance of $14t$ from the midpoint between the starting points of Melanie and Sherry. In the notation of the Claim, \\begin{align*} c = 80t, b = 60t, x = 14t \\implies a = \\frac{(80t)^2-(60t)^2}{2 \\cdot 14t}=\\frac{20^2}{4}\\cdot \\frac{16-9}{7}t = 100t.\\end{align*} Hence, \\begin{align*} AH = \\sqrt{BC^2-BH^2}= \\sqrt{(80t)^2-(50t+14t)^2}=16t \\cdot \\sqrt{5^2-4^2}= 48t = 264 \\implies t = 5.5.\\end{align*} \\[D = a = 100t = 550\\] vladimir.shelomovskii@gmail.com, vvsss",
"We have the following diagram: [asy] /* Made by MRENTHUSIASM */ size(350); pair A, B, C; A = (0,264); B = (-275,0); C = (275,0); draw((-300,0)--(300,0)^^(-300,264)--(300,264)^^A--B^^A--C,linewidth(2)); dot(\"Finish\",A,1.75*N,linewidth(5)); dot(\"Sherry\",B,1.75*S,linewidth(5)); dot(\"Melanie\",C,1.75*S,linewidth(5)); Label L1 = Label(\"$D$\", align=(0,0), position=MidPoint, filltype=Fill(3,0,white)); Label L2 = Label(\"$264$\", align=(0,0), position=MidPoint, filltype=Fill(0,3,white)); Label L3 = Label(\"Current $(14)$\", position=EndPoint, filltype=Fill(3,0,white)); Label L4 = Label(\"$y$\", align=(-1,0), position=Relative(0.4)); Label L5 = Label(\"$x$\", align=(0,1), position=Relative(0.4)); Label L6 = Label(\"$y$\", align=(1,0), position=Relative(0.4)); Label L7 = Label(\"$x$\", align=(0,1), position=Relative(0.4)); draw(B-(0,75)--C-(0,75), L=L1, arrow=Arrows(),bar=Bars(15)); draw((-350,0)--(-350,264), L=L2, arrow=Arrows(),bar=Bars(15)); draw((-300,-120)--(300,-120), L=L3, arrow=EndArrow()); draw(B--B+(0,48), L=L4, arrow=EndArrow()); draw(B+(0,48)--B+(50,48), L=L5, arrow=EndArrow()); draw(C--C+(0,48), L=L6, arrow=EndArrow()); draw(C+(0,48)--C+(-50,48), L=L7, arrow=EndArrow()); [/asy] Since Melanie and Sherry swim for the same distance and the same amount of time, they swim at the same net speed. Let $x$ and $y$ be some positive numbers. We have the following table: \\[\\begin{array}{c||c|c|c} & \\textbf{Net Velocity Vector (m/min)} & \\textbf{Natural Velocity Vector (m/min)} & \\textbf{Natural Speed (m/min)} \\\\ \\hline \\hline &&& \\\\ [-2.25ex] \\textbf{Melanie} & \\langle -x,y\\rangle & \\langle -x-14,y\\rangle & 80 \\\\ \\hline &&& \\\\ [-2.25ex] \\textbf{Sherry} & \\langle x,y\\rangle & \\langle x-14,y\\rangle & 60 \\end{array}\\] Recall that $|\\text{velocity}|=\\text{speed},$ so \\begin{align*} (-x-14)^2 + y^2 &= 80^2, &&(1) \\\\ (x-14)^2 + y^2 &= 60^2. &&(2) \\end{align*} We subtract $(2)$ from $(1)$ to get $56x=2800,$ from which $x=50.$ Substituting this into either equation, we have $y=48.$ It follows that Melanie and Sherry both swim for $264\\div y=5.5$ minutes. Therefore, the answer is \\[D=2x\\cdot5.5=550.\\] ~MRENTHUSIASM",
"We have the following diagram: [asy] /* Made by MRENTHUSIASM */ size(350); pair A, B, C; A = (0,264); B = (-275,0); C = (275,0); draw((-300,0)--(300,0)^^(-300,264)--(300,264)^^A--B^^A--C,linewidth(2)); dot(\"Finish\",A,1.75*N,linewidth(5)); dot(\"Sherry\",B,1.75*S,linewidth(5)); dot(\"Melanie\",C,1.75*S,linewidth(5)); Label L1 = Label(\"$D$\", align=(0,0), position=MidPoint, filltype=Fill(3,0,white)); Label L2 = Label(\"$264$\", align=(0,0), position=MidPoint, filltype=Fill(0,3,white)); Label L3 = Label(\"Current $(14)$\", position=EndPoint, filltype=Fill(3,0,white)); Label L4 = Label(\"$y$\", align=(-1,0), position=Relative(0.4)); Label L5 = Label(\"$x$\", align=(0,1), position=Relative(0.4)); Label L6 = Label(\"$y$\", align=(1,0), position=Relative(0.4)); Label L7 = Label(\"$x$\", align=(0,1), position=Relative(0.4)); draw(B-(0,75)--C-(0,75), L=L1, arrow=Arrows(),bar=Bars(15)); draw((-350,0)--(-350,264), L=L2, arrow=Arrows(),bar=Bars(15)); draw((-300,-120)--(300,-120), L=L3, arrow=EndArrow()); draw(B--B+(0,48), L=L4, arrow=EndArrow()); draw(B+(0,48)--B+(50,48), L=L5, arrow=EndArrow()); draw(C--C+(0,48), L=L6, arrow=EndArrow()); draw(C+(0,48)--C+(-50,48), L=L7, arrow=EndArrow()); [/asy] Since Melanie and Sherry swim for the same distance and the same amount of time, they swim at the same net speed. Let $x$ and $y$ be some positive numbers. We have the following table: \\[\\begin{array}{c||c|c|c} & \\textbf{Net Velocity Vector (m/min)} & \\textbf{Natural Velocity Vector (m/min)} & \\textbf{Natural Speed (m/min)} \\\\ \\hline \\hline &&& \\\\ [-2.25ex] \\textbf{Melanie} & \\langle -x,y\\rangle & \\langle -x-14,y\\rangle & 80 \\\\ \\hline &&& \\\\ [-2.25ex] \\textbf{Sherry} & \\langle x,y\\rangle & \\langle x-14,y\\rangle & 60 \\end{array}\\] Recall that $|\\text{velocity}|=\\text{speed},$ so \\begin{align*} (-x-14)^2 + y^2 &= 80^2, &&(1) \\\\ (x-14)^2 + y^2 &= 60^2. &&(2) \\end{align*} We subtract $(2)$ from $(1)$ to get $56x=2800,$ from which $x=50.$ Substituting this into either equation, we have $y=48.$ It follows that Melanie and Sherry both swim for $264\\div y=5.5$ minutes. Therefore, the answer is \\[D=2x\\cdot5.5=550.\\] ~MRENTHUSIASM",
"We can break down movement into two components: the $x$-component and the $y$-component. Suppose that Melanie travels a distance of $a$ in the $x$-direction and a distance of $c$ in the $y$-direction in one minute when there is no current. Similarly, suppose that Sherry travels a distance of $a$ in the $x$-direction but a distance of $b$ in the $y$-direction in one minute when there is no current. The current only affects the $x$-components because it goes in the $x$-direction. [asy] /* Ruthlessly plagiarized from MRENTHUSIASM by Curious_crow */ size(350); pair A, B, C; A = (0,264); B = (-275,0); C = (275,0); draw((-300,0)--(300,0)^^(-300,264)--(300,264)^^A--B^^A--C,linewidth(2)); dot(\"Finish\",A,1.75*N,linewidth(5)); dot(\"Sherry\",B,1.75*S,linewidth(5)); dot(\"Melanie\",C,1.75*S,linewidth(5)); Label L1 = Label(\"$D$\", align=(0,0), position=MidPoint, filltype=Fill(3,0,white)); Label L2 = Label(\"$264$\", align=(0,0), position=MidPoint, filltype=Fill(0,3,white)); Label L3 = Label(\"Current $(14)$\", position=EndPoint, filltype=Fill(3,0,white)); Label L4 = Label(\"$a$\", align=(-1,0), position=Relative(0.4)); Label L5 = Label(\"$b+14$\", align=(0,1), position=Relative(0.4)); Label L6 = Label(\"$a$\", align=(1,0), position=Relative(0.4)); Label L7 = Label(\"$c-14$\", align=(0,1), position=Relative(0.4)); draw(B-(0,75)--C-(0,75), L=L1, arrow=Arrows(),bar=Bars(15)); draw((-350,0)--(-350,264), L=L2, arrow=Arrows(),bar=Bars(15)); draw((-300,-120)--(300,-120), L=L3, arrow=EndArrow()); draw(B--B+(0,48), L=L4, arrow=EndArrow()); draw(B+(0,48)--B+(50,48), L=L5, arrow=EndArrow()); draw(C--C+(0,48), L=L6, arrow=EndArrow()); draw(C+(0,48)--C+(-50,48), L=L7, arrow=EndArrow()); [/asy] Now, note that $a^2 + b^2 = 60^2$ because Sherry travels 60 meters in a minute. Thus, $a^2 + c^2 = 80^2$ because Melanie travels 80 meters in a minute. Also, the distance they travel with the current must be the same in one minute because they reach the point equidistant from them at the same time. That means $b + 14 = c - 14$ or $b = c - 28$. So now we can plug that into the two equations to get: \\begin{align*} a^2 + c^2 &= 80^2, \\\\ a^2 + (c-28)^2 &= 60^2. \\end{align*} We can solve the system of equations to get $a = 48$ and $c = 64$. From this, we can figure out that it must've taken them $5.5$ minutes to get to the other side, because $264/48 = 5.5$. This means that there are $5.5$ lengths of $48$ in each person's travel. Also, $D$ must be equal to $11(b+14) = 11(c-14)$ because there are $(5.5)2 = 11$ lengths of $b-14$ between them, $5.5$ on each person's side. Since $c = 64$, we have $c-14 = 50$, so the answer is \\[D=11\\cdot50=550.\\] ~Curious_crow",
"We can break down movement into two components: the $x$-component and the $y$-component. Suppose that Melanie travels a distance of $a$ in the $x$-direction and a distance of $c$ in the $y$-direction in one minute when there is no current. Similarly, suppose that Sherry travels a distance of $a$ in the $x$-direction but a distance of $b$ in the $y$-direction in one minute when there is no current. The current only affects the $x$-components because it goes in the $x$-direction. [asy] /* Ruthlessly plagiarized from MRENTHUSIASM by Curious_crow */ size(350); pair A, B, C; A = (0,264); B = (-275,0); C = (275,0); draw((-300,0)--(300,0)^^(-300,264)--(300,264)^^A--B^^A--C,linewidth(2)); dot(\"Finish\",A,1.75*N,linewidth(5)); dot(\"Sherry\",B,1.75*S,linewidth(5)); dot(\"Melanie\",C,1.75*S,linewidth(5)); Label L1 = Label(\"$D$\", align=(0,0), position=MidPoint, filltype=Fill(3,0,white)); Label L2 = Label(\"$264$\", align=(0,0), position=MidPoint, filltype=Fill(0,3,white)); Label L3 = Label(\"Current $(14)$\", position=EndPoint, filltype=Fill(3,0,white)); Label L4 = Label(\"$a$\", align=(-1,0), position=Relative(0.4)); Label L5 = Label(\"$b+14$\", align=(0,1), position=Relative(0.4)); Label L6 = Label(\"$a$\", align=(1,0), position=Relative(0.4)); Label L7 = Label(\"$c-14$\", align=(0,1), position=Relative(0.4)); draw(B-(0,75)--C-(0,75), L=L1, arrow=Arrows(),bar=Bars(15)); draw((-350,0)--(-350,264), L=L2, arrow=Arrows(),bar=Bars(15)); draw((-300,-120)--(300,-120), L=L3, arrow=EndArrow()); draw(B--B+(0,48), L=L4, arrow=EndArrow()); draw(B+(0,48)--B+(50,48), L=L5, arrow=EndArrow()); draw(C--C+(0,48), L=L6, arrow=EndArrow()); draw(C+(0,48)--C+(-50,48), L=L7, arrow=EndArrow()); [/asy] Now, note that $a^2 + b^2 = 60^2$ because Sherry travels 60 meters in a minute. Thus, $a^2 + c^2 = 80^2$ because Melanie travels 80 meters in a minute. Also, the distance they travel with the current must be the same in one minute because they reach the point equidistant from them at the same time. That means $b + 14 = c - 14$ or $b = c - 28$. So now we can plug that into the two equations to get: \\begin{align*} a^2 + c^2 &= 80^2, \\\\ a^2 + (c-28)^2 &= 60^2. \\end{align*} We can solve the system of equations to get $a = 48$ and $c = 64$. From this, we can figure out that it must've taken them $5.5$ minutes to get to the other side, because $264/48 = 5.5$. This means that there are $5.5$ lengths of $48$ in each person's travel. Also, $D$ must be equal to $11(b+14) = 11(c-14)$ because there are $(5.5)2 = 11$ lengths of $b-14$ between them, $5.5$ on each person's side. Since $c = 64$, we have $c-14 = 50$, so the answer is \\[D=11\\cdot50=550.\\] ~Curious_crow"
] |
2022-I-6
| 2,022
| 6
|
Find the number of ordered pairs of integers $(a,b)$ such that the sequence \[3,4,5,a,b,30,40,50\] is strictly increasing and no set of four (not necessarily consecutive) terms forms an arithmetic progression.
| 228
|
I
|
[
"Since $3,4,5,a$ and $3,4,5,b$ cannot be an arithmetic progression, $a$ or $b$ can never be $6$. Since $b, 30, 40, 50$ and $a, 30, 40, 50$ cannot be an arithmetic progression, $a$ and $b$ can never be $20$. Since $a < b$, there are ${24 - 2 \\choose 2} = 231$ ways to choose $a$ and $b$ with these two restrictions in mind. However, there are still specific invalid cases counted in these $231$ pairs $(a,b)$. Since \\[3,5,a,b\\] cannot form an arithmetic progression, $\\underline{(a,b) \\neq (7,9)}$. \\[a,b,30,50\\] cannot be an arithmetic progression, so $(a,b) \\neq (-10,10)$; however, since this pair was not counted in our $231$, we do not need to subtract it off. \\[3,a,b,30\\] cannot form an arithmetic progression, so $\\underline{(a,b) \\neq (12,21)}$. \\[4, a, b, 40\\] cannot form an arithmetic progression, so $\\underline{(a,b) \\neq (16,28)}$. \\[5, a,b, 50\\] cannot form an arithmetic progression, $(a,b) \\neq 20, 35$; however, since this pair was not counted in our $231$ (since we disallowed $a$ or $b$ to be $20$), we do not to subtract it off. Also, the sequences $(3,a,b,40)$, $(3,a,b,50)$, $(4,a,b,30)$, $(4,a,b,50)$, $(5,a,b,30)$ and $(5,a,b,40)$ will never be arithmetic, since that would require $a$ and $b$ to be non-integers. So, we need to subtract off $3$ progressions from the $231$ we counted, to get our final answer of $228. ~ ihatemath123",
"We will follow the solution from earlier in a rigorous manner to show that there are no other cases missing. We recognize that an illegal sequence (defined as one that we subtract from our 231) can never have the numbers {3, 4} and {4,5} because we have not included a 6 in our count. Similarly, sequences with {30,40} and {40,50} will not give us any subtractions because those sequences must all include a 20. Let's stick with the lower ones for a minute: if we take them two at a time, then {3,5} will give us the subtraction of 1 sequence {3,5,7,9}. We have exhausted all pairs of numbers we can take, and if we take the triplet of single digit numbers, the only possible sequence must have a 6, which we already don't count. Therefore, we subtract $\\textbf{1}$ from the count of illegal sequences with any of the single-digit numbers and none of the numbers 30,40,50. (Note if we take only 1 at a time, there will have to be 3 of $a, b$, which is impossible.) If we have the sequence including {30,50}, we end up having negative values, so these do not give us any subtractions, and the triplet {30,40,50} gives us a 20. Hence by the same reasoning as earlier, we have 0 subtractions from the sequences with these numbers and none of the single digit numbers {3,4,5}. Finally, we count the sequences that are something like (one of 3,4,5,), $a, b$, (one of 30, 40, 50). If this is to be the case, then let $a$ be the starting value in the sequence. The sequence will be $a, a+d, a+2d, a+3d$; We see that if we subtract the largest term by the smallest term we have $3d$, so the subtraction of one of (30,40,50) and one of (3,4,5) must be divisible by 3. Therefore the only sequences possible are $3,a,b,30; 4,a,b,40; 5,a,b,50$. Of these, only the last is invalid because it gives $b = 35$, larger than our bounds $6<a<b<30$. Therefore, we subtract $\\textbf{2}$ from this case. Our final answer is $231 - 1 - 2 = 228 ~KingRavi",
"We will follow the solution from earlier in a rigorous manner to show that there are no other cases missing. We recognize that an illegal sequence (defined as one that we subtract from our 231) can never have the numbers {3, 4} and {4,5} because we have not included a 6 in our count. Similarly, sequences with {30,40} and {40,50} will not give us any subtractions because those sequences must all include a 20. Let's stick with the lower ones for a minute: if we take them two at a time, then {3,5} will give us the subtraction of 1 sequence {3,5,7,9}. We have exhausted all pairs of numbers we can take, and if we take the triplet of single digit numbers, the only possible sequence must have a 6, which we already don't count. Therefore, we subtract $\\textbf{1}$ from the count of illegal sequences with any of the single-digit numbers and none of the numbers 30,40,50. (Note if we take only 1 at a time, there will have to be 3 of $a, b$, which is impossible.) If we have the sequence including {30,50}, we end up having negative values, so these do not give us any subtractions, and the triplet {30,40,50} gives us a 20. Hence by the same reasoning as earlier, we have 0 subtractions from the sequences with these numbers and none of the single digit numbers {3,4,5}. Finally, we count the sequences that are something like (one of 3,4,5,), $a, b$, (one of 30, 40, 50). If this is to be the case, then let $a$ be the starting value in the sequence. The sequence will be $a, a+d, a+2d, a+3d$; We see that if we subtract the largest term by the smallest term we have $3d$, so the subtraction of one of (30,40,50) and one of (3,4,5) must be divisible by 3. Therefore the only sequences possible are $3,a,b,30; 4,a,b,40; 5,a,b,50$. Of these, only the last is invalid because it gives $b = 35$, larger than our bounds $6<a<b<30$. Therefore, we subtract $\\textbf{2}$ from this case. Our final answer is $231 - 1 - 2 = 228 ~KingRavi",
"Denote $S = \\left\\{ (a, b) : 6 \\leq a < b \\leq 29 \\right\\}$. Denote by $A$ a subset of $S$, such that there exists an arithmetic sequence that has 4 terms and includes $a$ but not $b$. Denote by $B$ a subset of $S$, such that there exists an arithmetic sequence that has 4 terms and includes $b$ but not $a$. Hence, $C$ is a subset of $S$, such that there exists an arithmetic sequence that has 4 terms and includes both $a$ and $b$. Hence, this problem asks us to compute \\[ | S | - \\left( | A | + | B | + | C | \\right) . \\] First, we compute $| S |$. We have $| S | = \\binom{29 - 6 + 1}{2} = \\binom{24}{2} = 276$. Second, we compute $| A |$. $\\textbf{Case 1}$: $a = 6$. We have $b = 8 , \\cdots , 19, 21, 22, \\cdots, 29$. Thus, the number of solutions is 22. $\\textbf{Case 2}$: $a = 20$. We have $b = 21, 22, \\cdots , 29$. Thus, the number of solutions is 9. Thus, $| A | = 22 + 9 = 31$. Third, we compute $| B |$. In $B$, we have $b = 6, 20$. However, because $6 \\leq a < b$, we have $b \\geq 7$. Thus, $b = 20$. This implies $a = 7, 8, 9, 11, 12, \\cdots , 19$. Note that $(a, b)=(10, 20)$ belongs in $C$. Thus, $| B | = 12$. Fourth, we compute $| C |$. $\\textbf{Case 1}$: In the arithmetic sequence, the two numbers beyond $a$ and $b$ are on the same side of $a$ and $b$. Hence, $(a, b) = (6 , 7), (7, 9) , (10, 20)$. Therefore, the number solutions in this case is 3. $\\textbf{Case 2}$: In the arithmetic sequence, the two numbers beyond $a$ and $b$ are on the opposite sides of $a$ and $b$. $\\textbf{Case 2.1}$: The arithmetic sequence is $3, a, b, 30$. Hence, $(a, b) = (12, 21)$. $\\textbf{Case 2.2}$: The arithmetic sequence is $4, a, b, 40$. Hence, $(a, b) = (16, 28)$. $\\textbf{Case 2.3}$: The arithmetic sequence is $5, a, b, 50$. Hence, $(a, b) = (20, 35)$. However, the sequence $... 20, 35, 30, 40, 50$ is not strictly increasing. Putting two cases together, $| C | = 65.$ Therefore, \\[| S | - \\left( | A | + | B | + | C | \\right) = 276 - \\left( 31 + 12 + 5 \\right) = 228.\\] ~Steven Chen (www.professorchenedu.com)",
"divide cases into $7\\leq a<20; 21\\leq a\\leq28$.(Notice that $a$ can't be equal to $6,20$, that's why I divide them into two parts. There are three cases that arithmetic sequence forms: $3,12,21,30;4,16,28,40;3,5,7,9$.(NOTICE that $5,20,35,50$ IS NOT A VALID SEQUENCE!) So when $7\\leq a<20$, there are $10+11+12+...+22-3-13=192$ possible ways( 3 means the arithmetic sequence and 13 means there are 13 \"a\" s and b cannot be 20) When $21\\leq a \\leq 28$, there are $1+2+\\cdots+8=36$ ways. In all, there are $192+36=228 possible sequences. ~bluesoul"
] |
2022-I-7
| 2,022
| 7
|
Let $a,b,c,d,e,f,g,h,i$ be distinct integers from $1$ to $9.$ The minimum possible positive value of \[\dfrac{a \cdot b \cdot c - d \cdot e \cdot f}{g \cdot h \cdot i}\] can be written as $\frac{m}{n},$ where $m$ and $n$ are relatively prime positive integers. Find $m+n.$
| 289
|
I
|
[
"To minimize a positive fraction, we minimize its numerator and maximize its denominator. It is clear that $\\frac{a \\cdot b \\cdot c - d \\cdot e \\cdot f}{g \\cdot h \\cdot i} \\geq \\frac{1}{7\\cdot8\\cdot9}.$ If we minimize the numerator, then $a \\cdot b \\cdot c - d \\cdot e \\cdot f = 1.$ Note that $a \\cdot b \\cdot c \\cdot d \\cdot e \\cdot f = (a \\cdot b \\cdot c) \\cdot (a \\cdot b \\cdot c - 1) \\geq 6! = 720,$ so $a \\cdot b \\cdot c \\geq 28.$ It follows that $a \\cdot b \\cdot c$ and $d \\cdot e \\cdot f$ are consecutive composites with prime factors no other than $2,3,5,$ and $7.$ The smallest values for $a \\cdot b \\cdot c$ and $d \\cdot e \\cdot f$ are $36$ and $35,$ respectively. So, we have $\\{a,b,c\\} = \\{2,3,6\\}, \\{d,e,f\\} = \\{1,5,7\\},$ and $\\{g,h,i\\} = \\{4,8,9\\},$ from which $\\frac{a \\cdot b \\cdot c - d \\cdot e \\cdot f}{g \\cdot h \\cdot i} = \\frac{1}{288}.$ If we do not minimize the numerator, then $a \\cdot b \\cdot c - d \\cdot e \\cdot f > 1.$ Note that $\\frac{a \\cdot b \\cdot c - d \\cdot e \\cdot f}{g \\cdot h \\cdot i} \\geq \\frac{2}{7\\cdot8\\cdot9} > \\frac{1}{288}.$ Together, we conclude that the minimum possible positive value of $\\frac{a \\cdot b \\cdot c - d \\cdot e \\cdot f}{g \\cdot h \\cdot i}$ is $\\frac{1}{288}.$ Therefore, the answer is $1+288=289 ~MRENTHUSIASM ~jgplay",
"To minimize a positive fraction, we minimize its numerator and maximize its denominator. It is clear that $\\frac{a \\cdot b \\cdot c - d \\cdot e \\cdot f}{g \\cdot h \\cdot i} \\geq \\frac{1}{7\\cdot8\\cdot9}.$ If we minimize the numerator, then $a \\cdot b \\cdot c - d \\cdot e \\cdot f = 1.$ Note that $a \\cdot b \\cdot c \\cdot d \\cdot e \\cdot f = (a \\cdot b \\cdot c) \\cdot (a \\cdot b \\cdot c - 1) \\geq 6! = 720,$ so $a \\cdot b \\cdot c \\geq 28.$ It follows that $a \\cdot b \\cdot c$ and $d \\cdot e \\cdot f$ are consecutive composites with prime factors no other than $2,3,5,$ and $7.$ The smallest values for $a \\cdot b \\cdot c$ and $d \\cdot e \\cdot f$ are $36$ and $35,$ respectively. So, we have $\\{a,b,c\\} = \\{2,3,6\\}, \\{d,e,f\\} = \\{1,5,7\\},$ and $\\{g,h,i\\} = \\{4,8,9\\},$ from which $\\frac{a \\cdot b \\cdot c - d \\cdot e \\cdot f}{g \\cdot h \\cdot i} = \\frac{1}{288}.$ If we do not minimize the numerator, then $a \\cdot b \\cdot c - d \\cdot e \\cdot f > 1.$ Note that $\\frac{a \\cdot b \\cdot c - d \\cdot e \\cdot f}{g \\cdot h \\cdot i} \\geq \\frac{2}{7\\cdot8\\cdot9} > \\frac{1}{288}.$ Together, we conclude that the minimum possible positive value of $\\frac{a \\cdot b \\cdot c - d \\cdot e \\cdot f}{g \\cdot h \\cdot i}$ is $\\frac{1}{288}.$ Therefore, the answer is $1+288=289 ~MRENTHUSIASM ~jgplay",
"Obviously, to find the correct answer, we need to get the largest denominator with the smallest numerator. To bash efficiently, we can start out with $7\\cdot8\\cdot9$ as our denominator. This, however, leaves us with the numbers $1, 2, 3, 4, 5,$ and $6$ left. The smallest we can make out of this is $1\\cdot5\\cdot6 - 2\\cdot3\\cdot4 = 30 - 24 = 6$. When simplified, it gives us $\\frac{1}{84}$, which gives a small answer of $85$. Obviously there are larger answers than this. After the first bash, we learn to bash even more efficiently, we can consider both the numerator and the denominator when guessing. We know the numerator has to be extremely small while still having a large denominator. When bashing, we soon find out the couple $(a,b,c)=(2,3,6)$ and $(d,e,f)=(1,5,7)$. This gives us a numerator of $36-35=1$, which is by far the smallest yet. With the remaining numbers $4, 8,$ and $9$, we get $\\frac{36-35}{4\\cdot8\\cdot9}=\\frac{1}{288}$. Finally, we add up our numerator and denominator: The answer is $1+288=289. ~orenbad",
"Obviously, to find the correct answer, we need to get the largest denominator with the smallest numerator. To bash efficiently, we can start out with $7\\cdot8\\cdot9$ as our denominator. This, however, leaves us with the numbers $1, 2, 3, 4, 5,$ and $6$ left. The smallest we can make out of this is $1\\cdot5\\cdot6 - 2\\cdot3\\cdot4 = 30 - 24 = 6$. When simplified, it gives us $\\frac{1}{84}$, which gives a small answer of $85$. Obviously there are larger answers than this. After the first bash, we learn to bash even more efficiently, we can consider both the numerator and the denominator when guessing. We know the numerator has to be extremely small while still having a large denominator. When bashing, we soon find out the couple $(a,b,c)=(2,3,6)$ and $(d,e,f)=(1,5,7)$. This gives us a numerator of $36-35=1$, which is by far the smallest yet. With the remaining numbers $4, 8,$ and $9$, we get $\\frac{36-35}{4\\cdot8\\cdot9}=\\frac{1}{288}$. Finally, we add up our numerator and denominator: The answer is $1+288=289. ~orenbad",
"To minimize the numerator, we must have $abc - def = 1$. Thus, one of these products must be odd and the other must be even. The odd product must consist of only odd numbers. The smallest such value $(d, e, f) = (1, 3, 5)$ cannot result in a difference of $1$, and the next smallest product, $(d, e, f) = (1, 3, 7)$ cannot either, but $(d, e, f) = (1, 5, 7)$ can if $(a, b, c) = (2, 3, 6)$. Thus, the denominator must be $(g, h, i) = (4, 8, 9)$, and the smallest fraction possible is $\\dfrac{36 - 35}{288} = \\dfrac{1}{288}$, making the answer $1 + 288 = 289. ~A_MatheMagician",
"To minimize the numerator, we must have $abc - def = 1$. Thus, one of these products must be odd and the other must be even. The odd product must consist of only odd numbers. The smallest such value $(d, e, f) = (1, 3, 5)$ cannot result in a difference of $1$, and the next smallest product, $(d, e, f) = (1, 3, 7)$ cannot either, but $(d, e, f) = (1, 5, 7)$ can if $(a, b, c) = (2, 3, 6)$. Thus, the denominator must be $(g, h, i) = (4, 8, 9)$, and the smallest fraction possible is $\\dfrac{36 - 35}{288} = \\dfrac{1}{288}$, making the answer $1 + 288 = 289. ~A_MatheMagician"
] |
2022-I-8
| 2,022
| 8
|
Equilateral triangle $\triangle ABC$ is inscribed in circle $\omega$ with radius $18.$ Circle $\omega_A$ is tangent to sides $\overline{AB}$ and $\overline{AC}$ and is internally tangent to $\omega.$ Circles $\omega_B$ and $\omega_C$ are defined analogously. Circles $\omega_A,$ $\omega_B,$ and $\omega_C$ meet in six points---two points for each pair of circles. The three intersection points closest to the vertices of $\triangle ABC$ are the vertices of a large equilateral triangle in the interior of $\triangle ABC,$ and the other three intersection points are the vertices of a smaller equilateral triangle in the interior of $\triangle ABC.$ The side length of the smaller equilateral triangle can be written as $\sqrt{a} - \sqrt{b},$ where $a$ and $b$ are positive integers. Find $a+b.$
| 378
|
I
|
[
"We can extend $AB$ and $AC$ to $B'$ and $C'$ respectively such that circle $\\omega_A$ is the incircle of $\\triangle AB'C'$. [asy] /* Made by MRENTHUSIASM */ size(300); pair A, B, C, B1, C1, W, WA, WB, WC, X, Y, Z; A = 18*dir(90); B = 18*dir(210); C = 18*dir(330); B1 = A+24*sqrt(3)*dir(B-A); C1 = A+24*sqrt(3)*dir(C-A); W = (0,0); WA = 6*dir(270); WB = 6*dir(30); WC = 6*dir(150); X = (sqrt(117)-3)*dir(270); Y = (sqrt(117)-3)*dir(30); Z = (sqrt(117)-3)*dir(150); filldraw(X--Y--Z--cycle,green,dashed); draw(Circle(WA,12)^^Circle(WB,12)^^Circle(WC,12),blue); draw(Circle(W,18)^^A--B--C--cycle); draw(B--B1--C1--C,dashed); dot(\"$A$\",A,1.5*dir(A),linewidth(4)); dot(\"$B$\",B,1.5*(-1,0),linewidth(4)); dot(\"$C$\",C,1.5*(1,0),linewidth(4)); dot(\"$B'$\",B1,1.5*dir(B1),linewidth(4)); dot(\"$C'$\",C1,1.5*dir(C1),linewidth(4)); dot(\"$O$\",W,1.5*dir(90),linewidth(4)); dot(\"$X$\",X,1.5*dir(X),linewidth(4)); dot(\"$Y$\",Y,1.5*dir(Y),linewidth(4)); dot(\"$Z$\",Z,1.5*dir(Z),linewidth(4)); [/asy] Since the diameter of the circle is the height of this triangle, the height of this triangle is $36$. We can use inradius or equilateral triangle properties to get the inradius of this triangle is $12$ (The incenter is also a centroid in an equilateral triangle, and the distance from a side to the centroid is a third of the height). Therefore, the radius of each of the smaller circles is $12$. Let $O=\\omega$ be the center of the largest circle. We will set up a coordinate system with $O$ as the origin. The center of $\\omega_A$ will be at $(0,-6)$ because it is directly beneath $O$ and is the length of the larger radius minus the smaller radius, or $18-12 = 6$. By rotating this point $120^{\\circ}$ around $O$, we get the center of $\\omega_B$. This means that the magnitude of vector $\\overrightarrow{O\\omega_B}$ is $6$ and is at a $30$ degree angle from the horizontal. Therefore, the coordinates of this point are $(3\\sqrt{3},3)$ and by symmetry the coordinates of the center of $\\omega_C$ is $(-3\\sqrt{3},3)$. The upper left and right circles intersect at two points, the lower of which is $X$. The equations of these two circles are: \\begin{align*} (x+3\\sqrt3)^2 + (y-3)^2 &= 12^2, \\\\ (x-3\\sqrt3)^2 + (y-3)^2 &= 12^2. \\end{align*} We solve this system by subtracting to get $x = 0$. Plugging back in to the first equation, we have $(3\\sqrt{3})^2 + (y-3)^2 = 144 \\implies (y-3)^2 = 117 \\implies y-3 = \\pm \\sqrt{117} \\implies y = 3 \\pm \\sqrt{117}$. Since we know $X$ is the lower solution, we take the negative value to get $X = (0,3-\\sqrt{117})$. We can solve the problem two ways from here. We can find $Y$ by rotation and use the distance formula to find the length, or we can be somewhat more clever. We notice that it is easier to find $OX$ as they lie on the same vertical, $\\angle XOY$ is $120$ degrees so we can make use of $30-60-90$ triangles, and $OX = OY$ because $O$ is the center of triangle $XYZ$. We can draw the diagram as such: [asy] /* Made by MRENTHUSIASM */ size(300); pair A, B, C, B1, C1, W, WA, WB, WC, X, Y, Z; A = 18*dir(90); B = 18*dir(210); C = 18*dir(330); B1 = A+24*sqrt(3)*dir(B-A); C1 = A+24*sqrt(3)*dir(C-A); W = (0,0); WA = 6*dir(270); WB = 6*dir(30); WC = 6*dir(150); X = (sqrt(117)-3)*dir(270); Y = (sqrt(117)-3)*dir(30); Z = (sqrt(117)-3)*dir(150); filldraw(X--Y--Z--cycle,green,dashed); draw(Circle(WA,12)^^Circle(WB,12)^^Circle(WC,12),blue); draw(Circle(W,18)^^A--B--C--cycle); draw(B--B1--C1--C^^W--X^^W--Y^^W--midpoint(X--Y),dashed); dot(\"$A$\",A,1.5*dir(A),linewidth(4)); dot(\"$B$\",B,1.5*(-1,0),linewidth(4)); dot(\"$C$\",C,1.5*(1,0),linewidth(4)); dot(\"$B'$\",B1,1.5*dir(B1),linewidth(4)); dot(\"$C'$\",C1,1.5*dir(C1),linewidth(4)); dot(\"$O$\",W,1.5*dir(90),linewidth(4)); dot(\"$X$\",X,1.5*dir(X),linewidth(4)); dot(\"$Y$\",Y,1.5*dir(Y),linewidth(4)); dot(\"$Z$\",Z,1.5*dir(Z),linewidth(4)); [/asy] Note that $OX = OY = \\sqrt{117} - 3$. It follows that \\begin{align*} XY &= 2 \\cdot \\frac{OX\\cdot\\sqrt{3}}{2} \\\\ &= OX \\cdot \\sqrt{3} \\\\ &= (\\sqrt{117}-3) \\cdot \\sqrt{3} \\\\ &= \\sqrt{351}-\\sqrt{27}. \\end{align*} Finally, the answer is $351+27 = 378. ~KingRavi",
"We can extend $AB$ and $AC$ to $B'$ and $C'$ respectively such that circle $\\omega_A$ is the incircle of $\\triangle AB'C'$. [asy] /* Made by MRENTHUSIASM */ size(300); pair A, B, C, B1, C1, W, WA, WB, WC, X, Y, Z; A = 18*dir(90); B = 18*dir(210); C = 18*dir(330); B1 = A+24*sqrt(3)*dir(B-A); C1 = A+24*sqrt(3)*dir(C-A); W = (0,0); WA = 6*dir(270); WB = 6*dir(30); WC = 6*dir(150); X = (sqrt(117)-3)*dir(270); Y = (sqrt(117)-3)*dir(30); Z = (sqrt(117)-3)*dir(150); filldraw(X--Y--Z--cycle,green,dashed); draw(Circle(WA,12)^^Circle(WB,12)^^Circle(WC,12),blue); draw(Circle(W,18)^^A--B--C--cycle); draw(B--B1--C1--C,dashed); dot(\"$A$\",A,1.5*dir(A),linewidth(4)); dot(\"$B$\",B,1.5*(-1,0),linewidth(4)); dot(\"$C$\",C,1.5*(1,0),linewidth(4)); dot(\"$B'$\",B1,1.5*dir(B1),linewidth(4)); dot(\"$C'$\",C1,1.5*dir(C1),linewidth(4)); dot(\"$O$\",W,1.5*dir(90),linewidth(4)); dot(\"$X$\",X,1.5*dir(X),linewidth(4)); dot(\"$Y$\",Y,1.5*dir(Y),linewidth(4)); dot(\"$Z$\",Z,1.5*dir(Z),linewidth(4)); [/asy] Since the diameter of the circle is the height of this triangle, the height of this triangle is $36$. We can use inradius or equilateral triangle properties to get the inradius of this triangle is $12$ (The incenter is also a centroid in an equilateral triangle, and the distance from a side to the centroid is a third of the height). Therefore, the radius of each of the smaller circles is $12$. Let $O=\\omega$ be the center of the largest circle. We will set up a coordinate system with $O$ as the origin. The center of $\\omega_A$ will be at $(0,-6)$ because it is directly beneath $O$ and is the length of the larger radius minus the smaller radius, or $18-12 = 6$. By rotating this point $120^{\\circ}$ around $O$, we get the center of $\\omega_B$. This means that the magnitude of vector $\\overrightarrow{O\\omega_B}$ is $6$ and is at a $30$ degree angle from the horizontal. Therefore, the coordinates of this point are $(3\\sqrt{3},3)$ and by symmetry the coordinates of the center of $\\omega_C$ is $(-3\\sqrt{3},3)$. The upper left and right circles intersect at two points, the lower of which is $X$. The equations of these two circles are: \\begin{align*} (x+3\\sqrt3)^2 + (y-3)^2 &= 12^2, \\\\ (x-3\\sqrt3)^2 + (y-3)^2 &= 12^2. \\end{align*} We solve this system by subtracting to get $x = 0$. Plugging back in to the first equation, we have $(3\\sqrt{3})^2 + (y-3)^2 = 144 \\implies (y-3)^2 = 117 \\implies y-3 = \\pm \\sqrt{117} \\implies y = 3 \\pm \\sqrt{117}$. Since we know $X$ is the lower solution, we take the negative value to get $X = (0,3-\\sqrt{117})$. We can solve the problem two ways from here. We can find $Y$ by rotation and use the distance formula to find the length, or we can be somewhat more clever. We notice that it is easier to find $OX$ as they lie on the same vertical, $\\angle XOY$ is $120$ degrees so we can make use of $30-60-90$ triangles, and $OX = OY$ because $O$ is the center of triangle $XYZ$. We can draw the diagram as such: [asy] /* Made by MRENTHUSIASM */ size(300); pair A, B, C, B1, C1, W, WA, WB, WC, X, Y, Z; A = 18*dir(90); B = 18*dir(210); C = 18*dir(330); B1 = A+24*sqrt(3)*dir(B-A); C1 = A+24*sqrt(3)*dir(C-A); W = (0,0); WA = 6*dir(270); WB = 6*dir(30); WC = 6*dir(150); X = (sqrt(117)-3)*dir(270); Y = (sqrt(117)-3)*dir(30); Z = (sqrt(117)-3)*dir(150); filldraw(X--Y--Z--cycle,green,dashed); draw(Circle(WA,12)^^Circle(WB,12)^^Circle(WC,12),blue); draw(Circle(W,18)^^A--B--C--cycle); draw(B--B1--C1--C^^W--X^^W--Y^^W--midpoint(X--Y),dashed); dot(\"$A$\",A,1.5*dir(A),linewidth(4)); dot(\"$B$\",B,1.5*(-1,0),linewidth(4)); dot(\"$C$\",C,1.5*(1,0),linewidth(4)); dot(\"$B'$\",B1,1.5*dir(B1),linewidth(4)); dot(\"$C'$\",C1,1.5*dir(C1),linewidth(4)); dot(\"$O$\",W,1.5*dir(90),linewidth(4)); dot(\"$X$\",X,1.5*dir(X),linewidth(4)); dot(\"$Y$\",Y,1.5*dir(Y),linewidth(4)); dot(\"$Z$\",Z,1.5*dir(Z),linewidth(4)); [/asy] Note that $OX = OY = \\sqrt{117} - 3$. It follows that \\begin{align*} XY &= 2 \\cdot \\frac{OX\\cdot\\sqrt{3}}{2} \\\\ &= OX \\cdot \\sqrt{3} \\\\ &= (\\sqrt{117}-3) \\cdot \\sqrt{3} \\\\ &= \\sqrt{351}-\\sqrt{27}. \\end{align*} Finally, the answer is $351+27 = 378. ~KingRavi",
"[asy] /* Made by MRENTHUSIASM */ /* Modified by isabelchen */ size(250); pair A, B, C, W, WA, WB, WC, X, Y, Z, D, E; A = 18*dir(90); B = 18*dir(210); C = 18*dir(330); W = (0,0); WA = 6*dir(270); WB = 6*dir(30); WC = 6*dir(150); X = (sqrt(117)-3)*dir(270); Y = (sqrt(117)-3)*dir(30); Z = (sqrt(117)-3)*dir(150); D = intersectionpoint(Circle(WA,12),A--C); E = intersectionpoints(Circle(WB,12),Circle(WC,12))[0]; filldraw(X--Y--Z--cycle,green,dashed); draw(Circle(WA,12)^^Circle(WB,12)^^Circle(WC,12),blue); draw(Circle(W,18)^^A--B--C--cycle); dot(\"$A$\",A,1.5*dir(A),linewidth(4)); dot(\"$B$\",B,1.5*dir(B),linewidth(4)); dot(\"$C$\",C,1.5*dir(C),linewidth(4)); dot(\"$\\omega$\",W,1.5*dir(270),linewidth(4)); dot(\"$\\omega_A$\",WA,1.5*dir(-WA),linewidth(4)); dot(\"$\\omega_B$\",WB,1.5*dir(-WB),linewidth(4)); dot(\"$\\omega_C$\",WC,1.5*dir(-WC),linewidth(4)); dot(\"$X$\",X,1.5*dir(X),linewidth(4)); dot(\"$Y$\",Y,1.5*dir(Y),linewidth(4)); dot(\"$Z$\",Z,1.5*dir(Z),linewidth(4)); dot(\"$E$\",E,1.5*dir(E),linewidth(4)); dot(\"$D$\",D,1.5*dir(D),linewidth(4)); draw(WC--WB^^WC--X^^WC--E^^WA--D^^A--X); [/asy] For equilateral triangle with side length $l$, height $h$, and circumradius $r$, there are relationships: $h = \\frac{\\sqrt{3}}{2} l$, $r = \\frac{2}{3} h = \\frac{\\sqrt{3}}{3} l$, and $l = \\sqrt{3}r$. There is a lot of symmetry in the figure. The radius of the big circle $\\odot \\omega$ is $R = 18$, let the radius of the small circles $\\odot \\omega_A$, $\\odot \\omega_B$, $\\odot \\omega_C$ be $r$. We are going to solve this problem in $3$ steps: $\\textbf{Step 1:}$ We have $\\triangle A \\omega_A D$ is a $30-60-90$ triangle, and $A \\omega_A = 2 \\cdot \\omega_A D$, $A \\omega_A = 2R-r$ ($\\odot \\omega$ and $\\odot \\omega_A$ are tangent), and $\\omega_A D = r$. So, we get $2R-r = 2r$ and $r = \\frac{2}{3} \\cdot R = 12$. Since $\\odot \\omega$ and $\\odot \\omega_A$ are tangent, we get $\\omega \\omega_A = R - r = \\frac{1}{3} \\cdot R = 6$. Note that $\\triangle \\omega_A \\omega_B \\omega_C$ is an equilateral triangle, and $\\omega$ is its center, so $\\omega_B \\omega_C = \\sqrt{3} \\cdot \\omega \\omega_A = 6 \\sqrt{3}$. $\\textbf{Step 2:}$ Note that $\\triangle \\omega_C E X$ is an isosceles triangle, so \\[EX = 2 \\sqrt{(\\omega_C E)^2 - \\left(\\frac{\\omega_B \\omega_C}{2}\\right)^2} = 2 \\sqrt{r^2 - \\left(\\frac{\\omega_B \\omega_C}{2}\\right)^2} = 2 \\sqrt{12^2 - (3 \\sqrt{3})^2} = 2 \\sqrt{117}.\\] $\\textbf{Step 3:}$ In $\\odot \\omega_C$, Power of a Point gives $\\omega X \\cdot \\omega E = r^2 - (\\omega_C \\omega)^2$ and $\\omega E = EX - \\omega X = 2\\sqrt{117} - \\omega X$. It follows that $\\omega X \\cdot (2\\sqrt{117} - \\omega X) = 12^2 - 6^2$. We solve this quadratic equation: $\\omega X = \\sqrt{117} - 3$. Since $\\omega X$ is the circumradius of equilateral $\\triangle XYZ$, we have $XY = \\sqrt{3} \\cdot \\omega X = \\sqrt{3} \\cdot (\\sqrt{117} - 3) = \\sqrt{351}-\\sqrt{27}$. Therefore, the answer is $351+27 = 378. ~isabelchen",
"[asy] /* Made by MRENTHUSIASM */ /* Modified by isabelchen */ size(250); pair A, B, C, W, WA, WB, WC, X, Y, Z, D, E; A = 18*dir(90); B = 18*dir(210); C = 18*dir(330); W = (0,0); WA = 6*dir(270); WB = 6*dir(30); WC = 6*dir(150); X = (sqrt(117)-3)*dir(270); Y = (sqrt(117)-3)*dir(30); Z = (sqrt(117)-3)*dir(150); D = intersectionpoint(Circle(WA,12),A--C); E = intersectionpoints(Circle(WB,12),Circle(WC,12))[0]; filldraw(X--Y--Z--cycle,green,dashed); draw(Circle(WA,12)^^Circle(WB,12)^^Circle(WC,12),blue); draw(Circle(W,18)^^A--B--C--cycle); dot(\"$A$\",A,1.5*dir(A),linewidth(4)); dot(\"$B$\",B,1.5*dir(B),linewidth(4)); dot(\"$C$\",C,1.5*dir(C),linewidth(4)); dot(\"$\\omega$\",W,1.5*dir(270),linewidth(4)); dot(\"$\\omega_A$\",WA,1.5*dir(-WA),linewidth(4)); dot(\"$\\omega_B$\",WB,1.5*dir(-WB),linewidth(4)); dot(\"$\\omega_C$\",WC,1.5*dir(-WC),linewidth(4)); dot(\"$X$\",X,1.5*dir(X),linewidth(4)); dot(\"$Y$\",Y,1.5*dir(Y),linewidth(4)); dot(\"$Z$\",Z,1.5*dir(Z),linewidth(4)); dot(\"$E$\",E,1.5*dir(E),linewidth(4)); dot(\"$D$\",D,1.5*dir(D),linewidth(4)); draw(WC--WB^^WC--X^^WC--E^^WA--D^^A--X); [/asy] For equilateral triangle with side length $l$, height $h$, and circumradius $r$, there are relationships: $h = \\frac{\\sqrt{3}}{2} l$, $r = \\frac{2}{3} h = \\frac{\\sqrt{3}}{3} l$, and $l = \\sqrt{3}r$. There is a lot of symmetry in the figure. The radius of the big circle $\\odot \\omega$ is $R = 18$, let the radius of the small circles $\\odot \\omega_A$, $\\odot \\omega_B$, $\\odot \\omega_C$ be $r$. We are going to solve this problem in $3$ steps: $\\textbf{Step 1:}$ We have $\\triangle A \\omega_A D$ is a $30-60-90$ triangle, and $A \\omega_A = 2 \\cdot \\omega_A D$, $A \\omega_A = 2R-r$ ($\\odot \\omega$ and $\\odot \\omega_A$ are tangent), and $\\omega_A D = r$. So, we get $2R-r = 2r$ and $r = \\frac{2}{3} \\cdot R = 12$. Since $\\odot \\omega$ and $\\odot \\omega_A$ are tangent, we get $\\omega \\omega_A = R - r = \\frac{1}{3} \\cdot R = 6$. Note that $\\triangle \\omega_A \\omega_B \\omega_C$ is an equilateral triangle, and $\\omega$ is its center, so $\\omega_B \\omega_C = \\sqrt{3} \\cdot \\omega \\omega_A = 6 \\sqrt{3}$. $\\textbf{Step 2:}$ Note that $\\triangle \\omega_C E X$ is an isosceles triangle, so \\[EX = 2 \\sqrt{(\\omega_C E)^2 - \\left(\\frac{\\omega_B \\omega_C}{2}\\right)^2} = 2 \\sqrt{r^2 - \\left(\\frac{\\omega_B \\omega_C}{2}\\right)^2} = 2 \\sqrt{12^2 - (3 \\sqrt{3})^2} = 2 \\sqrt{117}.\\] $\\textbf{Step 3:}$ In $\\odot \\omega_C$, Power of a Point gives $\\omega X \\cdot \\omega E = r^2 - (\\omega_C \\omega)^2$ and $\\omega E = EX - \\omega X = 2\\sqrt{117} - \\omega X$. It follows that $\\omega X \\cdot (2\\sqrt{117} - \\omega X) = 12^2 - 6^2$. We solve this quadratic equation: $\\omega X = \\sqrt{117} - 3$. Since $\\omega X$ is the circumradius of equilateral $\\triangle XYZ$, we have $XY = \\sqrt{3} \\cdot \\omega X = \\sqrt{3} \\cdot (\\sqrt{117} - 3) = \\sqrt{351}-\\sqrt{27}$. Therefore, the answer is $351+27 = 378. ~isabelchen",
"Let $O$ be the center, $R = 18$ be the radius, and $CC'$ be the diameter of $\\omega.$ Let $r$ be the radius, $E,D,F$ are the centers of $\\omega_A, \\omega_B,\\omega_C.$ Let $KGH$ be the desired triangle with side $x.$ We find $r$ using \\[CC' = 2R = C'K + KC = r + \\frac{r}{\\sin 30^\\circ} = 3r.\\] \\[r = \\frac{2R}{3} = 12.\\] \\[OE = R – r = 6.\\] Triangles $\\triangle DEF$ and $\\triangle KGH$ – are equilateral triangles with a common center $O,$ therefore in the triangle $OEH$ $OE = 6, \\angle EOH = 120^\\circ, OH = \\frac{x}{\\sqrt3}.$ We apply the Law of Cosines to $\\triangle OEH$ and get \\[OE^2 + OH^2 + OE \\cdot OH = EH^2.\\] \\[6^2 + \\frac{x^2}{3} + \\frac{6x}{\\sqrt3} = 12^2.\\] \\[x^2 + 6x \\sqrt{3} = 324\\] \\[x= \\sqrt{351} - \\sqrt{27} \\implies 351 + 27 = 378\\] vladimir.shelomovskii@gmail.com, vvsss",
"Let $O$ be the center, $R = 18$ be the radius, and $CC'$ be the diameter of $\\omega.$ Let $r$ be the radius, $E,D,F$ are the centers of $\\omega_A, \\omega_B,\\omega_C.$ Let $KGH$ be the desired triangle with side $x.$ We find $r$ using \\[CC' = 2R = C'K + KC = r + \\frac{r}{\\sin 30^\\circ} = 3r.\\] \\[r = \\frac{2R}{3} = 12.\\] \\[OE = R – r = 6.\\] Triangles $\\triangle DEF$ and $\\triangle KGH$ – are equilateral triangles with a common center $O,$ therefore in the triangle $OEH$ $OE = 6, \\angle EOH = 120^\\circ, OH = \\frac{x}{\\sqrt3}.$ We apply the Law of Cosines to $\\triangle OEH$ and get \\[OE^2 + OH^2 + OE \\cdot OH = EH^2.\\] \\[6^2 + \\frac{x^2}{3} + \\frac{6x}{\\sqrt3} = 12^2.\\] \\[x^2 + 6x \\sqrt{3} = 324\\] \\[x= \\sqrt{351} - \\sqrt{27} \\implies 351 + 27 = 378\\] vladimir.shelomovskii@gmail.com, vvsss",
"Let $O$ be the center of $\\omega$, $X$ be the intersection of $\\omega_B,\\omega_C$ further from $A$, and $O_A$ be the center of $\\omega_A$. Define $Y, Z, O_B, O_C$ similarly. It is well-known that the $A$-mixtilinear inradius $R_A$ is $\\tfrac{r}{\\cos^2\\left(\\frac{\\angle A}{2}\\right)} = \\tfrac{9}{\\cos^2\\left(30^{\\circ}\\right)} = 12$, so in particular this means that $OO_B = 18 - R_B = 6 = OO_C$. Since $\\angle O_BOO_C = \\angle BOC = 120^\\circ$, it follows by Law of Cosines on $\\triangle OO_BO_C$ that $O_BO_C = 6\\sqrt{3}$. Then the Pythagorean theorem gives that the altitude of $O_BO_CX$ is $\\sqrt{117}$, so $OY = OX = \\text{dist}(X, YZ) - \\text{dist}(O, YZ) = \\sqrt{117} - 3$ and $YZ = \\tfrac{O_BO_C\\cdot OY}{OO_B} = \\tfrac{6\\sqrt{3}(\\sqrt{117} - 3)}{6}=\\sqrt{351} - \\sqrt{27}$ so the answer is $351 + 27 = 378. ~Kagebaka",
"Let $O$ be the center of $\\omega$, $X$ be the intersection of $\\omega_B,\\omega_C$ further from $A$, and $O_A$ be the center of $\\omega_A$. Define $Y, Z, O_B, O_C$ similarly. It is well-known that the $A$-mixtilinear inradius $R_A$ is $\\tfrac{r}{\\cos^2\\left(\\frac{\\angle A}{2}\\right)} = \\tfrac{9}{\\cos^2\\left(30^{\\circ}\\right)} = 12$, so in particular this means that $OO_B = 18 - R_B = 6 = OO_C$. Since $\\angle O_BOO_C = \\angle BOC = 120^\\circ$, it follows by Law of Cosines on $\\triangle OO_BO_C$ that $O_BO_C = 6\\sqrt{3}$. Then the Pythagorean theorem gives that the altitude of $O_BO_CX$ is $\\sqrt{117}$, so $OY = OX = \\text{dist}(X, YZ) - \\text{dist}(O, YZ) = \\sqrt{117} - 3$ and $YZ = \\tfrac{O_BO_C\\cdot OY}{OO_B} = \\tfrac{6\\sqrt{3}(\\sqrt{117} - 3)}{6}=\\sqrt{351} - \\sqrt{27}$ so the answer is $351 + 27 = 378. ~Kagebaka"
] |
2022-I-9
| 2,022
| 9
|
Ellina has twelve blocks, two each of red ( $\textbf{R}$ ), blue ( $\textbf{B}$ ), yellow ( $\textbf{Y}$ ), green ( $\textbf{G}$ ), orange ( $\textbf{O}$ ), and purple ( $\textbf{P}$ ). Call an arrangement of blocks $\textit{even}$ if there is an even number of blocks between each pair of blocks of the same color. For example, the arrangement \[\textbf{R B B Y G G Y R O P P O}\] is even. Ellina arranges her blocks in a row in random order. The probability that her arrangement is even is $\frac{m}{n},$ where $m$ and $n$ are relatively prime positive integers. Find $m+n.$
| 247
|
I
|
[
"Consider this position chart: \\[\\textbf{1 2 3 4 5 6 7 8 9 10 11 12}\\] Since there has to be an even number of spaces between each pair of the same color, spots $1$, $3$, $5$, $7$, $9$, and $11$ contain some permutation of all $6$ colored balls. Likewise, so do the even spots, so the number of even configurations is $6! \\cdot 6!$ (after putting every pair of colored balls in opposite parity positions, the configuration can be shown to be even). This is out of $\\frac{12!}{(2!)^6}$ possible arrangements, so the probability is: \\[\\frac{6!\\cdot6!}{\\frac{12!}{(2!)^6}} = \\frac{6!\\cdot2^6}{7\\cdot8\\cdot9\\cdot10\\cdot11\\cdot12} = \\frac{2^4}{7\\cdot11\\cdot3} = \\frac{16}{231},\\] which is in simplest form. So, $m + n = 16 + 231 = 247. ~Oxymoronic15",
"We can simply use constructive counting. First, let us place the red blocks; choose the first slot in $12$ ways, and the second in $6$ ways, because the number is cut in half due to the condition in the problem. This gives $12 \\cdot 6$ ways to place the red blocks. Similarly, there are $10 \\cdot 5$ ways to place the blue blocks, and so on, until there are $2 \\cdot 1$ ways to place the purple blocks. Thus, the probability is \\[\\frac{12 \\cdot 6 \\cdot 10 \\cdot 5 \\cdot 8 \\cdot 4 \\cdot 6 \\cdot 3 \\cdot 4 \\cdot 2 \\cdot 2 \\cdot 1}{12!}=\\frac{16}{231},\\] and the desired answer is $16+231=247. ~A1001",
