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749be077979b0f39ac838b83e0f388fc | https://artofproblemsolving.com/wiki/index.php/2013_AIME_I_Problems/Problem_3 | Let $ABCD$ be a square, and let $E$ and $F$ be points on $\overline{AB}$ and $\overline{BC},$ respectively. The line through $E$ parallel to $\overline{BC}$ and the line through $F$ parallel to $\overline{AB}$ divide $ABCD$ into two squares and two nonsquare rectangles. The sum of the areas of the two squares is $\frac{9}{10}$ of the area of square $ABCD.$ Find $\frac{AE}{EB} + \frac{EB}{AE}.$ | After we get the polynomial $x^2 - 18x + 1,$ we want to find $x + \frac 1 {x}.$ Since the product of the roots of the polynomial is 1, the roots of the polynomial are simply $x, \frac 1 {x}.$ Hence $x + \frac 1 {x}$ is just $18$ by Vieta's formula, or $\boxed{018}$ | null | 018 |
749be077979b0f39ac838b83e0f388fc | https://artofproblemsolving.com/wiki/index.php/2013_AIME_I_Problems/Problem_3 | Let $ABCD$ be a square, and let $E$ and $F$ be points on $\overline{AB}$ and $\overline{BC},$ respectively. The line through $E$ parallel to $\overline{BC}$ and the line through $F$ parallel to $\overline{AB}$ divide $ABCD$ into two squares and two nonsquare rectangles. The sum of the areas of the two squares is $\frac{9}{10}$ of the area of square $ABCD.$ Find $\frac{AE}{EB} + \frac{EB}{AE}.$ | We have the equation $x^2 + y^2$ $\frac {9}{10} \cdot (x+y)^2$ . We get $x^2 + y^2 = 18xy$ . We rearrange to get $x^2 + y^2 - 18xy = 0$ . Since the problem only asks us for a ratio, we assume $x$ $1$ . We have $y^2 - 18y + 1$ $0$ . Solving the quadratic yields $9 + 4 \sqrt 5$ and $9 - 4 \sqrt 5$ . It doesn't really matter which one it is, since both of them are positive. We will use $9 + 4 \sqrt 5$
We have $9 + 4 \sqrt 5 + \frac {1}{9+4 \sqrt 5}$ . Rationalizing the denominator gives us $9 + 4 \sqrt 5 + \frac {9 - 4 \sqrt 5}{81-80} = (9 + 4 \sqrt 5) + (9 - 4 \sqrt 5) = 18$ . Our answer is $\boxed{018}$ | null | 018 |
ac69eba14932c34ebd68027d0be49515 | https://artofproblemsolving.com/wiki/index.php/2013_AIME_I_Problems/Problem_4 | In the array of 13 squares shown below, 8 squares are colored red, and the remaining 5 squares are colored blue. If one of all possible such colorings is chosen at random, the probability that the chosen colored array appears the same when rotated 90 degrees around the central square is $\frac{1}{n}$ , where is a positive integer. Find
[asy] draw((0,0)--(1,0)--(1,1)--(0,1)--(0,0)); draw((2,0)--(2,2)--(3,2)--(3,0)--(3,1)--(2,1)--(4,1)--(4,0)--(2,0)); draw((1,2)--(1,4)--(0,4)--(0,2)--(0,3)--(1,3)--(-1,3)--(-1,2)--(1,2)); draw((-1,1)--(-3,1)--(-3,0)--(-1,0)--(-2,0)--(-2,1)--(-2,-1)--(-1,-1)--(-1,1)); draw((0,-1)--(0,-3)--(1,-3)--(1,-1)--(1,-2)--(0,-2)--(2,-2)--(2,-1)--(0,-1)); size(100);[/asy] | When the array appears the same after a 90-degree rotation, the top formation must look the same as the right formation, which looks the same as the bottom one, which looks the same as the right one. There are four of the same configuration. There are not enough red squares for these to be all red, nor are there enough blue squares for there to be more than one blue square in each three-square formation. Thus there are 2 reds and 1 blue in each, and a blue in the center. There are 3 ways to choose which of the squares in the formation will be blue, leaving the other two red.
There are $\binom{13}{5}$ ways to have 5 blue squares in an array of 13.