"Use constructive counting, as per above. WLOG, place the red blocks first. There are 11 ways to place them with distance 0, 9 ways them to place with distance 2, so on, so the way to place red blocks is $11+9+7+5+3+1=36$. Then place any other block similarly, with $25$ ways (basic counting). You get then $6!^2$ ways to place the blocks evenly, and $12!/64$ ways to place the blocks in any way, so you get $\\frac{16}{231}=247$ by simplifying. -drag00n",
"We can divide the $12$ positions into odd and even positions. Each color needs one block in an odd position and one block in an even position. WLOG, we place the first block of the first pair into an odd position. This leaves $6$ even positions out of the $11$ remaining positions for the second block of the first pair. This results in a probability of $\\frac{6}{11}$ for the second block to fall into an even position. We can now place the first block of the second pair into another odd position, leaving $5$ even positions out of the $9$ remaining positions for the second block of the second pair. Continuing this pattern for the other $4$ pairs results in the product $\\frac{6\\cdot5\\cdot4\\cdot3\\cdot2\\cdot1}{11\\cdot9\\cdot7\\cdot5\\cdot3\\cdot1}=\\frac{16}{231}$. Thus, our answer is $16+231=247. ~Zhixing"
] |
2022-I-10
| 2,022
| 10
|
Three spheres with radii $11,$ $13,$ and $19$ are mutually externally tangent. A plane intersects the spheres in three congruent circles centered at $A,$ $B,$ and $C,$ respectively, and the centers of the spheres all lie on the same side of this plane. Suppose that $AB^2 = 560.$ Find $AC^2.$
| 756
|
I
|
[
"This solution refers to the Diagram section. We let $\\ell$ be the plane that passes through the spheres and $O_A$ and $O_B$ be the centers of the spheres with radii $11$ and $13$. We take a cross-section that contains $A$ and $B$, which contains these two spheres but not the third, as shown below: [asy] size(400); pair A, B, OA, OB; B = (0,0); A = (-23.6643191,0); OB = (0,8); OA = (-23.6643191,4); draw(circle(OB,13)); draw(circle(OA,11)); draw((-48,0)--(24,0)); label(\"$\\ell$\",(-42,0),S); label(\"$A$\",A,S); label(\"$B$\",B,S); label(\"$O_A$\",OA,N); label(\"$O_B$\",OB,N); draw(A--OA); draw(B--OB); draw(OA--OB); draw(OA--(0,4)); draw(OA--(-33.9112699,0)); draw(OB--(10.2469508,0)); label(\"$24$\",midpoint(OA--OB),N); label(\"$\\sqrt{560}$\",midpoint(A--B),S); label(\"$11$\",midpoint(OA--(-33.9112699,0)),NW); label(\"$13$\",midpoint(OB--(10.2469508,0)),NE); label(\"$r$\",midpoint(midpoint(A--B)--A),S); label(\"$r$\",midpoint(midpoint(A--B)--B),S); label(\"$r$\",midpoint(A--(-33.9112699,0)),S); label(\"$r$\",midpoint(B--(10.2469508,0)),S); label(\"$x$\",midpoint(midpoint(B--OB)--OB),W); label(\"$D$\",midpoint(B--OB),E); [/asy] Because the plane cuts out congruent circles, they have the same radius and from the given information, $AB = \\sqrt{560}$. Since $ABO_BO_A$ is a trapezoid, we can drop an altitude from $O_A$ to $BO_B$ to create a rectangle and triangle to use Pythagorean theorem. We know that the length of the altitude is $\\sqrt{560}$ and let the distance from $O_B$ to $D$ be $x$. Then we have $x^2 = 576-560 \\implies x = 4$. We have $AO_A = BD$ because of the rectangle, so $\\sqrt{11^2-r^2} = \\sqrt{13^2-r^2}-4$. Squaring, we have $121-r^2 = 169-r^2 + 16 - 8 \\cdot \\sqrt{169-r^2}$. Subtracting, we get $8 \\cdot \\sqrt{169-r^2} = 64 \\implies \\sqrt{169-r^2} = 8 \\implies 169-r^2 = 64 \\implies r^2 = 105$. We also notice that since we had $\\sqrt{169-r^2} = 8$ means that $BO_B = 8$ and since we know that $x = 4$, $AO_A = 4$. We take a cross-section that contains $A$ and $C$, which contains these two spheres but not the third, as shown below: [asy] size(400); pair A, C, OA, OC, M; C = (0,0); A = (-27.4954541697,0); OC = (0,16); OA = (-27.4954541697,4); M = midpoint(A--C); draw(circle(OC,19)); draw(circle(OA,11)); draw((-48,0)--(24,0)); label(\"$\\ell$\",(-42,0),S); label(\"$A$\",A,S); label(\"$C$\",C,S); label(\"$O_A$\",OA,N); label(\"$O_C$\",OC,N); draw(A--OA); draw(C--OC); draw(OA--OC); draw(OA--(0,4)); draw(OA--(-37.8877590151,0)); draw(OC--(10.2469508,0)); label(\"$30$\",midpoint(OA--OC),NW); label(\"$11$\",midpoint(OA--(-37.8877590151,0)),NW); label(\"$19$\",midpoint(OC--(10.2469508,0)),NE); label(\"$r$\",midpoint(midpoint(M--A)--A),S); label(\"$r$\",midpoint(midpoint(M--C)--C),S); label(\"$r$\",midpoint(A--(-37.8877590151,0)),S); label(\"$r$\",midpoint(C--(10.2469508,0)),S); label(\"$E$\",(0,4),E); [/asy] We have $CO_C = \\sqrt{19^2-r^2} = \\sqrt{361 - 105} = \\sqrt{256} = 16$. Since $AO_A = 4$, we have $EO_C = 16-4 = 12$. Using Pythagorean theorem, $O_AE = \\sqrt{30^2 - 12^2} = \\sqrt{900-144} = \\sqrt{756}$. Therefore, $O_AE^2 = AC^2 = 756. ~KingRavi",
"Let the distance between the center of the sphere to the center of those circular intersections as $a,b,c$ separately. According to the problem, we have $a^2-11^2=b^2-13^2=c^2-19^2; (11+13)^2-(b-a)^2=560.$ After solving we have $b-a=4,$ plug this back to $11^2-a^2=13^2-b^2,$ we have $a=4, b=8,$ and $c=16.$ The desired value is $(11+19)^2-(16-4)^2=756 ~bluesoul",
"Denote by $r$ the radius of three congruent circles formed by the cutting plane. Denote by $O_A$, $O_B$, $O_C$ the centers of three spheres that intersect the plane to get circles centered at $A$, $B$, $C$, respectively. Because three spheres are mutually tangent, $O_A O_B = 11 + 13 = 24$, $O_A O_C = 11 + 19 = 30$. We have $O_A A^2 = 11^2 - r^2$, $O_B B^2 = 13^2 - r^2$, $O_C C^2 = 19^2 - r^2$. Because $O_A A$ and $O_B B$ are perpendicular to the plane, $O_A AB O_B$ is a right trapezoid, with $\\angle O_A A B = \\angle O_B BA = 90^\\circ$. Hence, \\begin{align*} O_B B - O_A A & = \\sqrt{O_A O_B^2 - AB^2} \\\\ & = 4 . \\hspace{1cm} (1) \\end{align*} Recall that \\begin{align*} O_B B^2 - O_A A^2 & = \\left( 13^2 - r^2 \\right) - \\left( 11^2 - r^2 \\right) \\\\ & = 48 . \\hspace{1cm} (2) \\end{align*} Hence, taking $\\frac{(2)}{(1)}$, we get \\[ O_B B + O_A A = 12 . \\hspace{1cm} (3) \\] Solving (1) and (3), we get $O_B B = 8$ and $O_A A = 4$. Thus, $r^2 = 11^2 - O_A A^2 = 105$. Thus, $O_C C = \\sqrt{19^2 - r^2} = 16$. Because $O_A A$ and $O_C C$ are perpendicular to the plane, $O_A AC O_C$ is a right trapezoid, with $\\angle O_A A C = \\angle O_C CA = 90^\\circ$. Therefore, \\begin{align*} AC^2 & = O_A O_C^2 - \\left( O_C C - O_A A \\right)^2 \\\\ & = 756. ~Steven Chen (www.professorcheneeu.com) ~anonymous (minor edits)"
] |
2022-I-11
| 2,022
| 11
|
Let $ABCD$ be a parallelogram with $\angle BAD < 90^\circ.$ A circle tangent to sides $\overline{DA},$ $\overline{AB},$ and $\overline{BC}$ intersects diagonal $\overline{AC}$ at points $P$ and $Q$ with $AP < AQ,$ as shown. Suppose that $AP=3,$ $PQ=9,$ and $QC=16.$ Then the area of $ABCD$ can be expressed in the form $m\sqrt{n},$ where $m$ and $n$ are positive integers, and $n$ is not divisible by the square of any prime. Find $m+n.$ [asy] defaultpen(linewidth(0.6)+fontsize(11)); size(8cm); pair A,B,C,D,P,Q; A=(0,0); label("$A$", A, SW); B=(6,15); label("$B$", B, NW); C=(30,15); label("$C$", C, NE); D=(24,0); label("$D$", D, SE); P=(5.2,2.6); label("$P$", (5.8,2.6), N); Q=(18.3,9.1); label("$Q$", (18.1,9.7), W); draw(A--B--C--D--cycle); draw(C--A); draw(Circle((10.95,7.45), 7.45)); dot(A^^B^^C^^D^^P^^Q); [/asy]
| 150
|
I
|
[
"Let's redraw the diagram, but extend some helpful lines. [asy] size(10cm); pair A,B,C,D,EE,F,P,Q,O; A=(0,0); EE = (24,15); F = (30,0); O = (10.5,7.5); label(\"$A$\", A, SW); B=(6,15); label(\"$B$\", B, NW); C=(30,15); label(\"$C$\", C, NE); D=(24,0); label(\"$D$\", D, SE); P=(5.2,2.6); label(\"$P$\", (5.8,2.6), N); Q=(18.3,9.1); label(\"$Q$\", (18.1,9.7), W); draw(A--B--C--D--cycle); draw(C--A); draw(Circle((10.95,7.45), 7.45)); dot(A^^B^^C^^D^^P^^Q); dot(O); label(\"$O$\",O,W); draw((10.5,15)--(10.5,0)); draw(D--(24,15),dashed); draw(C--(30,0),dashed); draw(D--(30,0)); dot(EE); dot(F); label(\"$3$\", midpoint(A--P), S); label(\"$9$\", midpoint(P--Q), S); label(\"$16$\", midpoint(Q--C), S); label(\"$x$\", (5.5,13.75), W); label(\"$20$\", (20.25,15), N); label(\"$6$\", (5.25,0), S); label(\"$6$\", (1.5,3.75), W); label(\"$x$\", (8.25,15),N); label(\"$14+x$\", (17.25,0), S); label(\"$6-x$\", (27,15), N); label(\"$6+x$\", (27,7.5), S); label(\"$6\\sqrt{3}$\", (30,7.5), E); label(\"$T_1$\", (10.5,15), N); label(\"$T_2$\", (10.5,0), S); label(\"$T_3$\", (4.5,11.25), W); label(\"$E$\", EE, N); label(\"$F$\", F, S); [/asy] We obviously see that we must use power of a point since they've given us lengths in a circle and there are intersection points. Let $T_1, T_2, T_3$ be our tangents from the circle to the parallelogram. By the secant power of a point, the power of $A = 3 \\cdot (3+9) = 36$. Then $AT_2 = AT_3 = \\sqrt{36} = 6$. Similarly, the power of $C = 16 \\cdot (16+9) = 400$ and $CT_1 = \\sqrt{400} = 20$. We let $BT_3 = BT_1 = x$ and label the diagram accordingly. Notice that because $BC = AD, 20+x = 6+DT_2 \\implies DT_2 = 14+x$. Let $O$ be the center of the circle. Since $OT_1$ and $OT_2$ intersect $BC$ and $AD$, respectively, at right angles, we have $T_2T_1CD$ is a right-angled trapezoid and more importantly, the diameter of the circle is the height of the triangle. Therefore, we can drop an altitude from $D$ to $BC$ and $C$ to $AD$, and both are equal to $2r$. Since $T_1E = T_2D$, $20 - CE = 14+x \\implies CE = 6-x$. Since $CE = DF, DF = 6-x$ and $AF = 6+14+x+6-x = 26$. We can now use Pythagorean theorem on $\\triangle ACF$; we have $26^2 + (2r)^2 = (3+9+16)^2 \\implies 4r^2 = 784-676 \\implies 4r^2 = 108 \\implies 2r = 6\\sqrt{3}$ and $r^2 = 27$. We know that $CD = 6+x$ because $ABCD$ is a parallelogram. Using Pythagorean theorem on $\\triangle CDF$, $(6+x)^2 = (6-x)^2 + 108 \\implies (6+x)^2-(6-x)^2 = 108 \\implies 12 \\cdot 2x = 108 \\implies 2x = 9 \\implies x = \\frac{9}{2}$. Therefore, base $BC = 20 + \\frac{9}{2} = \\frac{49}{2}$. Thus the area of the parallelogram is the base times the height, which is $\\frac{49}{2} \\cdot 6\\sqrt{3} = 147\\sqrt{3}$ and the answer is $150 ~KingRavi",
"Let's redraw the diagram, but extend some helpful lines. [asy] size(10cm); pair A,B,C,D,EE,F,P,Q,O; A=(0,0); EE = (24,15); F = (30,0); O = (10.5,7.5); label(\"$A$\", A, SW); B=(6,15); label(\"$B$\", B, NW); C=(30,15); label(\"$C$\", C, NE); D=(24,0); label(\"$D$\", D, SE); P=(5.2,2.6); label(\"$P$\", (5.8,2.6), N); Q=(18.3,9.1); label(\"$Q$\", (18.1,9.7), W); draw(A--B--C--D--cycle); draw(C--A); draw(Circle((10.95,7.45), 7.45)); dot(A^^B^^C^^D^^P^^Q); dot(O); label(\"$O$\",O,W); draw((10.5,15)--(10.5,0)); draw(D--(24,15),dashed); draw(C--(30,0),dashed); draw(D--(30,0)); dot(EE); dot(F); label(\"$3$\", midpoint(A--P), S); label(\"$9$\", midpoint(P--Q), S); label(\"$16$\", midpoint(Q--C), S); label(\"$x$\", (5.5,13.75), W); label(\"$20$\", (20.25,15), N); label(\"$6$\", (5.25,0), S); label(\"$6$\", (1.5,3.75), W); label(\"$x$\", (8.25,15),N); label(\"$14+x$\", (17.25,0), S); label(\"$6-x$\", (27,15), N); label(\"$6+x$\", (27,7.5), S); label(\"$6\\sqrt{3}$\", (30,7.5), E); label(\"$T_1$\", (10.5,15), N); label(\"$T_2$\", (10.5,0), S); label(\"$T_3$\", (4.5,11.25), W); label(\"$E$\", EE, N); label(\"$F$\", F, S); [/asy] We obviously see that we must use power of a point since they've given us lengths in a circle and there are intersection points. Let $T_1, T_2, T_3$ be our tangents from the circle to the parallelogram. By the secant power of a point, the power of $A = 3 \\cdot (3+9) = 36$. Then $AT_2 = AT_3 = \\sqrt{36} = 6$. Similarly, the power of $C = 16 \\cdot (16+9) = 400$ and $CT_1 = \\sqrt{400} = 20$. We let $BT_3 = BT_1 = x$ and label the diagram accordingly. Notice that because $BC = AD, 20+x = 6+DT_2 \\implies DT_2 = 14+x$. Let $O$ be the center of the circle. Since $OT_1$ and $OT_2$ intersect $BC$ and $AD$, respectively, at right angles, we have $T_2T_1CD$ is a right-angled trapezoid and more importantly, the diameter of the circle is the height of the triangle. Therefore, we can drop an altitude from $D$ to $BC$ and $C$ to $AD$, and both are equal to $2r$. Since $T_1E = T_2D$, $20 - CE = 14+x \\implies CE = 6-x$. Since $CE = DF, DF = 6-x$ and $AF = 6+14+x+6-x = 26$. We can now use Pythagorean theorem on $\\triangle ACF$; we have $26^2 + (2r)^2 = (3+9+16)^2 \\implies 4r^2 = 784-676 \\implies 4r^2 = 108 \\implies 2r = 6\\sqrt{3}$ and $r^2 = 27$. We know that $CD = 6+x$ because $ABCD$ is a parallelogram. Using Pythagorean theorem on $\\triangle CDF$, $(6+x)^2 = (6-x)^2 + 108 \\implies (6+x)^2-(6-x)^2 = 108 \\implies 12 \\cdot 2x = 108 \\implies 2x = 9 \\implies x = \\frac{9}{2}$. Therefore, base $BC = 20 + \\frac{9}{2} = \\frac{49}{2}$. Thus the area of the parallelogram is the base times the height, which is $\\frac{49}{2} \\cdot 6\\sqrt{3} = 147\\sqrt{3}$ and the answer is $150 ~KingRavi",
"Let the circle tangent to $BC,AD,AB$ at $P,Q,M$ separately, denote that $\\angle{ABC}=\\angle{D}=\\alpha$ Using POP, it is very clear that $PC=20,AQ=AM=6$, let $BM=BP=x,QD=14+x$, using LOC in $\\triangle{ABP}$,$x^2+(x+6)^2-2x(x+6)\\cos\\alpha=36+PQ^2$, similarly, use LOC in $\\triangle{DQC}$, getting that $(14+x)^2+(6+x)^2-2(6+x)(14+x)\\cos\\alpha=400+PQ^2$. We use the second equation to minus the first equation, getting that $28x+196-(2x+12)\\times14\\times\\cos\\alpha=364$, we can get $\\cos\\alpha=\\frac{2x-12}{2x+12}$. Now applying LOC in $\\triangle{ADC}$, getting $(6+x)^2+(20+x)^2-2(6+x)\\times(20+x)\\times\\frac{2x-12}{2x+12}=(3+9+16)^2$, solving this equation to get $x=\\frac{9}{2}$, then $\\cos\\alpha=-\\frac{1}{7}$, $\\sin\\alpha=\\frac{4\\sqrt{3}}{7}$, the area is $\\frac{21}{2}\\cdot\\frac{49}{2}\\cdot\\frac{4\\sqrt{3}}{7}=147\\sqrt{3}$ leads to $150 ~bluesoul,HarveyZhang",
"Denote by $O$ the center of the circle. Denote by $r$ the radius of the circle. Denote by $E$, $F$, $G$ the points that the circle meets $AB$, $CD$, $AD$ at, respectively. Because the circle is tangent to $AD$, $CB$, $AB$, $OE = OF = OG = r$, $OE \\perp AD$, $OF \\perp CB$, $OG \\perp AB$. Because $AD \\parallel CB$, $E$, $O$, $F$ are collinear. Following from the power of a point, $AG^2 = AE^2 = AP \\cdot AQ$. Hence, $AG = AE = 6$. Following from the power of a point, $CF^2 = CQ \\cdot CP$. Hence, $CF = 20$. Denote $BG = x$. Because $DG$ and $DF$ are tangents to the circle, $BF = x$. Because $AEFB$ is a right trapezoid, $AB^2 = EF^2 + \\left( AE - BF \\right)^2$. Hence, $\\left( 6 + x \\right)^2 = 4 r^2 + \\left( 6 - x \\right)^2$. This can be simplified as \\[ 6 x = r^2 . \\hspace{1cm} (1) \\] In $\\triangle ACB$, by applying the law of cosines, we have \\begin{align*} AC^2 & = AB^2 + CB^2 - 2 AB \\cdot CB \\cos B \\\\ & = AB^2 + CB^2 + 2 AB \\cdot CB \\cos A \\\\ & = AB^2 + CB^2 + 2 AB \\cdot CB \\cdot \\frac{AE - BF}{AB} \\\\ & = AB^2 + CB^2 + 2 CB \\left( AE - BF \\right) \\\\ & = \\left( 6 + x \\right)^2 + \\left( 20 + x \\right)^2 + 2 \\left( 20 + x \\right) \\left( 6 - x \\right) \\\\ & = 24 x + 676 . \\end{align*} Because $AC = AP + PQ + QC = 28$, we get $x = \\frac{9}{2}$. Plugging this into Equation (1), we get $r = 3 \\sqrt{3}$. Therefore, \\begin{align*} {\\rm Area} \\ ABCD & = CB \\cdot EF \\\\ & = \\left( 20 + x \\right) \\cdot 2r \\\\ & = 147 \\sqrt{3} . \\end{align*} Therefore, the answer is $147 + 3 = \\textbf{(150) }. ~Steven Chen (www.professorchenedu.com)",
"Let $\\omega$ be the circle, let $r$ be the radius of $\\omega$, and let the points at which $\\omega$ is tangent to $AB$, $BC$, and $AD$ be $X$, $Y$, and $Z$, respectively. Note that PoP on $A$ and $C$ with respect to $\\omega$ yields $AX=6$ and $CY=20$. We can compute the area of $ABC$ in two ways: 1. By the half-base-height formula, $[ABC]=r(20+BX)$. 2. We can drop altitudes from the center $O$ of $\\omega$ to $AB$, $BC$, and $AC$, which have lengths $r$, $r$, and $\\sqrt{r^2-\\frac{81}{4}}$. Thus, $[ABC]=[OAB]+[OBC]+[OAC]=r(BX+13)+14\\sqrt{r^2-\\frac{81}{4}}$. Equating the two expressions for $[ABC]$ and solving for $r$ yields $r=3\\sqrt{3}$. Let $BX=BY=a$. By the Parallelogram Law, $(a+6)^2+(a+20)^2=38^2$. Solving for $a$ yields $a=9/2$. Thus, $[ABCD]=2[ABC]=2r(20+a)=147\\sqrt{3}$, for a final answer of $150. ~ Leo.Euler",
"Let $\\omega$ be the circle, let $r$ be the radius of $\\omega$, and let the points at which $\\omega$ is tangent to $AB$, $BC$, and $AD$ be $H$, $K$, and $T$, respectively. PoP on $A$ and $C$ with respect to $\\omega$ yields \\[AT=6, CK=20.\\] Let $TG = AC, CG||AT.$ In $\\triangle KGT$ $KT \\perp BC,$ $KT = \\sqrt{GT^2 – (KC + AT)^2} = 6 \\sqrt{3}=2r.$ $\\angle AOB = 90^{\\circ}, OH \\perp AB, OH = r = \\frac{KT}{2},$ \\[OH^2 = AH \\cdot BH \\implies BH = \\frac {9}{2}.\\] Area is \\[(BK + KC) \\cdot KT = (BH + KC) \\cdot 2r = \\frac{49}{2} \\cdot 6\\sqrt{3} = 147 \\sqrt{3} \\implies 147+3 = \\textbf{150}.\\] vladimir.shelomovskii@gmail.com, vvsss",
"Let $O$ be the center of the circle. Let points $M, N$ and $L$ be the tangent points of lines $BC, AD$ and $AB$ respectively to the circle. By Power of a Point, $({MC})^2=16\\cdot{25} \\Longrightarrow MC=20$. Similarly, $({AL})^2=3\\cdot{12} \\Longrightarrow AL=6$. Notice that $AL=AN=6$ since quadrilateral $LONA$ is symmetrical. Let $AC$ intersect $MN$ at $I$. Then, $\\bigtriangleup{IMC}$ is similar to $\\bigtriangleup{AIN}$. Therefore, $\\frac{CI}{MC}=\\frac{AI}{AN}$. Let the length of $PI=l$, then $\\frac{25-l}{20}=\\frac{3+l}{6}$. Solving we get $l=\\frac{45}{13}$. Doing the Pythagorean theorem on triangles $IMC$ and $AIN$ for sides $MI$ and $IN$ respectively, we obtain the equation $\\sqrt{(\\frac{280}{13})^2-400} +\\sqrt{(\\frac{84}{13})^2-36}=MN=2r_1$ where $r_1$ denotes the radius of the circle. Solving, we get $MN=6\\sqrt{3}$. Additionally, quadrilateral $OLBM$ is symmetrical so $OL=OM$. Let $OL=OM=x$ and extend a perpendicular foot from $B$ to $AD$ and call it $R$. Then, $\\bigtriangleup{ABR}$ is right with $AR=6-x$, $AB=6+x$, and $RB=2r_1=MN=6\\sqrt{3}$. Taking the difference of squares, we get $108=24x \\Longrightarrow x=\\frac{9}{2}$. The area of $ABCD$ is $MN\\cdot{BC}=(20+x)\\cdot{MN} \\Longrightarrow \\frac{49}{2}\\cdot{6\\sqrt{3}}=147\\sqrt{3}$. Therefore, the answer is $147+3=150 ~Magnetoninja",
"Let $O$ be the center of the circle. Let points $M, N$ and $L$ be the tangent points of lines $BC, AD$ and $AB$ respectively to the circle. By Power of a Point, $({MC})^2=16\\cdot{25} \\Longrightarrow MC=20$. Similarly, $({AL})^2=3\\cdot{12} \\Longrightarrow AL=6$. Notice that $AL=AN=6$ since quadrilateral $LONA$ is symmetrical. Let $AC$ intersect $MN$ at $I$. Then, $\\bigtriangleup{IMC}$ is similar to $\\bigtriangleup{AIN}$. Therefore, $\\frac{CI}{MC}=\\frac{AI}{AN}$. Let the length of $PI=l$, then $\\frac{25-l}{20}=\\frac{3+l}{6}$. Solving we get $l=\\frac{45}{13}$. Doing the Pythagorean theorem on triangles $IMC$ and $AIN$ for sides $MI$ and $IN$ respectively, we obtain the equation $\\sqrt{(\\frac{280}{13})^2-400} +\\sqrt{(\\frac{84}{13})^2-36}=MN=2r_1$ where $r_1$ denotes the radius of the circle. Solving, we get $MN=6\\sqrt{3}$. Additionally, quadrilateral $OLBM$ is symmetrical so $OL=OM$. Let $OL=OM=x$ and extend a perpendicular foot from $B$ to $AD$ and call it $R$. Then, $\\bigtriangleup{ABR}$ is right with $AR=6-x$, $AB=6+x$, and $RB=2r_1=MN=6\\sqrt{3}$. Taking the difference of squares, we get $108=24x \\Longrightarrow x=\\frac{9}{2}$. The area of $ABCD$ is $MN\\cdot{BC}=(20+x)\\cdot{MN} \\Longrightarrow \\frac{49}{2}\\cdot{6\\sqrt{3}}=147\\sqrt{3}$. Therefore, the answer is $147+3=150 ~Magnetoninja",
"Say that $BC$ is tangent to the circle at $X$ and $AD$ tangent at $Y$. Also, $H$ is the intersection of $XY$ (diameter) and $AC$ (diagonal). Then by power of a point with given info on $A$ and $C$ we get that $AY=6$ and $CX=20$. Note that $HAY \\sim HCX$, and since $\\frac{AY}{CX}=\\frac{3}{10}$ we note that \\[\\frac{AH}{CH} = \\frac{AP+PH}{CQ+QH} = \\frac{3+PH}{16+QH} =\\frac{AY}{CX}=\\frac{3}{10}\\]. Since $PH+HQ=9$, we get that $PH=\\frac{45}{13}$ and $QH=\\frac{72}{13}$. This is the length information within the circle. The same triangle similarity also means that $\\frac{YH}{XH}=\\frac{3}{10}$, so if the radius of the circle is $r$ then we have $XH=\\frac{20}{13}r$ and $YH = \\frac{6}{13}r$. By power of a point on H, we can figure out $r$: \\[XH\\cdot YH = PH \\cdot QG\\] \\[\\frac{20}{13}r \\cdot \\frac{6}{13}r = \\frac{45}{13} \\cdot \\frac{72}{13}\\] and we get that $r = 3 \\sqrt 3$. Thus, we have that the height of the parallelogram is $2r=6 \\sqrt 3$ and we want to find $BC$. If $AB$ is tangent to the circle at $E$, then set $a = BX = BE$. Using pythagorean theorem, $AO^2+BO^2=AB^2$ and we can plug in diagram values: \\[(AY^2+OY^2)+(BX^2+OX)^2=AB^2\\] \\[(6^2+(3 \\sqrt 3)^2) + (a^2+(3 \\sqrt 3)^2)=(a+6)^2.\\] Solving, we get $a=\\frac{9}{2}$ Finally, we have $[ABCD]=XY \\cdot BC = 6 \\sqrt 3 \\cdot (20+\\frac{9}{2}) \\rightarrow 150 ~ Brocolimanx",
"Say that $BC$ is tangent to the circle at $X$ and $AD$ tangent at $Y$. Also, $H$ is the intersection of $XY$ (diameter) and $AC$ (diagonal). Then by power of a point with given info on $A$ and $C$ we get that $AY=6$ and $CX=20$. Note that $HAY \\sim HCX$, and since $\\frac{AY}{CX}=\\frac{3}{10}$ we note that \\[\\frac{AH}{CH} = \\frac{AP+PH}{CQ+QH} = \\frac{3+PH}{16+QH} =\\frac{AY}{CX}=\\frac{3}{10}\\]. Since $PH+HQ=9$, we get that $PH=\\frac{45}{13}$ and $QH=\\frac{72}{13}$. This is the length information within the circle. The same triangle similarity also means that $\\frac{YH}{XH}=\\frac{3}{10}$, so if the radius of the circle is $r$ then we have $XH=\\frac{20}{13}r$ and $YH = \\frac{6}{13}r$. By power of a point on H, we can figure out $r$: \\[XH\\cdot YH = PH \\cdot QG\\] \\[\\frac{20}{13}r \\cdot \\frac{6}{13}r = \\frac{45}{13} \\cdot \\frac{72}{13}\\] and we get that $r = 3 \\sqrt 3$. Thus, we have that the height of the parallelogram is $2r=6 \\sqrt 3$ and we want to find $BC$. If $AB$ is tangent to the circle at $E$, then set $a = BX = BE$. Using pythagorean theorem, $AO^2+BO^2=AB^2$ and we can plug in diagram values: \\[(AY^2+OY^2)+(BX^2+OX)^2=AB^2\\] \\[(6^2+(3 \\sqrt 3)^2) + (a^2+(3 \\sqrt 3)^2)=(a+6)^2.\\] Solving, we get $a=\\frac{9}{2}$ Finally, we have $[ABCD]=XY \\cdot BC = 6 \\sqrt 3 \\cdot (20+\\frac{9}{2}) \\rightarrow 150 ~ Brocolimanx",
"Let $E$, $F$, $G$ be the circle's point of tangency with sides $AD$, $AB$, and $BC$, respectively. Let $O$ be the center of the inscribed circle. By Power of a Point, $AE^2 = AP \\cdot AQ = 3(3+9) = 36$, so $AE = 6$. Similarly, $GC^2 = CQ \\cdot CP = 16(16+9) = 400$, so $GC = 20$. Construct $GE$, and let $I$ be the point of intersection of $GE$ and $AC$. $GE \\perp BC$ and $GE \\perp AD$. By AA, $\\triangle IGC \\sim \\triangle IEA$, and we have $\\frac{AI}{IC} = \\frac{AE}{GC} = \\frac{3}{10}$. We also know $AI + IC = AC = 28$, so $AI = \\frac{84}{13}$ and $IC = \\frac{280}{13}$. Using Pythagorean Theorem on $\\triangle IEA$ and $\\triangle CIG$, we find that $EI = \\frac{18\\sqrt{3}}{13}$ and $IG = \\frac{60\\sqrt{3}}{13}$. Thus, $GE = EI + IG = 6\\sqrt{3}$, and the radius of the circle is $3\\sqrt{3}$. Construct $EF$, $FG$. $\\angle AFO = \\angle AEO = 90^{\\circ}$, so $AEOF$ is cyclic. Similarly, $BFOG$ is cyclic. Now, we attempt to set up Ptolemy. Using Pythagorean Theorem on $\\triangle AEO$, we find that $AO = 3\\sqrt{7}$. By Ptolemy's Theorem, $(AE)(FO) + (AF)(EO) = (AO)(FE)$, from which we have $(6)(3\\sqrt{3}) + (6)(3\\sqrt{3}) = (3\\sqrt{7})(FE)$ and $FE = 12\\frac{\\sqrt{3}}{\\sqrt{7}}$. From Thales' Circle, $\\triangle FGE$ is a right triangle, and $EF^2 + FG^2 = GE^2$, so $FG = \\frac{18}{\\sqrt{7}}$. Set $BF = BG = s$. $BO = \\sqrt{s^2 + (3\\sqrt{3})^2} = \\sqrt{s^2+27}$, so by Ptolemy's Theorem on $BFOG$, we have \\[(BF)(GO) + (BG)(FO) = (FG)(BO)\\] \\[(3\\sqrt{3})(s) + (3\\sqrt{3})(s) = (\\frac{18}{\\sqrt{7}})(\\sqrt{s^2+27})\\] Solving yields $s = \\frac{9}{2}$. We know that $BC = BG + GC = 20 + \\frac{9}{2} = \\frac{49}{2}$, so the area of $ABCD = (\\frac{49}{2})(6\\sqrt{3}) = 147\\sqrt{3}$. The requested answer is $147 + 3 = 150. ~ adam_zheng",
"Let $E$, $F$, $G$ be the circle's point of tangency with sides $AD$, $AB$, and $BC$, respectively. Let $O$ be the center of the inscribed circle. By Power of a Point, $AE^2 = AP \\cdot AQ = 3(3+9) = 36$, so $AE = 6$. Similarly, $GC^2 = CQ \\cdot CP = 16(16+9) = 400$, so $GC = 20$. Construct $GE$, and let $I$ be the point of intersection of $GE$ and $AC$. $GE \\perp BC$ and $GE \\perp AD$. By AA, $\\triangle IGC \\sim \\triangle IEA$, and we have $\\frac{AI}{IC} = \\frac{AE}{GC} = \\frac{3}{10}$. We also know $AI + IC = AC = 28$, so $AI = \\frac{84}{13}$ and $IC = \\frac{280}{13}$. Using Pythagorean Theorem on $\\triangle IEA$ and $\\triangle CIG$, we find that $EI = \\frac{18\\sqrt{3}}{13}$ and $IG = \\frac{60\\sqrt{3}}{13}$. Thus, $GE = EI + IG = 6\\sqrt{3}$, and the radius of the circle is $3\\sqrt{3}$. Construct $EF$, $FG$. $\\angle AFO = \\angle AEO = 90^{\\circ}$, so $AEOF$ is cyclic. Similarly, $BFOG$ is cyclic. Now, we attempt to set up Ptolemy. Using Pythagorean Theorem on $\\triangle AEO$, we find that $AO = 3\\sqrt{7}$. By Ptolemy's Theorem, $(AE)(FO) + (AF)(EO) = (AO)(FE)$, from which we have $(6)(3\\sqrt{3}) + (6)(3\\sqrt{3}) = (3\\sqrt{7})(FE)$ and $FE = 12\\frac{\\sqrt{3}}{\\sqrt{7}}$. From Thales' Circle, $\\triangle FGE$ is a right triangle, and $EF^2 + FG^2 = GE^2$, so $FG = \\frac{18}{\\sqrt{7}}$. Set $BF = BG = s$. $BO = \\sqrt{s^2 + (3\\sqrt{3})^2} = \\sqrt{s^2+27}$, so by Ptolemy's Theorem on $BFOG$, we have \\[(BF)(GO) + (BG)(FO) = (FG)(BO)\\] \\[(3\\sqrt{3})(s) + (3\\sqrt{3})(s) = (\\frac{18}{\\sqrt{7}})(\\sqrt{s^2+27})\\] Solving yields $s = \\frac{9}{2}$. We know that $BC = BG + GC = 20 + \\frac{9}{2} = \\frac{49}{2}$, so the area of $ABCD = (\\frac{49}{2})(6\\sqrt{3}) = 147\\sqrt{3}$. The requested answer is $147 + 3 = 150. ~ adam_zheng",
"Let points $E$, $F$, $G$ be the points where lines $BC$, $AB$, and $AD$ are tangent to the circle respectively. Then extend line $BC$ to point $H$ such that $AH$ is perpendicular to $BC$. According to Power of a Point, $CQ\\cdot CP=CE^2\\rightarrow 16\\cdot 25=CE^2\\rightarrow CE=20$. Similarly, $AP\\cdot AQ=AF^2\\rightarrow 3\\cdot 12=AF^2\\rightarrow AF=6$. (Note that $AG=AF=6$ because they are intersecting tangents from the same circle. Furthermore, $HE=AG=6$ due to a simple upwards translation.) Therefore, $HC=HE+EC=6+20=26$, and we already know that $AC=28$, meaning that we can use the Pythagorean Theorem on $\\Delta AHC$ to obtain: $AH=6\\sqrt{3}$. Let $HB=k$. Then $BE=6-HB=6-k$. Since they are intersecting tangents from the same circle, $BF=BE=6-k$. Therefore, $AB=AF+BF=6+6-k=12-k$. We have all the side lengths of $\\Delta ABH$, so applying the Pythagorean Theorem: $(6\\sqrt{3})^2+k^2=(12-k)^2\\rightarrow k=\\frac{3}{2}$. Thus, $BC=BE+EC=6-k+20=26-k=\\frac{49}{2}$. $[ABCD]=AH\\cdot BC=6\\sqrt{3}\\cdot \\frac{49}{2}=147\\sqrt{3}$, so the answer is $150. ~sid2012"
] |
2022-I-12
| 2,022
| 12
|
For any finite set $X,$ let $|X|$ denote the number of elements in $X.$ Define \[S_n = \sum |A \cap B|,\] where the sum is taken over all ordered pairs $(A,B)$ such that $A$ and $B$ are subsets of $\{1,2,3,\ldots,n\}$ with $|A|=|B|.$ For example, $S_2 = 4$ because the sum is taken over the pairs of subsets \[(A,B) \in \left\{(\emptyset,\emptyset),(\{1\},\{1\}),(\{1\},\{2\}),(\{2\},\{1\}),(\{2\},\{2\}),(\{1,2\},\{1,2\})\right\},\] giving $S_2 = 0+1+0+0+1+2=4.$ Let $\frac{S_{2022}}{S_{2021}} = \frac{p}{q},$ where $p$ and $q$ are relatively prime positive integers. Find the remainder when $p+q$ is divided by $1000.$
| 245
|
I
|
[
"Let's try out for small values of $n$ to get a feel for the problem. When $n=1, S_n$ is obviously $1$. The problem states that for $n=2, S_n$ is $4$. Let's try it out for $n=3$. Let's perform casework on the number of elements in $A, B$. $\\textbf{Case 1:} |A| = |B| = 1$ In this case, the only possible equivalencies will be if they are the exact same element, which happens $3$ times. $\\textbf{Case 2:} |A| = |B| = 2$ In this case, if they share both elements, which happens $3$ times, we will get $2$ for each time, and if they share only one element, which also happens $6$ times, we will get $1$ for each time, for a total of $12$ for this case. $\\textbf{Case 3:} |A| = |B| = 3$ In this case, the only possible scenario is that they both are the set $\\{1,2,3\\}$, and we have $3$ for this case. In total, $S_3 = 18$. Now notice, the number of intersections by each element $1 \\ldots 3$, or in general, $1 \\ldots n$ is equal for each element because of symmetry - each element when $n=3$ adds $6$ to the answer. Notice that $6 = \\binom{4}{2}$ - let's prove that $S_n = n \\cdot \\binom{2n-2}{n-1}$ (note that you can assume this and answer the problem if you're running short on time in the real test). Let's analyze the element $k$ - to find a general solution, we must count the number of these subsets that $k$ appears in. For $k$ to be in both $A$ and $B$, we need both sets to contain $k$ and another subset of $1$ through $n$ not including $k$. ($A = \\{k\\} \\cup A'| A' \\subset \\{1,2,\\ldots,n\\} \\land A' \\not \\subset \\{k\\}$ and $B = \\{k\\} \\cup B'| B' \\subset \\{1,2,\\ldots,n\\} \\land B' \\not \\subset \\{k\\}$) For any $0\\leq l \\leq n-1$ that is the size of both $A'$ and $B'$, the number of ways to choose the subsets $A'$ and $B'$ is $\\binom{n-1}{l}$ for both subsets, so the total number of ways to choose the subsets are $\\binom{n-1}{l}^2$. Now we sum this over all possible $l$'s to find the total number of ways to form sets $A$ and $B$ that contain $k$. This is equal to $\\sum_{l=0}^{n-1} \\binom{n-1}{l}^2$. This is a simplification of Vandermonde's identity, which states that $\\sum_{k=0}^{r} \\binom{m}{k} \\cdot \\binom{n}{r-k} = \\binom{m+n}{r}$. Here, $m$, $n$ and $r$ are all $n-1$, so this sum is equal to $\\binom{2n-2}{n-1}$. Finally, since we are iterating over all $k$'s for $n$ values of $k$, we have $S_n = n \\cdot \\binom{2n-2}{n-1}$, proving our claim. We now plug in $S_n$ to the expression we want to find. This turns out to be $\\frac{2022 \\cdot \\binom{4042}{2021}}{2021 \\cdot \\binom{4040}{2020}}$. Expanding produces $\\frac{2022 \\cdot 4042!\\cdot 2020! \\cdot 2020!}{2021 \\cdot 4040! \\cdot 2021! \\cdot 2021!}$. After cancellation, we have \\[\\frac{2022 \\cdot 4042 \\cdot 4041}{2021 \\cdot 2021 \\cdot 2021} \\implies \\frac{4044\\cdot 4041}{2021 \\cdot 2021}\\] $4044$ and $4041$ don't have any common factors with $2021$, so we're done with the simplification. We want to find $4044 \\cdot 4041 + 2021^2 \\pmod{1000} \\equiv 44 \\cdot 41 + 21^2 \\pmod{1000} \\equiv 1804+441 \\pmod{1000} \\equiv 2245 \\pmod{1000} \\equiv 245 ~KingRavi ~Edited by MY-2",
"Let's try out for small values of $n$ to get a feel for the problem. When $n=1, S_n$ is obviously $1$. The problem states that for $n=2, S_n$ is $4$. Let's try it out for $n=3$. Let's perform casework on the number of elements in $A, B$. $\\textbf{Case 1:} |A| = |B| = 1$ In this case, the only possible equivalencies will be if they are the exact same element, which happens $3$ times. $\\textbf{Case 2:} |A| = |B| = 2$ In this case, if they share both elements, which happens $3$ times, we will get $2$ for each time, and if they share only one element, which also happens $6$ times, we will get $1$ for each time, for a total of $12$ for this case. $\\textbf{Case 3:} |A| = |B| = 3$ In this case, the only possible scenario is that they both are the set $\\{1,2,3\\}$, and we have $3$ for this case. In total, $S_3 = 18$. Now notice, the number of intersections by each element $1 \\ldots 3$, or in general, $1 \\ldots n$ is equal for each element because of symmetry - each element when $n=3$ adds $6$ to the answer. Notice that $6 = \\binom{4}{2}$ - let's prove that $S_n = n \\cdot \\binom{2n-2}{n-1}$ (note that you can assume this and answer the problem if you're running short on time in the real test). Let's analyze the element $k$ - to find a general solution, we must count the number of these subsets that $k$ appears in. For $k$ to be in both $A$ and $B$, we need both sets to contain $k$ and another subset of $1$ through $n$ not including $k$. ($A = \\{k\\} \\cup A'| A' \\subset \\{1,2,\\ldots,n\\} \\land A' \\not \\subset \\{k\\}$ and $B = \\{k\\} \\cup B'| B' \\subset \\{1,2,\\ldots,n\\} \\land B' \\not \\subset \\{k\\}$) For any $0\\leq l \\leq n-1$ that is the size of both $A'$ and $B'$, the number of ways to choose the subsets $A'$ and $B'$ is $\\binom{n-1}{l}$ for both subsets, so the total number of ways to choose the subsets are $\\binom{n-1}{l}^2$. Now we sum this over all possible $l$'s to find the total number of ways to form sets $A$ and $B$ that contain $k$. This is equal to $\\sum_{l=0}^{n-1} \\binom{n-1}{l}^2$. This is a simplification of Vandermonde's identity, which states that $\\sum_{k=0}^{r} \\binom{m}{k} \\cdot \\binom{n}{r-k} = \\binom{m+n}{r}$. Here, $m$, $n$ and $r$ are all $n-1$, so this sum is equal to $\\binom{2n-2}{n-1}$. Finally, since we are iterating over all $k$'s for $n$ values of $k$, we have $S_n = n \\cdot \\binom{2n-2}{n-1}$, proving our claim. We now plug in $S_n$ to the expression we want to find. This turns out to be $\\frac{2022 \\cdot \\binom{4042}{2021}}{2021 \\cdot \\binom{4040}{2020}}$. Expanding produces $\\frac{2022 \\cdot 4042!\\cdot 2020! \\cdot 2020!}{2021 \\cdot 4040! \\cdot 2021! \\cdot 2021!}$. After cancellation, we have \\[\\frac{2022 \\cdot 4042 \\cdot 4041}{2021 \\cdot 2021 \\cdot 2021} \\implies \\frac{4044\\cdot 4041}{2021 \\cdot 2021}\\] $4044$ and $4041$ don't have any common factors with $2021$, so we're done with the simplification. We want to find $4044 \\cdot 4041 + 2021^2 \\pmod{1000} \\equiv 44 \\cdot 41 + 21^2 \\pmod{1000} \\equiv 1804+441 \\pmod{1000} \\equiv 2245 \\pmod{1000} \\equiv 245 ~KingRavi ~Edited by MY-2",
"We take cases based on the number of values in each of the subsets in the pair. Suppose we have $k$ elements in each of the subsets in a pair (for a total of n elements in the set). The expected number of elements in any random pair will be $n \\cdot \\frac{k}{n} \\cdot \\frac{k}{n}$ by linearity of expectation because for each of the $n$ elements, there is a $\\frac{k}{n}$ probability that the element will be chosen. To find the sum over all such values, we multiply this quantity by $\\binom{n}{k}^2$. Summing, we get \\[\\sum_{k=1}^{n} \\frac{k^2}{n} \\binom{n}{k}^2\\] Notice that we can rewrite this as \\[\\sum_{k=1}^{n} \\frac{1}{n} \\left(\\frac{k \\cdot n!}{(k)!(n - k)!}\\right)^2 = \\sum_{k=1}^{n} \\frac{1}{n} n^2 \\left(\\frac{(n-1)!}{(k - 1)!(n - k)!}\\right)^2 = n \\sum_{k=1}^{n} \\binom{n - 1}{k - 1}^2 = n \\sum_{k=1}^{n} \\binom{n - 1}{k - 1}\\binom{n - 1}{n - k}\\] We can simplify this using Vandermonde's identity to get $n \\binom{2n - 2}{n - 1}$. Evaluating this for $2022$ and $2021$ gives \\[\\frac{2022\\binom{4042}{2021}}{2021\\binom{4040}{2020}} = \\frac{2022 \\cdot 4042 \\cdot 4041}{2021^3} = \\frac{2022 \\cdot 2 \\cdot 4041}{2021^2}\\] Evaluating the numerators and denominators mod $1000$ gives $804 + 441 = 1245 - pi_is_3.14",
"We take cases based on the number of values in each of the subsets in the pair. Suppose we have $k$ elements in each of the subsets in a pair (for a total of n elements in the set). The expected number of elements in any random pair will be $n \\cdot \\frac{k}{n} \\cdot \\frac{k}{n}$ by linearity of expectation because for each of the $n$ elements, there is a $\\frac{k}{n}$ probability that the element will be chosen. To find the sum over all such values, we multiply this quantity by $\\binom{n}{k}^2$. Summing, we get \\[\\sum_{k=1}^{n} \\frac{k^2}{n} \\binom{n}{k}^2\\] Notice that we can rewrite this as \\[\\sum_{k=1}^{n} \\frac{1}{n} \\left(\\frac{k \\cdot n!}{(k)!(n - k)!}\\right)^2 = \\sum_{k=1}^{n} \\frac{1}{n} n^2 \\left(\\frac{(n-1)!}{(k - 1)!(n - k)!}\\right)^2 = n \\sum_{k=1}^{n} \\binom{n - 1}{k - 1}^2 = n \\sum_{k=1}^{n} \\binom{n - 1}{k - 1}\\binom{n - 1}{n - k}\\] We can simplify this using Vandermonde's identity to get $n \\binom{2n - 2}{n - 1}$. Evaluating this for $2022$ and $2021$ gives \\[\\frac{2022\\binom{4042}{2021}}{2021\\binom{4040}{2020}} = \\frac{2022 \\cdot 4042 \\cdot 4041}{2021^3} = \\frac{2022 \\cdot 2 \\cdot 4041}{2021^2}\\] Evaluating the numerators and denominators mod $1000$ gives $804 + 441 = 1245 - pi_is_3.14",
"For each element $i$, denote $x_i = \\left( x_{i, A}, x_{i, B} \\right) \\in \\left\\{ 0 , 1 \\right\\}^2$, where $x_{i, A} = \\Bbb I \\left\\{ i \\in A \\right\\}$ (resp. $x_{i, B} = \\Bbb I \\left\\{ i \\in B \\right\\}$). Denote $\\Omega = \\left\\{ (x_1, \\cdots , x_n): \\sum_{i = 1}^n x_{i, A} = \\sum_{i = 1}^n x_{i, B} \\right\\}$. Denote $\\Omega_{-j} = \\left\\{ (x_1, \\cdots , x_{j-1} , x_{j+1} , \\cdots , x_n): \\sum_{i \\neq j} x_{i, A} = \\sum_{i \\neq j} x_{i, B} \\right\\}$. Hence, \\begin{align*} S_n & = \\sum_{(x_1, \\cdots , x_n) \\in \\Omega} \\sum_{i = 1}^n \\Bbb I \\left\\{ x_{i, A} = x_{i, B} = 1 \\right\\} \\\\ & = \\sum_{i = 1}^n \\sum_{(x_1, \\cdots , x_n) \\in \\Omega} \\Bbb I \\left\\{ x_{i, A} = x_{i, B} = 1 \\right\\} \\\\ & = \\sum_{i = 1}^n \\sum_{(x_1, \\cdots , x_{i-1} , x_{i+1} , \\cdots , x_n) \\in \\Omega_{-i}} 1 \\\\ & = \\sum_{i = 1}^n \\sum_{j=0}^{n-1} \\left( \\binom{n-1}{j} \\right)^2 \\\\ & = n \\sum_{j=0}^{n-1} \\left( \\binom{n-1}{j} \\right)^2 \\\\ & = n \\sum_{j=0}^{n-1} \\binom{n-1}{j} \\binom{n-1}{n-1-j} \\\\ & = n \\binom{2n-2}{n-1} . \\end{align*} Therefore, \\begin{align*} \\frac{S_{2022}}{S_{2021}} & = \\frac{2022 \\binom{4042}{2021}}{2021 \\binom{4040}{2020}} \\\\ & = \\frac{4044 \\cdot 4041}{2021^2} . \\end{align*} This is in the lowest term. Therefore, modulo 1000, \\begin{align*} p + q & \\equiv 4044 \\cdot 4041 + 2021^2 \\\\ & \\equiv 44 \\cdot 41 + 21^2 \\\\ & \\equiv \\textbf{(245) } . \\end{align*} ~Steven Chen (www.professorchenedu.com)",
"For each element $i$, denote $x_i = \\left( x_{i, A}, x_{i, B} \\right) \\in \\left\\{ 0 , 1 \\right\\}^2$, where $x_{i, A} = \\Bbb I \\left\\{ i \\in A \\right\\}$ (resp. $x_{i, B} = \\Bbb I \\left\\{ i \\in B \\right\\}$). Denote $\\Omega = \\left\\{ (x_1, \\cdots , x_n): \\sum_{i = 1}^n x_{i, A} = \\sum_{i = 1}^n x_{i, B} \\right\\}$. Denote $\\Omega_{-j} = \\left\\{ (x_1, \\cdots , x_{j-1} , x_{j+1} , \\cdots , x_n): \\sum_{i \\neq j} x_{i, A} = \\sum_{i \\neq j} x_{i, B} \\right\\}$. Hence, \\begin{align*} S_n & = \\sum_{(x_1, \\cdots , x_n) \\in \\Omega} \\sum_{i = 1}^n \\Bbb I \\left\\{ x_{i, A} = x_{i, B} = 1 \\right\\} \\\\ & = \\sum_{i = 1}^n \\sum_{(x_1, \\cdots , x_n) \\in \\Omega} \\Bbb I \\left\\{ x_{i, A} = x_{i, B} = 1 \\right\\} \\\\ & = \\sum_{i = 1}^n \\sum_{(x_1, \\cdots , x_{i-1} , x_{i+1} , \\cdots , x_n) \\in \\Omega_{-i}} 1 \\\\ & = \\sum_{i = 1}^n \\sum_{j=0}^{n-1} \\left( \\binom{n-1}{j} \\right)^2 \\\\ & = n \\sum_{j=0}^{n-1} \\left( \\binom{n-1}{j} \\right)^2 \\\\ & = n \\sum_{j=0}^{n-1} \\binom{n-1}{j} \\binom{n-1}{n-1-j} \\\\ & = n \\binom{2n-2}{n-1} . \\end{align*} Therefore, \\begin{align*} \\frac{S_{2022}}{S_{2021}} & = \\frac{2022 \\binom{4042}{2021}}{2021 \\binom{4040}{2020}} \\\\ & = \\frac{4044 \\cdot 4041}{2021^2} . \\end{align*} This is in the lowest term. Therefore, modulo 1000, \\begin{align*} p + q & \\equiv 4044 \\cdot 4041 + 2021^2 \\\\ & \\equiv 44 \\cdot 41 + 21^2 \\\\ & \\equiv \\textbf{(245) } . \\end{align*} ~Steven Chen (www.professorchenedu.com)",
"Let's ask what the contribution of an element $k\\in \\{1,2,\\cdots,n\\}$ is to the sum $S_n = \\sum | A \\cap B |.$ The answer is given by the number of $(A,B)$ such that $|A|=|B|$ and $k \\in A\\cap B$, which is given by $\\binom{2n-2}{n-1}$ by the following construction: Write down 1 to $n$ except $k$ in a row. Do the same in a second row. Then choose $n-1$ numbers out of these $2n-2$ numbers. $k$ and the numbers chosen in the first row make up $A$. $k$ and the numbers not chosen in the second row make up $B$. This is a one-to-one correspondence between $(A,B)$ and the ways to choose $n-1$ numbers from $2n-2$ numbers. The contribution from all elements is therefore \\[S_n = n\\binom{2n-2}{n-1}.\\] For the rest please see Solution 1 or 2. ~qyang"
] |
2022-I-13
| 2,022
| 13
|
Let $S$ be the set of all rational numbers that can be expressed as a repeating decimal in the form $0.\overline{abcd},$ where at least one of the digits $a,$ $b,$ $c,$ or $d$ is nonzero. Let $N$ be the number of distinct numerators obtained when numbers in $S$ are written as fractions in lowest terms. For example, both $4$ and $410$ are counted among the distinct numerators for numbers in $S$ because $0.\overline{3636} = \frac{4}{11}$ and $0.\overline{1230} = \frac{410}{3333}.$ Find the remainder when $N$ is divided by $1000.$
| 392
|
I
|
[
"$0.\\overline{abcd}=\\frac{abcd}{9999} = \\frac{x}{y}$, $9999=9\\times 11\\times 101$. Then we need to find the number of positive integers $x$ that (with one of more $y$ such that $y|9999$) can meet the requirement $1 \\leq {x}\\cdot\\frac{9999}{y} \\leq 9999$. Make cases by factors of $x$. (A venn diagram of cases would be nice here.) Case $A$: $3 \\nmid x$ and $11 \\nmid x$ and $101 \\nmid x$, aka $\\gcd (9999, x)=1$. Euler's totient function counts these: \\[\\varphi \\left(3^2 \\cdot 11 \\cdot 101 \\right) = ((3-1)\\cdot 3)(11-1)(101-1)= \\bf{6000}\\] values (but it's enough to note that it's a multiple of 1000 and thus does not contribute to the final answer) Note: You don't need to know this formula. The remaining cases essentially re-derive the same computation for other factors of $9999$. This case isn't actually different. The remaining cases have $3$ (or $9$), $11$, and/or $101$ as factors of $abcd$, which cancel out part of $9999$. Note: Take care about when to use $3$ vs $9$. Case $B$: $3|x$, but $11 \\nmid x$ and $101 \\nmid x$. Then $abcd=9x$ to leave 3 uncancelled, and $x=3p$, so $x \\leq \\frac{9999}{9} = 1111$, giving: $x \\in 3 \\cdot \\{1, \\dots \\left\\lfloor \\frac{1111}{3}\\right\\rfloor\\}$, $x \\notin (3\\cdot 11) \\cdot \\{1 \\dots \\left\\lfloor \\frac{1111}{3\\cdot 11}\\right\\rfloor\\}$, $x \\notin (3 \\cdot 101) \\cdot \\{1 \\dots \\left\\lfloor \\frac{1111}{3 \\cdot 101}\\right\\rfloor\\}$, for a subtotal of $\\left\\lfloor \\frac{1111}{3}\\right\\rfloor - (\\left\\lfloor\\frac{1111}{3 \\cdot 11}\\right\\rfloor + \\left\\lfloor\\frac{1111}{3 \\cdot 101}\\right\\rfloor ) = 370 - (33+3) = \\bf{334}$ values. Case $C$: $11|x$, but $3 \\nmid x$ and $101 \\nmid x$. Much like previous case, $abcd$ is $11x$, so $x \\leq \\frac{9999}{11} = 909$, giving $\\left\\lfloor \\frac{909}{11}\\right\\rfloor - \\left(\\left\\lfloor\\frac{909}{11 \\cdot 3}\\right\\rfloor + \\left\\lfloor\\frac{909}{11 \\cdot 101}\\right\\rfloor \\right) = 82 - (27 + 0) = \\bf{55}$ values. Case $D$: $3|x$ and $11|x$ (so $33|x$), but $101 \\nmid x$. Here, $abcd$ is $99x$, so $x \\leq \\frac{9999}{99} = 101$, giving $\\left\\lfloor \\frac{101}{33}\\right\\rfloor - \\left\\lfloor \\frac{101}{33 \\cdot 101}\\right\\rfloor = 3-0 = \\bf{3}$ values. Case $E$: $101|x$. Here, $abcd$ is $101x$, so $x \\leq \\frac{9999}{101} = 99$, giving $\\left\\lfloor \\frac{99}{101}\\right\\rfloor = \\bf{0}$ values, so we don't need to account for multiples of $3$ and $11$. To sum up, the answer is \\[6000+334+55+3+0\\equiv392,\" it is a bit vague. The best way to clarify this is by this exact example - what is really meant is we need to divide by 9 first to achieve 1111, which has no multiple of 3; thus, given that the fraction x/y is the simplest form, x can be a multiple of 3. Similar explanations can be said when the solution divides 9999 by 11, 101, and uses that divided result in the PIE calculation rather than 9999. mathboy282"
] |
2022-I-14
| 2,022
| 14
|
Given $\triangle ABC$ and a point $P$ on one of its sides, call line $\ell$ the $\textit{splitting line}$ of $\triangle ABC$ through $P$ if $\ell$ passes through $P$ and divides $\triangle ABC$ into two polygons of equal perimeter. Let $\triangle ABC$ be a triangle where $BC = 219$ and $AB$ and $AC$ are positive integers. Let $M$ and $N$ be the midpoints of $\overline{AB}$ and $\overline{AC},$ respectively, and suppose that the splitting lines of $\triangle ABC$ through $M$ and $N$ intersect at $30^\circ.$ Find the perimeter of $\triangle ABC.$
| 459
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I
|
[
"Denote $BC = a$, $CA = b$, $AB = c$. Let the splitting line of $\\triangle ABC$ through $M$ (resp. $N$) crosses $\\triangle ABC$ at another point $X$ (resp. $Y$). WLOG, we assume $c \\leq b$. $\\textbf{Case 1}$: $a \\leq c \\leq b$. We extend segment $AB$ to $D$, such that $BD = a$. We extend segment $AC$ to $E$, such that $CE = a$. In this case, $X$ is the midpoint of $AE$, and $Y$ is the midpoint of $AD$. Because $M$ and $X$ are the midpoints of $AB$ and $AE$, respectively, $MX \\parallel BE$. Because $N$ and $Y$ are the midpoints of $AC$ and $AD$, respectively, $NY \\parallel CD$. Because $CB = CE$, $\\angle CBE =\\angle CEB = \\frac{\\angle ACB}{2}$. Because $BC = BD$, $\\angle BCD = \\angle BDC = \\frac{\\angle ABC}{2}$. Let $BE$ and $CD$ intersect at $O$. Because $MX \\parallel BE$ and $NY \\parallel CD$, the angle formed between lines $MX$ and $NY$ is congruent to $\\angle BOD$. Hence, $\\angle BOD = 30^\\circ$ or $150^\\circ$. We have \\begin{align*} \\angle BOD & = \\angle CBE + \\angle BCD \\\\ & = \\frac{\\angle ACB}{2} + \\frac{\\angle ABC}{2} \\\\ & = 90^\\circ - \\frac{\\angle A}{2} . \\end{align*} Hence, we must have $\\angle BOD = 30^\\circ$, not $150^\\circ$. Hence, $\\angle A = 120^\\circ$. This implies $a > b$ and $a >c$. This contradicts the condition specified for this case. Therefore, this case is infeasible. $\\textbf{Case 2}$: $c \\leq a \\leq b$. We extend segment $CB$ to $D$, such that $BD = c$. We extend segment $AC$ to $E$, such that $CE = a$. In this case, $X$ is the midpoint of $AE$, and $Y$ is the midpoint of $CD$. Because $M$ and $X$ are the midpoints of $AB$ and $AE$, respectively, $MX \\parallel BE$. Because $N$ and $Y$ are the midpoints of $AC$ and $CD$, respectively, $NY \\parallel AD$. Because $CB = CE$, $\\angle CBE =\\angle CEB = \\frac{\\angle ACB}{2}$. Because $BA = BD$, $\\angle BAD = \\angle BDA = \\frac{\\angle ABC}{2}$. Let $O$ be a point of $AC$, such that $BO \\parallel AD$. Hence, $\\angle OBC = \\angle BDA = \\frac{B}{2}$. Because $MX \\parallel BE$ and $NY \\parallel AD$ and $AD \\parallel BO$, the angle formed between lines $MX$ and $NY$ is congruent to $\\angle OBE$. Hence, $\\angle OBE = 30^\\circ$ or $150^\\circ$. We have \\begin{align*} \\angle OBE & = \\angle OBC + \\angle CBE \\\\ & = \\frac{\\angle ABC}{2} + \\frac{\\angle ACB}{2} \\\\ & = 90^\\circ - \\frac{\\angle A}{2} . \\end{align*} Hence, we must have $\\angle OBE = 30^\\circ$, not $150^\\circ$. Hence, $\\angle A = 120^\\circ$. This implies $a > b$ and $a >c$. This contradicts the condition specified for this case. Therefore, this case is infeasible. $\\textbf{Case 3}$: $c \\leq b \\leq a$. We extend segment $CB$ to $D$, such that $BD = c$. We extend segment $BC$ to $E$, such that $CE = b$. In this case, $X$ is the midpoint of $BE$, and $Y$ is the midpoint of $CD$. Because $M$ and $X$ are the midpoints of $AB$ and $BE$, respectively, $MX \\parallel AE$. Because $N$ and $Y$ are the midpoints of $AC$ and $CD$, respectively, $NY \\parallel AD$. Because $CA = CE$, $\\angle CAE =\\angle CEB = \\frac{\\angle ACB}{2}$. Because $BA = BD$, $\\angle BAD = \\angle BDA = \\frac{\\angle ABC}{2}$. Because $MX \\parallel AE$ and $NY \\parallel AD$, the angle formed between lines $MX$ and $NY$ is congruent to $\\angle DAE$. Hence, $\\angle DAE = 30^\\circ$ or $150^\\circ$. We have \\begin{align*} \\angle DAE & = \\angle BAD + \\angle CAE + \\angle BAC \\\\ & = \\frac{\\angle ABC}{2} + \\frac{\\angle ACB}{2} + \\angle BAC \\\\ & = 90^\\circ + \\frac{\\angle BAC}{2} . \\end{align*} Hence, we must have $\\angle OBE = 150^\\circ$, not $30^\\circ$. Hence, $\\angle BAC = 120^\\circ$. In $\\triangle ABC$, by applying the law of cosines, we have \\begin{align*} a^2 & = b^2 + c^2 - 2bc \\cos \\angle BAC\\\\ & = b^2 + c^2 - 2bc \\cos 120^\\circ \\\\ & = b^2 + c^2 + bc . \\end{align*} Because $a = 219$, we have \\[ b^2 + c^2 + bc = 219^2 . \\] Now, we find integer solution(s) of this equation with $c \\leq b$. Multiplying this equation by 4, we get \\[ \\left( 2 c + b \\right)^2 + 3 b^2 = 438^2 . \\hspace{1cm} (1) \\] Denote $d = 2 c + b$. Because $c \\leq b$, $b < d \\leq 3 b$. Because $438^2 - 3 b^2 \\equiv 0 \\pmod{3}$, $d^2 \\equiv 0 \\pmod{3}$. Thus, $d \\equiv 0 \\pmod{3}$. This implies $d^2 \\equiv 0 \\pmod{9}$. We also have $438^2 \\equiv 0 \\pmod{9}$. Hence, $3 b^2 \\equiv 0 \\pmod{9}$. This implies $b \\equiv 0 \\pmod{3}$. Denote $b = 3 p$ and $d = 3 q$. Hence, $p < q \\leq 3 p$. Hence, Equation (1) can be written as \\[ q^2 + 3 p^2 = 146^2 . \\hspace{1cm} (2) \\] Now, we solve this equation. First, we find an upper bound of $q$. We have $q^2 + 3 p^2 \\geq q^2 + 3 \\left( \\frac{q}{3} \\right)^2 = \\frac{4 q^2}{3}$. Hence, $\\frac{4 q^2}{3} \\leq 146^2$. Hence, $q \\leq 73 \\sqrt{3} < 73 \\cdot 1.8 = 131.4$. Because $q$ is an integer, we must have $q \\leq 131$. Second, we find a lower bound of $q$. We have $q^2 + 3 p^2 < q^2 + 3 q^2 = 4 q^2$. Hence, $4 q^2 > 146^2$. Hence, $q > 73$. Because $q$ is an integer, we must have $q \\geq 74$. Now, we find the integer solutions of $p$ and $q$ that satisfy Equation (2) with $74 \\leq q \\leq 131$. First, modulo 9, \\begin{align*} q^2 & \\equiv 146^2 - 3 p^2 \\\\ & \\equiv 4 - 3 \\cdot ( 0 \\mbox{ or } 1 ) \\\\ & \\equiv 4 \\mbox{ or } 1 . \\end{align*} Hence $q \\equiv \\pm 1, \\pm 2 \\pmod{9}$. Second, modulo 5, \\begin{align*} q^2 & \\equiv 146^2 - 3 p^2 \\\\ & \\equiv 1 + 2 p^2 \\\\ & \\equiv 1 + 2 \\cdot ( 0 \\mbox{ or } 1 \\mbox{ or } -1 ) \\\\ & \\equiv 1 \\mbox{ or } 3 \\mbox{ or } - 1 . \\end{align*} Because $q^2 \\equiv 0 \\mbox{ or } 1 \\mbox{ or } - 1$, we must have $q^2 \\equiv 1 \\mbox{ or } - 1$. Hence, $5 \\nmid q$. Third, modulo 7, \\begin{align*} q^2 & \\equiv 146^2 - 3 p^2 \\\\ & \\equiv 1 - 3 \\cdot ( 0 \\mbox{ or } 1 \\mbox{ or } 5 \\mbox{ or } 2 ) \\\\ & \\equiv 1 \\mbox{ or } 2 \\mbox{ or } 3 \\mbox{ or } 5 . \\end{align*} Because $q^2 \\equiv 0 \\mbox{ or } 1 \\mbox{ or } 2 \\mbox{ or } 4 \\pmod{ 7 }$, we must have $q^2 \\equiv 1 \\mbox{ or } 2 \\pmod{7}$. Hence, $q \\equiv 1, 3, 4, 6 \\pmod{7}$. Given all conditions above, the possible $q$ are 74, 83, 88, 92, 97, 101, 106, 109, 116, 118, 127. By testing all these numbers, we find that the only solution is $q = 97$. This implies $p = 63$. Hence, $b = 3p = 189$ and $d = 3q = 291$. Hence, $c = \\frac{d - b}{2} = 51$. Therefore, the perimeter of $\\triangle ABC$ is $b + c + a = 189 + 51 + 219 = \\textbf{(459) }. ~Steven Chen (www.professorchenedu.com)",
"Denote $BC = a$, $CA = b$, $AB = c$. Let the splitting line of $\\triangle ABC$ through $M$ (resp. $N$) crosses $\\triangle ABC$ at another point $X$ (resp. $Y$). WLOG, we assume $c \\leq b$. $\\textbf{Case 1}$: $a \\leq c \\leq b$. We extend segment $AB$ to $D$, such that $BD = a$. We extend segment $AC$ to $E$, such that $CE = a$. In this case, $X$ is the midpoint of $AE$, and $Y$ is the midpoint of $AD$. Because $M$ and $X$ are the midpoints of $AB$ and $AE$, respectively, $MX \\parallel BE$. Because $N$ and $Y$ are the midpoints of $AC$ and $AD$, respectively, $NY \\parallel CD$. Because $CB = CE$, $\\angle CBE =\\angle CEB = \\frac{\\angle ACB}{2}$. Because $BC = BD$, $\\angle BCD = \\angle BDC = \\frac{\\angle ABC}{2}$. Let $BE$ and $CD$ intersect at $O$. Because $MX \\parallel BE$ and $NY \\parallel CD$, the angle formed between lines $MX$ and $NY$ is congruent to $\\angle BOD$. Hence, $\\angle BOD = 30^\\circ$ or $150^\\circ$. We have \\begin{align*} \\angle BOD & = \\angle CBE + \\angle BCD \\\\ & = \\frac{\\angle ACB}{2} + \\frac{\\angle ABC}{2} \\\\ & = 90^\\circ - \\frac{\\angle A}{2} . \\end{align*} Hence, we must have $\\angle BOD = 30^\\circ$, not $150^\\circ$. Hence, $\\angle A = 120^\\circ$. This implies $a > b$ and $a >c$. This contradicts the condition specified for this case. Therefore, this case is infeasible. $\\textbf{Case 2}$: $c \\leq a \\leq b$. We extend segment $CB$ to $D$, such that $BD = c$. We extend segment $AC$ to $E$, such that $CE = a$. In this case, $X$ is the midpoint of $AE$, and $Y$ is the midpoint of $CD$. Because $M$ and $X$ are the midpoints of $AB$ and $AE$, respectively, $MX \\parallel BE$. Because $N$ and $Y$ are the midpoints of $AC$ and $CD$, respectively, $NY \\parallel AD$. Because $CB = CE$, $\\angle CBE =\\angle CEB = \\frac{\\angle ACB}{2}$. Because $BA = BD$, $\\angle BAD = \\angle BDA = \\frac{\\angle ABC}{2}$. Let $O$ be a point of $AC$, such that $BO \\parallel AD$. Hence, $\\angle OBC = \\angle BDA = \\frac{B}{2}$. Because $MX \\parallel BE$ and $NY \\parallel AD$ and $AD \\parallel BO$, the angle formed between lines $MX$ and $NY$ is congruent to $\\angle OBE$. Hence, $\\angle OBE = 30^\\circ$ or $150^\\circ$. We have \\begin{align*} \\angle OBE & = \\angle OBC + \\angle CBE \\\\ & = \\frac{\\angle ABC}{2} + \\frac{\\angle ACB}{2} \\\\ & = 90^\\circ - \\frac{\\angle A}{2} . \\end{align*} Hence, we must have $\\angle OBE = 30^\\circ$, not $150^\\circ$. Hence, $\\angle A = 120^\\circ$. This implies $a > b$ and $a >c$. This contradicts the condition specified for this case. Therefore, this case is infeasible. $\\textbf{Case 3}$: $c \\leq b \\leq a$. We extend segment $CB$ to $D$, such that $BD = c$. We extend segment $BC$ to $E$, such that $CE = b$. In this case, $X$ is the midpoint of $BE$, and $Y$ is the midpoint of $CD$. Because $M$ and $X$ are the midpoints of $AB$ and $BE$, respectively, $MX \\parallel AE$. Because $N$ and $Y$ are the midpoints of $AC$ and $CD$, respectively, $NY \\parallel AD$. Because $CA = CE$, $\\angle CAE =\\angle CEB = \\frac{\\angle ACB}{2}$. Because $BA = BD$, $\\angle BAD = \\angle BDA = \\frac{\\angle ABC}{2}$. Because $MX \\parallel AE$ and $NY \\parallel AD$, the angle formed between lines $MX$ and $NY$ is congruent to $\\angle DAE$. Hence, $\\angle DAE = 30^\\circ$ or $150^\\circ$. We have \\begin{align*} \\angle DAE & = \\angle BAD + \\angle CAE + \\angle BAC \\\\ & = \\frac{\\angle ABC}{2} + \\frac{\\angle ACB}{2} + \\angle BAC \\\\ & = 90^\\circ + \\frac{\\angle BAC}{2} . \\end{align*} Hence, we must have $\\angle OBE = 150^\\circ$, not $30^\\circ$. Hence, $\\angle BAC = 120^\\circ$. In $\\triangle ABC$, by applying the law of cosines, we have \\begin{align*} a^2 & = b^2 + c^2 - 2bc \\cos \\angle BAC\\\\ & = b^2 + c^2 - 2bc \\cos 120^\\circ \\\\ & = b^2 + c^2 + bc . \\end{align*} Because $a = 219$, we have \\[ b^2 + c^2 + bc = 219^2 . \\] Now, we find integer solution(s) of this equation with $c \\leq b$. Multiplying this equation by 4, we get \\[ \\left( 2 c + b \\right)^2 + 3 b^2 = 438^2 . \\hspace{1cm} (1) \\] Denote $d = 2 c + b$. Because $c \\leq b$, $b < d \\leq 3 b$. Because $438^2 - 3 b^2 \\equiv 0 \\pmod{3}$, $d^2 \\equiv 0 \\pmod{3}$. Thus, $d \\equiv 0 \\pmod{3}$. This implies $d^2 \\equiv 0 \\pmod{9}$. We also have $438^2 \\equiv 0 \\pmod{9}$. Hence, $3 b^2 \\equiv 0 \\pmod{9}$. This implies $b \\equiv 0 \\pmod{3}$. Denote $b = 3 p$ and $d = 3 q$. Hence, $p < q \\leq 3 p$. Hence, Equation (1) can be written as \\[ q^2 + 3 p^2 = 146^2 . \\hspace{1cm} (2) \\] Now, we solve this equation. First, we find an upper bound of $q$. We have $q^2 + 3 p^2 \\geq q^2 + 3 \\left( \\frac{q}{3} \\right)^2 = \\frac{4 q^2}{3}$. Hence, $\\frac{4 q^2}{3} \\leq 146^2$. Hence, $q \\leq 73 \\sqrt{3} < 73 \\cdot 1.8 = 131.4$. Because $q$ is an integer, we must have $q \\leq 131$. Second, we find a lower bound of $q$. We have $q^2 + 3 p^2 < q^2 + 3 q^2 = 4 q^2$. Hence, $4 q^2 > 146^2$. Hence, $q > 73$. Because $q$ is an integer, we must have $q \\geq 74$. Now, we find the integer solutions of $p$ and $q$ that satisfy Equation (2) with $74 \\leq q \\leq 131$. First, modulo 9, \\begin{align*} q^2 & \\equiv 146^2 - 3 p^2 \\\\ & \\equiv 4 - 3 \\cdot ( 0 \\mbox{ or } 1 ) \\\\ & \\equiv 4 \\mbox{ or } 1 . \\end{align*} Hence $q \\equiv \\pm 1, \\pm 2 \\pmod{9}$. Second, modulo 5, \\begin{align*} q^2 & \\equiv 146^2 - 3 p^2 \\\\ & \\equiv 1 + 2 p^2 \\\\ & \\equiv 1 + 2 \\cdot ( 0 \\mbox{ or } 1 \\mbox{ or } -1 ) \\\\ & \\equiv 1 \\mbox{ or } 3 \\mbox{ or } - 1 . \\end{align*} Because $q^2 \\equiv 0 \\mbox{ or } 1 \\mbox{ or } - 1$, we must have $q^2 \\equiv 1 \\mbox{ or } - 1$. Hence, $5 \\nmid q$. Third, modulo 7, \\begin{align*} q^2 & \\equiv 146^2 - 3 p^2 \\\\ & \\equiv 1 - 3 \\cdot ( 0 \\mbox{ or } 1 \\mbox{ or } 5 \\mbox{ or } 2 ) \\\\ & \\equiv 1 \\mbox{ or } 2 \\mbox{ or } 3 \\mbox{ or } 5 . \\end{align*} Because $q^2 \\equiv 0 \\mbox{ or } 1 \\mbox{ or } 2 \\mbox{ or } 4 \\pmod{ 7 }$, we must have $q^2 \\equiv 1 \\mbox{ or } 2 \\pmod{7}$. Hence, $q \\equiv 1, 3, 4, 6 \\pmod{7}$. Given all conditions above, the possible $q$ are 74, 83, 88, 92, 97, 101, 106, 109, 116, 118, 127. By testing all these numbers, we find that the only solution is $q = 97$. This implies $p = 63$. Hence, $b = 3p = 189$ and $d = 3q = 291$. Hence, $c = \\frac{d - b}{2} = 51$. Therefore, the perimeter of $\\triangle ABC$ is $b + c + a = 189 + 51 + 219 = \\textbf{(459) }. ~Steven Chen (www.professorchenedu.com)",
"We wish to solve the Diophantine equation $a^2+ab+b^2=3^2 \\cdot 73^2$. It can be shown that $3|a$ and $3|b$, so we make the substitution $a=3x$ and $b=3y$ to obtain $x^2+xy+y^2=73^2$ as our new equation to solve for. Notice that $r^2+r+1=(r-\\omega)(r-{\\omega}^2)$, where $\\omega=e^{i\\frac{2\\pi}{3}}$. Thus, \\[x^2+xy+y^2 = y^2((x/y)^2+(x/y)+1) = y^2 (\\frac{x}{y}-\\omega)(\\frac{x}{y}-{\\omega}^2) = (x-y\\omega)(x-y{\\omega}^2).\\] Note that $8^2+1^2+8 \\cdot 1=73$. Thus, $(8-\\omega)(8-{\\omega}^2)=73$. Squaring both sides yields \\begin{align} (8-\\omega)^2(8-{\\omega}^2)^2&=73^2\\\\ (63-17\\omega)(63-17{\\omega}^2)&=73^2. \\end{align} Thus, by $(2)$, $(63, 17)$ is a solution to $x^2+xy+y^2=73^2$. This implies that $a=189$ and $b=51$, so our final answer is $189+51+219=459. ~ Leo.Euler",
"We wish to solve the Diophantine equation $a^2+ab+b^2=3^2 \\cdot 73^2$. It can be shown that $3|a$ and $3|b$, so we make the substitution $a=3x$ and $b=3y$ to obtain $x^2+xy+y^2=73^2$ as our new equation to solve for. Notice that $r^2+r+1=(r-\\omega)(r-{\\omega}^2)$, where $\\omega=e^{i\\frac{2\\pi}{3}}$. Thus, \\[x^2+xy+y^2 = y^2((x/y)^2+(x/y)+1) = y^2 (\\frac{x}{y}-\\omega)(\\frac{x}{y}-{\\omega}^2) = (x-y\\omega)(x-y{\\omega}^2).\\] Note that $8^2+1^2+8 \\cdot 1=73$. Thus, $(8-\\omega)(8-{\\omega}^2)=73$. Squaring both sides yields \\begin{align} (8-\\omega)^2(8-{\\omega}^2)^2&=73^2\\\\ (63-17\\omega)(63-17{\\omega}^2)&=73^2. \\end{align} Thus, by $(2)$, $(63, 17)$ is a solution to $x^2+xy+y^2=73^2$. This implies that $a=189$ and $b=51$, so our final answer is $189+51+219=459. ~ Leo.Euler",
"We look at upper and middle diagrams and get $\\angle BAC = 120^\\circ$. Next we use only the lower Diagram. Let $I$ be incenter $\\triangle ABC$, E be midpoint of biggest arc $\\overset{\\Large\\frown} {BC}.$ Then bisector $AI$ cross circumcircle $\\triangle ABC$ at point $E$. Quadrilateral $ABEC$ is cyclic, so \\[\\angle BEC = 180^\\circ - \\angle ABC = 60^\\circ \\implies BE = CE = IE = BC.\\] \\[AE \\cdot BC = AB \\cdot CE + AC \\cdot BE \\implies AE = AB + AC\\] $\\implies AI +EI = AB + AC, \\hspace{10mm} AI = AB+ AC – BC$ is integer. \\[AI = \\frac {2AB \\cdot AC \\cdot cos \\angle CAI}{AB+AC + BC} = \\frac {AB \\cdot AC}{AB+AC + BC} =\\] \\[= AB + AC – BC \\implies AC^2 + AB^2 + AB \\cdot AC = BC^2.\\] A quick $(\\mod9)$ check gives that $3\\mid AC$ and $3\\mid AB$. \\[AI \\le A_0I_0 = EA_0 – EI_0 = \\frac{2 BC}{\\sqrt{3}} – BC = \\frac {2 - \\sqrt{3}}{\\sqrt{3}} BC = 33.88.\\] Denote $a= \\frac {BC}{3}= 73, b = \\frac {AC}{3}, c = \\frac {AB}{3}, l = \\frac {AI}{3} \\le 11.$ We have equations in integers $\\frac{bc}{a+b+c} = b + c – a = l \\le 11.$ The solution $(b > c)$ is \\[b = \\frac{a + l +\\sqrt{a^2 – 6al – 3l^2}}{2}, c = \\frac{a + l -\\sqrt{a^2 – 6al – 3l^2}}{2}.\\] Suppose, $a^2 – 6al – 3l^2 = (a – 3l – t)^2 \\implies \\frac {12l^2}{t} + t + 6l= 2a = 146.$ Now we check all possible $t = {2,3,4,6,12, ml}.$ Case $t = 2 \\implies 6l^2 + 6l = 146 – 2 \\implies l^2 + l = 24 \\implies \\O$ Case $t = 3 \\implies 4l^2 + 6l = 146 – 3 = 143\\implies \\O.$ Case $t = 4 \\implies 3l^2 + 6l = 146 – 4 =142 \\implies \\O.$ Case $t = 6 \\implies 2l^2 + 6l = 146 – 6 = 140 \\implies l = 7, b = 63, c = 17.$ Case $t = 12 \\implies l^2 + 6l = 146 – 12 = 134 \\implies \\O.$ Case $t = ml \\implies \\frac{12l}{m} + 6l + ml = 146 \\implies \\frac{12}{m} + 6 + m = \\frac{73 \\cdot 2}{l}\\implies \\O.$ vladimir.shelomovskii@gmail.com, vvsss",
"We look at upper and middle diagrams and get $\\angle BAC = 120^\\circ$. Next we use only the lower Diagram. Let $I$ be incenter $\\triangle ABC$, E be midpoint of biggest arc $\\overset{\\Large\\frown} {BC}.$ Then bisector $AI$ cross circumcircle $\\triangle ABC$ at point $E$. Quadrilateral $ABEC$ is cyclic, so \\[\\angle BEC = 180^\\circ - \\angle ABC = 60^\\circ \\implies BE = CE = IE = BC.\\] \\[AE \\cdot BC = AB \\cdot CE + AC \\cdot BE \\implies AE = AB + AC\\] $\\implies AI +EI = AB + AC, \\hspace{10mm} AI = AB+ AC – BC$ is integer. \\[AI = \\frac {2AB \\cdot AC \\cdot cos \\angle CAI}{AB+AC + BC} = \\frac {AB \\cdot AC}{AB+AC + BC} =\\] \\[= AB + AC – BC \\implies AC^2 + AB^2 + AB \\cdot AC = BC^2.\\] A quick $(\\mod9)$ check gives that $3\\mid AC$ and $3\\mid AB$. \\[AI \\le A_0I_0 = EA_0 – EI_0 = \\frac{2 BC}{\\sqrt{3}} – BC = \\frac {2 - \\sqrt{3}}{\\sqrt{3}} BC = 33.88.\\] Denote $a= \\frac {BC}{3}= 73, b = \\frac {AC}{3}, c = \\frac {AB}{3}, l = \\frac {AI}{3} \\le 11.$ We have equations in integers $\\frac{bc}{a+b+c} = b + c – a = l \\le 11.$ The solution $(b > c)$ is \\[b = \\frac{a + l +\\sqrt{a^2 – 6al – 3l^2}}{2}, c = \\frac{a + l -\\sqrt{a^2 – 6al – 3l^2}}{2}.\\] Suppose, $a^2 – 6al – 3l^2 = (a – 3l – t)^2 \\implies \\frac {12l^2}{t} + t + 6l= 2a = 146.$ Now we check all possible $t = {2,3,4,6,12, ml}.$ Case $t = 2 \\implies 6l^2 + 6l = 146 – 2 \\implies l^2 + l = 24 \\implies \\O$ Case $t = 3 \\implies 4l^2 + 6l = 146 – 3 = 143\\implies \\O.$ Case $t = 4 \\implies 3l^2 + 6l = 146 – 4 =142 \\implies \\O.$ Case $t = 6 \\implies 2l^2 + 6l = 146 – 6 = 140 \\implies l = 7, b = 63, c = 17.$ Case $t = 12 \\implies l^2 + 6l = 146 – 12 = 134 \\implies \\O.$ Case $t = ml \\implies \\frac{12l}{m} + 6l + ml = 146 \\implies \\frac{12}{m} + 6 + m = \\frac{73 \\cdot 2}{l}\\implies \\O.$ vladimir.shelomovskii@gmail.com, vvsss"
] |
2022-I-15
| 2,022
| 15
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Let $x,$ $y,$ and $z$ be positive real numbers satisfying the system of equations: \begin{align*} \sqrt{2x-xy} + \sqrt{2y-xy} &= 1 \\ \sqrt{2y-yz} + \sqrt{2z-yz} &= \sqrt2 \\ \sqrt{2z-zx} + \sqrt{2x-zx} &= \sqrt3. \end{align*} Then $\left[ (1-x)(1-y)(1-z) \right]^2$ can be written as $\frac{m}{n},$ where $m$ and $n$ are relatively prime positive integers. Find $m+n.$
| 33
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I
|
[
"First, let define a triangle with side lengths $\\sqrt{2x}$, $\\sqrt{2z}$, and $l$, with altitude from $l$'s equal to $\\sqrt{xz}$. $l = \\sqrt{2x - xz} + \\sqrt{2z - xz}$, the left side of one equation in the problem. Let $\\theta$ be angle opposite the side with length $\\sqrt{2x}$. Then the altitude has length $\\sqrt{2z} \\cdot \\sin(\\theta) = \\sqrt{xz}$ and thus $\\sin(\\theta) = \\sqrt{\\frac{x}{2}}$, so $x=2\\sin^2(\\theta)$ and the side length $\\sqrt{2x}$ is equal to $2\\sin(\\theta)$. We can symmetrically apply this to the two other equations/triangles. By law of sines, we have $\\frac{2\\sin(\\theta)}{\\sin(\\theta)} = 2R$, with $R=1$ as the circumradius, same for all 3 triangles. The circumcircle's central angle to a side is $2 \\arcsin(l/2)$, so the 3 triangles' $l=1, \\sqrt{2}, \\sqrt{3}$, have angles $120^{\\circ}, 90^{\\circ}, 60^{\\circ}$, respectively. This means that by half angle arcs, we see that we have in some order, $x=2\\sin^2(\\alpha)$, $y=2\\sin^2(\\beta)$, and $z=2\\sin^2(\\gamma)$ (not necessarily this order, but here it does not matter due to symmetry), satisfying that $\\alpha+\\beta=180^{\\circ}-\\frac{120^{\\circ}}{2}$, $\\beta+\\gamma=180^{\\circ}-\\frac{90^{\\circ}}{2}$, and $\\gamma+\\alpha=180^{\\circ}-\\frac{60^{\\circ}}{2}$. Solving, we get $\\alpha=\\frac{135^{\\circ}}{2}$, $\\beta=\\frac{105^{\\circ}}{2}$, and $\\gamma=\\frac{165^{\\circ}}{2}$. We notice that \\[[(1-x)(1-y)(1-z)]^2=[\\sin(2\\alpha)\\sin(2\\beta)\\sin(2\\gamma)]^2=[\\sin(135^{\\circ})\\sin(105^{\\circ})\\sin(165^{\\circ})]^2\\] \\[=\\left(\\frac{\\sqrt{2}}{2} \\cdot \\frac{\\sqrt{6}-\\sqrt{2}}{4} \\cdot \\frac{\\sqrt{6}+\\sqrt{2}}{4}\\right)^2 = \\left(\\frac{\\sqrt{2}}{8}\\right)^2=\\frac{1}{32} \\to 033. \\blacksquare\\] - kevinmathz",
"First, let define a triangle with side lengths $\\sqrt{2x}$, $\\sqrt{2z}$, and $l$, with altitude from $l$'s equal to $\\sqrt{xz}$. $l = \\sqrt{2x - xz} + \\sqrt{2z - xz}$, the left side of one equation in the problem. Let $\\theta$ be angle opposite the side with length $\\sqrt{2x}$. Then the altitude has length $\\sqrt{2z} \\cdot \\sin(\\theta) = \\sqrt{xz}$ and thus $\\sin(\\theta) = \\sqrt{\\frac{x}{2}}$, so $x=2\\sin^2(\\theta)$ and the side length $\\sqrt{2x}$ is equal to $2\\sin(\\theta)$. We can symmetrically apply this to the two other equations/triangles. By law of sines, we have $\\frac{2\\sin(\\theta)}{\\sin(\\theta)} = 2R$, with $R=1$ as the circumradius, same for all 3 triangles. The circumcircle's central angle to a side is $2 \\arcsin(l/2)$, so the 3 triangles' $l=1, \\sqrt{2}, \\sqrt{3}$, have angles $120^{\\circ}, 90^{\\circ}, 60^{\\circ}$, respectively. This means that by half angle arcs, we see that we have in some order, $x=2\\sin^2(\\alpha)$, $y=2\\sin^2(\\beta)$, and $z=2\\sin^2(\\gamma)$ (not necessarily this order, but here it does not matter due to symmetry), satisfying that $\\alpha+\\beta=180^{\\circ}-\\frac{120^{\\circ}}{2}$, $\\beta+\\gamma=180^{\\circ}-\\frac{90^{\\circ}}{2}$, and $\\gamma+\\alpha=180^{\\circ}-\\frac{60^{\\circ}}{2}$. Solving, we get $\\alpha=\\frac{135^{\\circ}}{2}$, $\\beta=\\frac{105^{\\circ}}{2}$, and $\\gamma=\\frac{165^{\\circ}}{2}$. We notice that \\[[(1-x)(1-y)(1-z)]^2=[\\sin(2\\alpha)\\sin(2\\beta)\\sin(2\\gamma)]^2=[\\sin(135^{\\circ})\\sin(105^{\\circ})\\sin(165^{\\circ})]^2\\] \\[=\\left(\\frac{\\sqrt{2}}{2} \\cdot \\frac{\\sqrt{6}-\\sqrt{2}}{4} \\cdot \\frac{\\sqrt{6}+\\sqrt{2}}{4}\\right)^2 = \\left(\\frac{\\sqrt{2}}{8}\\right)^2=\\frac{1}{32} \\to 033. \\blacksquare\\] - kevinmathz",
"(This eventually whittles down to the same concept as Solution 1) Note that in each equation in this system, it is possible to factor $\\sqrt{x}$, $\\sqrt{y}$, or $\\sqrt{z}$ from each term (on the left sides), since each of $x$, $y$, and $z$ are positive real numbers. After factoring out accordingly from each terms one of $\\sqrt{x}$, $\\sqrt{y}$, or $\\sqrt{z}$, the system should look like this: \\begin{align*} \\sqrt{x}\\cdot\\sqrt{2-y} + \\sqrt{y}\\cdot\\sqrt{2-x} &= 1 \\\\ \\sqrt{y}\\cdot\\sqrt{2-z} + \\sqrt{z}\\cdot\\sqrt{2-y} &= \\sqrt2 \\\\ \\sqrt{z}\\cdot\\sqrt{2-x} + \\sqrt{x}\\cdot\\sqrt{2-z} &= \\sqrt3. \\end{align*} This should give off tons of trigonometry vibes. To make the connection clear, $x = 2\\cos^2 \\alpha$, $y = 2\\cos^2 \\beta$, and $z = 2\\cos^2 \\theta$ is a helpful substitution: \\begin{align*} \\sqrt{2\\cos^2 \\alpha}\\cdot\\sqrt{2-2\\cos^2 \\beta} + \\sqrt{2\\cos^2 \\beta}\\cdot\\sqrt{2-2\\cos^2 \\alpha} &= 1 \\\\ \\sqrt{2\\cos^2 \\beta}\\cdot\\sqrt{2-2\\cos^2 \\theta} + \\sqrt{2\\cos^2 \\theta}\\cdot\\sqrt{2-2\\cos^2 \\beta} &= \\sqrt2 \\\\ \\sqrt{2\\cos^2 \\theta}\\cdot\\sqrt{2-2\\cos^2 \\alpha} + \\sqrt{2\\cos^2 \\alpha}\\cdot\\sqrt{2-2\\cos^2 \\theta} &= \\sqrt3. \\end{align*} From each equation $\\sqrt{2}^2$ can be factored out, and when every equation is divided by 2, we get: \\begin{align*} \\sqrt{\\cos^2 \\alpha}\\cdot\\sqrt{1-\\cos^2 \\beta} + \\sqrt{\\cos^2 \\beta}\\cdot\\sqrt{1-\\cos^2 \\alpha} &= \\frac{1}{2} \\\\ \\sqrt{\\cos^2 \\beta}\\cdot\\sqrt{1-\\cos^2 \\theta} + \\sqrt{\\cos^2 \\theta}\\cdot\\sqrt{1-\\cos^2 \\beta} &= \\frac{\\sqrt2}{2} \\\\ \\sqrt{\\cos^2 \\theta}\\cdot\\sqrt{1-\\cos^2 \\alpha} + \\sqrt{\\cos^2 \\alpha}\\cdot\\sqrt{1-\\cos^2 \\theta} &= \\frac{\\sqrt3}{2}. \\end{align*} which simplifies to (using the Pythagorean identity $\\sin^2 \\phi + \\cos^2 \\phi = 1 \\; \\forall \\; \\phi \\in \\mathbb{C}$): \\begin{align*} \\cos \\alpha\\cdot\\sin \\beta + \\cos \\beta\\cdot\\sin \\alpha &= \\frac{1}{2} \\\\ \\cos \\beta\\cdot\\sin \\theta + \\cos \\theta\\cdot\\sin \\beta &= \\frac{\\sqrt2}{2} \\\\ \\cos \\theta\\cdot\\sin \\alpha + \\cos \\alpha\\cdot\\sin \\theta &= \\frac{\\sqrt3}{2}. \\end{align*} which further simplifies to (using sine addition formula $\\sin(a + b) = \\sin a \\cos b + \\cos a \\sin b$): \\begin{align*} \\sin(\\alpha + \\beta) &= \\frac{1}{2} \\\\ \\sin(\\beta + \\theta) &= \\frac{\\sqrt2}{2} \\\\ \\sin(\\alpha + \\theta) &= \\frac{\\sqrt3}{2}. \\end{align*} Taking the inverse sine ($0\\leq\\theta\\frac{\\pi}{2}$) of each equation yields a simple system: \\begin{align*} \\alpha + \\beta &= \\frac{\\pi}{6} \\\\ \\beta + \\theta &= \\frac{\\pi}{4} \\\\ \\alpha + \\theta &= \\frac{\\pi}{3} \\end{align*} giving solutions: \\begin{align*} \\alpha &= \\frac{\\pi}{8} \\\\ \\beta &= \\frac{\\pi}{24} \\\\ \\theta &= \\frac{5\\pi}{24} \\end{align*} Since these unknowns are directly related to our original unknowns, there are consequent solutions for those: \\begin{align*} x &= 2\\cos^2\\left(\\frac{\\pi}{8}\\right) \\\\ y &= 2\\cos^2\\left(\\frac{\\pi}{24}\\right) \\\\ z &= 2\\cos^2\\left(\\frac{5\\pi}{24}\\right) \\end{align*} When plugging into the expression $\\left[ (1-x)(1-y)(1-z) \\right]^2$, noting that $-\\cos 2\\phi = 1 - 2\\cos^2 \\phi\\; \\forall \\; \\phi \\in \\mathbb{C}$ helps to simplify this expression into: \\begin{align*} \\left[ (-1)^3\\left(\\cos \\left(2\\cdot\\frac{\\pi}{8}\\right)\\cos \\left(2\\cdot\\frac{\\pi}{24}\\right)\\cos \\left(2\\cdot\\frac{5\\pi}{24}\\right)\\right)\\right]^2 \\\\ = \\left[ (-1)\\left(\\cos \\left(\\frac{\\pi}{4}\\right)\\cos \\left(\\frac{\\pi}{12}\\right)\\cos \\left(\\frac{5\\pi}{12}\\right)\\right)\\right]^2 \\end{align*} Now, all the cosines in here are fairly standard: \\begin{align*} \\cos \\frac{\\pi}{4} &= \\frac{\\sqrt{2}}{2} \\\\ \\cos \\frac{\\pi}{12} &=\\frac{\\sqrt{6} + \\sqrt{2}}{4} & (= \\cos{\\frac{\\frac{\\pi}{6}}{2}} ) \\\\ \\cos \\frac{5\\pi}{12} &= \\frac{\\sqrt{6} - \\sqrt{2}}{4} & (= \\cos\\left({\\frac{\\pi}{6} + \\frac{\\pi}{4}} \\right) ) \\end{align*} With some final calculations: \\begin{align*} &(-1)^2\\left(\\frac{\\sqrt{2}}{2}\\right)^2\\left(\\frac{\\sqrt{6} + \\sqrt{2}}{4}\\right)^2\\left(\\frac{\\sqrt{6} - \\sqrt{2}}{4}\\right)^2 \\\\ =& \\left(\\frac{1}{2}\\right) \\left(\\left(\\frac{\\sqrt{6} + \\sqrt{2}}{4}\\right)\\left(\\frac{\\sqrt{6} - \\sqrt{2}}{4}\\right)\\right)^2 \\\\ =&\\frac{1}{2} \\frac{4^2}{16^2} = \\frac{1}{32} \\end{align*} This is our answer in simplest form $\\frac{m}{n}$, so $m + n = 1 + 32 = 033. ~Oxymoronic15",
"(This eventually whittles down to the same concept as Solution 1) Note that in each equation in this system, it is possible to factor $\\sqrt{x}$, $\\sqrt{y}$, or $\\sqrt{z}$ from each term (on the left sides), since each of $x$, $y$, and $z$ are positive real numbers. After factoring out accordingly from each terms one of $\\sqrt{x}$, $\\sqrt{y}$, or $\\sqrt{z}$, the system should look like this: \\begin{align*} \\sqrt{x}\\cdot\\sqrt{2-y} + \\sqrt{y}\\cdot\\sqrt{2-x} &= 1 \\\\ \\sqrt{y}\\cdot\\sqrt{2-z} + \\sqrt{z}\\cdot\\sqrt{2-y} &= \\sqrt2 \\\\ \\sqrt{z}\\cdot\\sqrt{2-x} + \\sqrt{x}\\cdot\\sqrt{2-z} &= \\sqrt3. \\end{align*} This should give off tons of trigonometry vibes. To make the connection clear, $x = 2\\cos^2 \\alpha$, $y = 2\\cos^2 \\beta$, and $z = 2\\cos^2 \\theta$ is a helpful substitution: \\begin{align*} \\sqrt{2\\cos^2 \\alpha}\\cdot\\sqrt{2-2\\cos^2 \\beta} + \\sqrt{2\\cos^2 \\beta}\\cdot\\sqrt{2-2\\cos^2 \\alpha} &= 1 \\\\ \\sqrt{2\\cos^2 \\beta}\\cdot\\sqrt{2-2\\cos^2 \\theta} + \\sqrt{2\\cos^2 \\theta}\\cdot\\sqrt{2-2\\cos^2 \\beta} &= \\sqrt2 \\\\ \\sqrt{2\\cos^2 \\theta}\\cdot\\sqrt{2-2\\cos^2 \\alpha} + \\sqrt{2\\cos^2 \\alpha}\\cdot\\sqrt{2-2\\cos^2 \\theta} &= \\sqrt3. \\end{align*} From each equation $\\sqrt{2}^2$ can be factored out, and when every equation is divided by 2, we get: \\begin{align*} \\sqrt{\\cos^2 \\alpha}\\cdot\\sqrt{1-\\cos^2 \\beta} + \\sqrt{\\cos^2 \\beta}\\cdot\\sqrt{1-\\cos^2 \\alpha} &= \\frac{1}{2} \\\\ \\sqrt{\\cos^2 \\beta}\\cdot\\sqrt{1-\\cos^2 \\theta} + \\sqrt{\\cos^2 \\theta}\\cdot\\sqrt{1-\\cos^2 \\beta} &= \\frac{\\sqrt2}{2} \\\\ \\sqrt{\\cos^2 \\theta}\\cdot\\sqrt{1-\\cos^2 \\alpha} + \\sqrt{\\cos^2 \\alpha}\\cdot\\sqrt{1-\\cos^2 \\theta} &= \\frac{\\sqrt3}{2}. \\end{align*} which simplifies to (using the Pythagorean identity $\\sin^2 \\phi + \\cos^2 \\phi = 1 \\; \\forall \\; \\phi \\in \\mathbb{C}$): \\begin{align*} \\cos \\alpha\\cdot\\sin \\beta + \\cos \\beta\\cdot\\sin \\alpha &= \\frac{1}{2} \\\\ \\cos \\beta\\cdot\\sin \\theta + \\cos \\theta\\cdot\\sin \\beta &= \\frac{\\sqrt2}{2} \\\\ \\cos \\theta\\cdot\\sin \\alpha + \\cos \\alpha\\cdot\\sin \\theta &= \\frac{\\sqrt3}{2}. \\end{align*} which further simplifies to (using sine addition formula $\\sin(a + b) = \\sin a \\cos b + \\cos a \\sin b$): \\begin{align*} \\sin(\\alpha + \\beta) &= \\frac{1}{2} \\\\ \\sin(\\beta + \\theta) &= \\frac{\\sqrt2}{2} \\\\ \\sin(\\alpha + \\theta) &= \\frac{\\sqrt3}{2}. \\end{align*} Taking the inverse sine ($0\\leq\\theta\\frac{\\pi}{2}$) of each equation yields a simple system: \\begin{align*} \\alpha + \\beta &= \\frac{\\pi}{6} \\\\ \\beta + \\theta &= \\frac{\\pi}{4} \\\\ \\alpha + \\theta &= \\frac{\\pi}{3} \\end{align*} giving solutions: \\begin{align*} \\alpha &= \\frac{\\pi}{8} \\\\ \\beta &= \\frac{\\pi}{24} \\\\ \\theta &= \\frac{5\\pi}{24} \\end{align*} Since these unknowns are directly related to our original unknowns, there are consequent solutions for those: \\begin{align*} x &= 2\\cos^2\\left(\\frac{\\pi}{8}\\right) \\\\ y &= 2\\cos^2\\left(\\frac{\\pi}{24}\\right) \\\\ z &= 2\\cos^2\\left(\\frac{5\\pi}{24}\\right) \\end{align*} When plugging into the expression $\\left[ (1-x)(1-y)(1-z) \\right]^2$, noting that $-\\cos 2\\phi = 1 - 2\\cos^2 \\phi\\; \\forall \\; \\phi \\in \\mathbb{C}$ helps to simplify this expression into: \\begin{align*} \\left[ (-1)^3\\left(\\cos \\left(2\\cdot\\frac{\\pi}{8}\\right)\\cos \\left(2\\cdot\\frac{\\pi}{24}\\right)\\cos \\left(2\\cdot\\frac{5\\pi}{24}\\right)\\right)\\right]^2 \\\\ = \\left[ (-1)\\left(\\cos \\left(\\frac{\\pi}{4}\\right)\\cos \\left(\\frac{\\pi}{12}\\right)\\cos \\left(\\frac{5\\pi}{12}\\right)\\right)\\right]^2 \\end{align*} Now, all the cosines in here are fairly standard: \\begin{align*} \\cos \\frac{\\pi}{4} &= \\frac{\\sqrt{2}}{2} \\\\ \\cos \\frac{\\pi}{12} &=\\frac{\\sqrt{6} + \\sqrt{2}}{4} & (= \\cos{\\frac{\\frac{\\pi}{6}}{2}} ) \\\\ \\cos \\frac{5\\pi}{12} &= \\frac{\\sqrt{6} - \\sqrt{2}}{4} & (= \\cos\\left({\\frac{\\pi}{6} + \\frac{\\pi}{4}} \\right) ) \\end{align*} With some final calculations: \\begin{align*} &(-1)^2\\left(\\frac{\\sqrt{2}}{2}\\right)^2\\left(\\frac{\\sqrt{6} + \\sqrt{2}}{4}\\right)^2\\left(\\frac{\\sqrt{6} - \\sqrt{2}}{4}\\right)^2 \\\\ =& \\left(\\frac{1}{2}\\right) \\left(\\left(\\frac{\\sqrt{6} + \\sqrt{2}}{4}\\right)\\left(\\frac{\\sqrt{6} - \\sqrt{2}}{4}\\right)\\right)^2 \\\\ =&\\frac{1}{2} \\frac{4^2}{16^2} = \\frac{1}{32} \\end{align*} This is our answer in simplest form $\\frac{m}{n}$, so $m + n = 1 + 32 = 033. ~Oxymoronic15",
"Let $1-x=a;1-y=b;1-z=c$, rewrite those equations $\\sqrt{(1-a)(1+b)}+\\sqrt{(1+a)(1-b)}=1$; $\\sqrt{(1-b)(1+c)}+\\sqrt{(1+b)(1-c)}=\\sqrt{2}$ $\\sqrt{(1-a)(1+c)}+\\sqrt{(1-c)(1+a)}=\\sqrt{3}$ and solve for $m/n = (abc)^2 = a^2b^2c^2$ Square both sides and simplify, to get three equations: $2ab-1=2\\sqrt{(1-a^2)(1-b^2)}$ $2bc~ ~ ~ ~ ~ ~=2\\sqrt{(1-b^2)(1-c^2)}$ $2ac+1=2\\sqrt{(1-c^2)(1-a^2)}$ Square both sides again, and simplify to get three equations: $a^2+b^2-ab=\\frac{3}{4}$ $b^2+c^2~ ~ ~ ~ ~ ~=1$ $a^2+c^2+ac=\\frac{3}{4}$ Subtract first and third equation, getting $(b+c)(b-c)=a(b+c)$, $a=b-c$ Put it in first equation, getting $b^2-2bc+c^2+b^2-b(b-c)=b^2+c^2-bc=\\frac{3}{4}$, $bc=\\frac{1}{4}$ Since $a^2=b^2+c^2-2bc=\\frac{1}{2}$, $m/n = a^2b^2c^2 = a^2(bc)^2 = \\frac{1}{2}\\left(\\frac{1}{4}\\right)^2=\\frac{1}{32}$ and so the final answer is $033 ~bluesoul",
"Let $1-x=a;1-y=b;1-z=c$, rewrite those equations $\\sqrt{(1-a)(1+b)}+\\sqrt{(1+a)(1-b)}=1$; $\\sqrt{(1-b)(1+c)}+\\sqrt{(1+b)(1-c)}=\\sqrt{2}$ $\\sqrt{(1-a)(1+c)}+\\sqrt{(1-c)(1+a)}=\\sqrt{3}$ and solve for $m/n = (abc)^2 = a^2b^2c^2$ Square both sides and simplify, to get three equations: $2ab-1=2\\sqrt{(1-a^2)(1-b^2)}$ $2bc~ ~ ~ ~ ~ ~=2\\sqrt{(1-b^2)(1-c^2)}$ $2ac+1=2\\sqrt{(1-c^2)(1-a^2)}$ Square both sides again, and simplify to get three equations: $a^2+b^2-ab=\\frac{3}{4}$ $b^2+c^2~ ~ ~ ~ ~ ~=1$ $a^2+c^2+ac=\\frac{3}{4}$ Subtract first and third equation, getting $(b+c)(b-c)=a(b+c)$, $a=b-c$ Put it in first equation, getting $b^2-2bc+c^2+b^2-b(b-c)=b^2+c^2-bc=\\frac{3}{4}$, $bc=\\frac{1}{4}$ Since $a^2=b^2+c^2-2bc=\\frac{1}{2}$, $m/n = a^2b^2c^2 = a^2(bc)^2 = \\frac{1}{2}\\left(\\frac{1}{4}\\right)^2=\\frac{1}{32}$ and so the final answer is $033 ~bluesoul",
"Denote $u = 1 - x$, $v = 1 - y$, $w = 1 - z$. Hence, the system of equations given in the problem can be written as \\begin{align*} \\sqrt{(1-u)(1+v)} + \\sqrt{(1+u)(1-v)} & = 1 \\hspace{1cm} (1) \\\\ \\sqrt{(1-v)(1+w)} + \\sqrt{(1+v)(1-w)} & = \\sqrt{2} \\hspace{1cm} (2) \\\\ \\sqrt{(1-w)(1+u)} + \\sqrt{(1+w)(1-u)} & = \\sqrt{3} . \\hspace{1cm} (3) \\end{align*} Each equation above takes the following form: \\[ \\sqrt{(1-a)(1+b)} + \\sqrt{(1+a)(1-b)} = k . \\] Now, we simplify this equation by removing radicals. Denote $p = \\sqrt{(1-a)(1+b)}$ and $q = \\sqrt{(1+a)(1-b)}$. Hence, the equation above implies \\[ \\left\\{ \\begin{array}{l} p + q = k \\\\ p^2 = (1-a)(1+b) \\\\ q^2 = (1+a)(1-b) \\end{array} \\right.. \\] Hence, $q^2 - p^2 = (1+a)(1-b) - (1-a)(1+b) = 2 (a-b)$. Hence, $q - p = \\frac{q^2 - p^2}{p+q} = \\frac{2}{k} (a-b)$. Because $p + q = k$ and $q - p = \\frac{2}{k} (a-b)$, we get $q = \\frac{a-b}{k} + \\frac{k}{2}$. Plugging this into the equation $q^2 = (1+a)(1-b)$ and simplifying it, we get \\[ a^2 + \\left( k^2 - 2 \\right) ab + b^2 = k^2 - \\frac{k^4}{4} . \\] Therefore, the system of equations above can be simplified as \\begin{align*} u^2 - uv + v^2 & = \\frac{3}{4} \\\\ v^2 + w^2 & = 1 \\\\ w^2 + wu + u^2 & = \\frac{3}{4} . \\end{align*} Denote $w' = - w$. The system of equations above can be equivalently written as \\begin{align*} u^2 - uv + v^2 & = \\frac{3}{4} \\hspace{1cm} (1') \\\\ v^2 + w'^2 & = 1 \\hspace{1cm} (2') \\\\ w'^2 - w'u + u^2 & = \\frac{3}{4} \\hspace{1cm} (3') . \\end{align*} Taking $(1') - (3')$, we get \\[ (v - w') (v + w' - u) = 0 . \\] Thus, we have either $v - w' = 0$ or $v + w' - u = 0$. $\\textbf{Case 1}$: $v - w' = 0$. Equation (2') implies $v = w' = \\pm \\frac{1}{\\sqrt{2}}$. Plugging $v$ and $w'$ into Equation (2), we get contradiction. Therefore, this case is infeasible. $\\textbf{Case 2}$: $v + w' - u = 0$. Plugging this condition into (1') to substitute $u$, we get \\[ v^2 + v w' + w'^2 = \\frac{3}{4} \\hspace{1cm} (4) . \\] Taking $(4) - (2')$, we get \\[ v w' = - \\frac{1}{4} . \\hspace{1cm} (5) . \\] Taking (4) + (5), we get \\[ \\left( v + w' \\right)^2 = \\frac{1}{2} . \\] Hence, $u^2 = \\left( v + w' \\right)^2 = \\frac{1}{2}$. Therefore, \\begin{align*} \\left[ (1-x)(1-y)(1-z) \\right]^2 & = u^2 (vw)^2 \\\\ & = u^2 (vw')^2 \\\\ & = \\frac{1}{2} \\left( - \\frac{1}{4} \\right)^2 \\\\ & = \\frac{1}{32} . \\end{align*} Therefore, the answer is $1 + 32 = \\textbf{(033) }, and when making the substitution in each equation of the initial set of equations, we obtain a new equation in the form of the sine addition formula. ~ Leo.Euler",
"In given equations, $0 \\leq x,y,z \\leq 2,$ so we define some points: \\[\\bar {O} = (0, 0), \\bar {A} = (1, 0), \\bar{M} = \\left(\\frac {1}{\\sqrt{2}},\\frac {1}{\\sqrt{2}}\\right),\\] \\[\\bar {X} = \\left(\\sqrt {\\frac {x}{2}}, \\sqrt{1 – \\frac{x}{2}}\\right), \\bar {Y'} = \\left(\\sqrt {\\frac {y}{2}}, \\sqrt{1 – \\frac{y}{2}}\\right),\\] \\[\\bar {Y} = \\left(\\sqrt {1 – \\frac{y}{2}},\\sqrt{\\frac {y}{2}}\\right), \\bar {Z} = \\left(\\sqrt {1 – \\frac{z}{2}},\\sqrt{\\frac {z}{2}}\\right).\\] Notice, that \\[\\mid \\vec {AO} \\mid = \\mid \\vec {MO} \\mid = \\mid \\vec {XO} \\mid =\\mid \\vec {YO} \\mid = \\mid \\vec {Y'O} \\mid =\\mid \\vec {ZO} \\mid = 1\\] and each points lies in the first quadrant. We use given equations and get some scalar products: \\[(\\vec {XO} \\cdot \\vec {YO}) = \\frac {1}{2} = \\cos \\angle XOY \\implies \\angle XOY = 60 ^\\circ,\\] \\[(\\vec {XO} \\cdot \\vec {ZO}) = \\frac {\\sqrt{3}}{2} = \\cos \\angle XOZ \\implies \\angle XOZ = 30^\\circ,\\] \\[(\\vec {Y'O} \\cdot \\vec {ZO}) = \\frac {1}{\\sqrt{2}} = \\cos \\angle Y'OZ \\implies \\angle Y'OZ = 45^\\circ.\\] So $\\angle YOZ = \\angle XOY – \\angle XOZ = 60 ^\\circ – 30 ^\\circ = 30 ^\\circ, \\angle Y'OY = \\angle Y'OZ + \\angle YOZ = 45^\\circ + 30 ^\\circ = 75^\\circ.$ Points $Y$ and $Y'$ are symmetric with respect to $OM.$ Case 1 \\[\\angle YOA = \\frac{90^\\circ – 75^\\circ}{2} = 7.5^\\circ, \\angle ZOA = 30^\\circ + 7.5^\\circ = 37.5^\\circ, \\angle XOA = 60^\\circ + 7.5^\\circ = 67.5^\\circ .\\] \\[1 – x = \\left(\\sqrt{1 – \\frac{x}{2}} \\right)^2– \\left(\\sqrt{\\frac {x}{2}}\\right)^2 = \\sin^2 \\angle XOA – \\cos^2 \\angle XOA = –\\cos 2 \\angle XOA = –\\cos 135^\\circ,\\] \\[1 – y = \\cos 15^\\circ, 1 – z = \\cos 75^\\circ \\implies \\left[ (1–x)(1–y)(1–z) \\right]^2 = \\left[ \\sin 45^\\circ \\cdot \\cos 15^\\circ \\cdot \\sin 15^\\circ \\right]^2 =\\] \\[=\\left[ \\frac {\\sin 45^\\circ \\cdot \\sin 30^\\circ}{2} \\right]^2 = \\frac {1}{32} \\implies \\textbf{033}.\\] Case 2 \\[\\angle Y_1 OA = \\frac{90^\\circ + 75^\\circ}{2} = 82.5^\\circ, \\angle Z_1 OA = 82.5^\\circ – 30^\\circ = 52.5^\\circ, \\angle X_1 OA = 82.5^\\circ – 60^\\circ = 22.5^\\circ \\implies \\textbf{033}.\\] vladimir.shelomovskii@gmail.com, vvsss",
"In given equations, $0 \\leq x,y,z \\leq 2,$ so we define some points: \\[\\bar {O} = (0, 0), \\bar {A} = (1, 0), \\bar{M} = \\left(\\frac {1}{\\sqrt{2}},\\frac {1}{\\sqrt{2}}\\right),\\] \\[\\bar {X} = \\left(\\sqrt {\\frac {x}{2}}, \\sqrt{1 – \\frac{x}{2}}\\right), \\bar {Y'} = \\left(\\sqrt {\\frac {y}{2}}, \\sqrt{1 – \\frac{y}{2}}\\right),\\] \\[\\bar {Y} = \\left(\\sqrt {1 – \\frac{y}{2}},\\sqrt{\\frac {y}{2}}\\right), \\bar {Z} = \\left(\\sqrt {1 – \\frac{z}{2}},\\sqrt{\\frac {z}{2}}\\right).\\] Notice, that \\[\\mid \\vec {AO} \\mid = \\mid \\vec {MO} \\mid = \\mid \\vec {XO} \\mid =\\mid \\vec {YO} \\mid = \\mid \\vec {Y'O} \\mid =\\mid \\vec {ZO} \\mid = 1\\] and each points lies in the first quadrant. We use given equations and get some scalar products: \\[(\\vec {XO} \\cdot \\vec {YO}) = \\frac {1}{2} = \\cos \\angle XOY \\implies \\angle XOY = 60 ^\\circ,\\] \\[(\\vec {XO} \\cdot \\vec {ZO}) = \\frac {\\sqrt{3}}{2} = \\cos \\angle XOZ \\implies \\angle XOZ = 30^\\circ,\\] \\[(\\vec {Y'O} \\cdot \\vec {ZO}) = \\frac {1}{\\sqrt{2}} = \\cos \\angle Y'OZ \\implies \\angle Y'OZ = 45^\\circ.\\] So $\\angle YOZ = \\angle XOY – \\angle XOZ = 60 ^\\circ – 30 ^\\circ = 30 ^\\circ, \\angle Y'OY = \\angle Y'OZ + \\angle YOZ = 45^\\circ + 30 ^\\circ = 75^\\circ.$ Points $Y$ and $Y'$ are symmetric with respect to $OM.$ Case 1 \\[\\angle YOA = \\frac{90^\\circ – 75^\\circ}{2} = 7.5^\\circ, \\angle ZOA = 30^\\circ + 7.5^\\circ = 37.5^\\circ, \\angle XOA = 60^\\circ + 7.5^\\circ = 67.5^\\circ .\\] \\[1 – x = \\left(\\sqrt{1 – \\frac{x}{2}} \\right)^2– \\left(\\sqrt{\\frac {x}{2}}\\right)^2 = \\sin^2 \\angle XOA – \\cos^2 \\angle XOA = –\\cos 2 \\angle XOA = –\\cos 135^\\circ,\\] \\[1 – y = \\cos 15^\\circ, 1 – z = \\cos 75^\\circ \\implies \\left[ (1–x)(1–y)(1–z) \\right]^2 = \\left[ \\sin 45^\\circ \\cdot \\cos 15^\\circ \\cdot \\sin 15^\\circ \\right]^2 =\\] \\[=\\left[ \\frac {\\sin 45^\\circ \\cdot \\sin 30^\\circ}{2} \\right]^2 = \\frac {1}{32} \\implies \\textbf{033}.\\] Case 2 \\[\\angle Y_1 OA = \\frac{90^\\circ + 75^\\circ}{2} = 82.5^\\circ, \\angle Z_1 OA = 82.5^\\circ – 30^\\circ = 52.5^\\circ, \\angle X_1 OA = 82.5^\\circ – 60^\\circ = 22.5^\\circ \\implies \\textbf{033}.\\] vladimir.shelomovskii@gmail.com, vvsss"
] |
2022-II-1
| 2,022
| 1
|
Adults made up $\frac5{12}$ of the crowd of people at a concert. After a bus carrying $50$ more people arrived, adults made up $\frac{11}{25}$ of the people at the concert. Find the minimum number of adults who could have been at the concert after the bus arrived.
| 154
|
II
|
[
"Let $x$ be the number of people at the party before the bus arrives. We know that $x\\equiv 0\\pmod {12}$, as $\\frac{5}{12}$ of people at the party before the bus arrives are adults. Similarly, we know that $x + 50 \\equiv 0 \\pmod{25}$, as $\\frac{11}{25}$ of the people at the party are adults after the bus arrives. $x + 50 \\equiv 0 \\pmod{25}$ can be reduced to $x \\equiv 0 \\pmod{25}$, and since we are looking for the minimum amount of people, $x$ is $300$. That means there are $350$ people at the party after the bus arrives, and thus there are $350 \\cdot \\frac{11}{25} = 154 adults at the party. ~eamo",
"Since at the beginning, adults make up $\\frac{5}{12}$ of the concert, the amount of people must be a multiple of 12. Call the amount of people in the beginning $x$.Then $x$ must be divisible by 12, in other words: $x$ must be a multiple of 12. Since after 50 more people arrived, adults make up $\\frac{11}{25}$ of the concert, $x+50$ is a multiple of 25. This means $x+50$ must be a multiple of 5. Notice that if a number is divisible by 5, it must end with a 0 or 5. Since 5 is impossible (obviously, since multiples of 12 end in 2, 4, 6, 8, 0,...), $x$ must end in 0. Notice that the multiples of 12 that end in 0 are: 60, 120, 180, etc.. By trying out, you can clearly see that $x=300$ is the minimum number of people at the concert. So therefore, after 50 more people arrive, there are $300+50=350$ people at the concert, and the number of adults is $350*\\frac{11}{25}=154$. Therefore the answer is $154. I know this solution is kind of lame, but this is still pretty straightforward. This solution is very similar to the first one, though. ~hastapasta",
"Since at the beginning, adults make up $\\frac{5}{12}$ of the concert, the amount of people must be a multiple of 12. Call the amount of people in the beginning $x$.Then $x$ must be divisible by 12, in other words: $x$ must be a multiple of 12. Since after 50 more people arrived, adults make up $\\frac{11}{25}$ of the concert, $x+50$ is a multiple of 25. This means $x+50$ must be a multiple of 5. Notice that if a number is divisible by 5, it must end with a 0 or 5. Since 5 is impossible (obviously, since multiples of 12 end in 2, 4, 6, 8, 0,...), $x$ must end in 0. Notice that the multiples of 12 that end in 0 are: 60, 120, 180, etc.. By trying out, you can clearly see that $x=300$ is the minimum number of people at the concert. So therefore, after 50 more people arrive, there are $300+50=350$ people at the concert, and the number of adults is $350*\\frac{11}{25}=154$. Therefore the answer is $154. I know this solution is kind of lame, but this is still pretty straightforward. This solution is very similar to the first one, though. ~hastapasta",
"Let $a$ be the number of adults before the bus arrived and $x$ be the total number of people at the concert. So, $\\frac{a}{x}=\\frac{5}{12}$. Solving for $x$ in terms of $a$, $x = \\frac{12}{5}a$. After the bus arrives, let's say there are an additional $y$ adults out of the 50 more people who enter the concert. From that, we get $\\frac{a+y}{x+50}=\\frac{11}{25}$. Replacing $x$ with the value of $a$, the second equation becomes $\\frac{a+y}{\\frac{12}{5}a+50}=\\frac{11}{25}$. By cross-multiplying and simplifying, we get that $25(y-22)=\\frac{7a}{5}$. Observe that we must make sure $y-22$ is positive and divisible by $7$ to have an integer value of $a$. The smallest possible value of $y$ that satisfies this conditions is $29$. Plugging this into the equation, $a = 125$. The question asks for the minimum number of adults that are there after the bus arrives, which is $a+y$. Thus, the answer is simply $125+29=154. ~mathical8"
] |
2022-II-2
| 2,022
| 2
|
Azar, Carl, Jon, and Sergey are the four players left in a singles tennis tournament. They are randomly assigned opponents in the semifinal matches, and the winners of those matches play each other in the final match to determine the winner of the tournament. When Azar plays Carl, Azar will win the match with probability $\frac23$ . When either Azar or Carl plays either Jon or Sergey, Azar or Carl will win the match with probability $\frac34$ . Assume that outcomes of different matches are independent. The probability that Carl will win the tournament is $\frac{p}{q}$ , where $p$ and $q$ are relatively prime positive integers. Find $p+q$ .