$\frac{3}{\binom{13}{5}}$ $\frac{1}{429}$ , so $n$ $\boxed{429}$ | null | 429 |
9cb37c683cff209c9df4525efa57b5ca | https://artofproblemsolving.com/wiki/index.php/2013_AIME_I_Problems/Problem_5 | The real root of the equation $8x^3 - 3x^2 - 3x - 1 = 0$ can be written in the form $\frac{\sqrt[3]a + \sqrt[3]b + 1}{c}$ , where $a$ $b$ , and $c$ are positive integers. Find $a+b+c$ | We note that $8x^3 - 3x^2 - 3x - 1 = 9x^3 - x^3 - 3x^2 - 3x - 1 = 9x^3 - (x + 1)^3$ . Therefore, we have that $9x^3 = (x+1)^3$ , so it follows that $x\sqrt[3]{9} = x+1$ . Solving for $x$ yields $\frac{1}{\sqrt[3]{9}-1} = \frac{\sqrt[3]{81}+\sqrt[3]{9}+1}{8}$ , so the answer is $\boxed{98}$ | null | 98 |
9cb37c683cff209c9df4525efa57b5ca | https://artofproblemsolving.com/wiki/index.php/2013_AIME_I_Problems/Problem_5 | The real root of the equation $8x^3 - 3x^2 - 3x - 1 = 0$ can be written in the form $\frac{\sqrt[3]a + \sqrt[3]b + 1}{c}$ , where $a$ $b$ , and $c$ are positive integers. Find $a+b+c$ | Let $r$ be the real root of the given polynomial . Now define the cubic polynomial $Q(x)=-x^3-3x^2-3x+8$ . Note that $1/r$ must be a root of $Q$ . However we can simplify $Q$ as $Q(x)=9-(x+1)^3$ , so we must have that $(\frac{1}{r}+1)^3=9$ . Thus $\frac{1}{r}=\sqrt[3]{9}-1$ , and $r=\frac{1}{\sqrt[3]{9}-1}$ . We can then multiply the numerator and denominator of $r$ by $\sqrt[3]{81}+\sqrt[3]{9}+1$ to rationalize the denominator, and we therefore have $r=\frac{\sqrt[3]{81}+\sqrt[3]{9}+1}{8}$ , and the answer is $\boxed{98}$ | null | 98 |
9cb37c683cff209c9df4525efa57b5ca | https://artofproblemsolving.com/wiki/index.php/2013_AIME_I_Problems/Problem_5 | The real root of the equation $8x^3 - 3x^2 - 3x - 1 = 0$ can be written in the form $\frac{\sqrt[3]a + \sqrt[3]b + 1}{c}$ , where $a$ $b$ , and $c$ are positive integers. Find $a+b+c$ | It is clear that for the algebraic degree of $x$ to be $3$ that there exists some cubefree integer $p$ and positive integers $m,n$ such that $a = m^3p$ and $b = n^3p^2$ (it is possible that $b = n^3p$ , but then the problem wouldn't ask for both an $a$ and $b$ ). Let $f_1$ be the automorphism over $\mathbb{Q}[\sqrt[3]{a}][\omega]$ which sends $\sqrt[3]{a} \to \omega \sqrt[3]{a}$ and $f_2$ which sends $\sqrt[3]{a} \to \omega^2 \sqrt[3]{a}$ (note : $\omega$ is a cubic root of unity ).
Letting $r$ be the root, we clearly we have $r + f_1(r) + f_2(r) = \frac{3}{8}$ by Vieta's formulas. Thus it follows $c=8$ .