| 125
|
II
|
[
"Let $A$ be Azar, $C$ be Carl, $J$ be Jon, and $S$ be Sergey. The $4$ circles represent the $4$ players, and the arrow is from the winner to the loser with the winning probability as the label. This problem can be solved by using $2$ cases. $\\textbf{Case 1:}$ $C$'s opponent for the semifinal is $A$ The probability $C$'s opponent is $A$ is $\\frac13$. Therefore the probability $C$ wins the semifinal in this case is $\\frac13 \\cdot \\frac13$. The other semifinal game is played between $J$ and $S$, it doesn't matter who wins because $C$ has the same probability of winning either one. The probability of $C$ winning in the final is $\\frac34$, so the probability of $C$ winning the tournament in case 1 is $\\frac13 \\cdot \\frac13 \\cdot \\frac34$ $\\textbf{Case 2:}$ $C$'s opponent for the semifinal is $J$ or $S$ It doesn't matter if $C$'s opponent is $J$ or $S$ because $C$ has the same probability of winning either one. The probability $C$'s opponent is $J$ or $S$ is $\\frac23$. Therefore the probability $C$ wins the semifinal in this case is $\\frac23 \\cdot \\frac34$. The other semifinal game is played between $A$ and $J$ or $S$. In this case it matters who wins in the other semifinal game because the probability of $C$ winning $A$ and $J$ or $S$ is different. $\\textbf{Case 2.1:}$ $C$'s opponent for the final is $A$ For this to happen, $A$ must have won $J$ or $S$ in the semifinal, the probability is $\\frac34$. Therefore, the probability that $C$ won $A$ in the final is $\\frac34 \\cdot \\frac13$. $\\textbf{Case 2.2:}$ $C$'s opponent for the final is $J$ or $S$ For this to happen, $J$ or $S$ must have won $A$ in the semifinal, the probability is $\\frac14$. Therefore, the probability that $C$ won $J$ or $S$ in the final is $\\frac14 \\cdot \\frac34$. In Case 2 the probability of $C$ winning the tournament is $\\frac23 \\cdot \\frac34 \\cdot (\\frac34 \\cdot \\frac13 + \\frac14 \\cdot \\frac34)$ Adding case 1 and case 2 together we get $\\frac13 \\cdot \\frac13 \\cdot \\frac34 + \\frac23 \\cdot \\frac34 \\cdot (\\frac34 \\cdot \\frac13 + \\frac14 \\cdot \\frac34) = \\frac{29}{96},$ so the answer is $29 + 96 = \\textbf{125}. ~isabelchen"
] |
2022-II-3
| 2,022
| 3
|
A right square pyramid with volume $54$ has a base with side length $6.$ The five vertices of the pyramid all lie on a sphere with radius $\frac mn$ , where $m$ and $n$ are relatively prime positive integers. Find $m+n$ .
| 21
|
II
|
[
"Although I can't draw the exact picture of this problem, but it is quite easy to imagine that four vertices of the base of this pyramid is on a circle (Radius $\\frac{6}{\\sqrt{2}} = 3\\sqrt{2}$). Since all five vertices are on the sphere, the distances of the spherical center and the vertices are the same: $l$. Because of the symmetrical property of the pyramid, we can imagine that the line of the apex and the (sphere's) center will intersect the square at the (base's) center. Since the volume is $54 = \\frac{1}{3} \\cdot S \\cdot h = \\frac{1}{3} \\cdot 6^2 \\cdot h$, where $h=\\frac{9}{2}$ is the height of this pyramid, we have: $l^2=\\left(\\frac{9}{2}-l\\right)^2+\\left(3\\sqrt{2}\\right)^2$ according to the Pythagorean theorem. Solve this equation will give us $l = \\frac{17}{4},$ therefore $m+n=021",
"To start, we find the height of the pyramid. By the volume of a pyramid formula, we have \\[\\frac13 \\cdot 6^2 \\cdot h=54 \\implies h=\\frac92.\\] Next, let us find the length of the non-base sides of the pyramid. By the Pythagorean Theorem, noting that the distance from one vertex of the base to the center of the base is $\\frac12 \\cdot 6\\sqrt2=3\\sqrt2$, we have \\[x=\\sqrt{\\left(\\frac92\\right)^2+(3\\sqrt2)^2}=\\sqrt{\\frac{153}4}=\\frac{3\\sqrt{17}}2.\\] Taking the cross section of the pyramid and transforming the problem into $2$-d, it suffices to find the radius of the circumcircle of a triangle of side lengths $\\frac{3\\sqrt{17}}2$, $\\frac{3\\sqrt{17}}2$, $6\\sqrt2$. This turns out to be easy by the formula $R=\\frac{abc}{4A}$, and through computing this value (the work has been left out) we find that $R=\\frac{17}4$, so our answer is $\\textbf{021}. ~A1001",
"By the volume of a pyramid formula, we have that the height of the pyramid is $\\frac{9}{2}$. Since the base is a square with side length 6, the simplest way to place it in the coordinate plane is to put the center of the square at the origin and let the base be on the $xy$ plane. Then, the vertices of the base would be $(3,3,0), (3,-3,0), (-3,3,0), (-3,-3,0)$ in some order. Also, let the vertex be $(0,0,\\frac{9}{2})$. Recall that the formula for a sphere is $(x-a)^2+(y-b)^2+(z-c)^2=r^2$ where the center is $(a,b,c)$ and the radius is $r$. Symmetry gives that $a=b=0$. Plug in $(3,3,0)$ and $(0,0,\\frac{9}{2})$ and you get the system of equation $18+c^2=r^2$ $(\\frac{9}{2}-c)^2=r^2$ Solving gives $c=1/4$ and $r=17/4$, so our answer is $17+4=021.~Ddk001",
"By the volume of a pyramid formula, we have that the height of the pyramid is $\\frac{9}{2}$. Since the base is a square with side length 6, the simplest way to place it in the coordinate plane is to put the center of the square at the origin and let the base be on the $xy$ plane. Then, the vertices of the base would be $(3,3,0), (3,-3,0), (-3,3,0), (-3,-3,0)$ in some order. Also, let the vertex be $(0,0,\\frac{9}{2})$. Recall that the formula for a sphere is $(x-a)^2+(y-b)^2+(z-c)^2=r^2$ where the center is $(a,b,c)$ and the radius is $r$. Symmetry gives that $a=b=0$. Plug in $(3,3,0)$ and $(0,0,\\frac{9}{2})$ and you get the system of equation $18+c^2=r^2$ $(\\frac{9}{2}-c)^2=r^2$ Solving gives $c=1/4$ and $r=17/4$, so our answer is $17+4=021.~Ddk001",
"We know that the volume of a square pyramid is $\\frac{1}{3}\\cdot{s^2}\\cdot{h}$. The volume of the pyramid is $54$ and the square's side length is $6$. Plugging the information back into the formula, we get \\[\\frac{1}{3} \\cdot 6^2 \\cdot h=54 \\implies h=\\frac{9}{2}\\]. By Pythagorean theorem, we can say the diagonal of the square is \\[\\sqrt{6^2+6^2}=6\\sqrt{2}\\]. If we draw a line going through the poles of the sphere, we can see that the line perpendicularly bisects the square's diagonal. If we take the cross-section of the sphere making the problem $2$-d, and call the radius of the sphere $r$, by the power of point, we can say \\[\\frac{9}{2}\\cdot(2r-\\frac{9}{2})=3\\sqrt2\\cdot{3\\sqrt2}\\] solving this equation gives you \\[r=\\frac{17}{4}\\] so the answer is $17+4=021. ~Lentarot",
"We know that the volume of a square pyramid is $\\frac{1}{3}\\cdot{s^2}\\cdot{h}$. The volume of the pyramid is $54$ and the square's side length is $6$. Plugging the information back into the formula, we get \\[\\frac{1}{3} \\cdot 6^2 \\cdot h=54 \\implies h=\\frac{9}{2}\\]. By Pythagorean theorem, we can say the diagonal of the square is \\[\\sqrt{6^2+6^2}=6\\sqrt{2}\\]. If we draw a line going through the poles of the sphere, we can see that the line perpendicularly bisects the square's diagonal. If we take the cross-section of the sphere making the problem $2$-d, and call the radius of the sphere $r$, by the power of point, we can say \\[\\frac{9}{2}\\cdot(2r-\\frac{9}{2})=3\\sqrt2\\cdot{3\\sqrt2}\\] solving this equation gives you \\[r=\\frac{17}{4}\\] so the answer is $17+4=021. ~Lentarot"
] |
2022-II-4
| 2,022
| 4
|
There is a positive real number $x$ not equal to either $\tfrac{1}{20}$ or $\tfrac{1}{2}$ such that \[\log_{20x} (22x)=\log_{2x} (202x).\] The value $\log_{20x} (22x)$ can be written as $\log_{10} (\tfrac{m}{n})$ , where $m$ and $n$ are relatively prime positive integers. Find $m+n$ .
| 112
|
II
|
[
"Define $a$ to be $\\log_{20x} (22x) = \\log_{2x} (202x)$, what we are looking for. Then, by the definition of the logarithm, \\[\\begin{cases} (20x)^{a} &= 22x \\\\ (2x)^{a} &= 202x. \\end{cases}\\] Dividing the first equation by the second equation gives us $10^a = \\frac{11}{101}$, so by the definition of logs, $a = \\log_{10} \\frac{11}{101}$. This is what the problem asked for, so the fraction $\\frac{11}{101}$ gives us $m+n = 112. ~ihatemath123",
"We could assume a variable $v$ which equals to both $\\log_{20x} (22x)$ and $\\log_{2x} (202x)$. So that $(20x)^v=22x \\textcircled{1}$ and $(2x)^v=202x \\textcircled{2}$ Express $\\textcircled{1}$ as: $(20x)^v=(2x \\cdot 10)^v=(2x)^v \\cdot \\left(10^v\\right)=22x \\textcircled{3}$ Substitute $\\textcircled{{2}}$ to $\\textcircled{3}$: $202x \\cdot (10^v)=22x$ Thus, $v=\\log_{10} \\left(\\frac{22x}{202x}\\right)= \\log_{10} \\left(\\frac{11}{101}\\right)$, where $m=11$ and $n=101$. Therefore, $m+n = 112.",
"We have \\begin{align*} \\log_{20x} (22x) & = \\frac{\\log_k 22x}{\\log_k 20x} \\\\ & = \\frac{\\log_k x + \\log_k 22}{\\log_k x + \\log_k 20} . \\end{align*} We have \\begin{align*} \\log_{2x} (202x) & = \\frac{\\log_k 202x}{\\log_k 2x} \\\\ & = \\frac{\\log_k x + \\log_k 202 }{\\log_k x + \\log_k 2} . \\end{align*} Because $\\log_{20x} (22x)=\\log_{2x} (202x)$, we get \\[ \\frac{\\log_k x + \\log_k 22}{\\log_k x + \\log_k 20} = \\frac{\\log_k x + \\log_k 202 }{\\log_k x + \\log_k 2} . \\] We denote this common value as $\\lambda$. By solving the equality $\\frac{\\log_k x + \\log_k 22}{\\log_k x + \\log_k 20} = \\lambda$, we get $\\log_k x = \\frac{\\log_k 22 - \\lambda \\log_k 20}{\\lambda - 1}$. By solving the equality $\\frac{\\log_k x + \\log_k 202 }{\\log_k x + \\log_k 2} = \\lambda$, we get $\\log_k x = \\frac{\\log_k 202 - \\lambda \\log_k 2}{\\lambda - 1}$. By equating these two equations, we get \\[ \\frac{\\log_k 22 - \\lambda \\log_k 20}{\\lambda - 1} = \\frac{\\log_k 202 - \\lambda \\log_k 2}{\\lambda - 1} . \\] Therefore, \\begin{align*} \\log_{20x} (22x) & = \\lambda \\\\ & = \\frac{\\log_k 22 - \\log_k 202}{\\log_k 20 - \\log_k 2} \\\\ & = \\frac{\\log_k \\frac{11}{101}}{\\log_k 10} \\\\ & = \\log_{10} \\frac{11}{101} . \\end{align*} Therefore, the answer is $11 + 101 = \\textbf{112}. ~Steven Chen (www.professorchenedu.com)",
"Let $a$ be the exponent such that $(20x)^a = 22x$ and $(2x)^a = 202x$. Dividing, we get \\begin{align*} \\dfrac{(20x)^a}{(2x)^a} &= \\dfrac{22x}{202x}. \\\\ \\left(\\dfrac{20x}{2x}\\right)^a &= \\dfrac{22x}{202x}. \\\\ 10^a &= \\dfrac{11}{101}. \\\\ \\end{align*} Thus, we see that $\\log_{10} \\left(\\dfrac{11}{101}\\right) = a = \\log_{20x} 22x$, so the answer is $11 + 101 = 112. ~A_MatheMagician",
"Let $a$ be the exponent such that $(20x)^a = 22x$ and $(2x)^a = 202x$. Dividing, we get \\begin{align*} \\dfrac{(20x)^a}{(2x)^a} &= \\dfrac{22x}{202x}. \\\\ \\left(\\dfrac{20x}{2x}\\right)^a &= \\dfrac{22x}{202x}. \\\\ 10^a &= \\dfrac{11}{101}. \\\\ \\end{align*} Thus, we see that $\\log_{10} \\left(\\dfrac{11}{101}\\right) = a = \\log_{20x} 22x$, so the answer is $11 + 101 = 112. ~A_MatheMagician",
"By the change of base rule, we have $\\frac{\\log 22x}{\\log 20x}=\\frac{\\log 202x}{\\log 2x}$, or $\\frac{\\log 22 +\\log x}{\\log 20 +\\log x}=\\frac{\\log 202 +\\log x}{\\log 2 +\\log x}=k$. We also know that if $a/b=c/d$, then this also equals $\\frac{a-c}{b-d}$. We use this identity and find that $k=\\frac{\\log 202 -\\log 22}{\\log 2 -\\log 20}=-\\log\\frac{202}{22}=\\log\\frac{11}{101}$. The requested sum is $11+101=112 ~MathIsFun286",
"By change of base formula, \\[\\frac{\\log_{2x} 22x}{\\log_{2x} 20x} = \\frac{{\\log_{2x} 11} + 1}{{\\log_{2x} 10} + 1} = {\\log_{2x} 101} + 1\\] \\[\\log_{2x} 11 + 1 = (\\log_{2x} 10)(\\log_{2x} 101) + \\log_{2x} 10 + 1\\] \\[\\frac{\\log_{2x} \\frac{11}{1010}}{\\log_{2x} 10} = \\log_{2x} 101\\] \\[\\log_{10} {\\frac{11}{1010}} = \\log_{2x} 101\\] \\[\\log_{10} {\\frac{11}{1010}} + 1 = \\log_{2x} 101 + 1 = \\log_{2x} 202x = \\log_{20x} {22x}\\] Thus, \\[\\log_{20x} 22x = \\log_{10} \\left( \\frac{11}{1010} \\times 10 \\right) = \\log_{10} \\frac{11}{101}\\] The requested answer is $11 + 101 = 112. ~ adam_zheng"
] |
2022-II-5
| 2,022
| 5
|
Twenty distinct points are marked on a circle and labeled $1$ through $20$ in clockwise order. A line segment is drawn between every pair of points whose labels differ by a prime number. Find the number of triangles formed whose vertices are among the original $20$ points.
| 72
|
II
|
[
"Let $a$, $b$, and $c$ be the vertex of a triangle that satisfies this problem, where $a > b > c$. \\[a - b = p_1\\] \\[b - c = p_2\\] \\[a - c = p_3\\] $p_3 = a - c = a - b + b - c = p_1 + p_2$. Because $p_3$ is the sum of two primes, $p_1$ and $p_2$, $p_1$ or $p_2$ must be $2$. Let $p_1 = 2$, then $p_3 = p_2 + 2$. There are only $8$ primes less than $20$: $2, 3, 5, 7, 11, 13, 17, 19$. Only $3, 5, 11, 17$ plus $2$ equals another prime. $p_2 \\in \\{ 3, 5, 11, 17 \\}$. Once $a$ is determined, $a = b+2$ and $b = c + p_2$. There are $18$ values of $a$ where $b+2 \\le 20$, and $4$ values of $p_2$. Therefore the answer is $18 \\cdot 4 = \\textbf{072}.",
"As above, we must deduce that the sum of two primes must be equal to the third prime. Then, we can finish the solution using casework. If the primes are $2,3,5$, then the smallest number can range between $1$ and $15$. If the primes are $2,5,7$, then the smallest number can range between $1$ and $13$. If the primes are $2,11,13$, then the smallest number can range between $1$ and $7$. If the primes are $2,17,19$, then the smallest number can only be $1$. Adding all cases gets $15+13+7+1=36$. However, due to the commutative property, we must multiply this by 2. For example, in the $2,17,19$ case the numbers can be $1,3,20$ or $1,18,20$. Therefore the answer is $36\\cdot2=072."
] |
2022-II-6
| 2,022
| 6
|
Let $x_1\leq x_2\leq \cdots\leq x_{100}$ be real numbers such that $|x_1| + |x_2| + \cdots + |x_{100}| = 1$ and $x_1 + x_2 + \cdots + x_{100} = 0$ . Among all such $100$ -tuples of numbers, the greatest value that $x_{76} - x_{16}$ can achieve is $\tfrac mn$ , where $m$ and $n$ are relatively prime positive integers. Find $m+n$ .
| 841
|
II
|
[
"To find the greatest value of $x_{76} - x_{16}$, $x_{76}$ must be as large as possible, and $x_{16}$ must be as small as possible. If $x_{76}$ is as large as possible, $x_{76} = x_{77} = x_{78} = \\dots = x_{100} > 0$. If $x_{16}$ is as small as possible, $x_{16} = x_{15} = x_{14} = \\dots = x_{1} < 0$. The other numbers between $x_{16}$ and $x_{76}$ equal to $0$. Let $a = x_{76}$, $b = x_{16}$. Substituting $a$ and $b$ into $|x_1| + |x_2| + \\cdots + |x_{100}| = 1$ and $x_1 + x_2 + \\cdots + x_{100} = 0$ we get: \\[25a - 16b = 1\\] \\[25a + 16b = 0\\] $a = \\frac{1}{50}$, $b = -\\frac{1}{32}$ $x_{76} - x_{16} = a - b = \\frac{1}{50} + \\frac{1}{32} = \\frac{41}{800}$. $m+n = \\textbf{841} ~isabelchen",
"Define $s_N$ to be the sum of all the negatives, and $s_P$ to be the sum of all the positives. Since the sum of the absolute values of all the numbers is $1$, $|s_N|+|s_P|=1$. Since the sum of all the numbers is $0$, $s_N=-s_P\\implies |s_N|=|s_P|$. Therefore, $|s_N|=|s_P|=\\frac 12$, so $s_N=-\\frac 12$ and $s_P=\\frac 12$ since $s_N$ is negative and $s_P$ is positive. To maximize $x_{76}-x_{16}$, we need to make $x_{16}$ as small of a negative as possible, and $x_{76}$ as large of a positive as possible. Note that $x_{76}+x_{77}+\\cdots+x_{100}=\\frac 12$ is greater than or equal to $25x_{76}$ because the numbers are in increasing order. Similarly, $x_{1}+x_{2}+\\cdots+x_{16}=-\\frac 12$ is less than or equal to $16x_{16}$. So we now know that $\\frac 1{50}$ is the best we can do for $x_{76}$, and $-\\frac 1{32}$ is the least we can do for $x_{16}$. Finally, the maximum value of $x_{76}-x_{16}=\\frac 1{50}+\\frac 1{32}=\\frac{41}{800}$, so the answer is $841 works.) ~inventivedant",
"Because the absolute value sum of all the numbers is $1$, and the normal sum of all the numbers is $0$, the positive numbers must add to $\\frac12$ and negative ones must add to $-\\dfrac12$. To maximize $x_{76} - x_{16}$, we must make $x_{76}$ as big as possible and $x_{16}$ as small as possible. We can do this by making $x_1 + x_2 + x_3 \\dots x_{16} = -\\dfrac{1}{2}$, where $x_1 = x_2 = x_3 = \\dots = x_{16}$ (because that makes $x_{16}$ the smallest possible value), and $x_{76} + x_{77} + x_{78} + \\dots + x_{100} = \\dfrac{1}{2}$, where similarly $x_{76} = x_{77} = \\dots = x_{100}$ (because it makes $x_{76}$ its biggest possible value.) That means $16(x_{16}) = -\\dfrac{1}{2}$, and $25(x_{76}) = \\dfrac{1}{2}$. $x_{16} = -\\dfrac{1}{32}$ and $x_{76} = \\dfrac{1}{50}$, and subtracting them $\\dfrac{1}{50} - \\left( -\\dfrac{1}{32}\\right) = \\dfrac{41}{800}$. $41 + 800 = 841$. ~heheman"
] |
2022-II-7
| 2,022
| 7
|
A circle with radius $6$ is externally tangent to a circle with radius $24$ . Find the area of the triangular region bounded by the three common tangent lines of these two circles.
| 192
|
II
|
[
"[asy] //Created by isabelchen size(12cm, 12cm); draw(circle((0,0),24)); draw(circle((30,0),6)); draw((72/5, 96/5) -- (40,0)); draw((72/5, -96/5) -- (40,0)); draw((24, 12) -- (24, -12)); draw((0, 0) -- (40, 0)); draw((72/5, 96/5) -- (0,0)); draw((168/5, 24/5) -- (30,0)); draw((54/5, 72/5) -- (30,0)); dot((72/5, 96/5)); label(\"$A$\",(72/5, 96/5),NE); dot((168/5, 24/5)); label(\"$B$\",(168/5, 24/5),NE); dot((24,0)); label(\"$C$\",(24,0),NW); dot((40, 0)); label(\"$D$\",(40, 0),NE); dot((24, 12)); label(\"$E$\",(24, 12),NE); dot((24, -12)); label(\"$F$\",(24, -12),SE); dot((54/5, 72/5)); label(\"$G$\",(54/5, 72/5),NW); dot((0, 0)); label(\"$O_1$\",(0, 0),S); dot((30, 0)); label(\"$O_2$\",(30, 0),S); [/asy] $r_1 = O_1A = 24$, $r_2 = O_2B = 6$, $AG = BO_2 = r_2 = 6$, $O_1G = r_1 - r_2 = 24 - 6 = 18$, $O_1O_2 = r_1 + r_2 = 30$ $\\triangle O_2BD \\sim \\triangle O_1GO_2$, $\\frac{O_2D}{O_1O_2} = \\frac{BO_2}{GO_1}$, $\\frac{O_2D}{30} = \\frac{6}{18}$, $O_2D = 10$ $CD = O_2D + r_2 = 10 + 6 = 16$, $EF = 2EC = EA + EB = AB = GO_2 = \\sqrt{(O_1O_2)^2-(O_1G)^2} = \\sqrt{30^2-18^2} = 24$ $DEF = \\frac12 \\cdot EF \\cdot CD = \\frac12 \\cdot 24 \\cdot 16 = \\textbf{192} with our known values will lead us to the same solution. mathboy282",
"Let the center of the circle with radius $6$ be labeled $A$ and the center of the circle with radius $24$ be labeled $B$. Drop perpendiculars on the same side of line $AB$ from $A$ and $B$ to each of the tangents at points $C$ and $D$, respectively. Then, let line $AB$ intersect the two diagonal tangents at point $P$. Since $\\triangle{APC} \\sim \\triangle{BPD}$, we have \\[\\frac{AP}{AP+30}=\\frac14 \\implies AP=10.\\] Next, throw everything on a coordinate plane with $A=(0, 0)$ and $B = (30, 0)$. Then, $P = (-10, 0)$, and if $C = (x, y)$, we have \\[(x+10)^2+y^2=64,\\] \\[x^2+y^2=36.\\] Combining these and solving, we get $(x, y)=\\left(-\\frac{18}5, \\frac{24}5\\right)$. Notice now that $P$, $C$, and the intersections of the lines $x=6$ (the vertical tangent) with the tangent containing these points are collinear, and thus every slope between a pair of points will have the same slope, which in this case is $\\frac{-\\frac{18}5+10}{\\frac{24}5}=\\frac34$. Thus, the other two vertices of the desired triangle are $(6, 12)$ and $(6, -12)$. By the Shoelace Formula, the area of a triangle with coordinates $(-10, 0)$, $(6, 12)$, and $(6, -12)$ is \\[\\frac12|-120-0-72-72+0-120|=\\textbf{192}.\\] ~A1001",
"(Taking diagram names from Solution 1. Also say the line that passes through $O_1$ and is parallel to line EF, call the points of intersection of that line and the circumference of circle $O_1$ points $X$ and $Y$.) First notice that $DO_1$ is a straight line because $DXY$ is an isosceles triangle(or you can realize it by symmetry). That means, because $DO_1$ is a straight line, so angle $BDO_2$ = angle $ADO_1,$ triangle $ADO_1$ is similar to triangle $BDO_2$. Also name $DO_2 = x$. By our similar triangles, $\\frac{BO_2}{AO_1} = \\frac{1}{4} = \\frac{x}{x+30}$. Solving we get $x = 10 = DO_2$. Pythagorean Theorem on triangle $DBO_2$ shows $BD = \\sqrt{10^2 - 6^2} = 8$. By similar triangles, $DA = 4 \\cdot 8 = 32$ which means $AB = DA - DB = 32 - 8 = 24$. Because $BE = CE = AE, AB = 2 \\cdot BE = 24$. $BE = 12,$ which means $CE = 12$. $CD = DO_2$(its value found earlier in this solution) + $CO_2$ ($O_2$ 's radius) $= 10 + 6 = 16$. The area of $DEF$ is $\\frac{1}{2} \\cdot CD \\cdot EF = CD \\cdot CE$ (because $CE$ is $\\tfrac{1}{2}$ of $EF$) $= 16 \\cdot 12 = 192$. ~Professor Rat's solution, added by @heheman and edited by @megahertz13 and @Yrock for $\\LaTeX$.",
"[asy] //Created by isabelchen and edited by afly size(12cm, 12cm); draw(circle((0,0),24)); draw(circle((30,0),6)); draw((72/5, 96/5) -- (40,0)); draw((72/5, -96/5) -- (40,0)); draw((24, 12) -- (24, -12)); draw((0, 0) -- (40, 0)); draw((72/5, 96/5) -- (0,0)); draw((168/5, 24/5) -- (30,0)); draw((54/5, 72/5) -- (30,0)); draw((30,0)--(30,15/2)); dot((72/5, 96/5)); label(\"$A$\",(72/5, 96/5),NE); dot((168/5, 24/5)); label(\"$B$\",(168/5, 24/5),NE); dot((24,0)); label(\"$C$\",(24,0),NW); dot((40, 0)); label(\"$D$\",(40, 0),NE); dot((24, 12)); label(\"$E$\",(24, 12),NE); dot((24, -12)); label(\"$F$\",(24, -12),SE); dot((54/5, 72/5)); label(\"$G$\",(54/5, 72/5),NW); dot((0, 0)); label(\"$O_1$\",(0, 0),S); dot((30, 0)); label(\"$O_2$\",(30, 0),S); dot((30,15/2)); label(\"$H$\",(30,15/2),N); label(\"$x$\",(30,0)--(40,0),N); [/asy] First, we want to find $O_2D$. We know that $\\angle O_1AD = \\angle O_2BD = 90^{\\circ}$, so by AA similarity, $\\triangle O_1AD \\sim \\triangle O_2BD$. We want to find the length of $x$, and using the similar triangles, we write an equation: $\\frac{30 + x}{4} = x$. Solving, we get $x=10$. Therefore, $CD = 10 + 6 = 16$. Next, we find that using AA similarity, $\\triangle O_2BD \\sim \\triangle HO_2D \\sim \\triangle ECD$ and they are 3-4-5 triangles. We can quickly compute $EF = 2EC = 2 \\cdot \\left( \\frac{3}{4} \\cdot 16 \\right) = 2 \\cdot 12 = 24$. Therefore, the area is $\\frac{1}{2} \\cdot 16 \\cdot 24 = 192. ~afly",
"[asy] //Created by isabelchen and edited by afly size(12cm, 12cm); draw(circle((0,0),24)); draw(circle((30,0),6)); draw((72/5, 96/5) -- (40,0)); draw((72/5, -96/5) -- (40,0)); draw((24, 12) -- (24, -12)); draw((0, 0) -- (40, 0)); draw((72/5, 96/5) -- (0,0)); draw((168/5, 24/5) -- (30,0)); draw((54/5, 72/5) -- (30,0)); draw((30,0)--(30,15/2)); dot((72/5, 96/5)); label(\"$A$\",(72/5, 96/5),NE); dot((168/5, 24/5)); label(\"$B$\",(168/5, 24/5),NE); dot((24,0)); label(\"$C$\",(24,0),NW); dot((40, 0)); label(\"$D$\",(40, 0),NE); dot((24, 12)); label(\"$E$\",(24, 12),NE); dot((24, -12)); label(\"$F$\",(24, -12),SE); dot((54/5, 72/5)); label(\"$G$\",(54/5, 72/5),NW); dot((0, 0)); label(\"$O_1$\",(0, 0),S); dot((30, 0)); label(\"$O_2$\",(30, 0),S); dot((30,15/2)); label(\"$H$\",(30,15/2),N); label(\"$x$\",(30,0)--(40,0),N); [/asy] First, we want to find $O_2D$. We know that $\\angle O_1AD = \\angle O_2BD = 90^{\\circ}$, so by AA similarity, $\\triangle O_1AD \\sim \\triangle O_2BD$. We want to find the length of $x$, and using the similar triangles, we write an equation: $\\frac{30 + x}{4} = x$. Solving, we get $x=10$. Therefore, $CD = 10 + 6 = 16$. Next, we find that using AA similarity, $\\triangle O_2BD \\sim \\triangle HO_2D \\sim \\triangle ECD$ and they are 3-4-5 triangles. We can quickly compute $EF = 2EC = 2 \\cdot \\left( \\frac{3}{4} \\cdot 16 \\right) = 2 \\cdot 12 = 24$. Therefore, the area is $\\frac{1}{2} \\cdot 16 \\cdot 24 = 192. ~afly",
"The common internal tangent of the two circles intersects the two external tangents at their midpoints as the internal tangent is the radical axis of the two circles. Therefore, the length of the internal tangent is equal to the length of the external tangents which is $24$ by pythagorean theorem. Let $x$ denote the length of the extension of the external tangents such that they intersect. By similar triangles, we get \\[\\frac{x}{6} = \\frac{x+24}{24}\\] Therefore, $x = 8$. Our triangle has side lengths $8+12$, $24$, and $8+12$ making it a $20-20-24$ triangle. Dropping the altitude to the side with length $24$, we get the altitude has length $\\sqrt{20^2-12^2} = 16$. Therefore, the area of our triangle is $16\\cdot 12 = 192. ~ Vedoral",
"The common internal tangent of the two circles intersects the two external tangents at their midpoints as the internal tangent is the radical axis of the two circles. Therefore, the length of the internal tangent is equal to the length of the external tangents which is $24$ by pythagorean theorem. Let $x$ denote the length of the extension of the external tangents such that they intersect. By similar triangles, we get \\[\\frac{x}{6} = \\frac{x+24}{24}\\] Therefore, $x = 8$. Our triangle has side lengths $8+12$, $24$, and $8+12$ making it a $20-20-24$ triangle. Dropping the altitude to the side with length $24$, we get the altitude has length $\\sqrt{20^2-12^2} = 16$. Therefore, the area of our triangle is $16\\cdot 12 = 192. ~ Vedoral"
] |
2022-II-9
| 2,022
| 9
|
Let $\ell_A$ and $\ell_B$ be two distinct parallel lines. For positive integers $m$ and $n$ , distinct points $A_1, A_2, \allowbreak A_3, \allowbreak \ldots, \allowbreak A_m$ lie on $\ell_A$ , and distinct points $B_1, B_2, B_3, \ldots, B_n$ lie on $\ell_B$ . Additionally, when segments $\overline{A_iB_j}$ are drawn for all $i=1,2,3,\ldots, m$ and $j=1,\allowbreak 2,\allowbreak 3, \ldots, \allowbreak n$ , no point strictly between $\ell_A$ and $\ell_B$ lies on more than two of the segments. Find the number of bounded regions into which this figure divides the plane when $m=7$ and $n=5$ . The figure shows that there are 8 regions when $m=3$ and $n=2$ . [asy] import geometry; size(10cm); draw((-2,0)--(13,0)); draw((0,4)--(10,4)); label("$\ell_A$",(-2,0),W); label("$\ell_B$",(0,4),W); point A1=(0,0),A2=(5,0),A3=(11,0),B1=(2,4),B2=(8,4),I1=extension(B1,A2,A1,B2),I2=extension(B1,A3,A1,B2),I3=extension(B1,A3,A2,B2); draw(B1--A1--B2); draw(B1--A2--B2); draw(B1--A3--B2); label("$A_1$",A1,S); label("$A_2$",A2,S); label("$A_3$",A3,S); label("$B_1$",B1,N); label("$B_2$",B2,N); label("1",centroid(A1,B1,I1)); label("2",centroid(B1,I1,I3)); label("3",centroid(B1,B2,I3)); label("4",centroid(A1,A2,I1)); label("5",(A2+I1+I2+I3)/4); label("6",centroid(B2,I2,I3)); label("7",centroid(A2,A3,I2)); label("8",centroid(A3,B2,I2)); dot(A1); dot(A2); dot(A3); dot(B1); dot(B2); [/asy]
| 244
|
II
|
[
"We can use recursion to solve this problem: 1. Fix 7 points on $\\ell_A$, then put one point $B_1$ on $\\ell_B$. Now, introduce a function $f(x)$ that indicates the number of regions created, where x is the number of points on $\\ell_B$. For example, $f(1) = 6$ because there are 6 regions. 2. Now, put the second point $B_2$ on $\\ell_B$. Join $A_1~A_7$ and $B_2$ will create $7$ new regions (and we are not going to count them again), and split the existing regions. Let's focus on the spliting process: line segment formed between $B_2$ and $A_1$ intersect lines $\\overline{B_1A_2}$, $\\overline{B_1A_3}$, ..., $\\overline{B_1A_7}$ at $6$ points $\\Longrightarrow$ creating $6$ regions (we already count one region at first), then $5$ points $\\Longrightarrow$ creating $5$ regions (we already count one region at first), 4 points, etc. So, we have: \\[f(2) = f(1) + 7 + (6+5+...+1) = 34.\\] 3. If you still need one step to understand this: $A_1~A_7$ and $B_3$ will still create $7$ new regions. Intersecting \\[\\overline{A_2B_1}, \\overline{A_2B_2};\\] \\[\\overline{A_3B_1}, \\overline{A_3B_2};\\] \\[...\\] \\[\\overline{A_7B_1}, \\overline{A_7B_2}\\] at $12$ points, creating $12$ regions, etc. Thus, we have: \\[f(3) = f(2)+7+(12+10+8+...+2)=34+7+6\\cdot 7=83.\\] Yes, you might already notice that: \\[f(n+1) = f(n)+7+(6+5+...+1)\\cdot n = f(n) + 7 + 21n.\\] 5. Finally, we have $f(4) = 153$, and $f(5)=244$. Therefore, the answer is $244",
"We want to derive a general function $f(m,n)$ that indicates the number of bounded regions. Observing symmetry, we know this is a symmetric function about $m$ and $n$. Now let's focus on $f(m+1, n)-f(m, n)$, which is the difference caused by adding one point to the existing $m$ points of line $\\ell_A$. This new point, call it #m, when connected to point #1 on $\\ell_B$, crosses $m*(n-1)$ lines, thus making additional $m*(n-1)+1$ bounded regions; when connected to point #2 on $\\ell_B$, it crosses $m*(n-2)$ lines, thus making additional $m*(n-2)+1$ bounded regions; etc. By simple algebra/recursion methods, we see $f(m+1, n)-f(m, n)=m*\\frac{n(n-1)}{2} +n$ Notice $f(1,n)=n-1$. Not very difficult to figure out: $f(m, n)=\\frac{m(m-1)n(n-1)}{4} +mn-1$ The fact that $f(3,2)=8$ makes us more confident about the formula. Now plug in $m=5, n=7$, we get the final answer of $244.",
"Let some number of segments be constructed. We construct a new segment. We start from the straight line $l_B.$ WLOG from point $B_3.$ Segment will cross several existing segments (points $A,B,C,...$) and enter one of the points of the line $l_A (A_1).$ Each of these points adds exactly 1 new bounded region (yellow bounded regions). The exception is the only first segment $(A_1 B_1),$ which does not create any bounded region. Thus, the number of bounded regions is $1$ less than the number of points of intersection of the segments plus the number of points of arrival of the segments to $l_A.$ Each point of intersection of two segments is determined uniquely by the choice of pairs of points on each line. The number of such pairs is $\\dbinom{n}{2} \\cdot \\dbinom{m}{2}.$ Exactly one segment comes to each of the $n$ points of the line $l_A$ from each of the $m$ points of the line $l_B.$ The total number of arrivals is equal to $mn.$ Hence, the total number of bounded regions is $N = \\dbinom{n}{2} \\cdot \\dbinom{m}{2} + mn – 1.$ We plug in $m=5, n=7$, we get the final answer of $244. vladimir.shelomovskii@gmail.com, vvsss",
"When a new point is added to a line, the number of newly bounded regions it creates with each line segment will be one more than the number of intersection points the line makes with other lines. Case 1: If a new point $P$ is added to the right on a line when both lines have an equal amount of points. WLOG, let the point be on line $\\ell_A$. We consider the complement, where new lines don't intersect other line segments. Simply observing, we see that the only line segments that don't intersect with the new lines are lines attached to some point that a new line does not pass through. If we look at a series of points on line $\\ell_B$ from left to right and a line connects $P$ to an arbitrary point, then the lines formed with that point and with remaining points on the left of that point never intersect with the line with $P$. Let there be $s$ points on lines $\\ell_A$ and $\\ell_B$ before $P$ was added. For each of the $s$ points on $\\ell_B$, we subtract the total number of lines formed, which is $s^2$, not counting $P$. Considering all possible points on $\\ell_B$, we get $(s^2-s)+(s^2-2s)\\cdots(s^2-s^2)$ total intersections. However, for each of the lines, there is one more bounded region than number of intersections, so we add $s$. Simplifying, we get $s^3-s\\sum_{i=1}^{s}{i}+s\\Longrightarrow s(s^2-\\sum_{i=1}^{s}{i}+1)$. Note that this is only a recursion formula to find the number of new regions added for a new point $P$ added to $\\ell_A$. Case 2: If a new point $P$ is added to the right of a line that has one less point than the other line. Continuing on case one, let this point $P$ be on line $\\ell_B$. With similar reasoning, we see that the idea remains the same, except $s+1$ lines are formed with $P$ instead of just $s$ lines. Once again, each line from $P$ to a point on line $\\ell_A$ creates $s$ non-intersecting lines for that point and each point to its left. Subtracting from $s(s+1)$ lines and considering all possible lines created by $P$, we get $(s(s+1)-s)+(s(s+1)-2s)\\cdots(s(s+1)-s(s+1)$ intersections. However, the number of newly bounded regions is the number of intersections plus the number of points on line $\\ell_A$. Simplying, we get $s(s+1)^2-s\\sum_{i=1}^{s+1}{i}+(s+1)$ newly bounded regions. For the base case $s=2$ for both lines, there are $4$ bounded regions. Next, we plug in $s=2,3,4$ for both formulas and plug $s=5$ for the first formula to find the number of regions when $m=6$ and $n=5$. Notice that adding a final point on $\\ell_A$ is a variation of our Case 1. The only difference is for each of the $s$ lines formed by $P$, there are $s+1$ points that can form a non-intersecting line. Therefore, we are subtracting a factor of $s+1$ lines instead of $s$ lines from a total of $s(s+1)$ lines. However, the number of lines formed by $P$ remains the same so we still add $s$ at the end when considering intersection points. Thus, the recursive equation becomes $(s(s+1)-(s+1))+(s(s+1)-2(s+1))\\cdots(s(s+1)-s(s+1))+s\\Longrightarrow s^2(s+1)-(s+1)\\sum_{i=1}^{s}{i}+s$. Plugging $s=5$ into this formula and adding the values we obtained from the other formulas, the final answer is $4+4+9+12+22+28+45+55+65=244. ~Magnetoninja",
"When a new point is added to a line, the number of newly bounded regions it creates with each line segment will be one more than the number of intersection points the line makes with other lines. Case 1: If a new point $P$ is added to the right on a line when both lines have an equal amount of points. WLOG, let the point be on line $\\ell_A$. We consider the complement, where new lines don't intersect other line segments. Simply observing, we see that the only line segments that don't intersect with the new lines are lines attached to some point that a new line does not pass through. If we look at a series of points on line $\\ell_B$ from left to right and a line connects $P$ to an arbitrary point, then the lines formed with that point and with remaining points on the left of that point never intersect with the line with $P$. Let there be $s$ points on lines $\\ell_A$ and $\\ell_B$ before $P$ was added. For each of the $s$ points on $\\ell_B$, we subtract the total number of lines formed, which is $s^2$, not counting $P$. Considering all possible points on $\\ell_B$, we get $(s^2-s)+(s^2-2s)\\cdots(s^2-s^2)$ total intersections. However, for each of the lines, there is one more bounded region than number of intersections, so we add $s$. Simplifying, we get $s^3-s\\sum_{i=1}^{s}{i}+s\\Longrightarrow s(s^2-\\sum_{i=1}^{s}{i}+1)$. Note that this is only a recursion formula to find the number of new regions added for a new point $P$ added to $\\ell_A$. Case 2: If a new point $P$ is added to the right of a line that has one less point than the other line. Continuing on case one, let this point $P$ be on line $\\ell_B$. With similar reasoning, we see that the idea remains the same, except $s+1$ lines are formed with $P$ instead of just $s$ lines. Once again, each line from $P$ to a point on line $\\ell_A$ creates $s$ non-intersecting lines for that point and each point to its left. Subtracting from $s(s+1)$ lines and considering all possible lines created by $P$, we get $(s(s+1)-s)+(s(s+1)-2s)\\cdots(s(s+1)-s(s+1)$ intersections. However, the number of newly bounded regions is the number of intersections plus the number of points on line $\\ell_A$. Simplying, we get $s(s+1)^2-s\\sum_{i=1}^{s+1}{i}+(s+1)$ newly bounded regions. For the base case $s=2$ for both lines, there are $4$ bounded regions. Next, we plug in $s=2,3,4$ for both formulas and plug $s=5$ for the first formula to find the number of regions when $m=6$ and $n=5$. Notice that adding a final point on $\\ell_A$ is a variation of our Case 1. The only difference is for each of the $s$ lines formed by $P$, there are $s+1$ points that can form a non-intersecting line. Therefore, we are subtracting a factor of $s+1$ lines instead of $s$ lines from a total of $s(s+1)$ lines. However, the number of lines formed by $P$ remains the same so we still add $s$ at the end when considering intersection points. Thus, the recursive equation becomes $(s(s+1)-(s+1))+(s(s+1)-2(s+1))\\cdots(s(s+1)-s(s+1))+s\\Longrightarrow s^2(s+1)-(s+1)\\sum_{i=1}^{s}{i}+s$. Plugging $s=5$ into this formula and adding the values we obtained from the other formulas, the final answer is $4+4+9+12+22+28+45+55+65=244. ~Magnetoninja",
"We know the by Euler's Formula for planar graphs that $F-E+V=2$, where $F$ is the number of bounded faces, plus the outer region, $E$ is the number of edges, and $V$ is the number of vertices. Temporarily disregarding the intersections between the lines, we can easily calculate that: $V_{i}=7+5=12$ $E_{i}=6+4+7\\cdot5=45$ However, the resulting graph is not planar, as the edges clearly intersect. To account for this, we must turn all intersection points into vertices, and update our values accordingly. Observe that each intersection point can be mapped to two points on either line, and analogously, two points on either line can be mapped to one intersection point, uniquely. Thus, to count intersection points, we simply calculate: ${7 \\choose 2}{5 \\choose 2} = 210$ And thus, $V=V_{i}+210=222$ We must also account for the edges. Observe that each intersection point turns the two edges that make it into four, that is, each intersection point adds $2$ to the number of edges. Therefore, $E=E_{i}+2\\cdot210=465$ Plugging these into Euler's Formula we get: $F-E+V=2$ $F-465+222=2$ $F=245$ Disregarding the outer region, we conclude that our answer is $F-1=245-1=244 ~Shadowleafy",
"We know the by Euler's Formula for planar graphs that $F-E+V=2$, where $F$ is the number of bounded faces, plus the outer region, $E$ is the number of edges, and $V$ is the number of vertices. Temporarily disregarding the intersections between the lines, we can easily calculate that: $V_{i}=7+5=12$ $E_{i}=6+4+7\\cdot5=45$ However, the resulting graph is not planar, as the edges clearly intersect. To account for this, we must turn all intersection points into vertices, and update our values accordingly. Observe that each intersection point can be mapped to two points on either line, and analogously, two points on either line can be mapped to one intersection point, uniquely. Thus, to count intersection points, we simply calculate: ${7 \\choose 2}{5 \\choose 2} = 210$ And thus, $V=V_{i}+210=222$ We must also account for the edges. Observe that each intersection point turns the two edges that make it into four, that is, each intersection point adds $2$ to the number of edges. Therefore, $E=E_{i}+2\\cdot210=465$ Plugging these into Euler's Formula we get: $F-E+V=2$ $F-465+222=2$ $F=245$ Disregarding the outer region, we conclude that our answer is $F-1=245-1=244 ~Shadowleafy"
] |
2022-II-10
| 2,022
| 10
|
Find the remainder when \[\binom{\binom{3}{2}}{2} + \binom{\binom{4}{2}}{2} + \dots + \binom{\binom{40}{2}}{2}\] is divided by $1000$ .
| 4
|
II
|
[
"We first write the expression as a summation. \\begin{align*} \\sum_{i=3}^{40} \\binom{\\binom{i}{2}}{2} & = \\sum_{i=3}^{40} \\binom{\\frac{i \\left( i - 1 \\right)}{2}}{2} \\\\ & = \\sum_{i=3}^{40} \\frac{\\frac{i \\left( i - 1 \\right)}{2} \\left( \\frac{i \\left( i - 1 \\right)}{2}- 1 \\right)}{2} \\\\ & = \\frac{1}{8} \\sum_{i=3}^{40} i \\left( i - 1 \\right) \\left( i \\left( i - 1 \\right) - 2 \\right) \\\\ & = \\frac{1}{8} \\sum_{i=3}^{40} i(i - 1)(i^2-i-2) \\\\ & = \\frac{1}{8} \\sum_{i=3}^{40} i(i-1)(i+1)(i-2) \\\\ & = \\frac{1}{8}\\sum_{i=3}^{40} (i-2)(i-1)i(i+1) \\\\ & = \\frac{1}{40}\\sum_{i=3}^{40}[(i-2)(i-1)i(i+1)(i+2)-(i-3)(i-2)(i-1)i(i+1)]* \\\\ & = \\frac{38\\cdot39\\cdot40\\cdot41\\cdot42-0}{40}\\\\ & = 38 \\cdot 39 \\cdot 41 \\cdot 42 \\\\ & = \\left( 40 - 2 \\right) \\left( 40 - 1 \\right) \\left( 40 + 1 \\right) \\left( 40 + 2 \\right) \\\\ & = \\left( 40^2 - 2^2 \\right) \\left( 40^2 - 1^2 \\right) \\\\ & = \\left( 40^2 - 4 \\right) \\left( 40^2 - 1 \\right) \\\\ & = 40^4 - 40^2 \\cdot 5 + 4 \\\\ & \\equiv 004 is how we force the expression to telescope. ~qyang",
"We first write the expression as a summation. \\begin{align*} \\sum_{i=3}^{40} \\binom{\\binom{i}{2}}{2} & = \\sum_{i=3}^{40} \\binom{\\frac{i \\left( i - 1 \\right)}{2}}{2} \\\\ & = \\sum_{i=3}^{40} \\frac{\\frac{i \\left( i - 1 \\right)}{2} \\left( \\frac{i \\left( i - 1 \\right)}{2}- 1 \\right)}{2} \\\\ & = \\frac{1}{8} \\sum_{i=3}^{40} i \\left( i - 1 \\right) \\left( i \\left( i - 1 \\right) - 2 \\right) \\\\ & = \\frac{1}{8} \\sum_{i=3}^{40} i(i - 1)(i^2-i-2) \\\\ & = \\frac{1}{8} \\sum_{i=3}^{40} i(i-1)(i+1)(i-2) \\\\ & = \\frac{1}{8}\\sum_{i=3}^{40} (i-2)(i-1)i(i+1) \\\\ & = \\frac{1}{40}\\sum_{i=3}^{40}[(i-2)(i-1)i(i+1)(i+2)-(i-3)(i-2)(i-1)i(i+1)]* \\\\ & = \\frac{38\\cdot39\\cdot40\\cdot41\\cdot42-0}{40}\\\\ & = 38 \\cdot 39 \\cdot 41 \\cdot 42 \\\\ & = \\left( 40 - 2 \\right) \\left( 40 - 1 \\right) \\left( 40 + 1 \\right) \\left( 40 + 2 \\right) \\\\ & = \\left( 40^2 - 2^2 \\right) \\left( 40^2 - 1^2 \\right) \\\\ & = \\left( 40^2 - 4 \\right) \\left( 40^2 - 1 \\right) \\\\ & = 40^4 - 40^2 \\cdot 5 + 4 \\\\ & \\equiv 004 is how we force the expression to telescope. ~qyang",
"Doing simple algebra calculation will give the following equation: \\begin{align*} \\binom{\\binom{n}{2}}{2} &= \\frac{\\frac{n(n-1)}{2} \\cdot (\\frac{n(n-1)}{2}-1)}{2}\\\\ &= \\frac{n(n-1)(n^2-n-2)}{8}\\\\ &= \\frac{(n+1)n(n-1)(n-2)}{8}\\\\ &= \\frac{(n+1)!}{8\\cdot (n-3)!} = 3 \\cdot \\frac{(n+1)!}{4!\\cdot (n-3)!}\\\\ &= 3 \\binom{n+1}{4} \\end{align*} Next, by using Hockey-Stick Identity, we have: \\[3 \\cdot \\sum_{i=3}^{40} \\binom{i+1}{4} = 3 \\binom{42}{5} = 42 \\cdot 41 \\cdot 39 \\cdot 38\\] \\[=(40^2-2^2)(40^2-1^2) \\equiv 004 ~(\\text{mod}~ 1000)\\]",
"Doing simple algebra calculation will give the following equation: \\begin{align*} \\binom{\\binom{n}{2}}{2} &= \\frac{\\frac{n(n-1)}{2} \\cdot (\\frac{n(n-1)}{2}-1)}{2}\\\\ &= \\frac{n(n-1)(n^2-n-2)}{8}\\\\ &= \\frac{(n+1)n(n-1)(n-2)}{8}\\\\ &= \\frac{(n+1)!}{8\\cdot (n-3)!} = 3 \\cdot \\frac{(n+1)!}{4!\\cdot (n-3)!}\\\\ &= 3 \\binom{n+1}{4} \\end{align*} Next, by using Hockey-Stick Identity, we have: \\[3 \\cdot \\sum_{i=3}^{40} \\binom{i+1}{4} = 3 \\binom{42}{5} = 42 \\cdot 41 \\cdot 39 \\cdot 38\\] \\[=(40^2-2^2)(40^2-1^2) \\equiv 004 ~(\\text{mod}~ 1000)\\]",
"Since $40$ seems like a completely arbitrary number, we can use Engineer's Induction by listing out the first few sums. These are, in the order of how many terms there are starting from $1$ term: $3$, $18$, $63$, $168$, $378$, and $756$. Notice that these are just $3 \\cdot \\dbinom50$, $3 \\cdot \\dbinom61$, $3 \\cdot \\dbinom72$, $3 \\cdot \\dbinom83$, $3 \\cdot \\dbinom94$, $3 \\cdot \\dbinom{10}5$. It's clear that this pattern continues up to $38$ terms, noticing that the \"indexing\" starts with $\\dbinom32$ instead of $\\dbinom12$. Thus, the value of the sum is $3 \\cdot \\dbinom{42}{37}=2552004 \\equiv \\textbf{004}. ~A1001",
"As in solution 1, obtain $\\sum_{i=3}^{40} \\binom{\\binom{i}{2}}{2} = \\frac{1}{8} \\sum_{i=3}^{40} i^4-2i^3-i^2+2i.$ Write this as \\[\\frac{1}{8}\\left(\\sum_{i=3}^{40} i^4 - 2\\sum_{i=3}^{40}i^3 - \\sum_{i=3}^{40}i^2 + 2\\sum_{i=3}^{40}i\\right).\\] We can safely write this expression as $\\frac{1}{8}\\left(\\sum_{i=1}^{40} i^4 - 2\\sum_{i=1}^{40}i^3 - \\sum_{i=1}^{40}i^2 + 2\\sum_{i=1}^{40}i\\right)$, since plugging $i=1$ and $i=2$ into $i^4-2i^3-i^2+2i$ both equal $0,$ meaning they won't contribute to the sum. Use the sum of powers formulae. We obtain \\[\\frac{1}{8}\\left(\\frac{i(i+1)(2i+1)(3i^2+3i-1)}{30} - \\frac{i^2(i+1)^2}{2} - \\frac{i(i+1)(2i+1)}{6} + i(i+1)\\right) \\text{ where i = 40.}\\] We can factor the following expression as $\\frac{1}{8}\\left(\\frac{i(i+1)(2i+1)(3i^2+3i-6)}{30} - \\frac{i(i+1)}{2} (i(i+1)-2)\\right),$ and simplifying, we have \\[\\sum_{i=3}^{40} \\binom{\\binom{i}{2}}{2} = \\frac{i(i+1)(2i+1)(i^2+i-2)}{80}-\\frac{i^2(i+1)^2-2i(i+1)}{16} \\text{ where i = 40.}\\] Substituting $i=40$ and simplifying gets $41\\cdot 81\\cdot 819 - 5\\cdot 41\\cdot 819,$ so we would like to find $819\\cdot 76\\cdot 41 \\pmod{1000}.$ To do this, get $819\\cdot 76\\equiv 244 \\pmod{1000}.$ Next, $244\\cdot 41 \\equiv 004 -sirswagger21",
"To solve this problem, we need to use the following result: \\[ \\sum_{i=n}^m \\binom{i}{k} = \\binom{m+1}{k+1} - \\binom{n}{k+1} . \\] Now, we use this result to solve this problem. We have \\begin{align*} \\sum_{i=3}^{40} \\binom{\\binom{i}{2}}{2} & = \\sum_{i=3}^{40} \\binom{\\frac{i \\left( i - 1 \\right)}{2}}{2} \\\\ & = \\sum_{i=3}^{40} \\frac{\\frac{i \\left( i - 1 \\right)}{2} \\left( \\frac{i \\left( i - 1 \\right)}{2}- 1 \\right)}{2} \\\\ & = \\frac{1}{8} \\sum_{i=3}^{40} i \\left( i - 1 \\right) \\left( i \\left( i - 1 \\right) - 2 \\right) \\\\ & = \\frac{1}{8} \\sum_{i=3}^{40} i \\left( i - 1 \\right) \\left( \\left( i - 2 \\right) \\left( i - 3 \\right) + 4 \\left( i - 2 \\right) \\right) \\\\ & = 3 \\left( \\sum_{i=3}^{40} \\binom{i}{4} + \\sum_{i=3}^{40} \\binom{i}{3} \\right) \\\\ & = 3 \\left( \\binom{41}{5} - \\binom{3}{5} + \\binom{41}{4} - \\binom{3}{4} \\right) \\\\ & = 3 \\left( \\binom{41}{5} + \\binom{41}{4} \\right) \\\\ & = 3 \\cdot \\frac{41 \\cdot 40 \\cdot 39 \\cdot 38}{5!} \\left( 37 + 5 \\right) \\\\ & = 3 \\cdot 41 \\cdot 13 \\cdot 38 \\cdot 42 \\\\ & = 38 \\cdot 39 \\cdot 41 \\cdot 42 \\\\ & = \\left( 40 - 2 \\right) \\left( 40 - 1 \\right) \\left( 40 + 1 \\right) \\left( 40 + 2 \\right) \\\\ & = \\left( 40^2 - 2^2 \\right) \\left( 40^2 - 1^2 \\right) \\\\ & = \\left( 40^2 - 4 \\right) \\left( 40^2 - 1 \\right) \\\\ & = 40^4 - 40^2 \\cdot 5 + 4 . \\end{align*} Therefore, modulo 1000, $\\sum_{i=3}^{40} \\binom{\\binom{i}{2}}{2} \\equiv \\textbf{(004) }. ~Steven Chen (www.professorchenedu.com)",
"We examine the expression $\\binom{\\binom{n}{2}}{2}$. Imagine we have a set $S$ of $n$ integers. Then the expression can be translated to the number of pairs of $2$ element subsets of $S$. To count this, note that each pair of $2$ element subsets can either share $1$ value or $0$ values. In the former case, pick three integers $a$, $b$, and $c$. There are $\\binom{n}{3}$ ways to select these integers and $3$ ways to pick which one of the three is the shared integer. This gives $3\\cdot \\binom{n}{3}$. In the latter case, we pick $4$ integers $a$, $b$, $c$, and $d$ in a total of $\\binom{n}{4}$ ways. There are $\\frac{1}{2}\\binom{4}{2} = 3$ ways to split this up into $2$ sets of $2$ integers — $\\binom{4}{2}$ ways to pick which $2$ integers are together and dividing by $2$ to prevent overcounting. This gives $3\\cdot \\binom{n}{4}$. So we have \\[\\binom{\\binom{n}{2}}{2} = 3\\cdot \\binom{n}{3} + 3\\cdot \\binom{n}{4}\\] We use the Hockey Stick Identity to evaluate this sum: \\begin{align*} \\sum_{n=3}^{40} \\binom{\\binom{n}{2}}{2} &= \\sum_{n=3}^{40} \\left( 3 \\binom{n}{3} + 3 \\binom{n}{4} \\right) \\\\ &= 3 \\left( \\sum_{n=3}^{40} \\binom{n}{3} \\right) + 3\\left( \\sum_{n=4}^{40} \\binom{n}{4} \\right) \\\\ &= 3\\left( \\binom{41}{4} + \\binom{41}{5} \\right) \\end{align*} Evaluating while accounting for mod $1000$ gives the final answer to be $004. ~ GoatPotato",
"We examine the expression $\\binom{\\binom{n}{2}}{2}$. Imagine we have a set $S$ of $n$ integers. Then the expression can be translated to the number of pairs of $2$ element subsets of $S$. To count this, note that each pair of $2$ element subsets can either share $1$ value or $0$ values. In the former case, pick three integers $a$, $b$, and $c$. There are $\\binom{n}{3}$ ways to select these integers and $3$ ways to pick which one of the three is the shared integer. This gives $3\\cdot \\binom{n}{3}$. In the latter case, we pick $4$ integers $a$, $b$, $c$, and $d$ in a total of $\\binom{n}{4}$ ways. There are $\\frac{1}{2}\\binom{4}{2} = 3$ ways to split this up into $2$ sets of $2$ integers — $\\binom{4}{2}$ ways to pick which $2$ integers are together and dividing by $2$ to prevent overcounting. This gives $3\\cdot \\binom{n}{4}$. So we have \\[\\binom{\\binom{n}{2}}{2} = 3\\cdot \\binom{n}{3} + 3\\cdot \\binom{n}{4}\\] We use the Hockey Stick Identity to evaluate this sum: \\begin{align*} \\sum_{n=3}^{40} \\binom{\\binom{n}{2}}{2} &= \\sum_{n=3}^{40} \\left( 3 \\binom{n}{3} + 3 \\binom{n}{4} \\right) \\\\ &= 3 \\left( \\sum_{n=3}^{40} \\binom{n}{3} \\right) + 3\\left( \\sum_{n=4}^{40} \\binom{n}{4} \\right) \\\\ &= 3\\left( \\binom{41}{4} + \\binom{41}{5} \\right) \\end{align*} Evaluating while accounting for mod $1000$ gives the final answer to be $004. ~ GoatPotato"
] |
2022-II-11
| 2,022
| 11
|
Let $ABCD$ be a convex quadrilateral with $AB=2, AD=7,$ and $CD=3$ such that the bisectors of acute angles $\angle{DAB}$ and $\angle{ADC}$ intersect at the midpoint of $\overline{BC}.$ Find the square of the area of $ABCD.$
| 180
|
II
|
[
"[asy] defaultpen(fontsize(12)+0.6); size(300); pair A,B,C,D,M,H; real xb=71, xd=121; A=origin; D=(7,0); B=2*dir(xb); C=3*dir(xd)+D; M=(B+C)/2; H=foot(M,A,D); path c=CR(D,3); pair A1=bisectorpoint(D,A,B), D1=bisectorpoint(C,D,A), Bp=IP(CR(A,2),A--H), Cp=IP(CR(D,3),D--H); draw(B--A--D--C--B); draw(A--M--D^^M--H^^Bp--M--Cp, gray+0.4); draw(rightanglemark(A,H,M,5)); dot(\"$A$\",A,SW); dot(\"$D$\",D,SE); dot(\"$B$\",B,NW); dot(\"$C$\",C,NE); dot(\"$M$\",M,up); dot(\"$H$\",H,down); dot(\"$B'$\",Bp,down); dot(\"$C'$\",Cp,down); [/asy] According to the problem, we have $AB=AB'=2$, $DC=DC'=3$, $MB=MB'$, $MC=MC'$, and $B'C'=7-2-3=2$ Because $M$ is the midpoint of $BC$, we have $BM=MC$, so: \\[MB=MB'=MC'=MC.\\] Then, we can see that $\\bigtriangleup{MB'C'}$ is an isosceles triangle with $MB'=MC'$ Therefore, we could start our angle chasing: $\\angle{MB'C'}=\\angle{MC'B'}=180^\\circ-\\angle{MC'D}=180^\\circ-\\angle{MCD}$. This is when we found that points $M$, $C$, $D$, and $B'$ are on a circle. Thus, $\\angle{BMB'}=\\angle{CDC'} \\Rightarrow \\angle{B'MA}=\\angle{C'DM}$. This is the time we found that $\\bigtriangleup{AB'M} \\sim \\bigtriangleup{MC'D}$. Thus, $\\frac{AB'}{B'M}=\\frac{MC'}{C'D} \\Longrightarrow (B'M)^2=AB' \\cdot C'D = 6$ Point $H$ is the midpoint of $B'C'$, and $MH \\perp AD$. $B'H=HC'=1 \\Longrightarrow MH=\\sqrt{B'M^2-B'H^2}=\\sqrt{6-1}=\\sqrt{5}$. The area of this quadrilateral is the sum of areas of triangles: \\[S_{\\bigtriangleup{ABM}}+S_{\\bigtriangleup{AB'M}}+S_{\\bigtriangleup{CDM}}+S_{\\bigtriangleup{C'DM}}+S_{\\bigtriangleup{B'C'M}}\\] \\[=S_{\\bigtriangleup{AB'M}}\\cdot 2 + S_{\\bigtriangleup{B'C'M}} + S_{\\bigtriangleup{C'DM}}\\cdot 2\\] \\[=2 \\cdot \\frac{1}{2} \\cdot AB' \\cdot MH + \\frac{1}{2} \\cdot B'C' \\cdot MH + 2 \\cdot \\frac{1}{2} \\cdot C'D \\cdot MH\\] \\[=2\\sqrt{5}+\\sqrt{5}+3\\sqrt{5}=6\\sqrt{5}\\] Finally, the square of the area is $(6\\sqrt{5})^2=180",
"Denote by $M$ the midpoint of segment $BC$. Let points $P$ and $Q$ be on segment $AD$, such that $AP = AB$ and $DQ = DC$. Denote $\\angle DAM = \\alpha$, $\\angle BAD = \\beta$, $\\angle BMA = \\theta$, $\\angle CMD = \\phi$. Denote $BM = x$. Because $M$ is the midpoint of $BC$, $CM = x$. Because $AM$ is the angle bisector of $\\angle BAD$ and $AB = AP$, $\\triangle BAM \\cong \\triangle PAM$. Hence, $MP = MB$ and $\\angle AMP = \\theta$. Hence, $\\angle MPD = \\angle MAP + \\angle PMA = \\alpha + \\theta$. Because $DM$ is the angle bisector of $\\angle CDA$ and $DC = DQ$, $\\triangle CDM \\cong \\triangle QDM$. Hence, $MQ = MC$ and $\\angle DMQ = \\phi$. Hence, $\\angle MQA = \\angle MDQ + \\angle QMD = \\beta + \\phi$. Because $M$ is the midpoint of segment $BC$, $MB = MC$. Because $MP = MB$ and $MQ = MC$, $MP = MQ$. Thus, $\\angle MPD = \\angle MQA$. Thus, \\[ \\alpha + \\theta = \\beta + \\phi . \\hspace{1cm} (1) \\] In $\\triangle AMD$, $\\angle AMD = 180^\\circ - \\angle MAD - \\angle MDA = 180^\\circ - \\alpha - \\beta$. In addition, $\\angle AMD = 180^\\circ - \\angle BMA - \\angle CMD = 180^\\circ - \\theta - \\phi$. Thus, \\[ \\alpha + \\beta = \\theta + \\phi . \\hspace{1cm} (2) \\] Taking $(1) + (2)$, we get $\\alpha = \\phi$. Taking $(1) - (2)$, we get $\\beta = \\theta$. Therefore, $\\triangle ADM \\sim \\triangle AMB \\sim \\triangle MDC$. Hence, $\\frac{AD}{AM} = \\frac{AM}{AB}$ and $\\frac{AD}{DM} = \\frac{DM}{CD}$. Thus, $AM = \\sqrt{AD \\cdot AD} = \\sqrt{14}$ and $DM = \\sqrt{AD \\cdot CD} = \\sqrt{21}$. In $\\triangle ADM$, by applying the law of cosines, $\\cos \\angle AMD = \\frac{AM^2 + DM^2 - AD^2}{2 AM \\cdot DM} = - \\frac{1}{\\sqrt{6}}$. Hence, $\\sin \\angle AMD = \\sqrt{1 - \\cos^2 \\angle AMD} = \\frac{\\sqrt{5}}{\\sqrt{6}}$. Hence, ${\\rm Area} \\ \\triangle ADM = \\frac{1}{2} AM \\cdot DM \\dot \\sin \\angle AMD = \\frac{7 \\sqrt{5}}{2}$. Therefore, \\begin{align*} {\\rm Area} \\ ABCD & = {\\rm Area} \\ \\triangle AMD + {\\rm Area} \\ \\triangle ABM + {\\rm Area} \\ \\triangle MCD \\\\ & = {\\rm Area} \\ \\triangle AMD \\left( 1 + \\left( \\frac{AM}{AD} \\right)^2 + \\left( \\frac{MD}{AD} \\right)^2 \\right) \\\\ & = 6 \\sqrt{5} . \\end{align*} Therefore, the square of ${\\rm Area} \\ ABCD$ is $\\left( 6 \\sqrt{5} \\right)^2 = \\textbf{(180) }. ~Steven Chen (www.professorchenedu.com)",
"Claim In the triangle $ABC, AB = 2AC, M$ is the midpoint of $AB. D$ is the point of intersection of the circumcircle and the bisector of angle $A.$ Then $DM = BD.$ Proof Let $A = 2\\alpha.$ Then $\\angle DBC = \\angle DCB = \\alpha.$ Let $E$ be the intersection point of the perpendicular dropped from $D$ to $AB$ with the circle. Then the sum of arcs $\\overset{\\Large\\frown} {BE} + \\overset{\\Large\\frown}{AC} + \\overset{\\Large\\frown}{CD} = 180^\\circ.$ \\[\\overset{\\Large\\frown} {BE} = 180^\\circ – 2\\alpha – \\overset{\\Large\\frown}{AC}.\\] Let $E'$ be the point of intersection of the line $CM$ with the circle. $CM$ is perpendicular to $AD, \\angle AMC = 90^\\circ – \\alpha,$ the sum of arcs $\\overset{\\Large\\frown}{A}C + \\overset{\\Large\\frown}{BE'} = 180^\\circ – 2\\alpha \\implies E'$ coincides with $E.$ The inscribed angles $\\angle DEM = \\angle DEB, M$ is symmetric to $B$ with respect to $DE, DM = DB.$ Solution Let $AB' = AB, DC' = DC, B'$ and $C'$ on $AD.$ Then $AB' = 2, DC' = 3, B'C' = 2 = AB'.$ Quadrilateral $ABMC'$ is cyclic. Let $\\angle A = 2\\alpha.$ Then $\\angle MBC' = \\angle MC'B = \\alpha.$ Circle $BB'C'C$ centered at $M, BC$ is its diameter, $\\angle BC'C = 90^\\circ.$ $\\angle DMC' = \\angle MC'B,$ since they both complete $\\angle MC'C$ to $90^\\circ.$ $\\angle MB'A = \\angle MC'D,$ since they are the exterior angles of an isosceles $\\triangle MB'C'.$ $\\triangle AMB' \\sim \\triangle MDC'$ by two angles. $\\frac {AB'}{MC'} = \\frac {MB'}{DC'}, MC' =\\sqrt{AB' \\cdot C'D} = \\sqrt{6}.$ The height dropped from $M$ to $AD$ is $\\sqrt{MB'^2 - (\\frac{B'C'}{2})^2} =\\sqrt{6 - 1} = \\sqrt{5}.$ The areas of triangles $\\triangle AMB'$ and $\\triangle MC'B'$ are equal to $\\sqrt{5},$ area of $\\triangle MC'D$ is $\\frac{3}{2} \\sqrt{5}.$ \\[\\triangle AMB' = \\triangle AMB, \\triangle MC'D = \\triangle MCD \\implies\\] The area of $ABCD$ is $(1 + 2 + 3) \\sqrt{5} = 6\\sqrt{5} \\implies 6^2 \\cdot 5 = 180 vladimir.shelomovskii@gmail.com, vvsss",
"Claim In the triangle $ABC, AB = 2AC, M$ is the midpoint of $AB. D$ is the point of intersection of the circumcircle and the bisector of angle $A.$ Then $DM = BD.$ Proof Let $A = 2\\alpha.$ Then $\\angle DBC = \\angle DCB = \\alpha.$ Let $E$ be the intersection point of the perpendicular dropped from $D$ to $AB$ with the circle. Then the sum of arcs $\\overset{\\Large\\frown} {BE} + \\overset{\\Large\\frown}{AC} + \\overset{\\Large\\frown}{CD} = 180^\\circ.$ \\[\\overset{\\Large\\frown} {BE} = 180^\\circ – 2\\alpha – \\overset{\\Large\\frown}{AC}.\\] Let $E'$ be the point of intersection of the line $CM$ with the circle. $CM$ is perpendicular to $AD, \\angle AMC = 90^\\circ – \\alpha,$ the sum of arcs $\\overset{\\Large\\frown}{A}C + \\overset{\\Large\\frown}{BE'} = 180^\\circ – 2\\alpha \\implies E'$ coincides with $E.$ The inscribed angles $\\angle DEM = \\angle DEB, M$ is symmetric to $B$ with respect to $DE, DM = DB.$ Solution Let $AB' = AB, DC' = DC, B'$ and $C'$ on $AD.$ Then $AB' = 2, DC' = 3, B'C' = 2 = AB'.$ Quadrilateral $ABMC'$ is cyclic. Let $\\angle A = 2\\alpha.$ Then $\\angle MBC' = \\angle MC'B = \\alpha.$ Circle $BB'C'C$ centered at $M, BC$ is its diameter, $\\angle BC'C = 90^\\circ.$ $\\angle DMC' = \\angle MC'B,$ since they both complete $\\angle MC'C$ to $90^\\circ.$ $\\angle MB'A = \\angle MC'D,$ since they are the exterior angles of an isosceles $\\triangle MB'C'.$ $\\triangle AMB' \\sim \\triangle MDC'$ by two angles. $\\frac {AB'}{MC'} = \\frac {MB'}{DC'}, MC' =\\sqrt{AB' \\cdot C'D} = \\sqrt{6}.$ The height dropped from $M$ to $AD$ is $\\sqrt{MB'^2 - (\\frac{B'C'}{2})^2} =\\sqrt{6 - 1} = \\sqrt{5}.$ The areas of triangles $\\triangle AMB'$ and $\\triangle MC'B'$ are equal to $\\sqrt{5},$ area of $\\triangle MC'D$ is $\\frac{3}{2} \\sqrt{5}.$ \\[\\triangle AMB' = \\triangle AMB, \\triangle MC'D = \\triangle MCD \\implies\\] The area of $ABCD$ is $(1 + 2 + 3) \\sqrt{5} = 6\\sqrt{5} \\implies 6^2 \\cdot 5 = 180 vladimir.shelomovskii@gmail.com, vvsss",
"Extend $AB$ and $CD$ so they intersect at a point $X$. Then note that $M$ is the incenter of $\\triangle{XAD}$, implying that $M$ is on the angle bisector of $X$. Now because $XM$ is both an angle bisector and a median of $\\triangle{XBC}$, $\\triangle{XBC}$ is isosceles. Then we can start angle chasing: Let $\\angle{BAM}=a, \\angle{CDM}=b,$ and $\\angle{XBC}=c$. Then $\\angle{AMD}=\\pi-(a+b), \\angle{ABM}=\\pi-c, \\angle{DCM}=\\pi-c$, implying that $\\angle{BMA}+\\angle{CMD}=a+b$, implying that $2c-(a+b)=(a+b)$, or that $c=a+b$. Substituting this into the rest of the diagram, we find that $\\triangle{BMA} \\sim \\triangle{CDM} \\sim \\triangle{MDA}$. Then $\\frac{AB}{BM}=\\frac{MC}{CD}$, or $BM=CM=\\sqrt{6}$. Moreover, $\\frac{AB}{AM}=\\frac{AM}{AD}$, or $AM=\\sqrt{14}$. Similarly, $\\frac{CD}{MD}=\\frac{MD}{AD}$, or $DM=\\sqrt{21}$. Then using Law of Cosines on $\\triangle{AMD}$, to get that $cos\\angle{AMD}=-\\frac{\\sqrt{6}}{6}$, or $sin\\angle{AMD}=\\frac{\\sqrt{30}}{6}$. We finish by using the formula $K=\\frac{1}{2}absinC$, as follows: $[ABCD]=[ABM]+[CDM]+[ADM]=\\frac{\\frac{\\sqrt{30}}{6}(2\\sqrt{6}+3\\sqrt{6}+7\\sqrt{6})}{2}=6\\sqrt{5}$. $(6\\sqrt{5})^2=180. -dragoon",
"Extend $AB$ and $CD$ so they intersect at a point $X$. Then note that $M$ is the incenter of $\\triangle{XAD}$, implying that $M$ is on the angle bisector of $X$. Now because $XM$ is both an angle bisector and a median of $\\triangle{XBC}$, $\\triangle{XBC}$ is isosceles. Then we can start angle chasing: Let $\\angle{BAM}=a, \\angle{CDM}=b,$ and $\\angle{XBC}=c$. Then $\\angle{AMD}=\\pi-(a+b), \\angle{ABM}=\\pi-c, \\angle{DCM}=\\pi-c$, implying that $\\angle{BMA}+\\angle{CMD}=a+b$, implying that $2c-(a+b)=(a+b)$, or that $c=a+b$. Substituting this into the rest of the diagram, we find that $\\triangle{BMA} \\sim \\triangle{CDM} \\sim \\triangle{MDA}$. Then $\\frac{AB}{BM}=\\frac{MC}{CD}$, or $BM=CM=\\sqrt{6}$. Moreover, $\\frac{AB}{AM}=\\frac{AM}{AD}$, or $AM=\\sqrt{14}$. Similarly, $\\frac{CD}{MD}=\\frac{MD}{AD}$, or $DM=\\sqrt{21}$. Then using Law of Cosines on $\\triangle{AMD}$, to get that $cos\\angle{AMD}=-\\frac{\\sqrt{6}}{6}$, or $sin\\angle{AMD}=\\frac{\\sqrt{30}}{6}$. We finish by using the formula $K=\\frac{1}{2}absinC$, as follows: $[ABCD]=[ABM]+[CDM]+[ADM]=\\frac{\\frac{\\sqrt{30}}{6}(2\\sqrt{6}+3\\sqrt{6}+7\\sqrt{6})}{2}=6\\sqrt{5}$. $(6\\sqrt{5})^2=180. -dragoon",
"As shown in paragraph one of solution 4, extending $AB$ and $CD$ to $X$, we realize that $\\triangle{XBC}$ is isosceles, thus $XM \\perp BC$. Let $XB = XC = x$. And, midpoint $M$ is the incenter of $\\triangle{XAD}$. Construct perpendiculars $ME, MF, MG$ to sides $AD, AX, DX$ respectively (constructing the radii of the incircle). Let $EM = FM = GM = r$. The semiperimeter $s = \\frac{2x + 2 + 3 + 7}{2} = x+6$. Since $FX$ is the tangent off the incircle, $FX = s - AD = x-1$. So, $BF = BX - FX = 1$. Because $\\triangle{BFM} \\sim \\triangle{MFX}$, \\[\\frac{BF}{FM} = \\frac{FM}{FX} \\implies {FM}^2 = BF \\cdot FX = x - 1 \\implies r^2 = x - 1\\]. By Heron's formula and the inradius area formula, \\[(x+6)r = \\sqrt{(x+6)\\cdot 4 \\cdot 3 \\cdot (x-1)} \\implies (x+6)r^2 = 12(x-1) \\implies (x+6)(x-1) = 12(x-1) \\implies x=6\\] Then, $r^2 = x - 1 = 5 \\implies r = \\sqrt{5}$. Finally, \\[[ABCD] = [ABM] + [CDM] + [AMD] = \\frac{AB \\cdot FM}{2} + \\frac{CD \\cdot GM}{2} + \\frac{AD \\cdot EM}{2} = \\frac{2r}{2} + \\frac{3r}{2} + \\frac{7r}{2} = 6r = 6\\sqrt{5}\\] Thus, our answer is $(6\\sqrt{5})^{2} = 180). This may be useful, but I haven't looked into it. ~CrazyVideoGamez",
"As shown in paragraph one of solution 4, extending $AB$ and $CD$ to $X$, we realize that $\\triangle{XBC}$ is isosceles, thus $XM \\perp BC$. Let $XB = XC = x$. And, midpoint $M$ is the incenter of $\\triangle{XAD}$. Construct perpendiculars $ME, MF, MG$ to sides $AD, AX, DX$ respectively (constructing the radii of the incircle). Let $EM = FM = GM = r$. The semiperimeter $s = \\frac{2x + 2 + 3 + 7}{2} = x+6$. Since $FX$ is the tangent off the incircle, $FX = s - AD = x-1$. So, $BF = BX - FX = 1$. Because $\\triangle{BFM} \\sim \\triangle{MFX}$, \\[\\frac{BF}{FM} = \\frac{FM}{FX} \\implies {FM}^2 = BF \\cdot FX = x - 1 \\implies r^2 = x - 1\\]. By Heron's formula and the inradius area formula, \\[(x+6)r = \\sqrt{(x+6)\\cdot 4 \\cdot 3 \\cdot (x-1)} \\implies (x+6)r^2 = 12(x-1) \\implies (x+6)(x-1) = 12(x-1) \\implies x=6\\] Then, $r^2 = x - 1 = 5 \\implies r = \\sqrt{5}$. Finally, \\[[ABCD] = [ABM] + [CDM] + [AMD] = \\frac{AB \\cdot FM}{2} + \\frac{CD \\cdot GM}{2} + \\frac{AD \\cdot EM}{2} = \\frac{2r}{2} + \\frac{3r}{2} + \\frac{7r}{2} = 6r = 6\\sqrt{5}\\] Thus, our answer is $(6\\sqrt{5})^{2} = 180). This may be useful, but I haven't looked into it. ~CrazyVideoGamez",
"Let the midpoint of $BC$ be $M$. Angle-chase and observe that $\\Delta AMD~\\Delta ABM~\\Delta MCD$. Let $BM=CM=a$ and $AM=x$ and $DM=y$. As a result of this similarity, we write \\[\\dfrac2a=\\dfrac a3,\\] which gives $a=\\sqrt 6$. Similarly, we write \\[\\dfrac2x=\\dfrac x7\\] and \\[\\dfrac3y=\\dfrac y7\\] to get $x=\\sqrt{14}$ and $y=\\sqrt{21}$. We now have all required side lengths; we can find the area of $\\Delta AMD$ with Heron's formula. Doing so yields $\\dfrac72\\sqrt5$. We could also bash out the areas of the other two triangles since we know all their side lengths (this is what I did :sob:), but a more intelligent method is to recall the triangles' similarity. The ratio of similarlity between $\\Delta AMD$ and $\\Delta ABM$ is $\\dfrac{\\sqrt{14}}7=\\sqrt{\\dfrac27}$, and between $\\Delta AMD$ and $\\Delta MCD$ is $\\dfrac{\\sqrt{21}}7=\\sqrt{\\dfrac37}$. Thus, the area ratios are $\\dfrac27$ and $\\dfrac37$, respectively, so adding together we have $\\dfrac27+\\dfrac37=\\dfrac57$. Multiplying this by our $\\dfrac72\\sqrt5$, we have $\\dfrac52\\sqrt5$ as their total area. Adding this to our original area, we have $\\dfrac52\\sqrt5+\\dfrac72\\sqrt5=\\sqrt5\\left(\\dfrac52+\\dfrac72\\right)=\\sqrt5\\left(\\dfrac{12}2\\right)=6\\sqrt5$. The square of this is $180. ~~Technodoggo",
"Let the midpoint of $BC$ be $M$. Angle-chase and observe that $\\Delta AMD~\\Delta ABM~\\Delta MCD$. Let $BM=CM=a$ and $AM=x$ and $DM=y$. As a result of this similarity, we write \\[\\dfrac2a=\\dfrac a3,\\] which gives $a=\\sqrt 6$. Similarly, we write \\[\\dfrac2x=\\dfrac x7\\] and \\[\\dfrac3y=\\dfrac y7\\] to get $x=\\sqrt{14}$ and $y=\\sqrt{21}$. We now have all required side lengths; we can find the area of $\\Delta AMD$ with Heron's formula. Doing so yields $\\dfrac72\\sqrt5$. We could also bash out the areas of the other two triangles since we know all their side lengths (this is what I did :sob:), but a more intelligent method is to recall the triangles' similarity. The ratio of similarlity between $\\Delta AMD$ and $\\Delta ABM$ is $\\dfrac{\\sqrt{14}}7=\\sqrt{\\dfrac27}$, and between $\\Delta AMD$ and $\\Delta MCD$ is $\\dfrac{\\sqrt{21}}7=\\sqrt{\\dfrac37}$. Thus, the area ratios are $\\dfrac27$ and $\\dfrac37$, respectively, so adding together we have $\\dfrac27+\\dfrac37=\\dfrac57$. Multiplying this by our $\\dfrac72\\sqrt5$, we have $\\dfrac52\\sqrt5$ as their total area. Adding this to our original area, we have $\\dfrac52\\sqrt5+\\dfrac72\\sqrt5=\\sqrt5\\left(\\dfrac52+\\dfrac72\\right)=\\sqrt5\\left(\\dfrac{12}2\\right)=6\\sqrt5$. The square of this is $180. ~~Technodoggo",
"As in solution 4, let $X=AB\\cap CD$, so $M$ is the incenter of $ADX$ and $XB=XC$. Let $XB=XC=x$. Then the normalized barycentric coordinates of $B$, $C$, and $M$ with respect to $ADX$ are $\\left[\\frac{x}{x+2}:0:\\frac{2}{x+2}\\right]$, $\\left[0:\\frac{x}{x+3}:\\frac{3}{x+3}\\right]$, and $\\left[\\frac{x+3}{2x+12}:\\frac{x+2}{2x+12}:\\frac{7}{2x+12}\\right]$. So we have $\\frac{1}{2}\\frac{x}{x+2}=\\frac{x+3}{2x+12}$ giving $x=6$. The sidelengths of $ADX$ are thus $AD=7$, $AX=8$, and $DX=9$ giving $[ADX]=12\\sqrt 5$. Also, we have $[BCX]=\\frac{6}{8}\\cdot\\frac{6}{9}[ADX]=6\\sqrt 5$ so that $[ABCD]=[ADX]-[BCX]=6\\sqrt 5$. The area squared is thus $180. ~~ golue3120",
"As in solution 4, let $X=AB\\cap CD$, so $M$ is the incenter of $ADX$ and $XB=XC$. Let $XB=XC=x$. Then the normalized barycentric coordinates of $B$, $C$, and $M$ with respect to $ADX$ are $\\left[\\frac{x}{x+2}:0:\\frac{2}{x+2}\\right]$, $\\left[0:\\frac{x}{x+3}:\\frac{3}{x+3}\\right]$, and $\\left[\\frac{x+3}{2x+12}:\\frac{x+2}{2x+12}:\\frac{7}{2x+12}\\right]$. So we have $\\frac{1}{2}\\frac{x}{x+2}=\\frac{x+3}{2x+12}$ giving $x=6$. The sidelengths of $ADX$ are thus $AD=7$, $AX=8$, and $DX=9$ giving $[ADX]=12\\sqrt 5$. Also, we have $[BCX]=\\frac{6}{8}\\cdot\\frac{6}{9}[ADX]=6\\sqrt 5$ so that $[ABCD]=[ADX]-[BCX]=6\\sqrt 5$. The area squared is thus $180. ~~ golue3120"
] |
2022-II-12
| 2,022
| 12
|
Let $a, b, x,$ and $y$ be real numbers with $a>4$ and $b>1$ such that \[\frac{x^2}{a^2}+\frac{y^2}{a^2-16}=\frac{(x-20)^2}{b^2-1}+\frac{(y-11)^2}{b^2}=1.\] Find the least possible value of $a+b.$
| 23
|
II
|
[
"Denote $P = \\left( x , y \\right)$. Because $\\frac{x^2}{a^2}+\\frac{y^2}{a^2-16} = 1$, $P$ is on an ellipse whose center is $\\left( 0 , 0 \\right)$ and foci are $\\left( - 4 , 0 \\right)$ and $\\left( 4 , 0 \\right)$. Hence, the sum of distance from $P$ to $\\left( - 4 , 0 \\right)$ and $\\left( 4 , 0 \\right)$ is equal to twice the major axis of this ellipse, $2a$. Because $\\frac{(x-20)^2}{b^2-1}+\\frac{(y-11)^2}{b^2} = 1$, $P$ is on an ellipse whose center is $\\left( 20 , 11 \\right)$ and foci are $\\left( 20 , 10 \\right)$ and $\\left( 20 , 12 \\right)$. Hence, the sum of distance from $P$ to $\\left( 20 , 10 \\right)$ and $\\left( 20 , 12 \\right)$ is equal to twice the major axis of this ellipse, $2b$. Therefore, $2a + 2b$ is the sum of the distance from $P$ to four foci of these two ellipses. To make this minimized, $P$ is the intersection point of the line that passes through $\\left( - 4 , 0 \\right)$ and $\\left( 20 , 10 \\right)$, and the line that passes through $\\left( 4 , 0 \\right)$ and $\\left( 20 , 12 \\right)$. The distance between $\\left( - 4 , 0 \\right)$ and $\\left( 20 , 10 \\right)$ is $\\sqrt{\\left( 20 + 4 \\right)^2 + \\left( 10 - 0 \\right)^2} = 26$. The distance between $\\left( 4 , 0 \\right)$ and $\\left( 20 , 12 \\right)$ is $\\sqrt{\\left( 20 - 4 \\right)^2 + \\left( 12 - 0 \\right)^2} = 20$. Hence, $2 a + 2 b = 26 + 20 = 46$. Therefore, $a + b = \\textbf{(023) } ~Steven Chen (www.professorchenedu.com)"
] |
2022-II-13
| 2,022
| 13
|
There is a polynomial $P(x)$ with integer coefficients such that \[P(x)=\frac{(x^{2310}-1)^6}{(x^{105}-1)(x^{70}-1)(x^{42}-1)(x^{30}-1)}\] holds for every $0<x<1.$ Find the coefficient of $x^{2022}$ in $P(x)$ .
| 220
|
II
|
[
"Because $0 < x < 1$, we have \\begin{align*} P \\left( x \\right) & = \\sum_{a=0}^6 \\sum_{b=0}^\\infty \\sum_{c=0}^\\infty \\sum_{d=0}^\\infty \\sum_{e=0}^\\infty \\binom{6}{a} x^{2310a} \\left( - 1 \\right)^{6-a} x^{105b} x^{70c} x^{42d} x^{30e} \\\\ & = \\sum_{a=0}^6 \\sum_{b=0}^\\infty \\sum_{c=0}^\\infty \\sum_{d=0}^\\infty \\sum_{e=0}^\\infty \\left( - 1 \\right)^{6-a} x^{2310 a + 105 b + 70 c + 42 d + 30 e} . \\end{align*} Denote by $c_{2022}$ the coefficient of $P \\left( x \\right)$. Thus, \\begin{align*} c_{2022} & = \\sum_{a=0}^6 \\sum_{b=0}^\\infty \\sum_{c=0}^\\infty \\sum_{d=0}^\\infty \\sum_{e=0}^\\infty \\left( - 1 \\right)^{6-a} \\Bbb I \\left\\{ 2310 a + 105 b + 70 c + 42 d + 30 e = 2022 \\right\\} \\\\ & = \\sum_{b=0}^\\infty \\sum_{c=0}^\\infty \\sum_{d=0}^\\infty \\sum_{e=0}^\\infty \\left( - 1 \\right)^{6-0} \\Bbb I \\left\\{ 2310 \\cdot 0 + 105 b + 70 c + 42 d + 30 e = 2022 \\right\\} \\\\ & = \\sum_{b=0}^\\infty \\sum_{c=0}^\\infty \\sum_{d=0}^\\infty \\sum_{e=0}^\\infty \\Bbb I \\left\\{ 105 b + 70 c + 42 d + 30 e = 2022 \\right\\} . \\end{align*} Now, we need to find the number of nonnegative integer tuples $\\left( b , c , d , e \\right)$ that satisfy \\[ 105 b + 70 c + 42 d + 30 e = 2022 . \\hspace{1cm} (1) \\] Modulo 2 on Equation (1), we have $b \\equiv 0 \\pmod{2}$. Hence, we can write $b = 2 b'$. Plugging this into (1), the problem reduces to finding the number of nonnegative integer tuples $\\left( b' , c , d , e \\right)$ that satisfy \\[ 105 b' + 35 c + 21 d + 15 e = 1011 . \\hspace{1cm} (2) \\] Modulo 3 on Equation (2), we have $2 c \\equiv 0 \\pmod{3}$. Hence, we can write $c = 3 c'$. Plugging this into (2), the problem reduces to finding the number of nonnegative integer tuples $\\left( b' , c' , d , e \\right)$ that satisfy \\[ 35 b' + 35 c' + 7 d + 5 e = 337 . \\hspace{1cm} (3) \\] Modulo 5 on Equation (3), we have $2 d \\equiv 2 \\pmod{5}$. Hence, we can write $d = 5 d' + 1$. Plugging this into (3), the problem reduces to finding the number of nonnegative integer tuples $\\left( b' , c' , d' , e \\right)$ that satisfy \\[ 7 b' + 7 c' + 7 d' + e = 66 . \\hspace{1cm} (4) \\] Modulo 7 on Equation (4), we have $e \\equiv 3 \\pmod{7}$. Hence, we can write $e = 7 e' + 3$. Plugging this into (4), the problem reduces to finding the number of nonnegative integer tuples $\\left( b' , c' , d' , e' \\right)$ that satisfy \\[ b' + c' + d' + e' = 9 . \\hspace{1cm} (5) \\] The number of nonnegative integer solutions to Equation (5) is $\\binom{9 + 4 - 1}{4 - 1} = \\binom{12}{3} = \\textbf{(220) }. ~Steven Chen (www.professorchenedu.com)",
"Note that $2022 = 210\\cdot 9 +132$. Since the only way to express $132$ in terms of $105$, $70$, $42$, or $30$ is $132 = 30+30+30+42$, we are essentially just counting the number of ways to express $210*9$ in terms of these numbers. Since $210 = 2*105=3*70=5*42=7*30$, it can only be expressed as a sum in terms of only one of the numbers ($105$, $70$, $42$, or $30$). Thus, the answer is (by sticks and stones) \\[\\binom{12}{3} = \\textbf{(220)}\\] ~Bigbrain123",
"We know that $\\frac{a^n-b^n}{a-b}=\\sum_{i=0}^{n-1} a^{n-1-i}b^i$. Applying this, we see that \\[P(x)=(1+x^{105}+x^{210}+...)(1+x^{70}+x^{140}+...)(1+x^{42}+x^{84}+...)(1+x^{30}+x^{60}+...)(x^{4620}-2x^{2310}+1)\\] The last factor does not contribute to the $x^{2022}$ term, so we can ignore it. Thus we only have left to solve the equation $105b+70c+42d+30e=2022$, and we can proceed from here with Solution 1. ~MathIsFun286",
"Essentially we want the coefficient of $x^{2022}$ in the expansion \\[\\frac{1}{(1 - x^{105})(1 - x^{70})(1 - x^{42})(1 - x^{30})}\\] As $(x^{2310} -1)^{6}$ does not contribute to the expansion, we omit it. Notice $\\text{lcm}(105,70,42,30) = 210$, so we can rewrite the generating function as \\[\\frac{(1 + x^{105})(1 + x^{70} + x^{140})(1 + x^{42} + ... + x^{168})(1 + x^{30} + ... + x^{180})}{(1-x^{210})^{4}}\\] Notice to obtain a $x^{2022}$ value, the $x^{42}$ must be used, so we can reduce the problem to finding the coefficient of $x^{1980}$ in \\[\\frac{(1 + x^{105})(1 + x^{70} + x^{140})(1 + x^{30} + ... + x^{180})}{(1-x^{210})^{4}}\\] If $(1 - x^{210})^{-4}$ is expanded, it will only generate multiples of 210. To compensate this, notice $1980 \\equiv 90 \\pmod{210}$, which means we need terms with a power of a 90 to acheive what we want (300 and larger values can be shown to be impossible upon inspection). This implies that the first two brackets do not contribute and we are left with \\[\\frac{x^{90}}{(1-x^{210})^{4}}\\] This reduces to finding the coefficient of $x^{1890}$ for the expansion $(1 - x^{210})^{-4}$, which is $\\binom{12}{3} = 220",
"Essentially we want the coefficient of $x^{2022}$ in the expansion \\[\\frac{1}{(1 - x^{105})(1 - x^{70})(1 - x^{42})(1 - x^{30})}\\] As $(x^{2310} -1)^{6}$ does not contribute to the expansion, we omit it. Notice $\\text{lcm}(105,70,42,30) = 210$, so we can rewrite the generating function as \\[\\frac{(1 + x^{105})(1 + x^{70} + x^{140})(1 + x^{42} + ... + x^{168})(1 + x^{30} + ... + x^{180})}{(1-x^{210})^{4}}\\] Notice to obtain a $x^{2022}$ value, the $x^{42}$ must be used, so we can reduce the problem to finding the coefficient of $x^{1980}$ in \\[\\frac{(1 + x^{105})(1 + x^{70} + x^{140})(1 + x^{30} + ... + x^{180})}{(1-x^{210})^{4}}\\] If $(1 - x^{210})^{-4}$ is expanded, it will only generate multiples of 210. To compensate this, notice $1980 \\equiv 90 \\pmod{210}$, which means we need terms with a power of a 90 to acheive what we want (300 and larger values can be shown to be impossible upon inspection). This implies that the first two brackets do not contribute and we are left with \\[\\frac{x^{90}}{(1-x^{210})^{4}}\\] This reduces to finding the coefficient of $x^{1890}$ for the expansion $(1 - x^{210})^{-4}$, which is $\\binom{12}{3} = 220",
"Expand the numerator and divide each term in the denominator separately to get $(1+x^{30}+x^{60}+\\dots)\\cdot (1+x^{42}+x^{84}+\\dots)\\cdot(1+x^{70}+x^{140}+\\dots)$ $\\cdot (1+x^{105}+x^{210}+\\dots) (x^{2310}-1)^{2}$ Since the factor of $(x^{2310}-1)^{2}$ is too big so we choose the -1 from both the 2 factors to get a 1. So now we need to find the number of quadruples of non negative integers $(a,b,c,d)$ such that $\\textbf{105a + 70b + 42c + 30d = 2022}$. By considering modulo $2$, $3$, $5$, $7$ we can reduce the above expression to something that can be solved using stars-and-bars. $a \\equiv 0 \\pmod 2$ $b \\equiv 0 \\pmod 3$ $c \\equiv 1 \\pmod 5$ $d \\equiv 3 \\pmod7$. Now we put in $a=2x$, $b=3y$, $c=5z+1$ and $d=7w+3$ to get $2022 = 210(x+y+z+w) + 132$ $\\Rightarrow x+y+z+w=9. ~Lakshya Pamecha",
"Expand the numerator and divide each term in the denominator separately to get $(1+x^{30}+x^{60}+\\dots)\\cdot (1+x^{42}+x^{84}+\\dots)\\cdot(1+x^{70}+x^{140}+\\dots)$ $\\cdot (1+x^{105}+x^{210}+\\dots) (x^{2310}-1)^{2}$ Since the factor of $(x^{2310}-1)^{2}$ is too big so we choose the -1 from both the 2 factors to get a 1. So now we need to find the number of quadruples of non negative integers $(a,b,c,d)$ such that $\\textbf{105a + 70b + 42c + 30d = 2022}$. By considering modulo $2$, $3$, $5$, $7$ we can reduce the above expression to something that can be solved using stars-and-bars. $a \\equiv 0 \\pmod 2$ $b \\equiv 0 \\pmod 3$ $c \\equiv 1 \\pmod 5$ $d \\equiv 3 \\pmod7$. Now we put in $a=2x$, $b=3y$, $c=5z+1$ and $d=7w+3$ to get $2022 = 210(x+y+z+w) + 132$ $\\Rightarrow x+y+z+w=9. ~Lakshya Pamecha"
] |
2022-II-14
| 2,022
| 14
|
For positive integers $a$ , $b$ , and $c$ with $a < b < c$ , consider collections of postage stamps in denominations $a$ , $b$ , and $c$ cents that contain at least one stamp of each denomination. If there exists such a collection that contains sub-collections worth every whole number of cents up to $1000$ cents, let $f(a, b, c)$ be the minimum number of stamps in such a collection. Find the sum of the three least values of $c$ such that $f(a, b, c) = 97$ for some choice of $a$ and $b$ .
| 188
|
II
|
[
"Notice that we must have $a = 1$; otherwise $1$ cent stamp cannot be represented. At least $b-1$ numbers of $1$ cent stamps are needed to represent the values less than $b$. Using at most $c-1$ stamps of value $1$ and $b$, it can have all the values from $1$ to $c-1$ cents. Plus $\\lfloor \\frac{999}{c} \\rfloor$ stamps of value $c$, every value up to $1000$ can be represented. Correction: This should be $\\lfloor \\frac{1000}{c} \\rfloor$. The current function breaks when $c \\mid 1000$ and $b \\mid c$. Take $c = 200$ and $b = 20$. Then, we have $\\lfloor \\frac{999}{200} \\rfloor = 4$ stamps of value 200, $\\lfloor \\frac{199}{20} \\rfloor = 9$ stamps of value b, and 19 stamps of value 1. The maximum such a collection can give is $200 \\cdot 4 + 20 \\cdot 9 +19 \\cdot 1 = 999$, just shy of the needed 1000. As for the rest of solution, proceed similarly, except use $1000$ instead of $999$. Also, some explanation: $b-1$ one cent stamps cover all residues module $b$. Having $\\lfloor \\frac{c-1}{b} \\rfloor$ stamps of value b covers all residue classes modulo $c$. Finally, we just need $\\lfloor \\frac{1000}{c} \\rfloor$ to cover everything up to 1000. In addition, note that this function sometimes may not always minimize the number of stamps required. This is due to the fact that the stamps of value $b$ and of value $1$ have the capacity to cover values greater than or equal to $c$ (which occurs when $c-1$ has a remainder less than $b-1$ when divided by $b$). Thus, in certain cases, not all $\\lfloor \\frac{1000}{c} \\rfloor$ stamps of value c may be necessary, because the stamps of value $b$ and 1 can replace one $c$. ~CrazyVideoGamez Correction for the correction and the original solution: Actually, $f(a, b, c) = b + \\lceil \\frac{c}{b} \\rceil + \\lceil \\frac{1001-b \\lceil \\frac{c}{b} \\rceil}{c} \\rceil - 2$. This could be obtained by solving the minimum of $A+B+C$ for the system of inequalities $A \\geq b-14$, $A+Bb \\geq c$, and $A+Bb+Cc \\geq 1000$, (Using the inequality adjustment method) where $A$, $B$, and $C$ represent the number of $a$, $b$, and $c$ postage stamps used. It is possible to get that $A=b-1$, $B=\\lceil \\frac{c}{b} \\rceil-1$, $C=\\lceil \\frac{1001-b \\lceil \\frac{c}{b} \\rceil}{c} \\rceil$. $f(a,b,c)=A+B+C$ yields the above. Note that in $\\text{Case } 2$, $c=88$, for the original solution $a=1$, $b=86$, $c=88$, $f(a,b,c)=96$ if the above definition is used. When $A=85$, $B=1$, and $C=10$, you can get a collection that satisfies the conditions of the problem. $A+B+C=96$ here. Note that using the correction does not change that the original $f(1,87,88)=97$. Under the correction of the correction: The actual solution could be achieved by using inequalities on the left and right of $f(a,b,c)$ to get $b+\\frac{1001}{c}-3<f(a,b,c)<c+1+\\frac{1001}{c}$. Similar steps follow after substituting $f(a,b,c)=97$. However, $f(1,7,11)=f(1,87,88)=f(1,87,89)=97$. Therefore the answer is still $11+88+89=188 ~isabelchen ~edited by bobjoebilly"
] |
2022-II-15
| 2,022
| 15
|
Two externally tangent circles $\omega_1$ and $\omega_2$ have centers $O_1$ and $O_2$ , respectively. A third circle $\Omega$ passing through $O_1$ and $O_2$ intersects $\omega_1$ at $B$ and $C$ and $\omega_2$ at $A$ and $D$ , as shown. Suppose that $AB = 2$ , $O_1O_2 = 15$ , $CD = 16$ , and $ABO_1CDO_2$ is a convex hexagon. Find the area of this hexagon. [asy] import geometry; size(10cm); point O1=(0,0),O2=(15,0),B=9*dir(30); circle w1=circle(O1,9),w2=circle(O2,6),o=circle(O1,O2,B); point A=intersectionpoints(o,w2)[1],D=intersectionpoints(o,w2)[0],C=intersectionpoints(o,w1)[0]; filldraw(A--B--O1--C--D--O2--cycle,0.2*red+white,black); draw(w1); draw(w2); draw(O1--O2,dashed); draw(o); dot(O1); dot(O2); dot(A); dot(D); dot(C); dot(B); label("$\omega_1$",8*dir(110),SW); label("$\omega_2$",5*dir(70)+(15,0),SE); label("$O_1$",O1,W); label("$O_2$",O2,E); label("$B$",B,N+1/2*E); label("$A$",A,N+1/2*W); label("$C$",C,S+1/4*W); label("$D$",D,S+1/4*E); label("$15$",midpoint(O1--O2),N); label("$16$",midpoint(C--D),N); label("$2$",midpoint(A--B),S); label("$\Omega$",o.C+(o.r-1)*dir(270)); [/asy]
| 140
|
II
|
[
"First observe that $AO_2 = O_2D$ and $BO_1 = O_1C$. Let points $A'$ and $B'$ be the reflections of $A$ and $B$, respectively, about the perpendicular bisector of $\\overline{O_1O_2}$. Then quadrilaterals $ABO_1O_2$ and $B'A'O_2O_1$ are congruent, so hexagons $ABO_1CDO_2$ and $A'B'O_1CDO_2$ have the same area. Furthermore, triangles $DO_2A'$ and $B'O_1C$ are congruent, so $A'D = B'C$ and quadrilateral $A'B'CD$ is an isosceles trapezoid. [asy] import olympiad; size(180); defaultpen(linewidth(0.7)); pair Ap = dir(105), Bp = dir(75), O1 = dir(25), C = dir(320), D = dir(220), O2 = dir(175); draw(unitcircle^^Ap--Bp--O1--C--D--O2--cycle); label(\"$A'$\",Ap,dir(origin--Ap)); label(\"$B'$\",Bp,dir(origin--Bp)); label(\"$O_1$\",O1,dir(origin--O1)); label(\"$C$\",C,dir(origin--C)); label(\"$D$\",D,dir(origin--D)); label(\"$O_2$\",O2,dir(origin--O2)); draw(O2--O1,linetype(\"4 4\")); draw(Ap--D^^Bp--C,linetype(\"2 2\")); [/asy] Next, remark that $B'O_1 = DO_2$, so quadrilateral $B'O_1DO_2$ is also an isosceles trapezoid; in turn, $B'D = O_1O_2 = 15$, and similarly $A'C = 15$. Thus, Ptolemy's theorem on $A'B'CD$ yields $A'D\\cdot B'C + 2\\cdot 16 = 15^2$, whence $A'D = B'C = \\sqrt{193}$. Let $\\alpha = \\angle A'B'D$. The Law of Cosines on triangle $A'B'D$ yields \\[\\cos\\alpha = \\frac{15^2 + 2^2 - (\\sqrt{193})^2}{2\\cdot 2\\cdot 15} = \\frac{36}{60} = \\frac 35,\\] and hence $\\sin\\alpha = \\tfrac 45$. Thus the distance between bases $A’B’$ and $CD$ is $12$ (in fact, $\\triangle A'B'D$ is a $9-12-15$ triangle with a $7-12-\\sqrt{193}$ triangle removed), which implies the area of $A'B'CD$ is $\\tfrac12\\cdot 12\\cdot(2+16) = 108$. Now let $O_1C = O_2A' = r_1$ and $O_2D = O_1B' = r_2$; the tangency of circles $\\omega_1$ and $\\omega_2$ implies $r_1 + r_2 = 15$. Furthermore, angles $A'O_2D$ and $A'B'D$ are opposite angles in cyclic quadrilateral $B'A'O_2D$, which implies the measure of angle $A'O_2D$ is $180^\\circ - \\alpha$. Therefore, the Law of Cosines applied to triangle $\\triangle A'O_2D$ yields \\begin{align*} 193 &= r_1^2 + r_2^2 - 2r_1r_2(-\\tfrac 35) = (r_1^2 + 2r_1r_2 + r_2^2) - \\tfrac45r_1r_2\\\\ &= (r_1+r_2)^2 - \\tfrac45 r_1r_2 = 225 - \\tfrac45r_1r_2. \\end{align*} Thus $r_1r_2 = 40$, and so the area of triangle $A'O_2D$ is $\\tfrac12r_1r_2\\sin\\alpha = 16$. Thus, the area of hexagon $ABO_{1}CDO_{2}$ is $108 + 2\\cdot 16 = 140. ~djmathman Additional Graph for Better Understanding (Rearranging of the Pizza Slices) This is how the reflection mentioned above is actually done. The reflection is actually a reorganization of the red and blue triangles, creating a symmetric figure with a isosceles trapezoid without changing the area. Basically you rearrange them so that each side contains a red triangle above a blue one. Then you calculate the area of the trapezoid and the two congruent triangles beside. ~cassphe",
"Denote by $O$ the center of $\\Omega$. Denote by $r$ the radius of $\\Omega$. We have $O_1$, $O_2$, $A$, $B$, $C$, $D$ are all on circle $\\Omega$. Denote $\\angle O_1 O O_2 = 2 \\theta$. Denote $\\angle O_1 O B = \\alpha$. Denote $\\angle O_2 O A = \\beta$. Because $B$ and $C$ are on circles $\\omega_1$ and $\\Omega$, $BC$ is a perpendicular bisector of $O_1 O$. Hence, $\\angle O_1 O C = \\alpha$. Because $A$ and $D$ are on circles $\\omega_2$ and $\\Omega$, $AD$ is a perpendicular bisector of $O_2 O$. Hence, $\\angle O_2 O D = \\beta$. In $\\triangle O O_1 O_2$, \\[ O_1 O_2 = 2 r \\sin \\theta . \\] Hence, \\[ 2 r \\sin \\theta = 15 . \\] In $\\triangle O AB$, \\begin{align*} AB & = 2 r \\sin \\frac{2 \\theta - \\alpha - \\beta}{2} \\\\ & = 2 r \\sin \\theta \\cos \\frac{\\alpha + \\beta}{2} - 2 r \\cos \\theta \\sin \\frac{\\alpha + \\beta}{2} \\\\ & = 15 \\cos \\frac{\\alpha + \\beta}{2} - 2 r \\cos \\theta \\sin \\frac{\\alpha + \\beta}{2} . \\end{align*} Hence, \\[ 15 \\cos \\frac{\\alpha + \\beta}{2} - 2 r \\cos \\theta \\sin \\frac{\\alpha + \\beta}{2} = 2 . \\hspace{1cm} (1) \\] In $\\triangle O CD$, \\begin{align*} CD & = 2 r \\sin \\frac{360^\\circ - 2 \\theta - \\alpha - \\beta}{2} \\\\ & = 2 r \\sin \\left( \\theta + \\frac{\\alpha + \\beta}{2} \\right) \\\\ & = 2 r \\sin \\theta \\cos \\frac{\\alpha + \\beta}{2} + 2 r \\cos \\theta \\sin \\frac{\\alpha + \\beta}{2} \\\\ & = 15 \\cos \\frac{\\alpha + \\beta}{2} + 2 r \\cos \\theta \\sin \\frac{\\alpha + \\beta}{2} . \\end{align*} Hence, \\[ 15 \\cos \\frac{\\alpha + \\beta}{2} + 2 r \\cos \\theta \\sin \\frac{\\alpha + \\beta}{2} = 16 . \\hspace{1cm} (2) \\] Taking $\\frac{(1) + (2)}{30}$, we get $\\cos \\frac{\\alpha + \\beta}{2} = \\frac{3}{5}$. Thus, $\\sin \\frac{\\alpha + \\beta}{2} = \\frac{4}{5}$. Taking these into (1), we get $2 r \\cos \\theta = \\frac{35}{4}$. Hence, \\begin{align*} 2 r & = \\sqrt{ \\left( 2 r \\sin \\theta \\right)^2 + \\left( 2 r \\cos \\theta \\right)^2} \\\\ & = \\frac{5}{4} \\sqrt{193} . \\end{align*} Hence, $\\cos \\theta = \\frac{7}{\\sqrt{193}}$. In $\\triangle O O_1 B$, \\[ O_1 B = 2 r \\sin \\frac{\\alpha}{2} . \\] In $\\triangle O O_2 A$, by applying the law of sines, we get \\[ O_2 A = 2 r \\sin \\frac{\\beta}{2} . \\] Because circles $\\omega_1$ and $\\omega_2$ are externally tangent, $B$ is on circle $\\omega_1$, $A$ is on circle $\\omega_2$, \\begin{align*} O_1 O_2 & = O_1 B + O_2 A \\\\ & = 2 r \\sin \\frac{\\alpha}{2} + 2 r \\sin \\frac{\\beta}{2} \\\\ & = 2 r \\left( \\sin \\frac{\\alpha}{2} + \\sin \\frac{\\beta}{2} \\right) . \\end{align*} Thus, $\\sin \\frac{\\alpha}{2} + \\sin \\frac{\\beta}{2} = \\frac{12}{\\sqrt{193}}$. Now, we compute $\\sin \\alpha$ and $\\sin \\beta$. Recall $\\cos \\frac{\\alpha + \\beta}{2} = \\frac{3}{5}$ and $\\sin \\frac{\\alpha + \\beta}{2} = \\frac{4}{5}$. Thus, $e^{i \\frac{\\alpha}{2}} e^{i \\frac{\\beta}{2}} = e^{i \\frac{\\alpha + \\beta}{2}} = \\frac{3}{5} + i \\frac{4}{5}$. We also have \\begin{align*} \\sin \\frac{\\alpha}{2} + \\sin \\frac{\\beta}{2} & = \\frac{1}{2i} \\left( e^{i \\frac{\\alpha}{2}} - e^{-i \\frac{\\alpha}{2}} + e^{i \\frac{\\beta}{2}} - e^{-i \\frac{\\beta}{2}} \\right) \\\\ & = \\frac{1}{2i} \\left( 1 - \\frac{1}{e^{i \\frac{\\alpha + \\beta}{2}} } \\right) \\left( e^{i \\frac{\\alpha}{2}} + e^{i \\frac{\\beta}{2}} \\right) . \\end{align*} Thus, \\begin{align*} \\sin \\alpha + \\sin \\beta & = \\frac{1}{2i} \\left( e^{i \\alpha} - e^{-i \\alpha} + e^{i \\beta} - e^{-i \\beta} \\right) \\\\ & = \\frac{1}{2i} \\left( 1 - \\frac{1}{e^{i \\left( \\alpha + \\beta \\right)}} \\right) \\left( e^{i \\alpha} + e^{i \\beta} \\right) \\\\ & = \\frac{1}{2i} \\left( 1 - \\frac{1}{e^{i \\left( \\alpha + \\beta \\right)}} \\right) \\left( \\left( e^{i \\frac{\\alpha}{2}} + e^{i \\frac{\\beta}{2}} \\right)^2 - 2 e^{i \\frac{\\alpha + \\beta}{2}} \\right) \\\\ & = \\frac{1}{2i} \\left( 1 - \\frac{1}{e^{i \\left( \\alpha + \\beta \\right)}} \\right) \\left( \\left( \\frac{2 i \\left( \\sin \\frac{\\alpha}{2} + \\sin \\frac{\\beta}{2} \\right)}{1 - \\frac{1}{e^{i \\frac{\\alpha + \\beta}{2}} }} \\right)^2 - 2 e^{i \\frac{\\alpha + \\beta}{2}} \\right) \\\\ & = - \\frac{1}{i} \\left( e^{i \\frac{\\alpha + \\beta}{2}} - e^{-i \\frac{\\alpha + \\beta}{2}} \\right) \\left( \\frac{2 \\left( \\sin \\frac{\\alpha}{2} + \\sin \\frac{\\beta}{2} \\right)^2} {e^{i \\frac{\\alpha + \\beta}{2}} + e^{-i \\frac{\\alpha + \\beta}{2}} - 2 } + 1 \\right) \\\\ & = \\frac{167 \\cdot 8}{193 \\cdot 5 } . \\end{align*} Therefore, \\begin{align*} {\\rm Area} \\ ABO_1CDO_2 & = {\\rm Area} \\ \\triangle O_3 AB + {\\rm Area} \\ \\triangle O_3 BO_1 + {\\rm Area} \\ \\triangle O_3 O_1 C \\\\ & \\quad + {\\rm Area} \\ \\triangle O_3 C D + {\\rm Area} \\ \\triangle O_3 D O_2 + {\\rm Area} \\ \\triangle O_3 O_2 A \\\\ & = \\frac{1}{2} r^2 \\left( \\sin \\left( 2 \\theta - \\alpha - \\beta \\right) + \\sin \\alpha + \\sin \\alpha + \\sin \\left( 360^\\circ - 2 \\theta - \\alpha - \\beta \\right) + \\sin \\beta + \\sin \\beta \\right) \\\\ & = \\frac{1}{2} r^2 \\left( \\sin \\left( 2 \\theta - \\alpha - \\beta \\right) - \\sin \\left( 2 \\theta + \\alpha + \\beta \\right) + 2 \\sin \\alpha + 2 \\sin \\beta \\right) \\\\ & = r^2 \\left( - \\cos 2 \\theta \\sin \\left( \\alpha + \\beta \\right) + \\sin \\alpha + \\sin \\beta \\right) \\\\ & = r^2 \\left( \\left( 1 - 2 \\cos^2 \\theta \\right) 2 \\sin \\frac{\\alpha + \\beta}{2} \\cos \\frac{\\alpha + \\beta}{2} + \\sin \\alpha + \\sin \\beta \\right) \\\\ & = \\textbf{(140) } . \\end{align*} ~Steven Chen (www.professorchenedu.com)",
"Let points $A'$ and $B'$ be the reflections of $A$ and $B,$ respectively, about the perpendicular bisector of $O_1 O_2.$ \\[B'O_2 = BO_1 = O_1 P = O_1 C,\\] \\[A'O_1 = AO_2 = O_2 P = O_2 D.\\] We establish the equality of the arcs and conclude that the corresponding chords are equal \\[\\overset{\\Large\\frown} {CO_1} + \\overset{\\Large\\frown} {A'O_1} +\\overset{\\Large\\frown} {A'B'} = \\overset{\\Large\\frown} {B'O_2} +\\overset{\\Large\\frown} {A'O_1} +\\overset{\\Large\\frown} {A'B'} =\\overset{\\Large\\frown} {B'O_2} +\\overset{\\Large\\frown} {DO_2} +\\overset{\\Large\\frown} {A'B'}\\] \\[\\implies A'D = B'C = O_1 O_2 = 15.\\] Similarly $A'C = B'D \\implies \\triangle A'CO_1 = \\triangle B'DO_2.$ Ptolemy's theorem on $A'CDB'$ yields \\[B'D \\cdot A'C + A'B' \\cdot CD = A'D \\cdot B'C \\implies\\] \\[B'D^2 + 2 \\cdot 16 = 15^2 \\implies B'D = A'C = \\sqrt{193}.\\] The area of the trapezoid $A'CDB'$ is equal to the area of an isosceles triangle with sides $A'D = B'C = 15$ and $A'B' + CD = 18.$ The height of this triangle is $\\sqrt{15^2-9^2} = 12.$ The area of $A'CDB'$ is $108.$ \\[\\sin \\angle B'CD = \\frac{12}{15} = \\frac{4}{5},\\] \\[\\angle B'CD + \\angle B'O_2 D = 180^o \\implies \\sin \\angle B'O_2 D = \\frac{4}{5}.\\] Denote $\\angle B'O_2 D = 2\\alpha.$ $\\angle B'O_2 D > \\frac{\\pi}{2},$ hence $\\cos \\angle B'O_2 D = \\cos 2\\alpha = -\\frac{3}{5}.$ \\[\\tan \\alpha =\\frac { \\sin 2 \\alpha}{1+\\cos 2 \\alpha} = \\frac {4/5}{1 - 3/5}=2.\\] Semiperimeter of $\\triangle B'O_2 D$ is $s = \\frac {15 + \\sqrt{193}}{2}.$ The distance from the vertex $O_2$ to the tangent points of the inscribed circle of the triangle $B'O_2 D$ is equal $s – B'D = \\frac{15 – \\sqrt{193}}{2}.$ The radius of the inscribed circle is $r = (s – B'D) \\tan \\alpha.$ The area of triangle $B'O_2 D$ is $[B'O_2 D] = sr = s (s – B'D) \\tan \\alpha = \\frac {15^2 – 193}{2} = 16.$ The hexagon $ABO_1 CDO_2$ has the same area as hexagon $B'A'O_1 CDO_2.$ The area of hexagon $B'A'O_1 CDO_2$ is equal to the sum of the area of the trapezoid $A'CDB'$ and the areas of two equal triangles $B'O_2 D$ and $A'O_1 C,$ so the area of the hexagon $ABO_1 CDO_2$ is \\[108 + 16 + 16 = 140.\\] vladimir.shelomovskii@gmail.com, vvsss",
"Let circle $O_1$'s radius be $r$, then the radius of circle $O_2$ is $15-r$. Based on Brahmagupta's Formula, the hexagon's Area $= \\sqrt{14(1)(16-r)(1+r)} + \\sqrt{7(8)(23-r)(8+r)}$. Now we only need to find the $r$. Connect $O_1$ and $A$ , $O_1$ and $D$ , and let $X$ be the point of intersection between $O_1D$ and circle $O_2$ , based on the \" 2 Non-Congruent Triangles of 'SSA' Scenario \" , we can immediately see $O_1X = O_1A$ and therefore get an equation from the \"Power of A Point Theorem: $(O_1A)(O_1D) = r(15+15-r) = 15r + r(15-r)$ (1). Similarly, $(O_2B)(O_2C) = (15-r)(15+r) = 15(15-r) + r(15-r)$ (2). We can also get two other equations about these 4 segments from Ptolemy's Theorem: $(O_1A)(O_2B) = 30 + r(15-r)$ (3) $(O_1D)(O_2C) = 240 + r(15-r)$ (4) Multiply equations (1) and (2), and equations (3) and (4) respectively, we will get a very simple and nice equation of $r$: $2(15^2)r(15-r) = 7200 + 270r(15-r)$, then: $r(15-r) = 40$. This result is good enough for us to find the hexagon's area, which: $= \\sqrt{14(1)(16-r)(1+r)} + \\sqrt{7(8)(23-r)(8+r)}$ $= \\sqrt{14(1)(1+15-r)(1+r)} + \\sqrt{7(8)(8+15-r)(8+r)}$ $= \\sqrt{14(1)(1+15+40)} + \\sqrt{7(8)(64+8(15)+40)}$ $= 28 + 112 = \\textbf{140}. eJMaSc"
] |
2023-I-1
| 2,023
| 1
|
Five men and nine women stand equally spaced around a circle in random order. The probability that every man stands diametrically opposite a woman is $\frac{m}{n},$ where $m$ and $n$ are relatively prime positive integers. Find $m+n.$
| 191
|
I
|
[
"For simplicity purposes, we consider two arrangements different even if they only differ by rotations or reflections. In this way, there are $14!$ arrangements without restrictions. First, there are $\\binom{7}{5}$ ways to choose the man-woman diameters. Then, there are $10\\cdot8\\cdot6\\cdot4\\cdot2$ ways to place the five men each in a man-woman diameter. Finally, there are $9!$ ways to place the nine women without restrictions. Together, the requested probability is \\[\\frac{\\tbinom{7}{5}\\cdot(10\\cdot8\\cdot6\\cdot4\\cdot2)\\cdot9!}{14!} = \\frac{21\\cdot(10\\cdot8\\cdot6\\cdot4\\cdot2)}{14\\cdot13\\cdot12\\cdot11\\cdot10} = \\frac{48}{143},\\] from which the answer is $48+143 = 191 ~MRENTHUSIASM",
"We can simply just loop through each of the men and find the probability that the person opposite from him is a woman. Start by sitting down the $1$st man. The probability that the person opposite to him is a woman is $\\frac{9}{13}$ since out of the $13$ people who can sit opposite to him, $9$ can be a woman. With the $2$nd man, we can use the same logic: there are $11$ people who can sit opposite to him, but only $8$ of them are a woman, so the probability is $\\frac{8}{11}.$ We use the same logic for the $3$rd, $4$th and $5$th men to get probabilities of $\\frac{7}{9}$, $\\frac{6}{7}$ and $\\frac{5}{5},$ respectively. Multiplying these probabilities, we get a final answer of \\[\\frac{9}{13}\\cdot\\frac{8}{11}\\cdot\\frac{7}{9}\\cdot\\frac{6}{7}\\cdot\\frac{5}{5}=\\frac{48}{143}\\longrightarrow 48+143 = 191.\\] ~s214425 (Inspired by Math Jam)",
"This problem is equivalent to solving for the probability that no man is standing diametrically opposite to another man. We can simply just construct this. We first place the $1$st man anywhere on the circle, now we have to place the $2$nd man somewhere around the circle such that he is not diametrically opposite to the first man. This can happen with a probability of $\\frac{12}{13}$ because there are $13$ available spots, and $12$ of them are not opposite to the first man. We do the same thing for the $3$rd man, finding a spot for him such that he is not opposite to the other $2$ men, which would happen with a probability of $\\frac{10}{12}$ using similar logic. Doing this for the $4$th and $5$th men, we get probabilities of $\\frac{8}{11}$ and $\\frac{6}{10}$ respectively. Multiplying these probabilities, we get, \\[\\frac{12}{13}\\cdot\\frac{10}{12}\\cdot\\frac{8}{11}\\cdot\\frac{6}{10}=\\frac{48}{143}\\longrightarrow191.\\] ~s214425",
"Assume that rotations and reflections are distinct arrangements, and replace men and women with identical M's and W's, respectively. (We can do that because the number of ways to arrange $5$ men in a circle and the number of ways to arrange $9$ women in a circle, are constants.) The total number of ways to arrange $5$ M's and $9$ W's is $\\binom{14}{5} = 2002.$ To count the number of valid arrangements (i.e. arrangements where every M is diametrically opposite a W), we notice that exactly $2$ of the pairs of diametrically opposite positions must be occupied by $2$ W's. There are $\\binom{7}{2} = 21$ ways to choose these $2$ pairs. For the remaining $5$ pairs, we have to choose which position is occupied by an M and which is occupied by a W. This can be done in $2^{5} = 32$ ways. Therefore, there are $21*32 = 672$ valid arrangements. Therefore, the probability that an arrangement is valid is $\\frac{672}{2002} = \\frac{48}{143}$ for an answer of $191 ~pianoboy",
"To start off, we calculate the total amount of ways to organize all $14$ people irrespective of any constraints. This is simply ${14\\choose5} = 2002$, because we just count how many ways we can place all $5$ men in any of the $14$ slots. Since men cannot be diametrically opposite with each other, because of the constraints, placing down one man in any given spot will make another spot on the opposite side of the circle unable to hold any men. This means that placing down one man will effectively take away $2$ spots. There are $14$ possible slots the first man can be placed. Once that man was placed, the next man only have $12$ possible slots because the slot that the first man is in is taken and the diametrically opposite spot to the first man can't have any men. Similar logic applies for the third man, who has $10$ possible slots. The fourth man has $8$ possible slots, and the fifth man has $6$ possible slots. This means the number of ways you can place all $5$ men down is $14 \\cdot 12 \\cdot 10 \\cdot 8 \\cdot 6$. However, since the men are all indistinct from each other, you also have to divide that value by $5! = 120$, since there are $120$ ways to arrange the $5$ men in each possible positioning of the men on the circle. This means the total number of ways to arrange the men around the circle so that none of them are diametrically opposite of each other is: $\\frac{14 \\cdot 12 \\cdot 10 \\cdot 8 \\cdot 6}{5!} = 672$. The women simply fill in the rest of the available slots in each arrangement of men. Thus, the final probability is $\\frac{672}{2002} = \\frac{48}{143}$, meaning the answer is $48 + 143 = 191. ~ericshi1685",
"We will first assign seats to the men. The first man can be placed in any of the $14$ slots. The second man can be placed in any of the remaining $13$ seats, except for the one diametrically opposite to the first man. So, there are $13 - 1 = 12$ ways to seat him. With a similar argument, the third man can be seated in $10$ ways, the fourth man in $8$ ways and the last man in $6$ ways. So, the total number of ways to arrange the men is $14 \\cdot 12 \\cdot 10 \\cdot 8 \\cdot 6$. The women go to the remaining $9$ spots. Note that since none of the seats diametrically opposite to the men is occupied, each man is opposite a woman. The number of ways to arrange the women is therefore, simply $9!$, meaning that the total number of ways to arrange the people with restrictions is $14 \\cdot 12 \\cdot 10 \\cdot 8 \\cdot 6 \\cdot 9!$ In general, there are $14!$ ways to arrange the people without restrictions. So, the probability is \\[\\frac{14 \\cdot 12 \\cdot 10 \\cdot 8 \\cdot 6 \\cdot 9!}{14!} = \\frac{8 \\cdot 6}{13 \\cdot 11} = \\frac{48}{143}.\\] The answer is $48 + 143 = 191. ~baassid24",
"First pin one man on one seat (to ensure no rotate situations). Then there are $13!$ arrangements. Because $5$ men must have women at their opposite side, we consider the $2$nd man and the woman opposite as one group and name it $P_2.$ There are $4$ groups, $P_1, P_2, P_3, P_4$ except the first man pinned on the same point. And for the rest $4$ women, name them $P_5$ and $P_6.$ First to order $P_1, P_2, P_3, P_4, P_5, P_6,$ there are $6!$ ways. For the $1$st man, there are $9$ women to choose, $8$ for the $2$nd, $\\ldots,$ $5$ for the $5$th, and then for the $2$ women pairs $3$ and $1.$ Because every $2$ person in the group have chance to change their position, there are $2^6$ possibilities. So the possibility is \\[P=\\frac{6!\\cdot 9\\cdot 8\\cdot 7\\cdot 6\\cdot 5\\cdot 3\\cdot 1 \\cdot 2^6}{13!}=\\frac{48}{143}.\\] The answer is $48+143=191 ~PLASTA",
"We get around the condition that each man can't be opposite to another man by simply considering all $7$ diagonals, and choosing $5$ where there will be a single man. For each diagonal, the man can go on either side, and there are $\\binom{14}{5}$ ways to arrange the men and the women in total. Thus our answer is $\\frac{\\binom{7}{5}\\cdot 2^5}{\\binom{14}{5}} = \\frac{48}{143}.$ We get $48 + 143 = 191 ~AtharvNaphade",
"We can find the probability of one arrangement occurring, and multiply it by the total number of arrangements. The probability of a man being in any specific position is $\\frac{5}{14}.$ The probability of a woman being across from him is $\\frac{9}{13}.$ The probability of a man being in any valid position is now $\\frac{4}{12},$ and the probability of a woman being across from him is $\\frac{8}{11},$ and so forth. We stop when there are no more men left. Multiplying these probabilities together, \\[P(\\mathrm{One\\ successful\\ outcome})=\\frac{5}{14}\\cdot \\frac{9}{13}\\cdot \\frac{4}{12}\\cdot \\frac{8}{11}\\cdot \\frac{3}{10}\\cdot \\frac{7}{9}\\cdot \\frac{2}{8}\\cdot \\frac{1}{6}\\cdot \\frac{6}{7}\\cdot \\frac{5}{5} = \\frac{1}{2002}.\\] To find the total number of successful outcomes, we consider the diagonals; the total number of diagonals to be made is $\\binom75$, since there are $7$ total diagonals, and we want to choose $5$ of them to connect a man to a woman. For each of these diagonals, the man can be on either side of the diagonal. It follows that there are $2$ possibilities for each diagonal (man on one side, woman on the other, and vice versa). There are $5$ diagonals with a man and a woman, so there are $2^5$ different ways for these diagonals to appear. There are $\\binom75$ successful diagonals, and for each of these diagonals, there are $2^5$ ways to seat the men and the women, there are $\\binom75$ $\\cdot 2^5$ successful outcomes. Recall that \\[P(\\mathrm{Any\\ successful\\ outcome})=P(\\mathrm{One\\ successful\\ outcome})\\cdot P(\\mathrm{Total\\ number\\ of\\ successful\\ outcomes}).\\] Therefore, \\[P(\\mathrm{Any\\ successful\\ outcome}) = \\frac{1}{2002}\\cdot 2^5\\cdot \\binom75 = \\frac{1}{2002}\\cdot 2^5\\cdot 21 = \\frac{2^5\\cdot 21}{2002} = \\frac{48}{143}.\\] The requested sum is $48+143=191 -Benedict T (countmath1)"
] |
2023-I-2
| 2,023
| 2
|
Positive real numbers $b \not= 1$ and $n$ satisfy the equations \[\sqrt{\log_b n} = \log_b \sqrt{n} \qquad \text{and} \qquad b \cdot \log_b n = \log_b (bn).\] The value of $n$ is $\frac{j}{k},$ where $j$ and $k$ are relatively prime positive integers. Find $j+k.$
| 881
|
I
|
[
"Denote $x = \\log_b n$. Hence, the system of equations given in the problem can be rewritten as \\begin{align*} \\sqrt{x} & = \\frac{1}{2} x , \\\\ bx & = 1 + x . \\end{align*} Solving the system gives $x = 4$ and $b = \\frac{5}{4}$. Therefore, \\[n = b^x = \\frac{625}{256}.\\] Therefore, the answer is $625 + 256 = 881. ~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)",
"We can use the property that $\\log(xy) = \\log(x) + \\log(y)$ on the first equation. We get $\\log_b(bn) = 1 + \\log_b(n)$. Then, subtracting $\\log_b(n)$ from both sides, we get $(b-1) \\log_b(n) = 1$, therefore $\\log_b(n) = \\frac{1}{b-1}$. Substituting that into our first equation, we get $\\frac{1}{2b-2} = \\sqrt{\\frac{1}{b-1}}$. Squaring, reciprocating, and simplifying both sides, we get the quadratic $4b^2 - 9b + 5 = 0$. Solving for $b$, we get $\\frac{5}{4}$ and $1$. Since the problem said that $b \\neq 1$, $b = \\frac{5}{4}$. To solve for $n$, we can use the property that $\\log_b(n) = \\frac{1}{b-1}$. $\\log_\\frac{5}{4}(n) = 4$, so $n = \\frac{5^4}{4^4} = \\frac{625}{256}$. Adding these together, we get $881 ~idk12345678",
"We can let $n=b^{4x^2}$. Then, in the first equation, the LHS becomes $2x$ and the RHS becomes $2x^2$. Therefore, $x$ must be $1$ ($x$ can't be $0$). So now we know $n=b^4$. So we can plug this into the second equation to get $n=b^4$. This gives $b\\cdot4=5$, so $b= \\frac{5}{4}$ and $n= b^4=\\frac{625}{256}$. Adding the numerator and denominator gives $881. Honestly, this problem is kinda well placed. ~yrock",
"We can let $n=b^{4x^2}$. Then, in the first equation, the LHS becomes $2x$ and the RHS becomes $2x^2$. Therefore, $x$ must be $1$ ($x$ can't be $0$). So now we know $n=b^4$. So we can plug this into the second equation to get $n=b^4$. This gives $b\\cdot4=5$, so $b= \\frac{5}{4}$ and $n= b^4=\\frac{625}{256}$. Adding the numerator and denominator gives $881. Honestly, this problem is kinda well placed. ~yrock"
] |
2023-I-3
| 2,023
| 3
|
A plane contains $40$ lines, no $2$ of which are parallel. Suppose that there are $3$ points where exactly $3$ lines intersect, $4$ points where exactly $4$ lines intersect, $5$ points where exactly $5$ lines intersect, $6$ points where exactly $6$ lines intersect, and no points where more than $6$ lines intersect. Find the number of points where exactly $2$ lines intersect.