Now, note that $\sqrt[3]{a} + \sqrt[3]{b} + 1$ is a root of $x^3 - 3x^2 - 24x - 64 = 0$ . Thus $(x-1)^3 = 27x + 63$ so $(\sqrt[3]{a} + \sqrt[3]{b})^3 = 27(\sqrt[3]{a} + \sqrt[3]{b}) + 90$ . Checking the non-cubicroot dimension part, we get $a + b = 90$ so it follows that $a + b + c = \boxed{98}$ | null | 98 |
9cb37c683cff209c9df4525efa57b5ca | https://artofproblemsolving.com/wiki/index.php/2013_AIME_I_Problems/Problem_5 | The real root of the equation $8x^3 - 3x^2 - 3x - 1 = 0$ can be written in the form $\frac{\sqrt[3]a + \sqrt[3]b + 1}{c}$ , where $a$ $b$ , and $c$ are positive integers. Find $a+b+c$ | We have $cx-1=\sqrt[3]{a}+\sqrt[3]{b}.$ Therefore $(cx-1)^3=(\sqrt[3]{a}+\sqrt[3]{b})^3=a+b+3\sqrt[3]{ab}(\sqrt[3]{a}+\sqrt[3]{b})=a+b+3\sqrt[3]{ab}(cx-1).$ We have \[c^3x^3-3c^2x^2-(3c\sqrt[3]{ab}-3c)x-(a+b+1-3\sqrt[3]{ab})=0.\] We will find $a,b,c$ so that the equation is equivalent to the original one. Let $\dfrac{3c^2}{c^3}=\dfrac{3}{8}, \dfrac{3c\sqrt[3]{ab}-3c}{c^3}=\dfrac{3}{8}, \dfrac{a+b+1-3\sqrt[3]{ab}}{c^3}=\dfrac{1}{8}.$ Easily, $c=8, \sqrt[3]{ab}=9,$ and $a+b=90.$ So $a + b + c = 90+8=\boxed{98}$ | null | 98 |
b269ed565dfac209d003fe89fa081dcb | https://artofproblemsolving.com/wiki/index.php/2013_AIME_I_Problems/Problem_6 | Melinda has three empty boxes and $12$ textbooks, three of which are mathematics textbooks. One box will hold any three of her textbooks, one will hold any four of her textbooks, and one will hold any five of her textbooks. If Melinda packs her textbooks into these boxes in random order, the probability that all three mathematics textbooks end up in the same box can be written as $\frac{m}{n}$ , where $m$ and $n$ are relatively prime positive integers. Find $m+n$ | The total ways the textbooks can be arranged in the 3 boxes is $12\textbf{C}3\cdot 9\textbf{C}4$ , which is equivalent to $\frac{12\cdot 11\cdot 10\cdot 9\cdot 8\cdot 7\cdot 6}{144}=12\cdot11\cdot10\cdot7\cdot3$ . If all of the math textbooks are put into the box that can hold $3$ textbooks, there are $9!/(4!\cdot 5!)=9\textbf{C}4$ ways for the other textbooks to be arranged. If all of the math textbooks are put into the box that can hold $4$ textbooks, there are $9$ ways to choose the other book in that box, times $8\textbf{C}3$ ways for the other books to be arranged. If all of the math textbooks are put into the box with the capability of holding $5$ textbooks, there are $9\textbf{C}2$ ways to choose the other 2 textbooks in that box, times $7\textbf{C}3$ ways to arrange the other 7 textbooks. $9\textbf{C}4=9\cdot7\cdot2=126$ $9\cdot 8\textbf{C}3=9\cdot8\cdot7=504$ , and $9\textbf{C}2\cdot 7\textbf{C}3=9\cdot7\cdot5\cdot4=1260$ , so the total number of ways the math textbooks can all be placed into the same box is $126+504+1260=1890$ . So, the probability of this occurring is $\frac{(9\cdot7)(2+8+(4\cdot5))}{12\cdot11\cdot10\cdot7\cdot3}=\frac{1890}{27720}$ . If the numerator and denominator are both divided by $9\cdot7$ , we have $\frac{(2+8+(4\cdot5))}{4\cdot11\cdot10}=\frac{30}{440}$ . Simplifying the numerator yields $\frac{30}{10\cdot4\cdot11}$ , and dividing both numerator and denominator by $10$ results in $\frac{3}{44}$ . This fraction cannot be simplified any further, so $m=3$ and $n=44$ . Therefore, $m+n=3+44=\boxed{047}$ | null | 047 |
b269ed565dfac209d003fe89fa081dcb | https://artofproblemsolving.com/wiki/index.php/2013_AIME_I_Problems/Problem_6 | Melinda has three empty boxes and $12$ textbooks, three of which are mathematics textbooks. One box will hold any three of her textbooks, one will hold any four of her textbooks, and one will hold any five of her textbooks. If Melinda packs her textbooks into these boxes in random order, the probability that all three mathematics textbooks end up in the same box can be written as $\frac{m}{n}$ , where $m$ and $n$ are relatively prime positive integers. Find $m+n$ | Consider the books as either math or not-math where books in each category are indistiguishable from one another. Then, there are $\,_{12}C_{3}$ total distinguishable ways to pack the books. Now, in order to determine the desired propability, we must find the total number of ways the condition that all math books are in the same box can be satisfied. We proceed with casework for each box:
Case 1: The math books are placed into the smallest box. This can be done in $\binom{3}{3}$ ways.
Case 2: The math books are placed into the middle box. This can be done in $\binom{4}{3}$ ways.
Case 3: The math books are placed into the largest box. This can be done in $\binom{5}{3}$ ways.
So, the total ways the condition can be satisfied is $\binom{3}{3} + \binom{4}{3} + \binom{5}{3}$ . This can be simplified to $\binom{6}{4} = \binom{6}{2}$ by the Hockey Stick Identity .
Therefore, the desired probability is $\dfrac{\dbinom{6}{2} }{\dbinom{12}{3}}$ $\dfrac{3}{44}$ , and $m+n=3+44=\boxed{047}$ | null | 047 |
b269ed565dfac209d003fe89fa081dcb | https://artofproblemsolving.com/wiki/index.php/2013_AIME_I_Problems/Problem_6 | Melinda has three empty boxes and $12$ textbooks, three of which are mathematics textbooks. One box will hold any three of her textbooks, one will hold any four of her textbooks, and one will hold any five of her textbooks. If Melinda packs her textbooks into these boxes in random order, the probability that all three mathematics textbooks end up in the same box can be written as $\frac{m}{n}$ , where $m$ and $n$ are relatively prime positive integers. Find $m+n$ | There are three cases as follows. Note these are PERMUTATIONS, as the books are distinct!
1. Math books in the 3-size box. Probability is $\frac{3\cdot2\cdot1}{12\cdot11\cdot10}$ , because we choose one of the $3$ places for math book 1, then one of the $2$ for math book 2, and the last one. Total number of orders: $12\cdot11\cdot10=1320$
2. In the 4-size: same logic gets you $\frac{1}{55}$ , since we have $4$ places for math book 1, and so on.
3. In the 5-size: you get $\frac{1}{22}$ , for a sum of $\frac{3}{44}$ so your answer is $\boxed{047}$ | null | 047 |
b269ed565dfac209d003fe89fa081dcb | https://artofproblemsolving.com/wiki/index.php/2013_AIME_I_Problems/Problem_6 | Melinda has three empty boxes and $12$ textbooks, three of which are mathematics textbooks. One box will hold any three of her textbooks, one will hold any four of her textbooks, and one will hold any five of her textbooks. If Melinda packs her textbooks into these boxes in random order, the probability that all three mathematics textbooks end up in the same box can be written as $\frac{m}{n}$ , where $m$ and $n$ are relatively prime positive integers. Find $m+n$ | Assume that the $9$ other books are all distinct. The number of ways to place the other $9$ books and the $3$ is $\frac {12!}{3!}$ . The number of ways to put all the $3$ math books into the box that holds $3$ books is $9!$ . The number of ways to put all the $3$ math books into the box that holds $4$ books is $\binom {4}{3} \cdot 9!$ , and the number of ways to put all the $3$ math books into the box that holds $5$ books is $\binom {5}{3} \cdot 9!$ . The number of desired outcomes is $15 \cdot 9!$ , and the total number of outcomes is $\frac {12!}{6}$ . Simplifying, we get the answer is $\frac {3}{44}$ , so our answer is $\boxed{047}$ | null | 047 |
c2b07e118e3f600bb94a8152c4ef82c4 | https://artofproblemsolving.com/wiki/index.php/2013_AIME_I_Problems/Problem_7 | A rectangular box has width $12$ inches, length $16$ inches, and height $\frac{m}{n}$ inches, where $m$ and $n$ are relatively prime positive integers. Three faces of the box meet at a corner of the box. The center points of those three faces are the vertices of a triangle with an area of $30$ square inches. Find $m+n$ | Let the height of the box be $x$
After using the Pythagorean Theorem three times, we can quickly see that the sides of the triangle are 10, $\sqrt{\left(\frac{x}{2}\right)^2 + 64}$ , and $\sqrt{\left(\frac{x}{2}\right)^2 + 36}$ . Since the area of the triangle is $30$ , the altitude of the triangle from the base with length $10$ is $6$
Considering the two triangles created by the altitude, we use the Pythagorean theorem twice to find the lengths of the two line segments that make up the base of $10$