| 607
|
I
|
[
"In this solution, let $\\boldsymbol{n}$-line points be the points where exactly $n$ lines intersect. We wish to find the number of $2$-line points. There are $\\binom{40}{2}=780$ pairs of lines. Among them: The $3$-line points account for $3\\cdot\\binom32=9$ pairs of lines. The $4$-line points account for $4\\cdot\\binom42=24$ pairs of lines. The $5$-line points account for $5\\cdot\\binom52=50$ pairs of lines. The $6$-line points account for $6\\cdot\\binom62=90$ pairs of lines. It follows that the $2$-line points account for $780-9-24-50-90=607 pairs of lines, where each pair intersect at a single point. ~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com) ~MRENTHUSIASM"
] |
2023-I-4
| 2,023
| 4
|
The sum of all positive integers $m$ such that $\frac{13!}{m}$ is a perfect square can be written as $2^a3^b5^c7^d11^e13^f,$ where $a,b,c,d,e,$ and $f$ are positive integers. Find $a+b+c+d+e+f.$
| 12
|
I
|
[
"We first rewrite $13!$ as a prime factorization, which is $2^{10}\\cdot3^5\\cdot5^2\\cdot7\\cdot11\\cdot13.$ For the fraction to be a square, it needs each prime to be an even power. This means $m$ must contain $7\\cdot11\\cdot13$. Also, $m$ can contain any even power of $2$ up to $2^{10}$, any odd power of $3$ up to $3^{5}$, and any even power of $5$ up to $5^{2}$. The sum of $m$ is \\[(2^0+2^2+2^4+2^6+2^8+2^{10})(3^1+3^3+3^5)(5^0+5^2)(7^1)(11^1)(13^1) =\\] \\[1365\\cdot273\\cdot26\\cdot7\\cdot11\\cdot13 = 2\\cdot3^2\\cdot5\\cdot7^3\\cdot11\\cdot13^4.\\] Therefore, the answer is $1+2+1+3+1+4=012. ~chem1kall",
"The prime factorization of $13!$ is \\[2^{10} \\cdot 3^5 \\cdot 5^2 \\cdot 7 \\cdot 11 \\cdot 13.\\] To get $\\frac{13!}{m}$ a perfect square, we must have $m = 2^{2x} \\cdot 3^{1 + 2y} \\cdot 5^{2z} \\cdot 7 \\cdot 11 \\cdot 13$, where $x \\in \\left\\{ 0, 1, \\cdots , 5 \\right\\}$, $y \\in \\left\\{ 0, 1, 2 \\right\\}$, $z \\in \\left\\{ 0, 1 \\right\\}$. Hence, the sum of all feasible $m$ is \\begin{align*} \\sum_{x=0}^5 \\sum_{y=0}^2 \\sum_{z=0}^1 2^{2x} \\cdot 3^{1 + 2y} \\cdot 5^{2z} \\cdot 7 \\cdot 11 \\cdot 13 & = \\left( \\sum_{x=0}^5 2^{2x} \\right) \\left( \\sum_{y=0}^2 3^{1 + 2y} \\right) \\left( \\sum_{z=0}^1 5^{2z} \\right) 7 \\cdot 11 \\cdot 13 \\\\ & = \\frac{4^6 - 1}{4-1} \\cdot \\frac{3 \\cdot \\left( 9^3 - 1 \\right)}{9 - 1} \\cdot \\frac{25^2 - 1}{25 - 1} \\cdot 7 \\cdot 11 \\cdot 13 \\\\ & = 2 \\cdot 3^2 \\cdot 5 \\cdot 7^3 \\cdot 11 \\cdot 13^4 . \\end{align*} Therefore, the answer is \\begin{align*} 1 + 2 + 1 + 3 + 1 + 4 & = 012 . \\end{align*} ~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)",
"Try smaller cases. There is clearly only one $m$ that makes $\\frac{2!}{m}$ a square, and this is $m=2$. Here, the sum of the exponents in the prime factorization is just $1$. Furthermore, the only $m$ that makes $\\frac{3!}{m}$ a square is $m = 6 = 2^13^1$, and the sum of the exponents is $2$ here. Trying $\\frac{4!}{m}$ and $\\frac{5!}{m}$, the sums of the exponents are $3$ and $4$. Based on this, we (incorrectly!) conclude that, when we are given $\\frac{n!}{m}$, the desired sum is $n-1$. The problem gives us $\\frac{13!}{m}$, so the answer is $13-1 = 012 -\"fake\" warning by oinava",
"Try smaller cases. There is clearly only one $m$ that makes $\\frac{2!}{m}$ a square, and this is $m=2$. Here, the sum of the exponents in the prime factorization is just $1$. Furthermore, the only $m$ that makes $\\frac{3!}{m}$ a square is $m = 6 = 2^13^1$, and the sum of the exponents is $2$ here. Trying $\\frac{4!}{m}$ and $\\frac{5!}{m}$, the sums of the exponents are $3$ and $4$. Based on this, we (incorrectly!) conclude that, when we are given $\\frac{n!}{m}$, the desired sum is $n-1$. The problem gives us $\\frac{13!}{m}$, so the answer is $13-1 = 012 -\"fake\" warning by oinava"
] |
2023-I-5
| 2,023
| 5
|
Let $P$ be a point on the circle circumscribing square $ABCD$ that satisfies $PA \cdot PC = 56$ and $PB \cdot PD = 90.$ Find the area of $ABCD.$
| 106
|
I
|
[
"Ptolemy's theorem states that for cyclic quadrilateral $WXYZ$, $WX\\cdot YZ + XY\\cdot WZ = WY\\cdot XZ$. We may assume that $P$ is between $B$ and $C$. Let $PA = a$, $PB = b$, $PC = c$, $PD = d$, and $AB = s$. We have $a^2 + c^2 = AC^2 = 2s^2$, because $AC$ is a diameter of the circle. Similarly, $b^2 + d^2 = 2s^2$. Therefore, $(a+c)^2 = a^2 + c^2 + 2ac = 2s^2 + 2(56) = 2s^2 + 112$. Similarly, $(b+d)^2 = 2s^2 + 180$. By Ptolemy's Theorem on $PCDA$, $as + cs = ds\\sqrt{2}$, and therefore $a + c = d\\sqrt{2}$. By Ptolemy's on $PBAD$, $bs + ds = as\\sqrt{2}$, and therefore $b + d = a\\sqrt{2}$. By squaring both equations, we obtain \\begin{alignat*}{8} 2d^2 &= (a+c)^2 &&= 2s^2 + 112, \\\\ 2a^2 &= (b+d)^2 &&= 2s^2 + 180. \\end{alignat*} Thus, $a^2 = s^2 + 90$, and $d^2 = s^2 + 56$. Plugging these values into $a^2 + c^2 = b^2 + d^2 = 2s^2$, we obtain $c^2 = s^2 - 90$, and $b^2 = s^2 - 56$. Now, we can solve using $a$ and $c$ (though using $b$ and $d$ yields the same solution for $s$). \\begin{align*} ac = (\\sqrt{s^2 - 90})(\\sqrt{s^2 + 90}) &= 56 \\\\ (s^2 + 90)(s^2 - 90) &= 56^2 \\\\ s^4 &= 90^2 + 56^2 = 106^2 \\\\ s^2 &= 106. \\end{align*} ~mathboy100",
"Ptolemy's theorem states that for cyclic quadrilateral $WXYZ$, $WX\\cdot YZ + XY\\cdot WZ = WY\\cdot XZ$. We may assume that $P$ is between $B$ and $C$. Let $PA = a$, $PB = b$, $PC = c$, $PD = d$, and $AB = s$. We have $a^2 + c^2 = AC^2 = 2s^2$, because $AC$ is a diameter of the circle. Similarly, $b^2 + d^2 = 2s^2$. Therefore, $(a+c)^2 = a^2 + c^2 + 2ac = 2s^2 + 2(56) = 2s^2 + 112$. Similarly, $(b+d)^2 = 2s^2 + 180$. By Ptolemy's Theorem on $PCDA$, $as + cs = ds\\sqrt{2}$, and therefore $a + c = d\\sqrt{2}$. By Ptolemy's on $PBAD$, $bs + ds = as\\sqrt{2}$, and therefore $b + d = a\\sqrt{2}$. By squaring both equations, we obtain \\begin{alignat*}{8} 2d^2 &= (a+c)^2 &&= 2s^2 + 112, \\\\ 2a^2 &= (b+d)^2 &&= 2s^2 + 180. \\end{alignat*} Thus, $a^2 = s^2 + 90$, and $d^2 = s^2 + 56$. Plugging these values into $a^2 + c^2 = b^2 + d^2 = 2s^2$, we obtain $c^2 = s^2 - 90$, and $b^2 = s^2 - 56$. Now, we can solve using $a$ and $c$ (though using $b$ and $d$ yields the same solution for $s$). \\begin{align*} ac = (\\sqrt{s^2 - 90})(\\sqrt{s^2 + 90}) &= 56 \\\\ (s^2 + 90)(s^2 - 90) &= 56^2 \\\\ s^4 &= 90^2 + 56^2 = 106^2 \\\\ s^2 &= 106. \\end{align*} ~mathboy100",
"By the Inscribed Angle Theorem, we conclude that $\\triangle PAC$ and $\\triangle PBD$ are right triangles. Let the brackets denote areas. We are given that \\begin{alignat*}{8} 2[PAC] &= PA \\cdot PC &&= 56, \\\\ 2[PBD] &= PB \\cdot PD &&= 90. \\end{alignat*} Let $O$ be the center of the circle, $X$ be the foot of the perpendicular from $P$ to $\\overline{AC},$ and $Y$ be the foot of the perpendicular from $P$ to $\\overline{BD},$ as shown below: [asy] /* Made by MRENTHUSIASM */ size(200); pair A, B, C, D, O, P, X, Y; A = (-sqrt(106)/2,sqrt(106)/2); B = (-sqrt(106)/2,-sqrt(106)/2); C = (sqrt(106)/2,-sqrt(106)/2); D = (sqrt(106)/2,sqrt(106)/2); O = origin; path p; p = Circle(O,sqrt(212)/2); draw(p); P = intersectionpoints(Circle(A,4),p)[1]; X = foot(P,A,C); Y = foot(P,B,D); draw(A--B--C--D--cycle); draw(P--A--C--cycle,red); draw(P--B--D--cycle,blue); draw(P--X,red+dashed); draw(P--Y,blue+dashed); markscalefactor=0.075; draw(rightanglemark(A,P,C),red); draw(rightanglemark(P,X,C),red); draw(rightanglemark(B,P,D),blue); draw(rightanglemark(P,Y,D),blue); dot(\"$A$\", A, 1.5*NW, linewidth(4)); dot(\"$B$\", B, 1.5*SW, linewidth(4)); dot(\"$C$\", C, 1.5*SE, linewidth(4)); dot(\"$D$\", D, 1.5*NE, linewidth(4)); dot(\"$P$\", P, 1.5*dir(P), linewidth(4)); dot(\"$X$\", X, 1.5*dir(20), linewidth(4)); dot(\"$Y$\", Y, 1.5*dir(Y-P), linewidth(4)); dot(\"$O$\", O, 1.5*E, linewidth(4)); [/asy] Let $d$ be the diameter of $\\odot O.$ It follows that \\begin{alignat*}{8} 2[PAC] &= d\\cdot PX &&= 56, \\\\ 2[PBD] &= d\\cdot PY &&= 90. \\end{alignat*} Moreover, note that $OXPY$ is a rectangle. By the Pythagorean Theorem, we have \\[PX^2+PY^2=PO^2.\\] We rewrite this equation in terms of $d:$ \\[\\left(\\frac{56}{d}\\right)^2+\\left(\\frac{90}{d}\\right)^2=\\left(\\frac d2\\right)^2,\\] from which $d^2=212.$ Therefore, we get \\[[ABCD] = \\frac{d^2}{2} = 106.\\] ~MRENTHUSIASM",
"By the Inscribed Angle Theorem, we conclude that $\\triangle PAC$ and $\\triangle PBD$ are right triangles. Let the brackets denote areas. We are given that \\begin{alignat*}{8} 2[PAC] &= PA \\cdot PC &&= 56, \\\\ 2[PBD] &= PB \\cdot PD &&= 90. \\end{alignat*} Let $O$ be the center of the circle, $X$ be the foot of the perpendicular from $P$ to $\\overline{AC},$ and $Y$ be the foot of the perpendicular from $P$ to $\\overline{BD},$ as shown below: [asy] /* Made by MRENTHUSIASM */ size(200); pair A, B, C, D, O, P, X, Y; A = (-sqrt(106)/2,sqrt(106)/2); B = (-sqrt(106)/2,-sqrt(106)/2); C = (sqrt(106)/2,-sqrt(106)/2); D = (sqrt(106)/2,sqrt(106)/2); O = origin; path p; p = Circle(O,sqrt(212)/2); draw(p); P = intersectionpoints(Circle(A,4),p)[1]; X = foot(P,A,C); Y = foot(P,B,D); draw(A--B--C--D--cycle); draw(P--A--C--cycle,red); draw(P--B--D--cycle,blue); draw(P--X,red+dashed); draw(P--Y,blue+dashed); markscalefactor=0.075; draw(rightanglemark(A,P,C),red); draw(rightanglemark(P,X,C),red); draw(rightanglemark(B,P,D),blue); draw(rightanglemark(P,Y,D),blue); dot(\"$A$\", A, 1.5*NW, linewidth(4)); dot(\"$B$\", B, 1.5*SW, linewidth(4)); dot(\"$C$\", C, 1.5*SE, linewidth(4)); dot(\"$D$\", D, 1.5*NE, linewidth(4)); dot(\"$P$\", P, 1.5*dir(P), linewidth(4)); dot(\"$X$\", X, 1.5*dir(20), linewidth(4)); dot(\"$Y$\", Y, 1.5*dir(Y-P), linewidth(4)); dot(\"$O$\", O, 1.5*E, linewidth(4)); [/asy] Let $d$ be the diameter of $\\odot O.$ It follows that \\begin{alignat*}{8} 2[PAC] &= d\\cdot PX &&= 56, \\\\ 2[PBD] &= d\\cdot PY &&= 90. \\end{alignat*} Moreover, note that $OXPY$ is a rectangle. By the Pythagorean Theorem, we have \\[PX^2+PY^2=PO^2.\\] We rewrite this equation in terms of $d:$ \\[\\left(\\frac{56}{d}\\right)^2+\\left(\\frac{90}{d}\\right)^2=\\left(\\frac d2\\right)^2,\\] from which $d^2=212.$ Therefore, we get \\[[ABCD] = \\frac{d^2}{2} = 106.\\] ~MRENTHUSIASM",
"[asy] /* Made by MRENTHUSIASM */ size(200); pair A, B, C, D, O, P, X, Y; A = (-sqrt(106)/2,sqrt(106)/2); B = (-sqrt(106)/2,-sqrt(106)/2); C = (sqrt(106)/2,-sqrt(106)/2); D = (sqrt(106)/2,sqrt(106)/2); O = origin; path p; p = Circle(O,sqrt(212)/2); draw(p); P = intersectionpoints(Circle(A,4),p)[1]; X = foot(P,A,C); Y = foot(P,B,D); draw(A--B--C--D--cycle); draw(P--A--C--cycle,red); draw(P--B--D--cycle,blue); draw(P--X,red+dashed); draw(P--Y,blue+dashed); markscalefactor=0.075; draw(rightanglemark(A,P,C),red); draw(rightanglemark(P,X,C),red); draw(rightanglemark(B,P,D),blue); draw(rightanglemark(P,Y,D),blue); dot(\"$A$\", A, 1.5*NW, linewidth(4)); dot(\"$B$\", B, 1.5*SW, linewidth(4)); dot(\"$C$\", C, 1.5*SE, linewidth(4)); dot(\"$D$\", D, 1.5*NE, linewidth(4)); dot(\"$P$\", P, 1.5*dir(P), linewidth(4)); dot(\"$X$\", X, 1.5*dir(20), linewidth(4)); dot(\"$Y$\", Y, 1.5*dir(Y-P), linewidth(4)); dot(\"$O$\", O, 1.5*E, linewidth(4)); [/asy] Let the center of the circle be $O$, and the radius of the circle be $r$. Since $ABCD$ is a rhombus with diagonals $2r$ and $2r$, its area is $\\dfrac{1}{2}(2r)(2r) = 2r^2$. Since $AC$ and $BD$ are diameters of the circle, $\\triangle APC$ and $\\triangle BPD$ are right triangles. Let $X$ and $Y$ be the foot of the altitudes to $AC$ and $BD$, respectively. We have \\[[\\triangle APC] = \\frac{1}{2}(PA)(PC) = \\frac{1}{2}(PX)(AC),\\] so $PX = \\dfrac{(PA)(PC)}{AC} = \\dfrac{28}{r}$. Similarly, \\[[\\triangle BPD] = \\frac{1}{2}(PB)(PD) = \\frac{1}{2}(PY)(PB),\\] so $PY = \\dfrac{(PB)(PD)}{BD} = \\dfrac{45}{r}$. Since $\\triangle APX \\sim \\triangle PCX,$ \\[\\frac{AX}{PX} = \\frac{PX}{CX}\\] \\[\\frac{AO - XO}{PX} = \\frac{PX}{OC + XO}.\\] But $PXOY$ is a rectangle, so $PY = XO$, and our equation becomes \\[\\frac{r - PY}{PX} = \\frac{PX}{r + PY}.\\] Cross multiplying and rearranging gives us $r^2 = PX^2 + PY^2 = \\left(\\dfrac{28}{r}\\right)^2 + \\left(\\dfrac{45}{r}\\right)^2$, which rearranges to $r^4 = 2809$. Therefore $[ABCD] = 2r^2 = 106. ~Cantalon",
"[asy] /* Made by MRENTHUSIASM */ size(200); pair A, B, C, D, O, P, X, Y; A = (-sqrt(106)/2,sqrt(106)/2); B = (-sqrt(106)/2,-sqrt(106)/2); C = (sqrt(106)/2,-sqrt(106)/2); D = (sqrt(106)/2,sqrt(106)/2); O = origin; path p; p = Circle(O,sqrt(212)/2); draw(p); P = intersectionpoints(Circle(A,4),p)[1]; X = foot(P,A,C); Y = foot(P,B,D); draw(A--B--C--D--cycle); draw(P--A--C--cycle,red); draw(P--B--D--cycle,blue); draw(P--X,red+dashed); draw(P--Y,blue+dashed); markscalefactor=0.075; draw(rightanglemark(A,P,C),red); draw(rightanglemark(P,X,C),red); draw(rightanglemark(B,P,D),blue); draw(rightanglemark(P,Y,D),blue); dot(\"$A$\", A, 1.5*NW, linewidth(4)); dot(\"$B$\", B, 1.5*SW, linewidth(4)); dot(\"$C$\", C, 1.5*SE, linewidth(4)); dot(\"$D$\", D, 1.5*NE, linewidth(4)); dot(\"$P$\", P, 1.5*dir(P), linewidth(4)); dot(\"$X$\", X, 1.5*dir(20), linewidth(4)); dot(\"$Y$\", Y, 1.5*dir(Y-P), linewidth(4)); dot(\"$O$\", O, 1.5*E, linewidth(4)); [/asy] Let the center of the circle be $O$, and the radius of the circle be $r$. Since $ABCD$ is a rhombus with diagonals $2r$ and $2r$, its area is $\\dfrac{1}{2}(2r)(2r) = 2r^2$. Since $AC$ and $BD$ are diameters of the circle, $\\triangle APC$ and $\\triangle BPD$ are right triangles. Let $X$ and $Y$ be the foot of the altitudes to $AC$ and $BD$, respectively. We have \\[[\\triangle APC] = \\frac{1}{2}(PA)(PC) = \\frac{1}{2}(PX)(AC),\\] so $PX = \\dfrac{(PA)(PC)}{AC} = \\dfrac{28}{r}$. Similarly, \\[[\\triangle BPD] = \\frac{1}{2}(PB)(PD) = \\frac{1}{2}(PY)(PB),\\] so $PY = \\dfrac{(PB)(PD)}{BD} = \\dfrac{45}{r}$. Since $\\triangle APX \\sim \\triangle PCX,$ \\[\\frac{AX}{PX} = \\frac{PX}{CX}\\] \\[\\frac{AO - XO}{PX} = \\frac{PX}{OC + XO}.\\] But $PXOY$ is a rectangle, so $PY = XO$, and our equation becomes \\[\\frac{r - PY}{PX} = \\frac{PX}{r + PY}.\\] Cross multiplying and rearranging gives us $r^2 = PX^2 + PY^2 = \\left(\\dfrac{28}{r}\\right)^2 + \\left(\\dfrac{45}{r}\\right)^2$, which rearranges to $r^4 = 2809$. Therefore $[ABCD] = 2r^2 = 106. ~Cantalon",
"Drop a height from point $P$ to line $\\overline{AC}$ and line $\\overline{BC}$. Call these two points to be $X$ and $Y$, respectively. Notice that the intersection of the diagonals of $\\square ABCD$ meets at a right angle at the center of the circumcircle, call this intersection point $O$. Since $OXPY$ is a rectangle, $OX$ is the distance from $P$ to line $\\overline{BD}$. We know that $\\tan{\\angle{POX}} = \\frac{PX}{XO} = \\frac{28}{45}$ by triangle area and given information. Then, notice that the measure of $\\angle{OCP}$ is half of $\\angle{XOP}$. Using the half-angle formula for tangent, \\begin{align*} \\frac{(2 \\cdot \\tan{\\angle{OCP}})}{(1-\\tan^2{\\angle{OCP}})} = \\tan{\\angle{POX}} = \\frac{28}{45} \\\\ 14\\tan^2{\\angle{OCP}} + 45\\tan{\\angle{OCP}} - 14 = 0 \\end{align*} Solving the equation above, we get that $\\tan{\\angle{OCP}} = -7/2$ or $2/7$. Since this value must be positive, we pick $\\frac{2}{7}$. Then, $\\frac{PA}{PC} = 2/7$ (since $\\triangle CAP$ is a right triangle with line $\\overline{AC}$ the diameter of the circumcircle) and $PA * PC = 56$. Solving we get $PA = 4$, $PC = 14$, giving us a diagonal of length $\\sqrt{212}$ and area $106. ~Danielzh",
"Drop a height from point $P$ to line $\\overline{AC}$ and line $\\overline{BC}$. Call these two points to be $X$ and $Y$, respectively. Notice that the intersection of the diagonals of $\\square ABCD$ meets at a right angle at the center of the circumcircle, call this intersection point $O$. Since $OXPY$ is a rectangle, $OX$ is the distance from $P$ to line $\\overline{BD}$. We know that $\\tan{\\angle{POX}} = \\frac{PX}{XO} = \\frac{28}{45}$ by triangle area and given information. Then, notice that the measure of $\\angle{OCP}$ is half of $\\angle{XOP}$. Using the half-angle formula for tangent, \\begin{align*} \\frac{(2 \\cdot \\tan{\\angle{OCP}})}{(1-\\tan^2{\\angle{OCP}})} = \\tan{\\angle{POX}} = \\frac{28}{45} \\\\ 14\\tan^2{\\angle{OCP}} + 45\\tan{\\angle{OCP}} - 14 = 0 \\end{align*} Solving the equation above, we get that $\\tan{\\angle{OCP}} = -7/2$ or $2/7$. Since this value must be positive, we pick $\\frac{2}{7}$. Then, $\\frac{PA}{PC} = 2/7$ (since $\\triangle CAP$ is a right triangle with line $\\overline{AC}$ the diameter of the circumcircle) and $PA * PC = 56$. Solving we get $PA = 4$, $PC = 14$, giving us a diagonal of length $\\sqrt{212}$ and area $106. ~Danielzh",
"Denote by $x$ the half length of each side of the square. We put the square to the coordinate plane, with $A = \\left( x, x \\right)$, $B = \\left( - x , x \\right)$, $C = \\left( - x , - x \\right)$, $D = \\left( x , - x \\right)$. The radius of the circumcircle of $ABCD$ is $\\sqrt{2} x$. Denote by $\\theta$ the argument of point $P$ on the circle. Thus, the coordinates of $P$ are $P = \\left( \\sqrt{2} x \\cos \\theta , \\sqrt{2} x \\sin \\theta \\right)$. Thus, the equations $PA \\cdot PC = 56$ and $PB \\cdot PD = 90$ can be written as \\begin{align*} \\sqrt{\\left( \\sqrt{2} x \\cos \\theta - x \\right)^2 + \\left( \\sqrt{2} x \\sin \\theta - x \\right)^2} \\cdot \\sqrt{\\left( \\sqrt{2} x \\cos \\theta + x \\right)^2 + \\left( \\sqrt{2} x \\sin \\theta + x \\right)^2} & = 56 \\\\ \\sqrt{\\left( \\sqrt{2} x \\cos \\theta + x \\right)^2 + \\left( \\sqrt{2} x \\sin \\theta - x \\right)^2} \\cdot \\sqrt{\\left( \\sqrt{2} x \\cos \\theta - x \\right)^2 + \\left( \\sqrt{2} x \\sin \\theta + x \\right)^2} & = 90 \\end{align*} These equations can be reformulated as \\begin{align*} x^4 \\left( 4 - 2 \\sqrt{2} \\left( \\cos \\theta + \\sin \\theta \\right) \\right) \\left( 4 + 2 \\sqrt{2} \\left( \\cos \\theta + \\sin \\theta \\right) \\right) & = 56^2 \\\\ x^4 \\left( 4 + 2 \\sqrt{2} \\left( \\cos \\theta - \\sin \\theta \\right) \\right) \\left( 4 - 2 \\sqrt{2} \\left( \\cos \\theta - \\sin \\theta \\right) \\right) & = 90^2 \\end{align*} These equations can be reformulated as \\begin{align*} 2 x^4 \\left( 1 - 2 \\cos \\theta \\sin \\theta \\right) & = 28^2 \\hspace{1cm} (1) \\\\ 2 x^4 \\left( 1 + 2 \\cos \\theta \\sin \\theta \\right) & = 45^2 \\hspace{1cm} (2) \\end{align*} Taking $\\frac{(1)}{(2)}$, by solving the equation, we get \\[ 2 \\cos \\theta \\sin \\theta = \\frac{45^2 - 28^2}{45^2 + 28^2} . \\hspace{1cm} (3) \\] Plugging (3) into (1), we get \\begin{align*} {\\rm Area} \\ ABCD & = \\left( 2 x \\right)^2 \\\\ & = 4 \\sqrt{\\frac{28^2}{2 \\left( 1 - 2 \\cos \\theta \\sin \\theta \\right)}} \\\\ & = 2 \\sqrt{45^2 + 28^2} \\\\ & = 2 \\cdot 53 \\\\ & = \\textbf{(106) } . \\end{align*} ~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)",
"Denote by $x$ the half length of each side of the square. We put the square to the coordinate plane, with $A = \\left( x, x \\right)$, $B = \\left( - x , x \\right)$, $C = \\left( - x , - x \\right)$, $D = \\left( x , - x \\right)$. The radius of the circumcircle of $ABCD$ is $\\sqrt{2} x$. Denote by $\\theta$ the argument of point $P$ on the circle. Thus, the coordinates of $P$ are $P = \\left( \\sqrt{2} x \\cos \\theta , \\sqrt{2} x \\sin \\theta \\right)$. Thus, the equations $PA \\cdot PC = 56$ and $PB \\cdot PD = 90$ can be written as \\begin{align*} \\sqrt{\\left( \\sqrt{2} x \\cos \\theta - x \\right)^2 + \\left( \\sqrt{2} x \\sin \\theta - x \\right)^2} \\cdot \\sqrt{\\left( \\sqrt{2} x \\cos \\theta + x \\right)^2 + \\left( \\sqrt{2} x \\sin \\theta + x \\right)^2} & = 56 \\\\ \\sqrt{\\left( \\sqrt{2} x \\cos \\theta + x \\right)^2 + \\left( \\sqrt{2} x \\sin \\theta - x \\right)^2} \\cdot \\sqrt{\\left( \\sqrt{2} x \\cos \\theta - x \\right)^2 + \\left( \\sqrt{2} x \\sin \\theta + x \\right)^2} & = 90 \\end{align*} These equations can be reformulated as \\begin{align*} x^4 \\left( 4 - 2 \\sqrt{2} \\left( \\cos \\theta + \\sin \\theta \\right) \\right) \\left( 4 + 2 \\sqrt{2} \\left( \\cos \\theta + \\sin \\theta \\right) \\right) & = 56^2 \\\\ x^4 \\left( 4 + 2 \\sqrt{2} \\left( \\cos \\theta - \\sin \\theta \\right) \\right) \\left( 4 - 2 \\sqrt{2} \\left( \\cos \\theta - \\sin \\theta \\right) \\right) & = 90^2 \\end{align*} These equations can be reformulated as \\begin{align*} 2 x^4 \\left( 1 - 2 \\cos \\theta \\sin \\theta \\right) & = 28^2 \\hspace{1cm} (1) \\\\ 2 x^4 \\left( 1 + 2 \\cos \\theta \\sin \\theta \\right) & = 45^2 \\hspace{1cm} (2) \\end{align*} Taking $\\frac{(1)}{(2)}$, by solving the equation, we get \\[ 2 \\cos \\theta \\sin \\theta = \\frac{45^2 - 28^2}{45^2 + 28^2} . \\hspace{1cm} (3) \\] Plugging (3) into (1), we get \\begin{align*} {\\rm Area} \\ ABCD & = \\left( 2 x \\right)^2 \\\\ & = 4 \\sqrt{\\frac{28^2}{2 \\left( 1 - 2 \\cos \\theta \\sin \\theta \\right)}} \\\\ & = 2 \\sqrt{45^2 + 28^2} \\\\ & = 2 \\cdot 53 \\\\ & = \\textbf{(106) } . \\end{align*} ~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)",
"WLOG, let $P$ be on minor arc $\\overarc {AB}$. Let $r$ and $O$ be the radius and center of the circumcircle respectively, and let $\\theta = \\angle AOP$. By the Pythagorean Theorem, the area of the square is $2r^2$. We can use the Law of Cosines on isosceles triangles $\\triangle AOP, \\, \\triangle COP, \\, \\triangle BOP, \\, \\triangle DOP$ to get \\begin{align*} PA^2 &= 2r^2(1 - \\cos \\theta), \\\\ PC^2 &= 2r^2(1 - \\cos (180 - \\theta)) = 2r^2(1 + \\cos \\theta), \\\\ PB^2 &= 2r^2(1 - \\cos (90 - \\theta)) = 2r^2(1 - \\sin \\theta), \\\\ PD^2 &= 2r^2(1 - \\cos (90 + \\theta)) = 2r^2(1 + \\sin \\theta). \\end{align*} Taking the products of the first two and last two equations, respectively, \\[56^2 = (PA \\cdot PC)^2 = 4r^4(1 - \\cos \\theta)(1 + \\cos \\theta) = 4r^4(1 - \\cos^2 \\theta) = 4r^4 \\sin^2 \\theta,\\] and \\[90^2 = (PB \\cdot PD)^2 = 4r^4(1 - \\sin \\theta)(1 + \\sin \\theta) = 4r^4(1 - \\sin^2 \\theta) = 4r^4 \\cos^2 \\theta.\\] Adding these equations, \\[56^2 + 90^2 = 4r^4,\\] so \\[2r^2 = \\sqrt{56^2+90^2} = 2\\sqrt{28^2+45^2} = 2\\sqrt{2809} = 2 \\cdot 53 = 106.\\] ~OrangeQuail9",
"WLOG, let $P$ be on minor arc $\\overarc {AB}$. Let $r$ and $O$ be the radius and center of the circumcircle respectively, and let $\\theta = \\angle AOP$. By the Pythagorean Theorem, the area of the square is $2r^2$. We can use the Law of Cosines on isosceles triangles $\\triangle AOP, \\, \\triangle COP, \\, \\triangle BOP, \\, \\triangle DOP$ to get \\begin{align*} PA^2 &= 2r^2(1 - \\cos \\theta), \\\\ PC^2 &= 2r^2(1 - \\cos (180 - \\theta)) = 2r^2(1 + \\cos \\theta), \\\\ PB^2 &= 2r^2(1 - \\cos (90 - \\theta)) = 2r^2(1 - \\sin \\theta), \\\\ PD^2 &= 2r^2(1 - \\cos (90 + \\theta)) = 2r^2(1 + \\sin \\theta). \\end{align*} Taking the products of the first two and last two equations, respectively, \\[56^2 = (PA \\cdot PC)^2 = 4r^4(1 - \\cos \\theta)(1 + \\cos \\theta) = 4r^4(1 - \\cos^2 \\theta) = 4r^4 \\sin^2 \\theta,\\] and \\[90^2 = (PB \\cdot PD)^2 = 4r^4(1 - \\sin \\theta)(1 + \\sin \\theta) = 4r^4(1 - \\sin^2 \\theta) = 4r^4 \\cos^2 \\theta.\\] Adding these equations, \\[56^2 + 90^2 = 4r^4,\\] so \\[2r^2 = \\sqrt{56^2+90^2} = 2\\sqrt{28^2+45^2} = 2\\sqrt{2809} = 2 \\cdot 53 = 106.\\] ~OrangeQuail9",
"First draw a diagram. [asy] pair A, B, C, D, O, P; A = (0,sqrt(106)); B = (0,0); C = (sqrt(106),0); D = (sqrt(106),sqrt(106)); O = (sqrt(106)/2, sqrt(106)/2); P = intersectionpoint(circle(A, sqrt(212)*sin(atan(28/45)/2)), circle(O, sqrt(212)/2)); draw(A--B--C--D--cycle); draw(circle(O, sqrt(212)/2)); label(\"$A$\", A, NW); label(\"$B$\", B, SW); label(\"$C$\", C, SE); label(\"$D$\", D, NE); label(\"$P$\", P, NW); label(\"$O$\", O, 1.5*S); label(\"$\\theta$\", O, dir(120)*5); draw(P--A--C--cycle, red); draw(P--B--D--cycle, blue); draw(P--O); draw(anglemark(P,O,A,30)); dot(P); dot(O); [/asy] Let's say that the radius is $r$. Then the area of the $ABCD$ is $(\\sqrt2r)^2 = 2r^2$ Using the formula for the length of a chord subtended by an angle, we get \\[PA = 2r\\sin\\left(\\dfrac{\\theta}2\\right)\\] \\[PC = 2r\\sin\\left(\\dfrac{180-\\theta}2\\right) = 2r\\sin\\left(90 - \\dfrac{\\theta}2\\right) = 2r\\cos\\left(\\dfrac{\\theta}2\\right)\\] Multiplying and simplifying these 2 equations gives \\[PA \\cdot PC = 4r^2 \\sin \\left(\\dfrac{\\theta}2 \\right) \\cos \\left(\\dfrac{\\theta}2 \\right) = 2r^2 \\sin\\left(\\theta \\right) = 56\\] Similarly $PB = 2r\\sin\\left(\\dfrac{90 +\\theta}2\\right)$ and $PD =2r\\sin\\left(\\dfrac{90 -\\theta}2\\right)$. Again, multiplying gives \\[PB \\cdot PD = 4r^2 \\sin\\left(\\dfrac{90 +\\theta}2\\right) \\sin\\left(\\dfrac{90 -\\theta}2\\right) = 4r^2 \\sin\\left(90 -\\dfrac{90 -\\theta}2\\right) \\sin\\left(\\dfrac{90 -\\theta}2\\right)\\] \\[=4r^2 \\sin\\left(\\dfrac{90 -\\theta}2\\right) \\cos\\left(\\dfrac{90 -\\theta}2\\right) = 2r^2 \\sin\\left(90 - \\theta \\right) = 2r^2 \\cos\\left(\\theta \\right) = 90\\] Dividing $2r^2 \\sin \\left(\\theta \\right)$ by $2r^2 \\cos \\left( \\theta \\right)$ gives $\\tan \\left(\\theta \\right) = \\dfrac{28}{45}$, so $\\theta = \\tan^{-1} \\left(\\dfrac{28}{45} \\right)$. Pluging this back into one of the equations, gives \\[2r^2 = \\dfrac{90}{\\cos\\left(\\tan^{-1}\\left(\\dfrac{28}{45}\\right)\\right)}\\] If we imagine a $28$-$45$-$53$ right triangle, we see that if $28$ is opposite and $45$ is adjacent, $\\cos\\left(\\theta\\right) = \\dfrac{\\text{adj}}{\\text{hyp}} = \\dfrac{45}{53}$. Now we see that \\[2r^2 = \\dfrac{90}{\\frac{45}{53}} = 106.\\] ~Voldemort101",
"First draw a diagram. [asy] pair A, B, C, D, O, P; A = (0,sqrt(106)); B = (0,0); C = (sqrt(106),0); D = (sqrt(106),sqrt(106)); O = (sqrt(106)/2, sqrt(106)/2); P = intersectionpoint(circle(A, sqrt(212)*sin(atan(28/45)/2)), circle(O, sqrt(212)/2)); draw(A--B--C--D--cycle); draw(circle(O, sqrt(212)/2)); label(\"$A$\", A, NW); label(\"$B$\", B, SW); label(\"$C$\", C, SE); label(\"$D$\", D, NE); label(\"$P$\", P, NW); label(\"$O$\", O, 1.5*S); label(\"$\\theta$\", O, dir(120)*5); draw(P--A--C--cycle, red); draw(P--B--D--cycle, blue); draw(P--O); draw(anglemark(P,O,A,30)); dot(P); dot(O); [/asy] Let's say that the radius is $r$. Then the area of the $ABCD$ is $(\\sqrt2r)^2 = 2r^2$ Using the formula for the length of a chord subtended by an angle, we get \\[PA = 2r\\sin\\left(\\dfrac{\\theta}2\\right)\\] \\[PC = 2r\\sin\\left(\\dfrac{180-\\theta}2\\right) = 2r\\sin\\left(90 - \\dfrac{\\theta}2\\right) = 2r\\cos\\left(\\dfrac{\\theta}2\\right)\\] Multiplying and simplifying these 2 equations gives \\[PA \\cdot PC = 4r^2 \\sin \\left(\\dfrac{\\theta}2 \\right) \\cos \\left(\\dfrac{\\theta}2 \\right) = 2r^2 \\sin\\left(\\theta \\right) = 56\\] Similarly $PB = 2r\\sin\\left(\\dfrac{90 +\\theta}2\\right)$ and $PD =2r\\sin\\left(\\dfrac{90 -\\theta}2\\right)$. Again, multiplying gives \\[PB \\cdot PD = 4r^2 \\sin\\left(\\dfrac{90 +\\theta}2\\right) \\sin\\left(\\dfrac{90 -\\theta}2\\right) = 4r^2 \\sin\\left(90 -\\dfrac{90 -\\theta}2\\right) \\sin\\left(\\dfrac{90 -\\theta}2\\right)\\] \\[=4r^2 \\sin\\left(\\dfrac{90 -\\theta}2\\right) \\cos\\left(\\dfrac{90 -\\theta}2\\right) = 2r^2 \\sin\\left(90 - \\theta \\right) = 2r^2 \\cos\\left(\\theta \\right) = 90\\] Dividing $2r^2 \\sin \\left(\\theta \\right)$ by $2r^2 \\cos \\left( \\theta \\right)$ gives $\\tan \\left(\\theta \\right) = \\dfrac{28}{45}$, so $\\theta = \\tan^{-1} \\left(\\dfrac{28}{45} \\right)$. Pluging this back into one of the equations, gives \\[2r^2 = \\dfrac{90}{\\cos\\left(\\tan^{-1}\\left(\\dfrac{28}{45}\\right)\\right)}\\] If we imagine a $28$-$45$-$53$ right triangle, we see that if $28$ is opposite and $45$ is adjacent, $\\cos\\left(\\theta\\right) = \\dfrac{\\text{adj}}{\\text{hyp}} = \\dfrac{45}{53}$. Now we see that \\[2r^2 = \\dfrac{90}{\\frac{45}{53}} = 106.\\] ~Voldemort101",
"[asy] pair A,B,C,D,P; A=(-3,3); B=(3,3); C=(3,-3); D=(-3,-3); draw(A--B--C--D--cycle); label(A,\"$A$\",NW); label(B,\"$B$\",NE); label(C,\"$C$\",SE); label(D,\"$D$\",SW); draw(circle((0,0),4.24264068712)); P=(-1,4.12310562562); label(P,\"$P$\", NW); pen k=red+dashed; draw(P--A,k); draw(P--B,k); draw(P--C,k); draw(P--D,k); dot(P); [/asy] Let $P=(a,b)$ on the upper quarter of the circle, and let $k$ be the side length of the square. Hence, we want to find $k^2$. Let the center of the circle be $(0,0)$. The two equations would thus become: \\[\\left(\\left(a+\\dfrac{k}2\\right)^2+\\left(b-\\dfrac{k}2\\right)^2\\right)\\left(\\left(a-\\dfrac{k}2\\right)^2+\\left(b+\\dfrac{k}2\\right)^2\\right)=56^2\\] \\[\\left(\\left(a-\\dfrac{k}2\\right)^2+\\left(b-\\dfrac{k}2\\right)^2\\right)\\left(\\left(a+\\dfrac{k}2\\right)^2+\\left(b+\\dfrac{k}2\\right)^2\\right)=90^2\\] Now, let $m=\\left(a+\\dfrac{k}2\\right)^2$, $n=\\left(a-\\dfrac{k}2\\right)^2$, $o=\\left(b+\\dfrac{k}2\\right)^2$, and $p=\\left(b-\\dfrac{k}2\\right)^2$. Our equations now change to $(m+p)(n+o)=56^2=mn+op+mo+pn$ and $(n+p)(m+o)=90^2=mn+op+no+pm$. Subtracting the first from the second, we have $pm+no-mo-pn=p(m-n)-o(m-n)=(m-n)(p-o)=34\\cdot146$. Substituting back in and expanding, we have $2ak\\cdot-2bk=34\\cdot146$, so $abk^2=-17\\cdot73$. We now have one of our terms we need ($k^2$). Therefore, we only need to find $ab$ to find $k^2$. We now write the equation of the circle, which point $P$ satisfies: \\[a^2+b^2=\\left(\\dfrac{k\\sqrt{2}}{2}\\right)^2=\\dfrac{k^2}2\\] We can expand the second equation, yielding \\[\\left(a^2+b^2+\\dfrac{k^2}2+(ak+bk)\\right)\\left(a^2+b^2+\\dfrac{k^2}2-(ak+bk)\\right)=(k^2+k(a+b))(k^2-k(a+b))=8100.\\] Now, with difference of squares, we get $k^4-k^2\\cdot(a+b)^2=k^2(k^2-(a+b)^2)=8100$. We can add $2abk^2=-17\\cdot73\\cdot2=-2482$ to this equation, which we can factor into $k^2(k^2-(a+b)^2+2ab)=k^2(k^2-(a^2+b^2))=8100-2482$. We realize that $a^2+b^2$ is the same as the equation of the circle, so we plug its equation in: $k^2\\left(k^2-\\dfrac{k^2}2\\right)=5618$. We can combine like terms to get $k^2\\cdot\\dfrac{k^2}2=5618$, so $(k^2)^2=11236$. Since the answer is an integer, we know $11236$ is a perfect square. Since it is even, it is divisible by $4$, so we can factor $11236=2^2\\cdot2809$. With some testing with approximations and last-digit methods, we can find that $53^2=2809$. Therefore, taking the square root, we find that $k^2$, the area of square $ABCD$, is $2\\cdot53=106. ~wuwang2002",
"[asy] pair A,B,C,D,P; A=(-3,3); B=(3,3); C=(3,-3); D=(-3,-3); draw(A--B--C--D--cycle); label(A,\"$A$\",NW); label(B,\"$B$\",NE); label(C,\"$C$\",SE); label(D,\"$D$\",SW); draw(circle((0,0),4.24264068712)); P=(-1,4.12310562562); label(P,\"$P$\", NW); pen k=red+dashed; draw(P--A,k); draw(P--B,k); draw(P--C,k); draw(P--D,k); dot(P); [/asy] Let $P=(a,b)$ on the upper quarter of the circle, and let $k$ be the side length of the square. Hence, we want to find $k^2$. Let the center of the circle be $(0,0)$. The two equations would thus become: \\[\\left(\\left(a+\\dfrac{k}2\\right)^2+\\left(b-\\dfrac{k}2\\right)^2\\right)\\left(\\left(a-\\dfrac{k}2\\right)^2+\\left(b+\\dfrac{k}2\\right)^2\\right)=56^2\\] \\[\\left(\\left(a-\\dfrac{k}2\\right)^2+\\left(b-\\dfrac{k}2\\right)^2\\right)\\left(\\left(a+\\dfrac{k}2\\right)^2+\\left(b+\\dfrac{k}2\\right)^2\\right)=90^2\\] Now, let $m=\\left(a+\\dfrac{k}2\\right)^2$, $n=\\left(a-\\dfrac{k}2\\right)^2$, $o=\\left(b+\\dfrac{k}2\\right)^2$, and $p=\\left(b-\\dfrac{k}2\\right)^2$. Our equations now change to $(m+p)(n+o)=56^2=mn+op+mo+pn$ and $(n+p)(m+o)=90^2=mn+op+no+pm$. Subtracting the first from the second, we have $pm+no-mo-pn=p(m-n)-o(m-n)=(m-n)(p-o)=34\\cdot146$. Substituting back in and expanding, we have $2ak\\cdot-2bk=34\\cdot146$, so $abk^2=-17\\cdot73$. We now have one of our terms we need ($k^2$). Therefore, we only need to find $ab$ to find $k^2$. We now write the equation of the circle, which point $P$ satisfies: \\[a^2+b^2=\\left(\\dfrac{k\\sqrt{2}}{2}\\right)^2=\\dfrac{k^2}2\\] We can expand the second equation, yielding \\[\\left(a^2+b^2+\\dfrac{k^2}2+(ak+bk)\\right)\\left(a^2+b^2+\\dfrac{k^2}2-(ak+bk)\\right)=(k^2+k(a+b))(k^2-k(a+b))=8100.\\] Now, with difference of squares, we get $k^4-k^2\\cdot(a+b)^2=k^2(k^2-(a+b)^2)=8100$. We can add $2abk^2=-17\\cdot73\\cdot2=-2482$ to this equation, which we can factor into $k^2(k^2-(a+b)^2+2ab)=k^2(k^2-(a^2+b^2))=8100-2482$. We realize that $a^2+b^2$ is the same as the equation of the circle, so we plug its equation in: $k^2\\left(k^2-\\dfrac{k^2}2\\right)=5618$. We can combine like terms to get $k^2\\cdot\\dfrac{k^2}2=5618$, so $(k^2)^2=11236$. Since the answer is an integer, we know $11236$ is a perfect square. Since it is even, it is divisible by $4$, so we can factor $11236=2^2\\cdot2809$. With some testing with approximations and last-digit methods, we can find that $53^2=2809$. Therefore, taking the square root, we find that $k^2$, the area of square $ABCD$, is $2\\cdot53=106. ~wuwang2002",
"WLOG, let $P$ be on minor arc $AD.$ Draw in $AP$, $BP$, $CP$, $DP$ and let $\\angle ABP = x.$ We can see, by the inscribed angle theorem, that $\\angle APB = \\angle ACB = 45$, and $\\angle CPD = \\angle CAD = 45.$ Then, $\\angle PAB = 135-x$, $\\angle PCD = \\angle PAD = (135-x)-90 = 45-x$, and $\\angle PDC = 90+x.$ Letting $(PA, PB, PC, PD, AB) = (a,b,c,d,s)$, we can use the law of sines on triangles $PAB$ and $PCD$ to get \\[s\\sqrt{2} = \\frac{a}{\\sin(x)} = \\frac{b}{\\sin(135-x)} = \\frac{c}{\\sin(90+x)} = \\frac{d}{\\sin(45-x)}.\\] Making all the angles in the above equation acute gives \\[s\\sqrt{2} = \\frac{a}{\\sin(x)} = \\frac{b}{\\sin(45+x)} = \\frac{c}{\\sin(90-x)} = \\frac{d}{\\sin(45-x)}.\\] Note that we are looking for $s^{2}.$ We are given that $ac = 56$ and $bd = 90.$ This means that $s^{2}\\sin(x)\\sin(90-x) = 28$ and $s^{2}\\sin(45+x)\\sin(45-x) = 45.$ However, \\[\\sin(x)\\sin(90-x) = \\sin(x)\\cos(x) = \\frac{\\sin(2x)}{2}\\] and \\[\\sin(45+x)\\sin(45-x) = \\frac{(\\cos(x) + \\sin(x))(\\cos(x) - \\sin(x))}{2} = \\frac{\\cos^{2}(x) - \\sin^{2}(x)}{2} = \\frac{\\cos(2x)}{2}.\\] Therefore, $s^{2}\\sin(2x) = 56$ and $s^{2}\\cos(2x) = 90.$ Therefore, by the Pythagorean Identity, \\[s^{2} = \\sqrt{(s^{2}\\sin(2x))^{2} + (s^{2}\\cos(2x))^{2}} = \\sqrt{56^{2} + 90^{2}} = 106.\\] ~pianoboy",
"WLOG, let $P$ be on minor arc $AD.$ Draw in $AP$, $BP$, $CP$, $DP$ and let $\\angle ABP = x.$ We can see, by the inscribed angle theorem, that $\\angle APB = \\angle ACB = 45$, and $\\angle CPD = \\angle CAD = 45.$ Then, $\\angle PAB = 135-x$, $\\angle PCD = \\angle PAD = (135-x)-90 = 45-x$, and $\\angle PDC = 90+x.$ Letting $(PA, PB, PC, PD, AB) = (a,b,c,d,s)$, we can use the law of sines on triangles $PAB$ and $PCD$ to get \\[s\\sqrt{2} = \\frac{a}{\\sin(x)} = \\frac{b}{\\sin(135-x)} = \\frac{c}{\\sin(90+x)} = \\frac{d}{\\sin(45-x)}.\\] Making all the angles in the above equation acute gives \\[s\\sqrt{2} = \\frac{a}{\\sin(x)} = \\frac{b}{\\sin(45+x)} = \\frac{c}{\\sin(90-x)} = \\frac{d}{\\sin(45-x)}.\\] Note that we are looking for $s^{2}.$ We are given that $ac = 56$ and $bd = 90.$ This means that $s^{2}\\sin(x)\\sin(90-x) = 28$ and $s^{2}\\sin(45+x)\\sin(45-x) = 45.$ However, \\[\\sin(x)\\sin(90-x) = \\sin(x)\\cos(x) = \\frac{\\sin(2x)}{2}\\] and \\[\\sin(45+x)\\sin(45-x) = \\frac{(\\cos(x) + \\sin(x))(\\cos(x) - \\sin(x))}{2} = \\frac{\\cos^{2}(x) - \\sin^{2}(x)}{2} = \\frac{\\cos(2x)}{2}.\\] Therefore, $s^{2}\\sin(2x) = 56$ and $s^{2}\\cos(2x) = 90.$ Therefore, by the Pythagorean Identity, \\[s^{2} = \\sqrt{(s^{2}\\sin(2x))^{2} + (s^{2}\\cos(2x))^{2}} = \\sqrt{56^{2} + 90^{2}} = 106.\\] ~pianoboy",
"Similar to Solution 6, let $P$ be on minor arc $\\overarc {AB}$, $r$ and $O$ be the radius and center of the circumcircle respectively, and $\\theta = \\angle AOP$. Since $\\triangle APC$ is a right triangle, $PA \\cdot PC$ equals the hypotenuse, $2r$, times its altitude, which can be represented as $r \\sin \\theta$. Therefore, $2r^2 \\sin \\theta = 56$. Applying similar logic to $\\triangle BPD$, we get $2r^2 \\sin (90^\\circ - \\theta) = 2r^2 \\cos \\theta = 90$. Dividing the two equations, we have \\begin{align*} \\frac{\\sin \\theta}{\\cos \\theta} &= \\frac{56}{90} \\\\ 56 \\cos \\theta &= 90 \\sin \\theta \\\\ (56 \\cos \\theta)^2 &= (90 \\sin \\theta)^2. \\end{align*} Adding $(56 \\sin \\theta)^2$ to both sides allows us to get rid of $\\cos \\theta$: \\begin{align*} (56 \\cos \\theta)^2 + (56 \\sin \\theta)^2 &= (90 \\sin \\theta)^2 + (56 \\sin \\theta)^2 \\\\ 56^2 &= (90^2 + 56^2)(\\sin \\theta)^2 \\\\ \\frac{56^2}{90^2 + 56^2} &= (\\sin \\theta)^2 \\\\ \\frac{28}{53} &= \\sin \\theta. \\end{align*} Therefore, we have $2r^2\\left(\\frac{28}{53}\\right) = 56$, and since the area of the square can be represented as $2r^2$, the answer is $56 \\cdot \\frac{53}{28} = 106. ~phillipzeng",
"Similar to Solution 6, let $P$ be on minor arc $\\overarc {AB}$, $r$ and $O$ be the radius and center of the circumcircle respectively, and $\\theta = \\angle AOP$. Since $\\triangle APC$ is a right triangle, $PA \\cdot PC$ equals the hypotenuse, $2r$, times its altitude, which can be represented as $r \\sin \\theta$. Therefore, $2r^2 \\sin \\theta = 56$. Applying similar logic to $\\triangle BPD$, we get $2r^2 \\sin (90^\\circ - \\theta) = 2r^2 \\cos \\theta = 90$. Dividing the two equations, we have \\begin{align*} \\frac{\\sin \\theta}{\\cos \\theta} &= \\frac{56}{90} \\\\ 56 \\cos \\theta &= 90 \\sin \\theta \\\\ (56 \\cos \\theta)^2 &= (90 \\sin \\theta)^2. \\end{align*} Adding $(56 \\sin \\theta)^2$ to both sides allows us to get rid of $\\cos \\theta$: \\begin{align*} (56 \\cos \\theta)^2 + (56 \\sin \\theta)^2 &= (90 \\sin \\theta)^2 + (56 \\sin \\theta)^2 \\\\ 56^2 &= (90^2 + 56^2)(\\sin \\theta)^2 \\\\ \\frac{56^2}{90^2 + 56^2} &= (\\sin \\theta)^2 \\\\ \\frac{28}{53} &= \\sin \\theta. \\end{align*} Therefore, we have $2r^2\\left(\\frac{28}{53}\\right) = 56$, and since the area of the square can be represented as $2r^2$, the answer is $56 \\cdot \\frac{53}{28} = 106. ~phillipzeng",
"First, we define a few points. Let $O$ be the center of the circle, let $E$ be the intersection of diameter $AC$ and chord $PD$, and let $F$ be the intersection of diameter $BD$ and chord $PC$. We know that $A$, $B$, $C$, and $D$ are four corners of a square. Therefore, the arcs $AD$, $DC$, and $CB$ are all $90$ degrees. By inscribed angles, angle $APD$, angle $DPC$, and angle $CPB$ are $45$ degrees each. Let the measure of angle $PAC$ be $a$. Similarly, let the measure of angle $PBD$ be $b$. Angle chasing will lead us to the fact that $a + b = 135$, or rather, $b = 135-a$. Let the diameter of the circle be $d$. Given by the problem, $d^2\\sin a \\cos a = 56$. Also, $d^2\\sin b \\cos b = 90$. Using the trigonometric identity $\\sin 2x = 2\\sin x \\cos x$, we can rewrite these as $d^2\\sin 2a = 112$ and $d^2\\sin 2b = 180$. Since we determined that $b = \\frac{3\\pi}{4}-a$, this can be substituted into the second equation. Then, we divide the two equations to get $\\frac{\\sin (\\frac{3\\pi}{2}-2a)}{\\sin 2a} = \\frac{45}{28}$. By using the trigonometric difference-of-angle identity, this simplifies to $\\frac{-\\cos 2a}{\\sin 2a} = \\frac{45}{28}$. By the definition of the tangent function, $\\tan 2a = -\\frac{28}{45}$ Considering this hypothetical right triangle with legs of $28$ and $45$, the hypotenuse is $\\sqrt{45^2+28^2} = 53$. Since $\\sin 2a$ must be positive (since $a$ is acute), $\\sin 2a = \\frac{28}{53}$. Substituting this into the first of the equations, $\\frac{28}{53}d^2 = 112$. From this, $d^2 = 212$. The area of square $ABCD$ is half of the square of its diagonal, which is $d$. Thus, the answer is $\\frac{d^2}{2} = 106. ~Curious_crow",
"First, we define a few points. Let $O$ be the center of the circle, let $E$ be the intersection of diameter $AC$ and chord $PD$, and let $F$ be the intersection of diameter $BD$ and chord $PC$. We know that $A$, $B$, $C$, and $D$ are four corners of a square. Therefore, the arcs $AD$, $DC$, and $CB$ are all $90$ degrees. By inscribed angles, angle $APD$, angle $DPC$, and angle $CPB$ are $45$ degrees each. Let the measure of angle $PAC$ be $a$. Similarly, let the measure of angle $PBD$ be $b$. Angle chasing will lead us to the fact that $a + b = 135$, or rather, $b = 135-a$. Let the diameter of the circle be $d$. Given by the problem, $d^2\\sin a \\cos a = 56$. Also, $d^2\\sin b \\cos b = 90$. Using the trigonometric identity $\\sin 2x = 2\\sin x \\cos x$, we can rewrite these as $d^2\\sin 2a = 112$ and $d^2\\sin 2b = 180$. Since we determined that $b = \\frac{3\\pi}{4}-a$, this can be substituted into the second equation. Then, we divide the two equations to get $\\frac{\\sin (\\frac{3\\pi}{2}-2a)}{\\sin 2a} = \\frac{45}{28}$. By using the trigonometric difference-of-angle identity, this simplifies to $\\frac{-\\cos 2a}{\\sin 2a} = \\frac{45}{28}$. By the definition of the tangent function, $\\tan 2a = -\\frac{28}{45}$ Considering this hypothetical right triangle with legs of $28$ and $45$, the hypotenuse is $\\sqrt{45^2+28^2} = 53$. Since $\\sin 2a$ must be positive (since $a$ is acute), $\\sin 2a = \\frac{28}{53}$. Substituting this into the first of the equations, $\\frac{28}{53}d^2 = 112$. From this, $d^2 = 212$. The area of square $ABCD$ is half of the square of its diagonal, which is $d$. Thus, the answer is $\\frac{d^2}{2} = 106. ~Curious_crow"
] |
2023-I-6
| 2,023
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Alice knows that $3$ red cards and $3$ black cards will be revealed to her one at a time in random order. Before each card is revealed, Alice must guess its color. If Alice plays optimally, the expected number of cards she will guess correctly is $\frac{m}{n},$ where $m$ and $n$ are relatively prime positive integers. Find $m+n.$
| 51
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I
|
[
"We break the problem into stages, one for each card revealed, then further into cases based on the number of remaining unrevealed cards of each color. Since expected value is linear, the expected value of the total number of correct card color guesses across all stages is the sum of the expected values of the number of correct card color guesses at each stage; that is, we add the probabilities of correctly guessing the color at each stage to get the final answer (See https://brilliant.org/wiki/linearity-of-expectation/) At any stage, if there are $a$ unrevealed cards of one color and $b$ of the other color, and $a \\geq b$, then the optimal strategy is to guess the color with $a$ unrevealed cards, which succeeds with probability $\\frac{a}{a+b}.$ Stage 1: There are always $3$ unrevealed cards of each color, so the probability of guessing correctly is $\\frac{1}{2}$. Stage 2: There is always a $3$-$2$ split ($3$ unrevealed cards of one color and $2$ of the other color), so the probability of guessing correctly is $\\frac{3}{5}$. Stage 3: There are now $2$ cases: The guess from Stage 2 was correct, so there is now a $2$-$2$ split of cards and a $\\frac{1}{2}$ probability of guessing the color of the third card correctly. The guess from Stage 2 was incorrect, so the split is $3$-$1$ and the probability of guessing correctly is $\\frac{3}{4}$. Thus, the overall probability of guessing correctly is $\\frac{3}{5} \\cdot \\frac{1}{2} + \\frac{2}{5} \\cdot \\frac{3}{4} = \\frac{3}{5}$. Stage 4: This stage has $2$ cases as well: The guesses from both Stage 2 and Stage 3 were incorrect. This occurs with probability $\\frac{2}{5} \\cdot \\frac{1}{4} = \\frac{1}{10}$ and results in a $3$-$0$ split and a certain correct guess at this stage. Otherwise, there must be a $2$-$1$ split and a $\\frac{2}{3}$ probability of guessing correctly. The probability of guessing the fourth card correctly is therefore $\\frac{1}{10} \\cdot 1 + \\frac{9}{10} \\cdot \\frac{2}{3} = \\frac{7}{10}$. Stage 5: Yet again, there are $2$ cases: In Stage 4, there was a $2$-$1$ split and the guess was correct. This occurs with probability $\\frac{9}{10} \\cdot \\frac{2}{3} = \\frac{3}{5}$ and results in a $1$-$1$ split with a $\\frac{1}{2}$ chance of a correct guess here. Otherwise, there must be a $2$-$0$ split, making a correct guess certain. In total, the fifth card can be guessed correctly with probability $\\frac{3}{5} \\cdot \\frac{1}{2} + \\frac{2}{5} \\cdot 1 = \\frac{7}{10}$. Stage 6: At this point, only $1$ card remains, so the probability of guessing its color correctly is $1$. In conclusion, the expected value of the number of cards guessed correctly is \\[\\frac{1}{2} + \\frac{3}{5} + \\frac{3}{5} + \\frac{7}{10} + \\frac{7}{10} + 1 = \\frac{5+6+6+7+7+10}{10} = \\frac{41}{10},\\] so the answer is $41 + 10 = 051 ~OrangeQuail9",