We find: \[10 = \sqrt{\left(28+x^2/4\right)}+x/2\]
Solving for $x$ gives us $x=\frac{36}{5}$ . Since this fraction is simplified: \[m+n=\boxed{041}\] | null | 041 |
c2b07e118e3f600bb94a8152c4ef82c4 | https://artofproblemsolving.com/wiki/index.php/2013_AIME_I_Problems/Problem_7 | A rectangular box has width $12$ inches, length $16$ inches, and height $\frac{m}{n}$ inches, where $m$ and $n$ are relatively prime positive integers. Three faces of the box meet at a corner of the box. The center points of those three faces are the vertices of a triangle with an area of $30$ square inches. Find $m+n$ | We may use vectors. Let the height of the box be $2h$ . Without loss of generality, let the front bottom left corner of the box be $(0,0,0)$ . Let the center point of the bottom face be $P_1$ , the center of the left face be $P_2$ and the center of the front face be $P_3$
We are given that the area of the triangle $\triangle P_1 P_2 P_3$ is $30$ . Thus, by a well known formula, we note that $\frac{1}{2}|\vec{P_1P_2} \text{x} \vec{P_1P_3}|=30$ We quickly attain that $\vec{P_1P_2}=<-6,0,h>$ and $\vec{P_1P_3}=<0,-8,h>$ (We can arbitrarily assign the long and short ends due to symmetry)
Computing the cross product, we find: \[\vec{P_1P_2} x \vec{P_1P_3}=-<6h,8h,48>\]
Thus: \[\sqrt{(6h)^2+(8h)^2+48^2}=2*30=60\] \[h=3.6\] \[2h=7.2\]
\[2h=36/5\]
\[m+n=\boxed{041}\] | null | 041 |
c2b07e118e3f600bb94a8152c4ef82c4 | https://artofproblemsolving.com/wiki/index.php/2013_AIME_I_Problems/Problem_7 | A rectangular box has width $12$ inches, length $16$ inches, and height $\frac{m}{n}$ inches, where $m$ and $n$ are relatively prime positive integers. Three faces of the box meet at a corner of the box. The center points of those three faces are the vertices of a triangle with an area of $30$ square inches. Find $m+n$ | Let the height of the box be $x$
After using the Pythagorean Theorem three times, we can quickly see that the sides of the triangle are 10, $\sqrt{(x/2)^2 + 64}$ , and $\sqrt{(x/2)^2 + 36}$ . Therefore, we can use Heron's formula to set up an equation for the area of the triangle.
The semiperimeter is $\left(10 + \sqrt{(x/2)^2 + 64} + \sqrt{(x/2)^2 + 36}\right)/2$ . Therefore, when we square Heron's formula, we find
\begin{align*}900 &= \frac{1}{2}\left(\left(10 + \sqrt{(x/2)^2 + 64} + \sqrt{(x/2)^2 + 36}\right)/2\right)\times\left(\left(10 + \sqrt{(x/2)^2 + 64} + \sqrt{(x/2)^2 + 36}\right)/2 - 10\right)\\&\qquad\times\left(\left(10 + \sqrt{(x/2)^2 + 64} + \sqrt{(x/2)^2 + 36}\right)/2 - \sqrt{(x/2)^2 + 64}\right)\\&\qquad\times\left(\left(10 + \sqrt{(x/2)^2 + 64} + \sqrt{(x/2)^2 + 36}\right)/2 - \sqrt{(x/2)^2 + 36}\right).\end{align*}
Solving, we get $\boxed{041}$ | null | 041 |
c2b07e118e3f600bb94a8152c4ef82c4 | https://artofproblemsolving.com/wiki/index.php/2013_AIME_I_Problems/Problem_7 | A rectangular box has width $12$ inches, length $16$ inches, and height $\frac{m}{n}$ inches, where $m$ and $n$ are relatively prime positive integers. Three faces of the box meet at a corner of the box. The center points of those three faces are the vertices of a triangle with an area of $30$ square inches. Find $m+n$ | Let half the height be $a$ (we want to find $2a$ ), then we see that the three sides of the triangle are (by Pyth Theorem) $10, \sqrt{a^2+36}, \sqrt{a^2+64}$ . Using the Law of Sines with the angle as the one included between the square roots, we see that this angle's cosine is $\frac{a^2}{\sqrt{(a^2+36)(a^2+64)}}$ by the Law of Cosines, meaning that its sine is $\frac{\sqrt{100a^2+2304}}{\sqrt{(a^2+36)(a^2+64)}}$ . Finally, multiply the two square-rooted sides by this sine and one-half, and equate to 30.