"We break the problem into stages, one for each card revealed, then further into cases based on the number of remaining unrevealed cards of each color. Since expected value is linear, the expected value of the total number of correct card color guesses across all stages is the sum of the expected values of the number of correct card color guesses at each stage; that is, we add the probabilities of correctly guessing the color at each stage to get the final answer (See https://brilliant.org/wiki/linearity-of-expectation/) At any stage, if there are $a$ unrevealed cards of one color and $b$ of the other color, and $a \\geq b$, then the optimal strategy is to guess the color with $a$ unrevealed cards, which succeeds with probability $\\frac{a}{a+b}.$ Stage 1: There are always $3$ unrevealed cards of each color, so the probability of guessing correctly is $\\frac{1}{2}$. Stage 2: There is always a $3$-$2$ split ($3$ unrevealed cards of one color and $2$ of the other color), so the probability of guessing correctly is $\\frac{3}{5}$. Stage 3: There are now $2$ cases: The guess from Stage 2 was correct, so there is now a $2$-$2$ split of cards and a $\\frac{1}{2}$ probability of guessing the color of the third card correctly. The guess from Stage 2 was incorrect, so the split is $3$-$1$ and the probability of guessing correctly is $\\frac{3}{4}$. Thus, the overall probability of guessing correctly is $\\frac{3}{5} \\cdot \\frac{1}{2} + \\frac{2}{5} \\cdot \\frac{3}{4} = \\frac{3}{5}$. Stage 4: This stage has $2$ cases as well: The guesses from both Stage 2 and Stage 3 were incorrect. This occurs with probability $\\frac{2}{5} \\cdot \\frac{1}{4} = \\frac{1}{10}$ and results in a $3$-$0$ split and a certain correct guess at this stage. Otherwise, there must be a $2$-$1$ split and a $\\frac{2}{3}$ probability of guessing correctly. The probability of guessing the fourth card correctly is therefore $\\frac{1}{10} \\cdot 1 + \\frac{9}{10} \\cdot \\frac{2}{3} = \\frac{7}{10}$. Stage 5: Yet again, there are $2$ cases: In Stage 4, there was a $2$-$1$ split and the guess was correct. This occurs with probability $\\frac{9}{10} \\cdot \\frac{2}{3} = \\frac{3}{5}$ and results in a $1$-$1$ split with a $\\frac{1}{2}$ chance of a correct guess here. Otherwise, there must be a $2$-$0$ split, making a correct guess certain. In total, the fifth card can be guessed correctly with probability $\\frac{3}{5} \\cdot \\frac{1}{2} + \\frac{2}{5} \\cdot 1 = \\frac{7}{10}$. Stage 6: At this point, only $1$ card remains, so the probability of guessing its color correctly is $1$. In conclusion, the expected value of the number of cards guessed correctly is \\[\\frac{1}{2} + \\frac{3}{5} + \\frac{3}{5} + \\frac{7}{10} + \\frac{7}{10} + 1 = \\frac{5+6+6+7+7+10}{10} = \\frac{41}{10},\\] so the answer is $41 + 10 = 051 ~OrangeQuail9",
"At any point in the game, Alice should guess whichever color has come up less frequently thus far (although if both colors have come up equally often, she may guess whichever she likes); using this strategy, her probability of guessing correctly is at least $\\frac{1}{2}$ on any given card, as desired. There are ${6 \\choose 3} = 20$ possible orderings of cards, all equally likely (since any of the $6! = 720$ permutations of the cards is equally likely, and each ordering covers $3!^2 = 6^2 = 36$ permutations). Each of the $10$ orderings that start with red cards corresponds with one that starts with a black card; the problem is symmetrical with respect to red and black cards, so we can, without loss of generality, consider only the orderings that start with red cards. We then generate a tally table showing whether Alice's guesses are correct for each ordering; for a given card, she guesses correctly if fewer than half the previously shown cards were the same color, guesses incorrectly if more than half were the same color, and guesses correctly with probability $\\frac{1}{2}$ if exactly half were the same color. In this table, $\\mid$ denotes a correct guess, $\\--$ denotes an incorrect guess, and $/$ denotes a guess with $\\frac{1}{2}$ probability of being correct. Now we sum the tallies across orderings, obtaining $41$, and finally divide by the number of orderings ($10$) to obtain the expected number of correct guesses, $\\frac{41}{10}$, which yields an answer of $41 + 10 = 051 ~IndigoEagle108",
"At any point in the game, Alice should guess whichever color has come up less frequently thus far (although if both colors have come up equally often, she may guess whichever she likes); using this strategy, her probability of guessing correctly is at least $\\frac{1}{2}$ on any given card, as desired. There are ${6 \\choose 3} = 20$ possible orderings of cards, all equally likely (since any of the $6! = 720$ permutations of the cards is equally likely, and each ordering covers $3!^2 = 6^2 = 36$ permutations). Each of the $10$ orderings that start with red cards corresponds with one that starts with a black card; the problem is symmetrical with respect to red and black cards, so we can, without loss of generality, consider only the orderings that start with red cards. We then generate a tally table showing whether Alice's guesses are correct for each ordering; for a given card, she guesses correctly if fewer than half the previously shown cards were the same color, guesses incorrectly if more than half were the same color, and guesses correctly with probability $\\frac{1}{2}$ if exactly half were the same color. In this table, $\\mid$ denotes a correct guess, $\\--$ denotes an incorrect guess, and $/$ denotes a guess with $\\frac{1}{2}$ probability of being correct. Now we sum the tallies across orderings, obtaining $41$, and finally divide by the number of orderings ($10$) to obtain the expected number of correct guesses, $\\frac{41}{10}$, which yields an answer of $41 + 10 = 051 ~IndigoEagle108",
"Denote by $N \\left( a, b \\right)$ the optimal expected number of cards that Alice guesses correctly, where the number of red and black cards are $a$ and $b$, respectively. Thus, for $a, b \\geq 1$, we have \\begin{align*} N \\left( a, b \\right) & = \\max \\left\\{ \\frac{a}{a+b} \\left( 1 + N \\left( a - 1 , b \\right) \\right) + \\frac{b}{a+b} N \\left( a , b - 1 \\right) , \\right. \\\\ & \\hspace{1cm} \\left. \\frac{a}{a+b} N \\left( a - 1 , b \\right) + \\frac{b}{a+b} \\left( 1 + N \\left( a , b - 1 \\right) \\right) \\right\\} . \\end{align*} For $a = 0$, Alice always guesses black. So $N \\left( 0 , b \\right) = b$. For $b = 0$, Alice always guesses red. So $N \\left( a , 0 \\right) = a$. To solve this dynamic program, we can also exploit its symmetry that $N \\left( a , b \\right) = N \\left( b , a \\right)$. By solving this dynamic program, we get $N \\left( 1, 1 \\right) = \\frac{3}{2}$, $N \\left( 1, 2 \\right) = \\frac{7}{3}$, $N \\left( 1 , 3 \\right) = \\frac{13}{4}$, $N \\left( 2 , 2 \\right) = \\frac{17}{6}$, $N \\left( 2 , 3 \\right) = \\frac{18}{5}$, $N \\left( 3, 3 \\right) = \\frac{41}{10}$. Therefore, the answer is $41 + 10 = \\textbf{(051) }. ~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)",
"Denote by $N \\left( a, b \\right)$ the optimal expected number of cards that Alice guesses correctly, where the number of red and black cards are $a$ and $b$, respectively. Thus, for $a, b \\geq 1$, we have \\begin{align*} N \\left( a, b \\right) & = \\max \\left\\{ \\frac{a}{a+b} \\left( 1 + N \\left( a - 1 , b \\right) \\right) + \\frac{b}{a+b} N \\left( a , b - 1 \\right) , \\right. \\\\ & \\hspace{1cm} \\left. \\frac{a}{a+b} N \\left( a - 1 , b \\right) + \\frac{b}{a+b} \\left( 1 + N \\left( a , b - 1 \\right) \\right) \\right\\} . \\end{align*} For $a = 0$, Alice always guesses black. So $N \\left( 0 , b \\right) = b$. For $b = 0$, Alice always guesses red. So $N \\left( a , 0 \\right) = a$. To solve this dynamic program, we can also exploit its symmetry that $N \\left( a , b \\right) = N \\left( b , a \\right)$. By solving this dynamic program, we get $N \\left( 1, 1 \\right) = \\frac{3}{2}$, $N \\left( 1, 2 \\right) = \\frac{7}{3}$, $N \\left( 1 , 3 \\right) = \\frac{13}{4}$, $N \\left( 2 , 2 \\right) = \\frac{17}{6}$, $N \\left( 2 , 3 \\right) = \\frac{18}{5}$, $N \\left( 3, 3 \\right) = \\frac{41}{10}$. Therefore, the answer is $41 + 10 = \\textbf{(051) }. ~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)",
"Denote by $N_{i,j}$ the optimal expected number of cards that Alice guesses correctly, where the number of cards are $i$ and $j \\ge i.$ If $i = 0$ then Alice guesses correctly all cards, so $N_{0,j} = j.$ If $j = i$ then Alice guesses next card with probability $\\frac {1}{2} \\implies N_{i,i} = \\frac {1}{2} + N_{i-1,i}.$ If $j = i+1$ then Alice guesses next card with probability $\\frac {i+1}{2i+1} \\implies N_{i,i+1} = \\frac {i+1}{2i+1} (1+ N_{i,i}) + \\frac{i}{2i+1} N_{i-1,i+1}.$ If $j = i+2$ then Alice guesses next card with probability $\\frac {i+2}{2i+2} \\implies N_{i,i+2} = \\frac {i+2}{2i+2} (1+ N_{i,i+1}) + \\frac{i}{2i+2} N_{i-1,i+2}.$ One can find consistently: $N_{1,1} = \\frac {1}{2} + N_{0,1} = \\frac {3}{2},$ \\[N_{1,2} = \\frac {2}{3} (1 + N_{1,1}) + \\frac {1}{3} N_{0,2} = \\frac {7}{3}.\\] \\[N_{2,2} = \\frac {1}{2} + N_{1,2} = \\frac {17}{6}.\\] \\[N_{1,3} = \\frac {3}{4} (1 + N_{1,2}) + \\frac {1}{4} N_{0,3} = \\frac {13}{4}.\\] \\[N_{2,3} = \\frac {3}{5} (1 + N_{2,2}) + \\frac {2}{5} N_{1,3} = \\frac {18}{5}.\\] \\[N_{3,3} = \\frac {1}{2} + N_{2,3} = \\frac {41}{10}.\\] Therefore, the answer is $41 + 10 = \\textbf{(051) }. vladimir.shelomovskii@gmail.com, vvsss",
"Denote by $N_{i,j}$ the optimal expected number of cards that Alice guesses correctly, where the number of cards are $i$ and $j \\ge i.$ If $i = 0$ then Alice guesses correctly all cards, so $N_{0,j} = j.$ If $j = i$ then Alice guesses next card with probability $\\frac {1}{2} \\implies N_{i,i} = \\frac {1}{2} + N_{i-1,i}.$ If $j = i+1$ then Alice guesses next card with probability $\\frac {i+1}{2i+1} \\implies N_{i,i+1} = \\frac {i+1}{2i+1} (1+ N_{i,i}) + \\frac{i}{2i+1} N_{i-1,i+1}.$ If $j = i+2$ then Alice guesses next card with probability $\\frac {i+2}{2i+2} \\implies N_{i,i+2} = \\frac {i+2}{2i+2} (1+ N_{i,i+1}) + \\frac{i}{2i+2} N_{i-1,i+2}.$ One can find consistently: $N_{1,1} = \\frac {1}{2} + N_{0,1} = \\frac {3}{2},$ \\[N_{1,2} = \\frac {2}{3} (1 + N_{1,1}) + \\frac {1}{3} N_{0,2} = \\frac {7}{3}.\\] \\[N_{2,2} = \\frac {1}{2} + N_{1,2} = \\frac {17}{6}.\\] \\[N_{1,3} = \\frac {3}{4} (1 + N_{1,2}) + \\frac {1}{4} N_{0,3} = \\frac {13}{4}.\\] \\[N_{2,3} = \\frac {3}{5} (1 + N_{2,2}) + \\frac {2}{5} N_{1,3} = \\frac {18}{5}.\\] \\[N_{3,3} = \\frac {1}{2} + N_{2,3} = \\frac {41}{10}.\\] Therefore, the answer is $41 + 10 = \\textbf{(051) }. vladimir.shelomovskii@gmail.com, vvsss",
"Denote $E_n$ the expected number of cards Alice guesses correctly given $n$ red cards and $n$ black cards. We want to find $E_3$. Alice has a $\\frac{1}{2}$ chance of guessing the first card. WLOG assume the first card color is red. For the next card, Alice has a $\\frac{3}{5}$ chance of guessing the card if she chooses black; if they guess right, there's one less red and black card, so the expected number of cards Alice guesses from here is $E_2$. If Alice does not guess correctly (which occurs with probability $\\frac{2}{5}$), this means that there's 3 black cards and 1 red card left, so Alice should guess black next with a $\\frac{3}{4}$ chance of being right. If Alice is wrong (with probability $\\frac{1}{4}$), there are only 3 black cards left, so Alice can guess these with certainty; if Alice is right, there are 2 blacks and 1 red left, so Alice should again guess black. If Alice is right (with probability $\\frac{2}{3}$), there is now 1 black and red card each, so the expected number of cards guessed is $E_1$; if she is wrong (with probability $\\frac{1}{3}$), there are 2 black cards left, so Alice can guess these with certainty. Summing this up into a formula: \\[E_3 = \\frac{1}{2} + \\frac{3}{5} \\left(1 + E_2 \\right) + \\frac{2}{5} \\left( \\frac{1}{4}(3) + \\frac{3}{4}\\left(1 + \\frac{2}{3}\\left(1 + E_1 \\right) + \\frac{1}{3}(2)\\right) \\right)\\] We can apply similar logic to compute $E_2$ and get \\[E_2 = \\frac{1}{2} + \\frac{2}{3}(1 + E_1) + \\frac{1}{3}(2)\\] To compute $E_1$, we know that Alice can guess the last card with certainty, and there's a $\\frac{1}{2}$ chance they get the first card as well, so $E_1 = \\frac{3}{2}$. Thus, $E_2 = \\frac{17}{6}$, and after long computation, we get $E_3 = \\frac{41}{10}$. The requested answer is $41 + 10 = 51. ~ adam_zheng",
"Denote $E_n$ the expected number of cards Alice guesses correctly given $n$ red cards and $n$ black cards. We want to find $E_3$. Alice has a $\\frac{1}{2}$ chance of guessing the first card. WLOG assume the first card color is red. For the next card, Alice has a $\\frac{3}{5}$ chance of guessing the card if she chooses black; if they guess right, there's one less red and black card, so the expected number of cards Alice guesses from here is $E_2$. If Alice does not guess correctly (which occurs with probability $\\frac{2}{5}$), this means that there's 3 black cards and 1 red card left, so Alice should guess black next with a $\\frac{3}{4}$ chance of being right. If Alice is wrong (with probability $\\frac{1}{4}$), there are only 3 black cards left, so Alice can guess these with certainty; if Alice is right, there are 2 blacks and 1 red left, so Alice should again guess black. If Alice is right (with probability $\\frac{2}{3}$), there is now 1 black and red card each, so the expected number of cards guessed is $E_1$; if she is wrong (with probability $\\frac{1}{3}$), there are 2 black cards left, so Alice can guess these with certainty. Summing this up into a formula: \\[E_3 = \\frac{1}{2} + \\frac{3}{5} \\left(1 + E_2 \\right) + \\frac{2}{5} \\left( \\frac{1}{4}(3) + \\frac{3}{4}\\left(1 + \\frac{2}{3}\\left(1 + E_1 \\right) + \\frac{1}{3}(2)\\right) \\right)\\] We can apply similar logic to compute $E_2$ and get \\[E_2 = \\frac{1}{2} + \\frac{2}{3}(1 + E_1) + \\frac{1}{3}(2)\\] To compute $E_1$, we know that Alice can guess the last card with certainty, and there's a $\\frac{1}{2}$ chance they get the first card as well, so $E_1 = \\frac{3}{2}$. Thus, $E_2 = \\frac{17}{6}$, and after long computation, we get $E_3 = \\frac{41}{10}$. The requested answer is $41 + 10 = 51. ~ adam_zheng",
"There's a beautiful observation that for any particular game path, the expected number of correct guesses is purely dependent on how often we are in a neutral state (i.e. there are the same number of red and black cards remaining in the deck). For example, given path $BRRBBR$ we are in a neutral state once in the beginning when no cards have been flipped, after the second card has been flipped, after the 4th card has been flipped, and finally a 4th time at the very end of the game. Call the number of times a given path is in the neutral state $n$. Then the expected number of correct guesses equals: \\[E = 3 + \\frac{n-1}{2}\\] Since this problem is small, we could just could just count them (noting that the problem is symmetric between reds and blacks): \\[ \\begin{array}{|c|c|c|} \\hline \\textbf{Path} & \\textbf{Neutral States} & \\textbf{E}\\\\ \\hline \\text{RRRBBB} & 2 & 3.5\\\\ \\text{RRBRBB} & 2 & 3.5 \\\\ \\text{RRBRBB} & 3 & 4 \\\\ \\text{RRBBBR} & 3 & 4 \\\\ \\text{RBRRBB} & 3 & 4 \\\\ \\text{RBRBRB} & 4 & 4.5 \\\\ \\text{RBRBBR} & 4 & 4.5 \\\\ \\text{RBBRRB} & 4 & 4.5 \\\\ \\text{RBBRBR} & 4 & 4.5 \\\\ \\text{RBBBRR} & 3 & 4 \\\\ \\hline \\end{array} \\] Using linearity of expectation, the expected number of correct guesses is: \\[E=\\frac{3.5+3.5+4+4+4+4.5+4.5+4.5+4.5+4}{10}=\\frac{41}{10}, \\text{ thus } 41 + 10 = 51 gives us: \\[ \\sum_{m=0}^{3} \\binom{2m}{m} \\cdot \\binom{6-2m}{3-m} = \\] \\[ \\binom{0}{0} \\cdot \\binom{6}{3} + \\binom{2}{1} \\cdot \\binom{4}{2} + \\binom{4}{2} \\cdot \\binom{2}{1} + \\binom{6}{3} \\cdot \\binom{0}{0} \\] \\[ = 1 \\cdot 20 + 2 \\cdot 6 + 6 \\cdot 2 + 20 \\cdot 1 \\] \\[ = 20 + 12 + 12 + 20 = 64 \\] and as before: \\[E=3 + \\frac{\\frac{N}{20}-1}{2}=3 + \\frac{\\frac{64}{20}-1}{2}=3+\\frac{11}{10}=\\frac{41}{10}\\] ~proloto",
"There's a beautiful observation that for any particular game path, the expected number of correct guesses is purely dependent on how often we are in a neutral state (i.e. there are the same number of red and black cards remaining in the deck). For example, given path $BRRBBR$ we are in a neutral state once in the beginning when no cards have been flipped, after the second card has been flipped, after the 4th card has been flipped, and finally a 4th time at the very end of the game. Call the number of times a given path is in the neutral state $n$. Then the expected number of correct guesses equals: \\[E = 3 + \\frac{n-1}{2}\\] Since this problem is small, we could just could just count them (noting that the problem is symmetric between reds and blacks): \\[ \\begin{array}{|c|c|c|} \\hline \\textbf{Path} & \\textbf{Neutral States} & \\textbf{E}\\\\ \\hline \\text{RRRBBB} & 2 & 3.5\\\\ \\text{RRBRBB} & 2 & 3.5 \\\\ \\text{RRBRBB} & 3 & 4 \\\\ \\text{RRBBBR} & 3 & 4 \\\\ \\text{RBRRBB} & 3 & 4 \\\\ \\text{RBRBRB} & 4 & 4.5 \\\\ \\text{RBRBBR} & 4 & 4.5 \\\\ \\text{RBBRRB} & 4 & 4.5 \\\\ \\text{RBBRBR} & 4 & 4.5 \\\\ \\text{RBBBRR} & 3 & 4 \\\\ \\hline \\end{array} \\] Using linearity of expectation, the expected number of correct guesses is: \\[E=\\frac{3.5+3.5+4+4+4+4.5+4.5+4.5+4.5+4}{10}=\\frac{41}{10}, \\text{ thus } 41 + 10 = 51 gives us: \\[ \\sum_{m=0}^{3} \\binom{2m}{m} \\cdot \\binom{6-2m}{3-m} = \\] \\[ \\binom{0}{0} \\cdot \\binom{6}{3} + \\binom{2}{1} \\cdot \\binom{4}{2} + \\binom{4}{2} \\cdot \\binom{2}{1} + \\binom{6}{3} \\cdot \\binom{0}{0} \\] \\[ = 1 \\cdot 20 + 2 \\cdot 6 + 6 \\cdot 2 + 20 \\cdot 1 \\] \\[ = 20 + 12 + 12 + 20 = 64 \\] and as before: \\[E=3 + \\frac{\\frac{N}{20}-1}{2}=3 + \\frac{\\frac{64}{20}-1}{2}=3+\\frac{11}{10}=\\frac{41}{10}\\] ~proloto"
] |
2023-I-7
| 2,023
| 7
|
Call a positive integer $n$ extra-distinct if the remainders when $n$ is divided by $2, 3, 4, 5,$ and $6$ are distinct. Find the number of extra-distinct positive integers less than $1000$ .
| 49
|
I
|
[
"$n$ can either be $0$ or $1$ (mod $2$). Case 1: $n \\equiv 0 \\pmod{2}$ Then, $n \\equiv 2 \\pmod{4}$, which implies $n \\equiv 1 \\pmod{3}$ and $n \\equiv 4 \\pmod{6}$, and therefore $n \\equiv 3 \\pmod{5}$. Using CRT, we obtain $n \\equiv 58 \\pmod{60}$, which gives $16$ values for $n$. Case 2: $n \\equiv 1 \\pmod{2}$ $n$ is then $3 \\pmod{4}$. If $n \\equiv 0 \\pmod{3}$, $n \\equiv 3 \\pmod{6}$, a contradiction. Thus, $n \\equiv 2 \\pmod{3}$, which implies $n \\equiv 5 \\pmod{6}$. $n$ can either be $0 \\pmod{5}$, which implies that $n \\equiv 35 \\pmod{60}$ by CRT, giving $17$ cases; or $4 \\pmod{5}$, which implies that $n \\equiv 59 \\pmod{60}$ by CRT, giving $16$ cases. The total number of extra-distinct numbers is thus $16 + 16 + 17 = 049. ~mathboy100",
"Because the LCM of all of the numbers we are dividing by is $60$, we know that all of the remainders are $0$ again at $60$, meaning that we have a cycle that repeats itself every $60$ numbers. After listing all of the remainders up to $60$, we find that $35$, $58$, and $59$ are extra-distinct. So, we have $3$ numbers every $60$ which are extra-distinct. $60\\cdot16 = 960$ and $3\\cdot16 = 48$, so we have $48$ extra-distinct numbers in the first $960$ numbers. Because of our pattern, we know that the numbers from $961$ thru $1000$ will have the same remainders as $1$ thru $40$, so we have $1$ other extra-distinct number ($35$). $48 + 1 = 049. ~Algebraik",
"Because the LCM of all of the numbers we are dividing by is $60$, we know that all of the remainders are $0$ again at $60$, meaning that we have a cycle that repeats itself every $60$ numbers. After listing all of the remainders up to $60$, we find that $35$, $58$, and $59$ are extra-distinct. So, we have $3$ numbers every $60$ which are extra-distinct. $60\\cdot16 = 960$ and $3\\cdot16 = 48$, so we have $48$ extra-distinct numbers in the first $960$ numbers. Because of our pattern, we know that the numbers from $961$ thru $1000$ will have the same remainders as $1$ thru $40$, so we have $1$ other extra-distinct number ($35$). $48 + 1 = 049. ~Algebraik",
"$\\textbf{Case 0: } {\\rm Rem} \\ \\left( n, 6 \\right) = 0$. We have ${\\rm Rem} \\ \\left( n, 2 \\right) = 0$. This violates the condition that $n$ is extra-distinct. Therefore, this case has no solution. $\\textbf{Case 1: } {\\rm Rem} \\ \\left( n, 6 \\right) = 1$. We have ${\\rm Rem} \\ \\left( n, 2 \\right) = 1$. This violates the condition that $n$ is extra-distinct. Therefore, this case has no solution. $\\textbf{Case 2: } {\\rm Rem} \\ \\left( n, 6 \\right) = 2$. We have ${\\rm Rem} \\ \\left( n, 3 \\right) = 2$. This violates the condition that $n$ is extra-distinct. Therefore, this case has no solution. $\\textbf{Case 3: } {\\rm Rem} \\ \\left( n, 6 \\right) = 3$. The condition ${\\rm Rem} \\ \\left( n, 6 \\right) = 3$ implies ${\\rm Rem} \\ \\left( n, 2 \\right) = 1$, ${\\rm Rem} \\ \\left( n, 3 \\right) = 0$. Because $n$ is extra-distinct, ${\\rm Rem} \\ \\left( n, l \\right)$ for $l \\in \\left\\{ 2, 3, 4 \\right\\}$ is a permutation of $\\left\\{ 0, 1 ,2 \\right\\}$. Thus, ${\\rm Rem} \\ \\left( n, 4 \\right) = 2$. However, ${\\rm Rem} \\ \\left( n, 4 \\right) = 2$ conflicts ${\\rm Rem} \\ \\left( n, 2 \\right) = 1$. Therefore, this case has no solution. $\\textbf{Case 4: } {\\rm Rem} \\ \\left( n, 6 \\right) = 4$. The condition ${\\rm Rem} \\ \\left( n, 6 \\right) = 4$ implies ${\\rm Rem} \\ \\left( n, 2 \\right) = 0$ and ${\\rm Rem} \\ \\left( n, 3 \\right) = 1$. Because $n$ is extra-distinct, ${\\rm Rem} \\ \\left( n, l \\right)$ for $l \\in \\left\\{ 2, 3, 4 , 5 \\right\\}$ is a permutation of $\\left\\{ 0, 1 ,2 , 3 \\right\\}$. Because ${\\rm Rem} \\ \\left( n, 2 \\right) = 0$, we must have ${\\rm Rem} \\ \\left( n, 4 \\right) = 2$. Hence, ${\\rm Rem} \\ \\left( n, 5 \\right) = 3$. Hence, $n \\equiv -2 \\pmod{{\\rm lcm} \\left( 4, 5 , 6 \\right)}$. Hence, $n \\equiv - 2 \\pmod{60}$. We have $1000 = 60 \\cdot 16 + 40$. Therefore, the number extra-distinct $n$ in this case is 16. $\\textbf{Case 5: } {\\rm Rem} \\ \\left( n, 6 \\right) = 5$. The condition ${\\rm Rem} \\ \\left( n, 6 \\right) = 5$ implies ${\\rm Rem} \\ \\left( n, 2 \\right) = 1$ and ${\\rm Rem} \\ \\left( n, 3 \\right) = 2$. Because $n$ is extra-distinct, ${\\rm Rem} \\ \\left( n, 4 \\right)$ and ${\\rm Rem} \\ \\left( n, 5 \\right)$ are two distinct numbers in $\\left\\{ 0, 3, 4 \\right\\}$. Because ${\\rm Rem} \\ \\left( n, 4 \\right) \\leq 3$ and $n$ is odd, we have ${\\rm Rem} \\ \\left( n, 4 \\right) = 3$. Hence, ${\\rm Rem} \\ \\left( n, 5 \\right) = 0$ or 4. $\\textbf{Case 5.1: } {\\rm Rem} \\ \\left( n, 6 \\right) = 5$, ${\\rm Rem} \\ \\left( n, 4 \\right) = 3$, ${\\rm Rem} \\ \\left( n, 5 \\right) = 0$. We have $n \\equiv 35 \\pmod{60}$. We have $1000 = 60 \\cdot 16 + 40$. Therefore, the number extra-distinct $n$ in this subcase is 17. $\\textbf{Case 5.2: } {\\rm Rem} \\ \\left( n, 6 \\right) = 5$, ${\\rm Rem} \\ \\left( n, 4 \\right) = 3$, ${\\rm Rem} \\ \\left( n, 5 \\right) = 4$. $n \\equiv - 1 \\pmod{60}$. We have $1000 = 60 \\cdot 16 + 40$. Therefore, the number extra-distinct $n$ in this subcase is 16. Putting all cases together, the total number of extra-distinct $n$ is $16 + 17 + 16 = 049. ~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)",
"We need to find that $35$, $58$, and $59$ are all extra-distinct numbers smaller than $61.$ Let $k \\in \\left\\{2, 3, 4, 5, 6 \\right\\}.$ Denote the remainder in the division of $a$ by $b$ as ${\\rm Rem} \\ \\left( a, b \\right).$ ${\\rm Rem} \\ \\left( -1, k \\right) = k - 1 \\implies {\\rm Rem} \\ \\left( 59, k \\right) = k - 1 = \\left\\{1, 2, 3, 4, 5 \\right\\}\\implies 59$ is extra-distinct. ${\\rm Rem} \\ \\left( -2, k \\right) = k - 2 \\implies {\\rm Rem} \\ \\left( 58, k \\right) = k - 2 = \\left\\{0, 1, 2, 3, 4 \\right\\} \\implies 58$ is extra-distinct. \\[{\\rm Rem} \\ \\left( x + 12y, k \\right) = {\\rm Rem} \\ \\left(x, k \\right) + \\left\\{0, 0, 0, {\\rm Rem} \\ \\left(12y, k \\right), 0 \\right\\}.\\] We need to check all of the remainders up to $12 - 3 = 9$ and remainders \\[{\\rm Rem} \\ \\left( 59 - 12, k \\right) = {\\rm Rem} \\ \\left( 59 - 36, k \\right) = \\left\\{1, 2, 3, 3, 5 \\right\\}, {\\rm Rem} \\ \\left( 59 - 48, k \\right) = \\left\\{1, 2, 3, 1, 5 \\right\\},\\] ${\\rm Rem} \\ \\left( 59 - 24, k \\right) ={\\rm Rem} \\ \\left(35, k \\right) = \\left\\{1, 2, 3, 0, 5 \\right\\} \\implies 35$ is extra-distinct. $58 - 12 = 46 \\implies {\\rm Rem} \\ \\left( 46, 5 \\right) = 1 = {\\rm Rem} \\ \\left( 46, 3 \\right),$ $58 - 24 = 34 \\implies {\\rm Rem} \\ \\left( 34, 5 \\right) = 4 = {\\rm Rem} \\ \\left( 34, 6 \\right),$ $58 - 36 = 22 \\implies {\\rm Rem} \\ \\left( 22, 5 \\right) = 2 = {\\rm Rem} \\ \\left( 22, 4 \\right),$ $58 - 48 = 10 \\implies {\\rm Rem} \\ \\left( 10, 5 \\right) = 0 = {\\rm Rem} \\ \\left( 10, 2 \\right).$ vladimir.shelomovskii@gmail.com, vvsss",
"We need to find that $35$, $58$, and $59$ are all extra-distinct numbers smaller than $61.$ Let $k \\in \\left\\{2, 3, 4, 5, 6 \\right\\}.$ Denote the remainder in the division of $a$ by $b$ as ${\\rm Rem} \\ \\left( a, b \\right).$ ${\\rm Rem} \\ \\left( -1, k \\right) = k - 1 \\implies {\\rm Rem} \\ \\left( 59, k \\right) = k - 1 = \\left\\{1, 2, 3, 4, 5 \\right\\}\\implies 59$ is extra-distinct. ${\\rm Rem} \\ \\left( -2, k \\right) = k - 2 \\implies {\\rm Rem} \\ \\left( 58, k \\right) = k - 2 = \\left\\{0, 1, 2, 3, 4 \\right\\} \\implies 58$ is extra-distinct. \\[{\\rm Rem} \\ \\left( x + 12y, k \\right) = {\\rm Rem} \\ \\left(x, k \\right) + \\left\\{0, 0, 0, {\\rm Rem} \\ \\left(12y, k \\right), 0 \\right\\}.\\] We need to check all of the remainders up to $12 - 3 = 9$ and remainders \\[{\\rm Rem} \\ \\left( 59 - 12, k \\right) = {\\rm Rem} \\ \\left( 59 - 36, k \\right) = \\left\\{1, 2, 3, 3, 5 \\right\\}, {\\rm Rem} \\ \\left( 59 - 48, k \\right) = \\left\\{1, 2, 3, 1, 5 \\right\\},\\] ${\\rm Rem} \\ \\left( 59 - 24, k \\right) ={\\rm Rem} \\ \\left(35, k \\right) = \\left\\{1, 2, 3, 0, 5 \\right\\} \\implies 35$ is extra-distinct. $58 - 12 = 46 \\implies {\\rm Rem} \\ \\left( 46, 5 \\right) = 1 = {\\rm Rem} \\ \\left( 46, 3 \\right),$ $58 - 24 = 34 \\implies {\\rm Rem} \\ \\left( 34, 5 \\right) = 4 = {\\rm Rem} \\ \\left( 34, 6 \\right),$ $58 - 36 = 22 \\implies {\\rm Rem} \\ \\left( 22, 5 \\right) = 2 = {\\rm Rem} \\ \\left( 22, 4 \\right),$ $58 - 48 = 10 \\implies {\\rm Rem} \\ \\left( 10, 5 \\right) = 0 = {\\rm Rem} \\ \\left( 10, 2 \\right).$ vladimir.shelomovskii@gmail.com, vvsss"
] |
2023-I-8
| 2,023
| 8
|
Rhombus $ABCD$ has $\angle BAD < 90^\circ.$ There is a point $P$ on the incircle of the rhombus such that the distances from $P$ to the lines $DA,AB,$ and $BC$ are $9,5,$ and $16,$ respectively. Find the perimeter of $ABCD.$
| 125
|
I
|
[
"This solution refers to the Diagram section. Let $O$ be the incenter of $ABCD$ for which $\\odot O$ is tangent to $\\overline{DA},\\overline{AB},$ and $\\overline{BC}$ at $X,Y,$ and $Z,$ respectively. Moreover, suppose that $R,S,$ and $T$ are the feet of the perpendiculars from $P$ to $\\overleftrightarrow{DA},\\overleftrightarrow{AB},$ and $\\overleftrightarrow{BC},$ respectively, such that $\\overline{RT}$ intersects $\\odot O$ at $P$ and $Q.$ We obtain the following diagram: [asy] /* Made by MRENTHUSIASM; inspired by Math Jams. */ size(300); pair A, B, C, D, O, P, R, S, T, X, Y, Z, Q; A = origin; B = (125/4,0); C = B + 125/4 * dir((3,4)); D = A + 125/4 * dir((3,4)); O = (25,25/2); P = (15,5); R = foot(P,A,D); S = foot(P,A,B); T = foot(P,B,C); X = (15,20); Y = (25,0); Z = (35,5); Q = intersectionpoints(Circle(O,25/2),R--T)[1]; fill(R--T--Z--X--cycle,cyan); markscalefactor=0.15; draw(rightanglemark(P,R,D)^^rightanglemark(P,S,B)^^rightanglemark(P,T,C),red); draw(Circle(O,25/2)^^A--B--C--D--cycle^^B--T); draw(P--R^^P--S^^P--T,red+dashed); draw(O--X^^O--Y^^O--Z); dot(\"$A$\",A,1.5*dir(225),linewidth(4.5)); dot(\"$B$\",B,1.5*dir(-45),linewidth(4.5)); dot(\"$C$\",C,1.5*dir(45),linewidth(4.5)); dot(\"$D$\",D,1.5*dir(135),linewidth(4.5)); dot(\"$P$\",P,1.5*dir(60),linewidth(4.5)); dot(\"$R$\",R,1.5*dir(135),linewidth(4.5)); dot(\"$S$\",S,1.5*dir(-90),linewidth(4.5)); dot(\"$T$\",T,1.5*dir(-45),linewidth(4.5)); dot(\"$O$\",O,1.5*dir(45),linewidth(4.5)); dot(\"$X$\",X,1.5*dir(135),linewidth(4.5)); dot(\"$Y$\",Y,1.5*dir(-90),linewidth(4.5)); dot(\"$Z$\",Z,1.5*dir(-45),linewidth(4.5)); dot(\"$Q$\",Q,1.5*dir(60),linewidth(4.5)); label(\"$9$\",midpoint(P--R),dir(A-D),red); label(\"$5$\",midpoint(P--S),dir(180),red); label(\"$16$\",midpoint(P--T),dir(A-D),red); [/asy] Note that $\\angle RXZ = \\angle TZX = 90^\\circ$ by the properties of tangents, so $RTZX$ is a rectangle. It follows that the diameter of $\\odot O$ is $XZ = RT = 25.$ Let $x=PQ$ and $y=RX=TZ.$ We apply the Power of a Point Theorem to $R$ and $T:$ \\begin{align*} y^2 &= 9(9+x), \\\\ y^2 &= 16(16-x). \\end{align*} We solve this system of equations to get $x=7$ and $y=12.$ Alternatively, we can find these results by the symmetry on rectangle $RTZX$ and semicircle $\\widehat{XPZ}.$ We extend $\\overline{SP}$ beyond $P$ to intersect $\\odot O$ and $\\overleftrightarrow{CD}$ at $E$ and $F,$ respectively, where $E\\neq P.$ So, we have $EF=SP=5$ and $PE=25-SP-EF=15.$ On the other hand, we have $PX=15$ by the Pythagorean Theorem on right $\\triangle PRX.$ Together, we conclude that $E=X.$ Therefore, points $S,P,$ and $X$ must be collinear. Let $G$ be the foot of the perpendicular from $D$ to $\\overline{AB}.$ Note that $\\overline{DG}\\parallel\\overline{XP},$ as shown below: [asy] /* Made by MRENTHUSIASM; inspired by Math Jams. */ size(300); pair A, B, C, D, O, P, R, S, T, X, Y, Z, Q, G; A = origin; B = (125/4,0); C = B + 125/4 * dir((3,4)); D = A + 125/4 * dir((3,4)); O = (25,25/2); P = (15,5); R = foot(P,A,D); S = foot(P,A,B); T = foot(P,B,C); X = (15,20); Y = (25,0); Z = (35,5); Q = intersectionpoints(Circle(O,25/2),R--T)[1]; G = foot(D,A,B); fill(D--A--G--cycle,green); fill(P--R--X--cycle,yellow); markscalefactor=0.15; draw(rightanglemark(P,R,D)^^rightanglemark(D,G,A),red); draw(Circle(O,25/2)^^A--B--C--D--cycle^^X--P^^D--G); draw(P--R,red+dashed); dot(\"$A$\",A,1.5*dir(225),linewidth(4.5)); dot(\"$B$\",B,1.5*dir(-45),linewidth(4.5)); dot(\"$C$\",C,1.5*dir(45),linewidth(4.5)); dot(\"$D$\",D,1.5*dir(135),linewidth(4.5)); dot(\"$P$\",P,1.5*dir(60),linewidth(4.5)); dot(\"$R$\",R,1.5*dir(135),linewidth(4.5)); dot(\"$O$\",O,1.5*dir(45),linewidth(4.5)); dot(\"$X$\",X,1.5*dir(135),linewidth(4.5)); dot(\"$G$\",G,1.5*dir(-90),linewidth(4.5)); draw(P--X,MidArrow(0.3cm,Fill(red))); draw(G--D,MidArrow(0.3cm,Fill(red))); label(\"$9$\",midpoint(P--R),dir(A-D),red); label(\"$12$\",midpoint(R--X),dir(135),red); label(\"$15$\",midpoint(X--P),dir(0),red); label(\"$25$\",midpoint(G--D),dir(0),red); [/asy] As $\\angle PRX = \\angle AGD = 90^\\circ$ and $\\angle PXR = \\angle ADG$ by the AA Similarity, we conclude that $\\triangle PRX \\sim \\triangle AGD.$ The ratio of similitude is \\[\\frac{PX}{AD} = \\frac{RX}{GD}.\\] We get $\\frac{15}{AD} = \\frac{12}{25},$ from which $AD = \\frac{125}{4}.$ Finally, the perimeter of $ABCD$ is $4AD = 125 ~MRENTHUSIASM (inspired by awesomeming327. and WestSuburb)",
"This solution refers to the Diagram section. Define points $O,R,S,$ and $T$ as Solution 1 does. Moreover, let $H$ be the foot of the perpendicular from $P$ to $\\overleftrightarrow{CD},$ $M$ be the foot of the perpendicular from $O$ to $\\overleftrightarrow{HS},$ and $N$ be the foot of the perpendicular from $O$ to $\\overleftrightarrow{RT}.$ We obtain the following diagram: [asy] /* Made by MRENTHUSIASM; inspired by Math Jams. */ size(300); pair A, B, C, D, O, P, R, S, T, H, M, N; A = origin; B = (125/4,0); C = B + 125/4 * dir((3,4)); D = A + 125/4 * dir((3,4)); O = (25,25/2); P = (15,5); R = foot(P,A,D); S = foot(P,A,B); T = foot(P,B,C); H = foot(S,C,D); M = foot(O,S,H); N = foot(O,R,T); fill(O--M--P--cycle,yellow); fill(O--N--P--cycle,green); markscalefactor=0.15; draw(rightanglemark(P,R,D)^^rightanglemark(P,S,B)^^rightanglemark(P,T,C)^^rightanglemark(O,M,P)^^rightanglemark(O,N,P)^^rightanglemark(S,H,D),red); draw(Circle(O,25/2)^^A--B--C--D--cycle^^B--T^^D--H^^O--M^^O--N^^O--P); draw(P--R^^P--S^^P--T^^P--H,red+dashed); dot(\"$A$\",A,1.5*dir(225),linewidth(4.5)); dot(\"$B$\",B,1.5*dir(-45),linewidth(4.5)); dot(\"$C$\",C,1.5*dir(45),linewidth(4.5)); dot(\"$D$\",D,1.5*dir(90),linewidth(4.5)); dot(\"$P$\",P,1.5*dir(60),linewidth(4.5)); dot(\"$R$\",R,1.5*dir(135),linewidth(4.5)); dot(\"$S$\",S,1.5*dir(-90),linewidth(4.5)); dot(\"$T$\",T,1.5*dir(-45),linewidth(4.5)); dot(\"$O$\",O,1.5*dir(45),linewidth(4.5)); dot(\"$H$\",H,1.5*dir(90),linewidth(4.5)); dot(\"$M$\",M,1.5*dir(180),linewidth(4.5)); dot(\"$N$\",N,1.5*dir(15),linewidth(4.5)); label(\"$9$\",midpoint(P--R),dir(A-D),red); label(\"$5$\",midpoint(P--S),dir(180),red); label(\"$16$\",midpoint(P--T),dir(A-D),red); [/asy] Note that the diameter of $\\odot O$ is $HS=RT=25,$ so $OP=\\frac{25}{2}.$ It follows that: In right $\\triangle OMP,$ we have $MP=\\frac{HS}{2}-PS=\\frac{15}{2}$ by symmetry, from which $OM=10$ by the Pythagorean Theorem. In right $\\triangle ONP,$ we have $NP=\\frac{RT}{2}-RP=\\frac{7}{2}$ by symmetry, from which $ON=12$ by the Pythagorean Theorem. Since $\\overline{MO}\\parallel\\overline{AB}$ and $\\overline{ON}\\parallel\\overline{DA},$ we conclude that $\\angle A = \\angle MON.$ We apply the Sine of a Sum Formula: \\begin{align*} \\sin\\angle A &= \\sin\\angle MON \\\\ &= \\sin(\\angle MOP + \\angle PON) \\\\ &= \\sin\\angle MOP \\cos\\angle PON + \\cos\\angle MOP \\sin\\angle PON \\\\ &= \\frac{3}{5}\\cdot\\frac{24}{25} + \\frac{4}{5}\\cdot\\frac{7}{25} \\\\ &= \\frac{4}{5}. \\end{align*} Note that \\[\\sin\\angle A = \\frac{HS}{DA},\\] from which $\\frac{4}{5} = \\frac{25}{DA}.$ We solve this equation to get $DA=\\frac{125}{4}.$ Finally, the perimeter of $ABCD$ is $4DA = 125 ~MRENTHUSIASM (credit given to TheAMCHub)",
"Label the points of the rhombus to be $X$, $Y$, $Z$, and $W$ and the center of the incircle to be $O$ so that $9$, $5$, and $16$ are the distances from point $P$ to side $ZW$, side $WX$, and $XY$ respectively. Through this, we know that the distance from the two pairs of opposite lines of rhombus $XYZW$ is $25$ and circle $O$ has radius $\\frac{25}{2}$. Call the feet of the altitudes from $P$ to side $ZW$, side $WX$, and side $XY$ to be $A$, $B$, and $C$ respectively. Additionally, call the feet of the altitudes from $O$ to side $ZW$, side $WX$, and side $XY$ to be $D$, $E$, and $F$ respectively. Draw a line segment from $P$ to $\\overline{OD}$ so that it is perpendicular to $\\overline{OD}$. Notice that this segment length is equal to $AD$ and is $\\sqrt{\\left(\\frac{25}{2}\\right)^2-\\left(\\frac{7}{2}\\right)^2}=12$ by Pythagorean Theorem. Similarly, perform the same operations with perpendicular from $P$ to $\\overline{OE}$ to get $BE=10$. By equal tangents, $WD=WE$. Now, label the length of segment $WA=n$ and $WB=n+2$. Using Pythagorean Theorem again, we get \\begin{align*} WA^2+PA^2&=WB^2+PB^2 \\\\ n^2+9^2&=(n+2)^2+5^2 \\\\ n&=13. \\end{align*} Which also gives us $\\tan{\\angle{OWX}}=\\frac{1}{2}$ and $OW=\\frac{25\\sqrt{5}}{2}$. Since the diagonals of the rhombus intersect at $O$ and are angle bisectors and are also perpendicular to each other, we can get that \\begin{align*} \\frac{OX}{OW}&=\\tan{\\angle{OWX}} \\\\ OX&=\\frac{25\\sqrt{5}}{4} \\\\ WX^2&=OW^2+OX^2 \\\\ WX&=\\frac{125}{4} \\\\ 4WX&=125. \\end{align*} ~Danielzh",
"Denote by $O$ the center of $ABCD$. We drop an altitude from $O$ to $AB$ that meets $AB$ at point $H$. We drop altitudes from $P$ to $AB$ and $AD$ that meet $AB$ and $AD$ at $E$ and $F$, respectively. We denote $\\theta = \\angle BAC$. We denote the side length of $ABCD$ as $d$. Because the distances from $P$ to $BC$ and $AD$ are $16$ and $9$, respectively, and $BC \\parallel AD$, the distance between each pair of two parallel sides of $ABCD$ is $16 + 9 = 25$. Thus, $OH = \\frac{25}{2}$ and $d \\sin \\theta = 25$. We have \\begin{align*} \\angle BOH & = 90^\\circ - \\angle HBO \\\\ & = 90^\\circ - \\angle HBD \\\\ & = 90^\\circ - \\frac{180^\\circ - \\angle C}{2} \\\\ & = 90^\\circ - \\frac{180^\\circ - \\theta}{2} \\\\ & = \\frac{\\theta}{2} . \\end{align*} Thus, $BH = OH \\tan \\angle BOH = \\frac{25}{2} \\tan \\frac{\\theta}{2}$. In $FAEP$, we have $\\overrightarrow{FA} + \\overrightarrow{AE} + \\overrightarrow{EP} + \\overrightarrow{PF} = 0$. Thus, \\[ AF + AE e^{i \\left( \\pi - \\theta \\right)} + EP e^{i \\left( \\frac{3 \\pi}{2} - \\theta \\right)} - PF i . \\] Taking the imaginary part of this equation and plugging $EP = 5$ and $PF = 9$ into this equation, we get \\[ AE = \\frac{9 + 5 \\cos \\theta}{\\sin \\theta} . \\] We have \\begin{align*} OP^2 & = \\left( OH - EP \\right)^2 + \\left( AH - AE \\right)^2 \\\\ & = \\left( \\frac{25}{2} - 5 \\right)^2 + \\left( d - \\frac{25}{2} \\tan \\frac{\\theta}{2} - \\frac{9 + 5 \\cos \\theta}{\\sin \\theta} \\right) \\\\ & = \\left( \\frac{15}{2} \\right)^2 + \\left( \\frac{25}{\\sin \\theta} - \\frac{25}{2} \\tan \\frac{\\theta}{2} - \\frac{9 + 5 \\cos \\theta}{\\sin \\theta} \\right) . \\hspace{1cm} (\\bigstar) \\end{align*} Because $P$ is on the incircle of $ABCD$, $OP = \\frac{25}{2}$. Plugging this into $(\\bigstar)$, we get the following equation \\[ 20 \\sin \\theta - 15 \\cos \\theta = 7 . \\] By solving this equation, we get $\\sin \\theta = \\frac{4}{5}$ and $\\cos \\theta = \\frac{3}{5}$. Therefore, $d = \\frac{25}{\\sin \\theta} = \\frac{125}{4}$. Therefore, the perimeter of $ABCD$ is $4d = 125. ~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)",
"The center of the incircle is $O.$ Denote the points in which the incircle meets $\\overline{AB},$ $\\overline{BC},$ $\\overline{CD},$ and $\\overline{DA}$ as $W,$ $X,$ $Y,$ and $Z,$ respectively. Next, also denote the base of the perpendicular from $P$ to $\\overline{AB},$ $\\overline{AD},$ $\\overline{OW},$ and $\\overline{OZ}$ as $M,$ $N,$ $S,$ and $T,$ respectively. We can easily see that the radius of the circle is $\\frac{25}{2}.$ Using this and Pythagorus on right $\\triangle OSP$ and $\\triangle OTP,$ we find that $MW = PS = 10$ and $NZ = PT = 12.$ Since $AW = AZ$ by properties of circle tangents, we can deduce by the above information that $AM = AN+2.$ Doing Pythagorus on right $\\triangle AMP$ and $\\triangle ANP$ we find that $a^2 = b^2 + 56$ (because $a^2+25=b^2+81.$) From solving the $2$ just derived equations, we find that $AM=15$ and $AN=13.$ Next, we use Pythagorus on right $\\triangle AOB$ (we can see it's right because of properties of rhombuses.) We get \\[AB^2 = AO^2 + BO^2.\\] We know $AB = AW + WB = 25 + WB.$ By Pythagorus on $\\triangle AWO$ and $\\triangle BWO,$ we also know $AO^2 = 25^2+\\left(\\frac{25}{2}\\right)^2$ and $BO^2=WB^2+\\left(\\frac{25}{2}\\right)^2.$ Substituting these in, we have \\[25^2 + 50WB + WB^2 = 25^2+\\left(\\frac{25}{2}\\right)^2+\\left(\\frac{25}{2}\\right)^2+WB^2.\\] Solving for $WB,$ we get $WB = \\frac{25}{4}.$ Now we find that each side of the rhombus $=AB=25+\\frac{25}{4}=\\frac{125}{4}.$ The perimeter of the rhombus would be that times $4.$ Our final answer is \\[\\frac{125}{4}\\cdot4=125.\\] ~s214425",
"Notation is shown on diagram, $RT \\perp AD, FG \\perp AB, E = AD \\cap \\omega, E' = FG \\cap AD.$ $RT = 9 + 16 = 25 = FG$ as hights of rhombus. \\[RP = QT = 9, PQ = 16 - 9 = 7, GE' = PF = 5,\\] \\[PE' = 25 - 5 - 5 = 15, RE = \\sqrt{RP \\cdot RQ} = \\sqrt{9 \\cdot 16} = 12.\\] \\[PE = \\sqrt{RP^2 + RE^2} = 15 \\implies E = E'.\\] \\[\\sin \\alpha = \\frac {RE}{PE} = \\frac {GF}{AD} \\implies AD = \\frac {15 \\cdot 25}{12} = \\frac {125}{4}.\\] The perimeter of $ABCD$ is $\\frac{125}{4}\\cdot4=125 vladimir.shelomovskii@gmail.com, vvsss"
] |
2023-I-9
| 2,023
| 9
|
Find the number of cubic polynomials $p(x) = x^3 + ax^2 + bx + c,$ where $a, b,$ and $c$ are integers in $\{-20,-19,-18,\ldots,18,19,20\},$ such that there is a unique integer $m \not= 2$ with $p(m) = p(2).$
| 738
|
I
|
[
"Plugging $2$ and $m$ into the expression for $p(x)$ and equating them, we get $8+4a+2b+c = m^3+am^2+bm+c$. Rearranging, we have \\[(m^3-8) + (m^2 - 4)a + (m-2)b = 0.\\] Note that the value of $c$ won't matter as it can be anything in the provided range, giving a total of $41$ possible choices for $c.$ So we only need to find the number of ordered pairs $(a, b)$ that work, and multiply it by $41.$ We can start by first dividing both sides by $m-2.$ (Note that this is valid since $m\\neq2:$ \\[m^2 + 2m + 4 + (m+2)a + b = 0.\\] We can rearrange this so it is a quadratic in $m$: \\[m^2 + (a+2)m + (4 + 2a + b) = 0.\\] Remember that $m$ has to be unique and not equal to $2.$ We can split this into two cases: case $1$ being that $m$ has exactly one solution, and it isn't equal to $2$; case $2$ being that $m$ has two solutions, one being $2,$ but the other is a unique solution not equal to $2.$ $\\textbf{Case 1:}$ There is exactly one solution for $m,$ and that solution is not $2.$ This means that the discriminant of the quadratic equation is $0,$ using that, we have $(a+2)^2 = 4(4 + 2a + b)$. Rearranging, we have \\[(a-2)^2 = 4(4 + b)\\Longrightarrow a = 2\\pm2\\sqrt{4+b}.\\] Using the fact that $4+b$ must be a perfect square, we can easily see that the values for $b$ can be $-4, -3, 0, 5,$ and $12.$ Also since it's a \"$\\pm$\" there will usually be $2$ solutions for $a$ for each value of $b.$ The two exceptions for this would be if $b = -4$ and $b = 12.$ For $b=-4$ because it would be a $\\pm0,$ which only gives one solution, instead of two. For $b=12$, because then $a = -6$, the solution for $m$ would equal $2,$, and we don't want this. (We can know this by replacing the solutions into the quadratic formula). So we have $5$ solutions for $b,$ each of which gives $2$ values for $a,$ except for $2$ and $b = -4,$ both of which only give one. So in total, there are $5*2 - 2 = 8$ ordered pairs of $(a,b)$ in this case. $\\textbf{Case 2:}$ $m$ has two solutions, but exactly one of them isn't equal to $2.$ This ensures that $1$ of the solutions is equal to $2.$ Let $r$ be the other value of $m$ that isn't $2.$ By Vieta: \\begin{align*} r+2 &= -a-2\\\\ 2r &= 4+2a+b. \\end{align*} From the first equation, we subtract both sides by $2$ and double both sides to get $2r = -2a - 8$ which also equals to $4+2a+b$ from the second equation. Equating both, we have $4a + b + 12 = 0.$ We can easily count that there would be $11$ ordered pairs $(a,b)$ that satisfy that. However, there's an outlier case in which $r$ happens to also equal $2,$ and we don't want that. We can reverse engineer and find out that $r=2$ when $(a,b) = (-6, 12),$ which we overcounted. So we subtract by one and we conclude that there are $10$ ordered pairs of $(a,b)$ that satisfy this case. This all shows that there is a total of $8+10 = 18$ amount of ordered pairs $(a,b).$ Multiplying this by $41$ (the amount of values for $c$) we get $18\\cdot41=738 as our final answer. ~s214425",
"$p(x)-p(2)$ is a cubic with at least two integral real roots, therefore it has three real roots, which are all integers. There are exactly two distinct roots, so either $p(x)=p(2)+(x-2)^2(x-m)$ or $p(x)=p(2)+(x-2)(x-m)^2$, with $m\\neq 2$. In the first case $p(x)=x^3-(4+m)x^2+(4+4m)x-4m+p(2)$, with $|4+4m|\\leq 20$ (which entails $|4+m|\\leq 20$), so $m$ can be $-6,-5,-4,-3,-2,-1,0,1, (\\textbf{not 2}!), 3,4$ and $-4m+p(2)$ can be any integer from $-20$ to $20$, giving $410$ polynomials (since the coefficients are given by linear functions of $m$ and thus are distinct). In the second case $p(x)=x^3-(2+2m)x^2+(4m+m^2)x-2m^2+p(2)$, and $m$ can be $-6,-5,-4,-3,-2,-1,0,1$ and $-4m+p(2)$ can be any integer from $-20$ to $20$, giving $328$ polynomials. The total is $738. ~EVIN-",
"$p(x)-p(2)$ is a cubic with at least two integral real roots, therefore it has three real roots, which are all integers. There are exactly two distinct roots, so either $p(x)=p(2)+(x-2)^2(x-m)$ or $p(x)=p(2)+(x-2)(x-m)^2$, with $m\\neq 2$. In the first case $p(x)=x^3-(4+m)x^2+(4+4m)x-4m+p(2)$, with $|4+4m|\\leq 20$ (which entails $|4+m|\\leq 20$), so $m$ can be $-6,-5,-4,-3,-2,-1,0,1, (\\textbf{not 2}!), 3,4$ and $-4m+p(2)$ can be any integer from $-20$ to $20$, giving $410$ polynomials (since the coefficients are given by linear functions of $m$ and thus are distinct). In the second case $p(x)=x^3-(2+2m)x^2+(4m+m^2)x-2m^2+p(2)$, and $m$ can be $-6,-5,-4,-3,-2,-1,0,1$ and $-4m+p(2)$ can be any integer from $-20$ to $20$, giving $328$ polynomials. The total is $738. ~EVIN-"
] |
2023-I-10
| 2,023
| 10
|
There exists a unique positive integer $a$ for which the sum \[U=\sum_{n=1}^{2023}\left\lfloor\dfrac{n^{2}-na}{5}\right\rfloor\] is an integer strictly between $-1000$ and $1000$ . For that unique $a$ , find $a+U$ . (Note that $\lfloor x\rfloor$ denotes the greatest integer that is less than or equal to $x$ .)
| 944
|
I
|
[
"Define $\\left\\{ x \\right\\} = x - \\left\\lfloor x \\right\\rfloor$. First, we bound $U$. We establish an upper bound of $U$. We have \\begin{align*} U & \\leq \\sum_{n=1}^{2023} \\frac{n^2 - na}{5} \\\\ & = \\frac{1}{5} \\sum_{n=1}^{2023} n^2 - \\frac{a}{5} \\sum_{n=1}^{2023} n \\\\ & = \\frac{1012 \\cdot 2023}{5} \\left( 1349 - a \\right) \\\\ & \\triangleq UB . \\end{align*} We establish a lower bound of $U$. We have \\begin{align*} U & = \\sum_{n=1}^{2023} \\left( \\frac{n^2 - na}{5} - \\left\\{ \\frac{n^2 - na}{5} \\right\\} \\right) \\\\ & = \\sum_{n=1}^{2023} \\frac{n^2 - na}{5} - \\sum_{n=1}^{2023} \\left\\{ \\frac{n^2 - na}{5} \\right\\} \\\\ & = UB - \\sum_{n=1}^{2023} \\left\\{ \\frac{n^2 - na}{5} \\right\\} \\\\ & \\geq UB - \\sum_{n=1}^{2023} \\mathbf 1 \\left\\{ \\frac{n^2 - na}{5} \\notin \\Bbb Z \\right\\} . \\end{align*} We notice that if $5 | n$, then $\\frac{n^2 - na}{5} \\in \\Bbb Z$. Thus, \\begin{align*} U & \\geq UB - \\sum_{n=1}^{2023} \\mathbf 1 \\left\\{ \\frac{n^2 - na}{5} \\notin \\Bbb Z \\right\\} \\\\ & \\geq UB - \\sum_{n=1}^{2023} \\mathbf 1 \\left\\{ 5 \\nmid n \\right\\} \\\\ & = UB - \\left( 2023 - \\left\\lfloor \\frac{2023}{5} \\right\\rfloor \\right) \\\\ & = UB - 1619 \\\\ & \\triangleq LB . \\end{align*} Because $U \\in \\left[ - 1000, 1000 \\right]$ and $UB - LB = 1619 < \\left( 1000 - \\left( - 1000 \\right) \\right)$, we must have either $UB \\in \\left[ - 1000, 1000 \\right]$ or $LB \\in \\left[ - 1000, 1000 \\right]$. For $UB \\in \\left[ - 1000, 1000 \\right]$, we get a unique $a = 1349$. For $LB \\in \\left[ - 1000, 1000 \\right]$, there is no feasible $a$. Therefore, $a = 1349$. Thus $UB = 0$. Next, we compute $U$. Let $n = 5 q + r$, where $r = {\\rm Rem} \\ \\left( n, 5 \\right)$. We have \\begin{align*} \\left\\{ \\frac{n^2 - na}{5} \\right\\} & = \\left\\{ \\frac{\\left( 5 q + r \\right)^2 - \\left( 5 q + r \\right)\\left( 1350 - 1 \\right)}{5} \\right\\} \\\\ & = \\left\\{ 5 q^2 + 2 q r - \\left( 5 q + r \\right) 270 + q + \\frac{r^2 + r}{5} \\right\\} \\\\ & = \\left\\{\\frac{r^2 + r}{5} \\right\\} \\\\ & = \\left\\{ \\begin{array}{ll} 0 & \\mbox{ if } r = 0, 4 \\\\ \\frac{2}{5} & \\mbox{ if } r = 1, 3 \\\\ \\frac{1}{5} & \\mbox{ if } r = 2 \\end{array} \\right. . \\end{align*} Therefore, \\begin{align*} U & = \\sum_{n=1}^{2023} \\left( \\frac{n^2 - na}{5} - \\left\\{ \\frac{n^2 - na}{5} \\right\\} \\right) \\\\ & = UB - \\sum_{n=1}^{2023} \\left\\{ \\frac{n^2 - na}{5} \\right\\} \\\\ & = - \\sum_{n=1}^{2023} \\left\\{ \\frac{n^2 - na}{5} \\right\\} \\\\ & = - \\sum_{q=0}^{404} \\sum_{r=0}^4 \\left\\{\\frac{r^2 + r}{5} \\right\\} + \\left\\{ \\frac{0^2 - 0 \\cdot a}{5} \\right\\} + \\left\\{ \\frac{2024^2 - 2024a}{5} \\right\\} \\\\ & = - \\sum_{q=0}^{404} \\left( 0 + 0 + \\frac{2}{5} + \\frac{2}{5} + \\frac{1}{5} \\right) + 0 + 0 \\\\ & = - 405 . \\end{align*} Therefore, $a + U = 1349 - 405 = \\textbf{944}. ~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com) ~minor edits by KevinChen_Yay",
"Define $\\left\\{ x \\right\\} = x - \\left\\lfloor x \\right\\rfloor$. First, we bound $U$. We establish an upper bound of $U$. We have \\begin{align*} U & \\leq \\sum_{n=1}^{2023} \\frac{n^2 - na}{5} \\\\ & = \\frac{1}{5} \\sum_{n=1}^{2023} n^2 - \\frac{a}{5} \\sum_{n=1}^{2023} n \\\\ & = \\frac{1012 \\cdot 2023}{5} \\left( 1349 - a \\right) \\\\ & \\triangleq UB . \\end{align*} We establish a lower bound of $U$. We have \\begin{align*} U & = \\sum_{n=1}^{2023} \\left( \\frac{n^2 - na}{5} - \\left\\{ \\frac{n^2 - na}{5} \\right\\} \\right) \\\\ & = \\sum_{n=1}^{2023} \\frac{n^2 - na}{5} - \\sum_{n=1}^{2023} \\left\\{ \\frac{n^2 - na}{5} \\right\\} \\\\ & = UB - \\sum_{n=1}^{2023} \\left\\{ \\frac{n^2 - na}{5} \\right\\} \\\\ & \\geq UB - \\sum_{n=1}^{2023} \\mathbf 1 \\left\\{ \\frac{n^2 - na}{5} \\notin \\Bbb Z \\right\\} . \\end{align*} We notice that if $5 | n$, then $\\frac{n^2 - na}{5} \\in \\Bbb Z$. Thus, \\begin{align*} U & \\geq UB - \\sum_{n=1}^{2023} \\mathbf 1 \\left\\{ \\frac{n^2 - na}{5} \\notin \\Bbb Z \\right\\} \\\\ & \\geq UB - \\sum_{n=1}^{2023} \\mathbf 1 \\left\\{ 5 \\nmid n \\right\\} \\\\ & = UB - \\left( 2023 - \\left\\lfloor \\frac{2023}{5} \\right\\rfloor \\right) \\\\ & = UB - 1619 \\\\ & \\triangleq LB . \\end{align*} Because $U \\in \\left[ - 1000, 1000 \\right]$ and $UB - LB = 1619 < \\left( 1000 - \\left( - 1000 \\right) \\right)$, we must have either $UB \\in \\left[ - 1000, 1000 \\right]$ or $LB \\in \\left[ - 1000, 1000 \\right]$. For $UB \\in \\left[ - 1000, 1000 \\right]$, we get a unique $a = 1349$. For $LB \\in \\left[ - 1000, 1000 \\right]$, there is no feasible $a$. Therefore, $a = 1349$. Thus $UB = 0$. Next, we compute $U$. Let $n = 5 q + r$, where $r = {\\rm Rem} \\ \\left( n, 5 \\right)$. We have \\begin{align*} \\left\\{ \\frac{n^2 - na}{5} \\right\\} & = \\left\\{ \\frac{\\left( 5 q + r \\right)^2 - \\left( 5 q + r \\right)\\left( 1350 - 1 \\right)}{5} \\right\\} \\\\ & = \\left\\{ 5 q^2 + 2 q r - \\left( 5 q + r \\right) 270 + q + \\frac{r^2 + r}{5} \\right\\} \\\\ & = \\left\\{\\frac{r^2 + r}{5} \\right\\} \\\\ & = \\left\\{ \\begin{array}{ll} 0 & \\mbox{ if } r = 0, 4 \\\\ \\frac{2}{5} & \\mbox{ if } r = 1, 3 \\\\ \\frac{1}{5} & \\mbox{ if } r = 2 \\end{array} \\right. . \\end{align*} Therefore, \\begin{align*} U & = \\sum_{n=1}^{2023} \\left( \\frac{n^2 - na}{5} - \\left\\{ \\frac{n^2 - na}{5} \\right\\} \\right) \\\\ & = UB - \\sum_{n=1}^{2023} \\left\\{ \\frac{n^2 - na}{5} \\right\\} \\\\ & = - \\sum_{n=1}^{2023} \\left\\{ \\frac{n^2 - na}{5} \\right\\} \\\\ & = - \\sum_{q=0}^{404} \\sum_{r=0}^4 \\left\\{\\frac{r^2 + r}{5} \\right\\} + \\left\\{ \\frac{0^2 - 0 \\cdot a}{5} \\right\\} + \\left\\{ \\frac{2024^2 - 2024a}{5} \\right\\} \\\\ & = - \\sum_{q=0}^{404} \\left( 0 + 0 + \\frac{2}{5} + \\frac{2}{5} + \\frac{1}{5} \\right) + 0 + 0 \\\\ & = - 405 . \\end{align*} Therefore, $a + U = 1349 - 405 = \\textbf{944}. ~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com) ~minor edits by KevinChen_Yay",
"We define $U' = \\sum^{2023}_{n=1} {\\frac{n^2-na}{5}}$. Since for any real number $x$, $\\lfloor x \\rfloor \\le x \\le \\lfloor x \\rfloor + 1$, we have $U \\le U' \\le U + 2023$. Now, since $-1000 \\le U \\le 1000$, we have $-1000 \\le U' \\le 3023$. Now, we can solve for $U'$ in terms of $a$. We have: \\begin{align*} U' &= \\sum^{2023}_{n=1} {\\frac{n^2-na}{5}} \\\\ &= \\sum^{2023}_{n=1} {\\frac{n^2}{5} - \\frac{na}{5}} \\\\ &= \\sum^{2023}_{n=1} {\\frac{n^2}{5}} - \\sum^{2023}_{n=1} {\\frac{na}{5}} \\\\ &= \\frac{\\sum^{2023}_{n=1} {{n^2}} - \\sum^{2023}_{n=1} {na}}{5} \\\\ &= \\frac{\\frac{2023(2023+1)(2023 \\cdot 2 + 1)}{6} - \\frac{a \\cdot 2023(2023+1)}{2} }{5} \\\\ &= \\frac{2023(2024)(4047-3a)}{30} \\\\ \\end{align*} So, we have $U' = \\frac{2023(2024)(4047-3a)}{30}$, and $-1000 \\le U' \\le 3023$, so we have $-1000 \\le \\frac{2023(2024)(4047-3a)}{30} \\le 3023$, or $-30000 \\le 2023(2024)(4047-3a) \\le 90690$. Now, $2023 \\cdot 2024$ is much bigger than $90690$ or $30000$, and since $4047-3a$ is an integer, to satsify the inequalities, we must have $4047 - 3a = 0$, or $a = 1349$, and $U' = 0$. Now, we can find $U - U'$. We have: \\begin{align*} U - U' &= \\sum^{2023}_{n=1} {\\lfloor \\frac{n^2-1349n}{5} \\rfloor} - \\sum^{2023}_{n=1} {\\frac{n^2-1349n}{5}} \\\\ &= \\sum^{2023}_{n=1} {\\lfloor \\frac{n^2-1349n}{5} \\rfloor - \\frac{n^2-1349n}{5}} \\end{align*}. Now, if $n^2-1349n \\equiv 0 \\text{ (mod 5)}$, then $\\lfloor \\frac{n^2-1349n}{5} \\rfloor - \\frac{n^2-1349n}{5} = 0$, and if $n^2-1349n \\equiv 1 \\text{ (mod 5)}$, then $\\lfloor \\frac{n^2-1349n}{5} \\rfloor - \\frac{n^2-1349n}{5} = -\\frac{1}{5}$, and so on. Testing with $n \\equiv 0,1,2,3,4, \\text{ (mod 5)}$, we get $n^2-1349n \\equiv 0,2,1,2,0 \\text{ (mod 5)}$ respectively. From 1 to 2023, there are 405 numbers congruent to 1 mod 5, 405 numbers congruent to 2 mod 5, 405 numbers congruent to 3 mod 5, 404 numbers congruent to 4 mod 5, and 404 numbers congruent to 0 mod 5. So, solving for $U - U'$, we get: \\begin{align*} U - U' &= \\sum^{2023}_{n=1} {\\lfloor \\frac{n^2-1349n}{5} \\rfloor - \\frac{n^2-1349n}{5}} \\\\ &= 404 \\cdot 0 - 405 \\cdot \\frac{2}{5} - 405 \\cdot \\frac{1}{5} - 405 \\cdot \\frac{2}{5} - 404 \\cdot 0 \\\\ &= -405(\\frac{2}{5}+\\frac{1}{5}+\\frac{2}{5}) \\\\ &= -405 \\end{align*} Since $U' = 0$, this gives $U = -405$, and we have $a + U = 1349-405 = 944. ~ genius_007",
"We can view the floor function in this problem as simply subtracting the remainder of $n^2 - na$ (mod $5$) from the numerator of $\\frac{n^2-na}{5}$. For example, $\\left\\lfloor \\frac{7}{5} \\right\\rfloor = \\frac{7-2}{5} = 1$. Note that the congruence of $n^2 - na$ (mod $5$) loops every time $n$ increases by 5. Also, note that the congruence of $a$ (mod $5$) determines the set of congruences of $n^2 - na$ for each congruence of $n$ (mod $5$). For example, if $a \\equiv 1$ (mod $5$), the set of remainders is $(0, 2, 1, 2, 0)$ for $n \\equiv 1,2,3,4,0$ (mod $5$). Let the sum of these elements be $s$. Note that for each “loop” of the numerator (mod $5$), each element of the set will be subtracted exactly once, meaning $s$ is subtracted once for each loop. The value of the numerator will loop $404$ times (mod $5$) throughout the sum, as $5 \\cdot 404=2020$. Then $U \\approx \\frac {\\left( \\frac {n(n+1)(2n+1)}{6} - \\frac{(a)(n)(n+1)}{2} -404s \\right)}{5}$ Where $n=2023$. Note that since $5 \\cdot 404=2020$, this is an approximation for $U$ because the equation disregards the remainder (mod $5$) when $n=2021, 2022$, and $2023$ so we must subtract the first 3 terms of our set of congruences one extra time to get the exact value of $U$ (*). However, we will find that this is a negligible error when it comes to the inequality $-1000<U<1000$, so we can proceed with this approximation to solve for $a$. Factoring our approximation gives $U \\approx \\frac {\\frac{(n)(n+1)(2n+1 - 3a)}{6}-404s}{5}$ We set $a= \\frac{(2n+1)}{3} = 1349$ to make $\\frac{(n)(n+1)(2n+1 - 3a)}{6}=0$, accordingly minimizing $|U|$, yielding $U \\approx \\frac{-404s}{5}$ If $a$ increases or decreases by $1$, then $U$ changes by $\\frac {(n)(n+1)}{2 \\cdot 5} = \\frac {2023 \\cdot 2024}{10}$ which clearly breaks the inequality on $U$. Therefore $a=1349 \\equiv 4$ (mod $5$) giving the set of remainders $(2,1,2,0,0)$, so $s=5$ and our approximation yields $U \\approx -404$. However, we must subtract 2, 1, and 2 (*) giving us $U = - 404 - \\frac{(2+1+2)}{5} = - 405$, giving an answer of $1349-405= 944 ~Spencer Danese",
"We can view the floor function in this problem as simply subtracting the remainder of $n^2 - na$ (mod $5$) from the numerator of $\\frac{n^2-na}{5}$. For example, $\\left\\lfloor \\frac{7}{5} \\right\\rfloor = \\frac{7-2}{5} = 1$. Note that the congruence of $n^2 - na$ (mod $5$) loops every time $n$ increases by 5. Also, note that the congruence of $a$ (mod $5$) determines the set of congruences of $n^2 - na$ for each congruence of $n$ (mod $5$). For example, if $a \\equiv 1$ (mod $5$), the set of remainders is $(0, 2, 1, 2, 0)$ for $n \\equiv 1,2,3,4,0$ (mod $5$). Let the sum of these elements be $s$. Note that for each “loop” of the numerator (mod $5$), each element of the set will be subtracted exactly once, meaning $s$ is subtracted once for each loop. The value of the numerator will loop $404$ times (mod $5$) throughout the sum, as $5 \\cdot 404=2020$. Then $U \\approx \\frac {\\left( \\frac {n(n+1)(2n+1)}{6} - \\frac{(a)(n)(n+1)}{2} -404s \\right)}{5}$ Where $n=2023$. Note that since $5 \\cdot 404=2020$, this is an approximation for $U$ because the equation disregards the remainder (mod $5$) when $n=2021, 2022$, and $2023$ so we must subtract the first 3 terms of our set of congruences one extra time to get the exact value of $U$ (*). However, we will find that this is a negligible error when it comes to the inequality $-1000<U<1000$, so we can proceed with this approximation to solve for $a$. Factoring our approximation gives $U \\approx \\frac {\\frac{(n)(n+1)(2n+1 - 3a)}{6}-404s}{5}$ We set $a= \\frac{(2n+1)}{3} = 1349$ to make $\\frac{(n)(n+1)(2n+1 - 3a)}{6}=0$, accordingly minimizing $|U|$, yielding $U \\approx \\frac{-404s}{5}$ If $a$ increases or decreases by $1$, then $U$ changes by $\\frac {(n)(n+1)}{2 \\cdot 5} = \\frac {2023 \\cdot 2024}{10}$ which clearly breaks the inequality on $U$. Therefore $a=1349 \\equiv 4$ (mod $5$) giving the set of remainders $(2,1,2,0,0)$, so $s=5$ and our approximation yields $U \\approx -404$. However, we must subtract 2, 1, and 2 (*) giving us $U = - 404 - \\frac{(2+1+2)}{5} = - 405$, giving an answer of $1349-405= 944 ~Spencer Danese",
"Consider the integral \\[\\int_{0}^{2023} \\dfrac{n^2-na}{5} \\, dn.\\] We hope this will give a good enough appoximation of $U$ to find $a.$ However, this integral can be easily evaluated to be \\[\\dfrac{1}{15}2023^3-\\dfrac{a}{10}2023^2=2023^2\\left(\\dfrac{2023}{15}-\\dfrac{a}{10}\\right).\\] Because we want this to be as close to $0$ as possible, we find that $a$ should equal $1349.$ Then, evaluating the sum becomes trivial. Set \\[U'=\\sum_{n=1}^{2023}\\dfrac{n^2-1349n}{5}\\] and \\[U''=\\sum_{n=1}^{2023}\\{\\dfrac{n^2-1349n}{5}\\}.\\] Then $U=U'-U''.$ We can evaluate $U'$ to be $0$ and $U''$ to be $-405$ (using some basic number theory). Thus, $U=-405$ and the answer is \\[1349+(-405)=944.\\] ~BS2012",
"Consider the integral \\[\\int_{0}^{2023} \\dfrac{n^2-na}{5} \\, dn.\\] We hope this will give a good enough appoximation of $U$ to find $a.$ However, this integral can be easily evaluated to be \\[\\dfrac{1}{15}2023^3-\\dfrac{a}{10}2023^2=2023^2\\left(\\dfrac{2023}{15}-\\dfrac{a}{10}\\right).\\] Because we want this to be as close to $0$ as possible, we find that $a$ should equal $1349.$ Then, evaluating the sum becomes trivial. Set \\[U'=\\sum_{n=1}^{2023}\\dfrac{n^2-1349n}{5}\\] and \\[U''=\\sum_{n=1}^{2023}\\{\\dfrac{n^2-1349n}{5}\\}.\\] Then $U=U'-U''.$ We can evaluate $U'$ to be $0$ and $U''$ to be $-405$ (using some basic number theory). Thus, $U=-405$ and the answer is \\[1349+(-405)=944.\\] ~BS2012"
] |
2023-I-11
| 2,023
| 11
|
Find the number of subsets of $\{1,2,3,\ldots,10\}$ that contain exactly one pair of consecutive integers. Examples of such subsets are $\{\mathbf{1},\mathbf{2},5\}$ and $\{1,3,\mathbf{6},\mathbf{7},10\}.$
| 235
|
I
|
[
"Define $f(x)$ to be the number of subsets of $\\{1, 2, 3, 4, \\ldots x\\}$ that have $0$ consecutive element pairs, and $f'(x)$ to be the number of subsets that have $1$ consecutive pair. Using casework on where the consecutive element pair is, there is a unique consecutive element pair that satisfies the conditions. It is easy to see that \\[f'(10) = 2f(7) + 2f(6) + 2f(1)f(5) + 2f(2)f(4) + f(3)^2.\\] We see that $f(1) = 2$, $f(2) = 3$, and $f(n) = f(n-1) + f(n-2)$. This is because if the element $n$ is included in our subset, then there are $f(n-2)$ possibilities for the rest of the elements (because $n-1$ cannot be used), and otherwise there are $f(n-1)$ possibilities. Thus, by induction, $f(n)$ is the $n+1$th Fibonacci number. This means that $f'(10) = 2(34) + 2(21) + 2(2)(13) + 2(3)(8) + 5^2 = 235. ~mathboy100",
"Define $f(x)$ to be the number of subsets of $\\{1, 2, 3, 4, \\ldots x\\}$ that have $0$ consecutive element pairs, and $f'(x)$ to be the number of subsets that have $1$ consecutive pair. Using casework on where the consecutive element pair is, there is a unique consecutive element pair that satisfies the conditions. It is easy to see that \\[f'(10) = 2f(7) + 2f(6) + 2f(1)f(5) + 2f(2)f(4) + f(3)^2.\\] We see that $f(1) = 2$, $f(2) = 3$, and $f(n) = f(n-1) + f(n-2)$. This is because if the element $n$ is included in our subset, then there are $f(n-2)$ possibilities for the rest of the elements (because $n-1$ cannot be used), and otherwise there are $f(n-1)$ possibilities. Thus, by induction, $f(n)$ is the $n+1$th Fibonacci number. This means that $f'(10) = 2(34) + 2(21) + 2(2)(13) + 2(3)(8) + 5^2 = 235. ~mathboy100",
"We can solve this problem using casework, with one case for each possible pair of consecutive numbers. $\\textbf{Case 1: (1,2)}$ If we have (1,2) as our pair, we are left with the numbers from 3-10 as elements that can be added to our subset. So, we must compute how many ways we can pick these numbers so that the set has no consecutive numbers other than (1,2). Our first option is to pick no more numbers, giving us $8 \\choose {0}$. We can also pick one number, giving us $7 \\choose {1}$ because 3 cannot be picked. Another choice is to pick two numbers and in order to make sure they are not consecutive we must fix one number in between them, giving us $6 \\choose {2}$. This pattern continues for each amount of numbers, yielding $5 \\choose {3}$ for 3 numbers and $4 \\choose {4}$ for four numbers. Adding these up, we have $8 \\choose {0}$ + $7 \\choose {1}$ + $6 \\choose {2}$ + $5 \\choose {3}$ + $4 \\choose {4}$ = $\\textbf{34}$. $\\textbf{Case 2: (2,3)}$ If we have (2,3) as our pair, everything works the same as with (1,2), because 1 is still unusable as it is consecutive with 2. The only difference is we now have only 4-10 to work with. Using the same pattern as before, we have $7 \\choose {0}$ + $6 \\choose {1}$ + $5 \\choose {2}$ + $4 \\choose {3}$ = $\\textbf{21}$. $\\textbf{Case 3: (3,4)}$ This case remains pretty much the same except we now have an option of whether or not to include 1. If we want to represent this like we have with our other choices, we would say $2 \\choose {0}$ for choosing no numbers and $1 \\choose {1}$ for choosing 1, leaving us with $2 \\choose {0}$ + $1 \\choose {1}$ = 2 choices (either including the number 1 in our subset or not including it). As far as the numbers from 5-10, our pattern from previous cases still holds. We have $6 \\choose {0}$ + $5 \\choose {1}$ + $4 \\choose {2}$ + $3 \\choose {3}$ = 13. With 2 choices on one side and 13 choices on the other side, we have $2\\cdot13$ = $\\textbf{26}$ combinations in all. $\\textbf{Case 4: (4,5)}$ Following the patterns we have already created in our previous cases, for the numbers 1-3 we have $3 \\choose {0}$ + $2 \\choose {1}$ = 3 choices (1, 2, or neither) and for the numbers 6-10 we have $5 \\choose {0}$ + $4 \\choose {1}$ + $3 \\choose {2}$ = 8 choices. With 3 choices on one side and 8 choices on the other side, we have $3\\cdot8$ = $\\textbf{24}$ combinations in all. $\\textbf{Case 5: (5,6)}$ Again following the patterns we have already created in our previous cases, for the numbers 1-4 we have $4 \\choose {0}$ + $3 \\choose {1}$ + $2 \\choose {2}$ = 5 choices and for the numbers 5-10 we have the same $4 \\choose {0}$ + $3 \\choose {1}$ + $2 \\choose {2}$ = 5 choices. $5\\cdot5$ = $\\textbf{25}$ combinations in all. $\\textbf{Rest of the cases}$ By symmetry, the case with (6,7) will act the same as case 4 with (4,5). This goes the same for (7,8) and case 3, (8.9) and case 2, and (9,10) and case 1. Now, we simply add up all of the possibilities for each case to get our final answer. 34 + 21 + 26 + 24 + 25 + 24 + 26 + 21 + 34 = $\\textbf{(235)} -Algebraik",
"Denote by $N_1 \\left( m \\right)$ the number of subsets of a set $S$ that consists of $m$ consecutive integers, such that each subset contains exactly one pair of consecutive integers. Denote by $N_0 \\left( m \\right)$ the number of subsets of a set $S$ that consists of $m$ consecutive integers, such that each subset does not contain any consecutive integers. Denote by $a$ the smallest number in set $S$. First, we compute $N_1 \\left( m \\right)$. Consider $m \\geq 3$. We do casework analysis. Case 1: A subset does not contain $a$. The number of subsets that has exactly one pair of consecutive integers is $N_1 \\left( m - 1 \\right)$. Case 2: A subset contains $a$ but does not contain $a + 1$. The number of subsets that has exactly one pair of consecutive integers is $N_1 \\left( m - 2 \\right)$. Case 3: A subset contains $a$ and $a + 1$. To have exactly one pair of consecutive integers, this subset cannot have $a + 2$, and cannot have consecutive integers in $\\left\\{ a+3, a+4, \\cdots , a + m - 1 \\right\\}$. Thus, the number of subsets that has exactly one pair of consecutive integers is $N_0 \\left( m - 3 \\right)$. Therefore, for $m \\geq 3$, \\[N_1 \\left( m \\right) = N_1 \\left( m - 1 \\right) + N_1 \\left( m - 2 \\right) + N_0 \\left( m - 3 \\right) .\\] For $m = 1$, we have $N_1 \\left( 1 \\right) = 0$. For $m = 2$, we have $N_1 \\left( 2 \\right) = 1$. Second, we compute $N_0 \\left( m \\right)$. Consider $m \\geq 2$. We do casework analysis. Case 1: A subset does not contain $a$. The number of subsets that has no consecutive integers is $N_0 \\left( m - 1 \\right)$. Case 2: A subset contains $a$. To avoid having consecutive integers, the subset cannot have $a + 1$. Thus, the number of subsets that has no consecutive integers is $N_0 \\left( m - 2 \\right)$. Therefore, for $m \\geq 2$, \\[N_0 \\left( m \\right) = N_0 \\left( m - 1 \\right) + N_0 \\left( m - 2 \\right) .\\] For $m = 0$, we have $N_0 \\left( 0 \\right) = 1$. For $m = 1$, we have $N_0 \\left( 1 \\right) = 2$. By solving the recursive equations above, we get $N_1 \\left( 10 \\right) = \\textbf{(235) }. ~ Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)",
"Denote by $N_1 \\left( m \\right)$ the number of subsets of a set $S$ that consists of $m$ consecutive integers, such that each subset contains exactly one pair of consecutive integers. Denote by $N_0 \\left( m \\right)$ the number of subsets of a set $S$ that consists of $m$ consecutive integers, such that each subset does not contain any consecutive integers. Denote by $a$ the smallest number in set $S$. First, we compute $N_1 \\left( m \\right)$. Consider $m \\geq 3$. We do casework analysis. Case 1: A subset does not contain $a$. The number of subsets that has exactly one pair of consecutive integers is $N_1 \\left( m - 1 \\right)$. Case 2: A subset contains $a$ but does not contain $a + 1$. The number of subsets that has exactly one pair of consecutive integers is $N_1 \\left( m - 2 \\right)$. Case 3: A subset contains $a$ and $a + 1$. To have exactly one pair of consecutive integers, this subset cannot have $a + 2$, and cannot have consecutive integers in $\\left\\{ a+3, a+4, \\cdots , a + m - 1 \\right\\}$. Thus, the number of subsets that has exactly one pair of consecutive integers is $N_0 \\left( m - 3 \\right)$. Therefore, for $m \\geq 3$, \\[N_1 \\left( m \\right) = N_1 \\left( m - 1 \\right) + N_1 \\left( m - 2 \\right) + N_0 \\left( m - 3 \\right) .\\] For $m = 1$, we have $N_1 \\left( 1 \\right) = 0$. For $m = 2$, we have $N_1 \\left( 2 \\right) = 1$. Second, we compute $N_0 \\left( m \\right)$. Consider $m \\geq 2$. We do casework analysis. Case 1: A subset does not contain $a$. The number of subsets that has no consecutive integers is $N_0 \\left( m - 1 \\right)$. Case 2: A subset contains $a$. To avoid having consecutive integers, the subset cannot have $a + 1$. Thus, the number of subsets that has no consecutive integers is $N_0 \\left( m - 2 \\right)$. Therefore, for $m \\geq 2$, \\[N_0 \\left( m \\right) = N_0 \\left( m - 1 \\right) + N_0 \\left( m - 2 \\right) .\\] For $m = 0$, we have $N_0 \\left( 0 \\right) = 1$. For $m = 1$, we have $N_0 \\left( 1 \\right) = 2$. By solving the recursive equations above, we get $N_1 \\left( 10 \\right) = \\textbf{(235) }. ~ Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)",
"Let $a_n$ be the number of subsets of the set $\\{1,2,3,\\ldots,n\\}$ such that there exists exactly 1 pair of consecutive elements. Let $b_n$ be the number of subsets of the set $\\{1, 2, 3\\ldots, n\\}$ such that there doesn't exist any pair of consecutive elements. First, lets see how we can construct $a_n.$ For each subset $S$ counted in $a_n,$ either: 1. $\\{n-1, n\\}\\subseteq S,$ 2. $n\\not\\in S$, or 3. $n-1 \\not\\in S$ and $n\\in S.$ The first case counts $b_{n-3}$ subsets (as $n-1$ cannot be included and the rest cannot have any consecutive elements), The second counts $a_{n-1},$ and the third counts $a_{n-2}.$ Thus, \\[a_n = a_{n-1} + a_{n-2} + b_{n-3}.\\] Next, Lets try to construct $b_n.$ For each subset $T$ counted in $b_n,$ either: 1.$n \\not\\in T,$ or 2.$n \\in T.$ The first case counts $b_{n-1}$ subsets and the second counts $b_{n-2}.$ Thus, \\[b_n = b_{n-1} + b_{n-2}.\\] Since $b_1 = 2$ and $b_2 = 3,$ we have that $b_n = F_{n+1},$ so $a_n = a_{n-1} + a_{n-2} + F_{n-2}.$ (The $F_i$ is the $i$th Fibonacci number). From here, we can construct a table of the values of $a_n$ until $n = 10.$ By listing out possibilities, we can solve for our first 3 values. \\[\\begin{array}{r|l} n & a_n \\\\ \\hline 1 & 0 \\\\ 2 & 1\\\\ 3 & 2\\\\ 4 & 2 + 1 + F_2 = 5\\\\5 & 5 + 2 + F_3 = 10\\\\6 & 10 + 5 + F_4 = 20 \\\\ 7 & 20 + 10 + F_5 = 38\\\\ 8 & 38 + 20 + F_6 = 71 \\\\ 9& 71 + 38 + F_7 = 130\\\\ 10 & 130 + 71 + F_8 = 235 \\end{array}\\] Our answer is $a_{10} = 235 ~AtharvNaphade",
"Let $a_n$ be the number of subsets of the set $\\{1,2,3,\\ldots,n\\}$ such that there exists exactly 1 pair of consecutive elements. Let $b_n$ be the number of subsets of the set $\\{1, 2, 3\\ldots, n\\}$ such that there doesn't exist any pair of consecutive elements. First, lets see how we can construct $a_n.$ For each subset $S$ counted in $a_n,$ either: 1. $\\{n-1, n\\}\\subseteq S,$ 2. $n\\not\\in S$, or 3. $n-1 \\not\\in S$ and $n\\in S.$ The first case counts $b_{n-3}$ subsets (as $n-1$ cannot be included and the rest cannot have any consecutive elements), The second counts $a_{n-1},$ and the third counts $a_{n-2}.$ Thus, \\[a_n = a_{n-1} + a_{n-2} + b_{n-3}.\\] Next, Lets try to construct $b_n.$ For each subset $T$ counted in $b_n,$ either: 1.$n \\not\\in T,$ or 2.$n \\in T.$ The first case counts $b_{n-1}$ subsets and the second counts $b_{n-2}.$ Thus, \\[b_n = b_{n-1} + b_{n-2}.\\] Since $b_1 = 2$ and $b_2 = 3,$ we have that $b_n = F_{n+1},$ so $a_n = a_{n-1} + a_{n-2} + F_{n-2}.$ (The $F_i$ is the $i$th Fibonacci number). From here, we can construct a table of the values of $a_n$ until $n = 10.$ By listing out possibilities, we can solve for our first 3 values. \\[\\begin{array}{r|l} n & a_n \\\\ \\hline 1 & 0 \\\\ 2 & 1\\\\ 3 & 2\\\\ 4 & 2 + 1 + F_2 = 5\\\\5 & 5 + 2 + F_3 = 10\\\\6 & 10 + 5 + F_4 = 20 \\\\ 7 & 20 + 10 + F_5 = 38\\\\ 8 & 38 + 20 + F_6 = 71 \\\\ 9& 71 + 38 + F_7 = 130\\\\ 10 & 130 + 71 + F_8 = 235 \\end{array}\\] Our answer is $a_{10} = 235 ~AtharvNaphade",
"Note: This is a very common stars and bars application. Casework on number of terms, let the number of terms be $n$. We can come up with a generalized formula for the number of subsets with n terms. Let $d_1, d_2, ..., d_{n-1}$ be the differences between the n terms. For example, in the set {2, 3, 6}, $d_1 = 1; d_2 = 3; d_3 = 5$ Let the range of the set be k for now, $d_1 + ... + d_{n-1} = k$. We select one pair of terms to be consecutive by selecting one of the (n-1) terms to be 1. WLOG, let $d_1 = 1$. $1 + d_2 + ... + d_{n-1} = k$. To ensure the other $d_2, d_3, ..., d_{n-1}$ are greater than 1 such that no two other terms are consecutive or the same, let $D_2 = d_2 + 1; D_3 = d_3 + 1; ...$. $(n-2) + 1 + D_2 + ... + D_{n-1} = k$ where $D_2, D_3, ..., D_{n-1}$ are positive integers. Finally, we add in $D_0$, the distance between 0 and the first term of the set, and $D_n$, the distance between the last term and 11. This way, the \"distance\" from 0 to 11 is \"bridged\" by $D_0$, k, and $D_N$. \\[D_0 + k + D_n = D_0 + ((n-1) + D_2 + ... + D_{n-1}) + D_N = 11\\] \\[D_0 + D_2 + D_3 + ... + D_n = 12 - n\\] There are n positive terms, by Balls and Urns, there are ${(12-n)-1 \\choose (n)-1} = {11-n \\choose n-1}$ ways of doing this. However, recall that there were $(n-1)$ ways, and we had used a WLOG to choose which two digits are consecutive. The final formula for the number of valid n-element subsets is hence $(n-1){11-n \\choose n-1}$ for $n > 2$. Case 1: Two terms $n = 2$, so $(2-1){11-2 \\choose 2-1} = 1{9 \\choose 1} = 9$ Case 2: Three terms $n = 3$, so $(3-1){11-3 \\choose 3-1} = 2{8 \\choose 2} = 56$ Case 3: Four terms $n = 4$, so $(4-1){11-4 \\choose 4-1} = 3{7 \\choose 3} = 105$ Case 4: Five terms $n = 5$, so $(5-1){11-5 \\choose 5-1} = 4{6 \\choose 4} = 60$ Case 5: Six terms $n = 6$, so $(6-1){11-6 \\choose 6-1} = 5{5 \\choose 5} = 5$ We can check by the Pigeonhole principle that there cannot be more than six terms, so the answer is $9+56+105+60+5=235. ~Mathandski",
"Note: This is a very common stars and bars application. Casework on number of terms, let the number of terms be $n$. We can come up with a generalized formula for the number of subsets with n terms. Let $d_1, d_2, ..., d_{n-1}$ be the differences between the n terms. For example, in the set {2, 3, 6}, $d_1 = 1; d_2 = 3; d_3 = 5$ Let the range of the set be k for now, $d_1 + ... + d_{n-1} = k$. We select one pair of terms to be consecutive by selecting one of the (n-1) terms to be 1. WLOG, let $d_1 = 1$. $1 + d_2 + ... + d_{n-1} = k$. To ensure the other $d_2, d_3, ..., d_{n-1}$ are greater than 1 such that no two other terms are consecutive or the same, let $D_2 = d_2 + 1; D_3 = d_3 + 1; ...$. $(n-2) + 1 + D_2 + ... + D_{n-1} = k$ where $D_2, D_3, ..., D_{n-1}$ are positive integers. Finally, we add in $D_0$, the distance between 0 and the first term of the set, and $D_n$, the distance between the last term and 11. This way, the \"distance\" from 0 to 11 is \"bridged\" by $D_0$, k, and $D_N$. \\[D_0 + k + D_n = D_0 + ((n-1) + D_2 + ... + D_{n-1}) + D_N = 11\\] \\[D_0 + D_2 + D_3 + ... + D_n = 12 - n\\] There are n positive terms, by Balls and Urns, there are ${(12-n)-1 \\choose (n)-1} = {11-n \\choose n-1}$ ways of doing this. However, recall that there were $(n-1)$ ways, and we had used a WLOG to choose which two digits are consecutive. The final formula for the number of valid n-element subsets is hence $(n-1){11-n \\choose n-1}$ for $n > 2$. Case 1: Two terms $n = 2$, so $(2-1){11-2 \\choose 2-1} = 1{9 \\choose 1} = 9$ Case 2: Three terms $n = 3$, so $(3-1){11-3 \\choose 3-1} = 2{8 \\choose 2} = 56$ Case 3: Four terms $n = 4$, so $(4-1){11-4 \\choose 4-1} = 3{7 \\choose 3} = 105$ Case 4: Five terms $n = 5$, so $(5-1){11-5 \\choose 5-1} = 4{6 \\choose 4} = 60$ Case 5: Six terms $n = 6$, so $(6-1){11-6 \\choose 6-1} = 5{5 \\choose 5} = 5$ We can check by the Pigeonhole principle that there cannot be more than six terms, so the answer is $9+56+105+60+5=235. ~Mathandski",
"Note that there are $F_{n+2}$ subsets of a set of $n$ consecutive integers that contains no two consecutive integers. (This can be proven by induction.) Now, notice that if we take $i$ and $i+1$ as the consecutive integers in our subset, we need to make a subset of the remaining integers such that it doesn't contain any two consecutive integers. Clearly, $i-1$ and $i+2$ cannot be chosen, and since $i-2$ and $i+3$ are sufficiently far apart, it is obvious we do not need to be concerned that an element of the set $\\{1, 2, ... , i-2 \\}$ is consecutive with any element of the set $\\{ i+3, i+4, ... , 10 \\}$ Thus, we can count the number of ways to choose a subset from the first set without any two elements being consecutive and multiply this by the number of ways to choose a subset from the second set without any two elements being consecutive. From above, and noting that the first set has $i-2$ consecutive integer elements and the second set has $8-i$ consecutive integer elements, we know that this is $F_i F_{10-i}.$ Summing this over for all $1 \\leq i \\leq 9$ yields \\[\\sum_{i=1}^9 F_i F_{10-i} = F_1F_9 + F_2F_8 + F_3F_7 + F_4F_6 + F_5 F_5 + F_6 F_4 + F_7 F_3 + F_8 F_2 + F_9 F_1 = 2(F_1F_9 + F_2F_8 + F_3F_7+F_4F_6) + F_5^2 = 2(1 \\cdot 34 + 1 \\cdot 21 + 2 \\cdot 13 + 3 \\cdot 8) + 5^2 = 2 \\cdot 105+25 = 235.\\] ~lpieleanu",
"Note that there are $F_{n+2}$ subsets of a set of $n$ consecutive integers that contains no two consecutive integers. (This can be proven by induction.) Now, notice that if we take $i$ and $i+1$ as the consecutive integers in our subset, we need to make a subset of the remaining integers such that it doesn't contain any two consecutive integers. Clearly, $i-1$ and $i+2$ cannot be chosen, and since $i-2$ and $i+3$ are sufficiently far apart, it is obvious we do not need to be concerned that an element of the set $\\{1, 2, ... , i-2 \\}$ is consecutive with any element of the set $\\{ i+3, i+4, ... , 10 \\}$ Thus, we can count the number of ways to choose a subset from the first set without any two elements being consecutive and multiply this by the number of ways to choose a subset from the second set without any two elements being consecutive. From above, and noting that the first set has $i-2$ consecutive integer elements and the second set has $8-i$ consecutive integer elements, we know that this is $F_i F_{10-i}.$ Summing this over for all $1 \\leq i \\leq 9$ yields \\[\\sum_{i=1}^9 F_i F_{10-i} = F_1F_9 + F_2F_8 + F_3F_7 + F_4F_6 + F_5 F_5 + F_6 F_4 + F_7 F_3 + F_8 F_2 + F_9 F_1 = 2(F_1F_9 + F_2F_8 + F_3F_7+F_4F_6) + F_5^2 = 2(1 \\cdot 34 + 1 \\cdot 21 + 2 \\cdot 13 + 3 \\cdot 8) + 5^2 = 2 \\cdot 105+25 = 235.\\] ~lpieleanu",
"The problem is the same as laying out a line of polynomoes to cover spots $0,1,...10$: 1 triomino ($RGG$), $n$ dominoes ($RG$), and $8-2n$ monominoes ($R$). The $G$ spots cover the members of the subset. The total number spots is 11, because one $R$ spot always covers the 0, and the other spots cover 1 through 10. There are 5 ways to choose polyomino sets, and many ways to order each set: $R + RG + RGG =$ Polyominoes $\\rightarrow$ Orderings $0 + 4 + 1 = 5 \\rightarrow 5! / 0!4!1! = ~~~5$ $2 + 3 + 1 = 6 \\rightarrow 6! / 2!3!1! = ~60$ $4 + 2 + 1 = 7 \\rightarrow 7! / 4!2!1! = 105$ $6 + 1 + 1 = 8 \\rightarrow 8! / 6!1!1! = ~56$ $8 + 0 + 1 = 9 \\rightarrow 9! / 8!0!1! = ~~~9$ The sum is $235. ~BraveCobra22aops",
"The problem is the same as laying out a line of polynomoes to cover spots $0,1,...10$: 1 triomino ($RGG$), $n$ dominoes ($RG$), and $8-2n$ monominoes ($R$). The $G$ spots cover the members of the subset. The total number spots is 11, because one $R$ spot always covers the 0, and the other spots cover 1 through 10. There are 5 ways to choose polyomino sets, and many ways to order each set: $R + RG + RGG =$ Polyominoes $\\rightarrow$ Orderings $0 + 4 + 1 = 5 \\rightarrow 5! / 0!4!1! = ~~~5$ $2 + 3 + 1 = 6 \\rightarrow 6! / 2!3!1! = ~60$ $4 + 2 + 1 = 7 \\rightarrow 7! / 4!2!1! = 105$ $6 + 1 + 1 = 8 \\rightarrow 8! / 6!1!1! = ~56$ $8 + 0 + 1 = 9 \\rightarrow 9! / 8!0!1! = ~~~9$ The sum is $235. ~BraveCobra22aops",
"Let $dp(i)$ be the number of subsets of a set $i$ consecutive integers such that the maximum value in the set is $i$ and there exists exactly one pair of consecutive integers. Define $dp2(i)$ similarly, but without any pair of consecutive integers. The base cases are $dp2(1)=dp2(2)=1$, $dp(1)=0$, and $dp(2)=1$. The transitions are: \\[dp2(i)=\\sum_{j=1}^{i-2}(dp2(j))+1\\] \\[dp(i)=dp2(i-1)+\\sum_{j=1}^{i-2}dp(j)\\] Note that $dp2$ is the Fibonacci numbers. \\[\\begin{array}{rcl} i & dp & dp2\\\\ \\hline 1 & 0 & 1 \\\\ 2 & 1 & 1 \\\\ 3 & 1 & 2 \\\\ 4 & 3 & 3 \\\\ 5 & 5 & 5 \\\\ 6 & 10 & 8 \\\\ 7 & 18 & 13 \\\\ 8 & 33 & 21 \\\\ 9 & 59 & 34 \\\\ 10 & 105 & 55 \\\\ \\end{array}\\] Summing over $dp$ yields $1+1+3+5+10+18+33+59+105=235 ~Mathenthus",
"Let $dp(i)$ be the number of subsets of a set $i$ consecutive integers such that the maximum value in the set is $i$ and there exists exactly one pair of consecutive integers. Define $dp2(i)$ similarly, but without any pair of consecutive integers. The base cases are $dp2(1)=dp2(2)=1$, $dp(1)=0$, and $dp(2)=1$. The transitions are: \\[dp2(i)=\\sum_{j=1}^{i-2}(dp2(j))+1\\] \\[dp(i)=dp2(i-1)+\\sum_{j=1}^{i-2}dp(j)\\] Note that $dp2$ is the Fibonacci numbers. \\[\\begin{array}{rcl} i & dp & dp2\\\\ \\hline 1 & 0 & 1 \\\\ 2 & 1 & 1 \\\\ 3 & 1 & 2 \\\\ 4 & 3 & 3 \\\\ 5 & 5 & 5 \\\\ 6 & 10 & 8 \\\\ 7 & 18 & 13 \\\\ 8 & 33 & 21 \\\\ 9 & 59 & 34 \\\\ 10 & 105 & 55 \\\\ \\end{array}\\] Summing over $dp$ yields $1+1+3+5+10+18+33+59+105=235 ~Mathenthus"
] |
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