You get $\sqrt{25a^2+576} = 30$ , giving $a=\frac{18}{5}$ , so our answer is $\boxed{041}$ | null | 041 |
c2b07e118e3f600bb94a8152c4ef82c4 | https://artofproblemsolving.com/wiki/index.php/2013_AIME_I_Problems/Problem_7 | A rectangular box has width $12$ inches, length $16$ inches, and height $\frac{m}{n}$ inches, where $m$ and $n$ are relatively prime positive integers. Three faces of the box meet at a corner of the box. The center points of those three faces are the vertices of a triangle with an area of $30$ square inches. Find $m+n$ | Let $x$ be $\frac12$ the height of the box. We will solve for $x$ and then multiply by $2$ at the end. The three side lengths of the triangle, by the Pythagorean Theorem, are $\sqrt{x^2+6^2},\sqrt{x^2+8^2},$ and $10$
Heron's formula states that the area of the triangle is $\sqrt{s(s-a)(s-b)(s-c)}$ where $s$ is the semiperimeter. Applying difference of squares to make the formula less computational, we get $\frac{\sqrt{(a+b+c)(a+b-c)(a-b+c)(-a+b+c)}}{4}=\frac{\sqrt{(a^2+b^2+2ab-c^2)(c^2-b^2-a^2+2ab)}}4=\frac{\sqrt{4a^2b^2-(a^2+b^2-c^2)^2}}4$
Now, we plug in $\sqrt{x^2+6^2}$ for $a$ $\sqrt{x^2+8^2}$ for $b$ , and $10$ for $c$ . This gives us \[\frac{\sqrt{4(x^2+6^2)(x^2+8^2)-(x^2+6^2+x^2+8^2-10^2)^2}}4=30\] \[\sqrt{4x^4+400x^2+4\cdot48^2-4x^4}=120\] \[400x^2+4\cdot48^2=120^2\] \[25x^2+24^2=30^2\] \[25x^2=6^2(5^2-4^2)\] \[25x^2=18^2\] \[5x=18\] \[x=\frac{18}5\]
Now multiplying by $2$ , we get that the height is $\frac{36}5$ , and $m+n=36+5=\boxed{041}$ | null | 041 |
2c3ac594d1d18f074f87e0d98d997956 | https://artofproblemsolving.com/wiki/index.php/2013_AIME_I_Problems/Problem_8 | The domain of the function $f(x) = \arcsin(\log_{m}(nx))$ is a closed interval of length $\frac{1}{2013}$ , where $m$ and $n$ are positive integers and $m>1$ . Find the remainder when the smallest possible sum $m+n$ is divided by 1000. | We start with the same method as above. The domain of the arcsin function is $[-1, 1]$ , so $-1 \le \log_{m}(nx) \le 1$
\[\frac{1}{m} \le nx \le m\] \[\frac{1}{mn} \le x \le \frac{m}{n}\] \[\frac{m}{n} - \frac{1}{mn} = \frac{1}{2013}\] \[n = 2013m - \frac{2013}{m}\]
For $n$ to be an integer, $m$ must divide $2013$ , and $m > 1$ . To minimize $n$ $m$ should be as small as possible because increasing $m$ will decrease $\frac{2013}{m}$ , the amount you are subtracting, and increase $2013m$ , the amount you are adding; this also leads to a small $n$ which clearly minimizes $m+n$
We let $m$ equal $3$ , the smallest factor of $2013$ that isn't $1$ . Then we have $n = 2013*3 - \frac{2013}{3} = 6039 - 671 = 5368$
$m + n = 5371$ , so the answer is $\boxed{371}$ | null | 371 |
2c3ac594d1d18f074f87e0d98d997956 | https://artofproblemsolving.com/wiki/index.php/2013_AIME_I_Problems/Problem_8 | The domain of the function $f(x) = \arcsin(\log_{m}(nx))$ is a closed interval of length $\frac{1}{2013}$ , where $m$ and $n$ are positive integers and $m>1$ . Find the remainder when the smallest possible sum $m+n$ is divided by 1000. | Note that we need $-1\le f(x)\le 1$ , and this eventually gets to $\frac{m^2-1}{mn}=\frac{1}{2013}$ . From there, break out the quadratic formula and note that \[m= \frac{n+\sqrt{n^2+4026^2}}{2013\times 2}.\] Then we realize that the square root, call it $a$ , must be an integer. Then $(a-n)(a+n)=4026^2.$
Observe carefully that $4026^2 = 2\times 2\times 3\times 3\times 11\times 11\times 61\times 61$ ! It is not difficult to see that to minimize the sum, we want to minimize $n$ as much as possible. Seeing that $2a$ is even, we note that a $2$ belongs in each factor. Now, since we want to minimize $a$ to minimize $n$ , we want to distribute the factors so that their ratio is as small as possible (sum is thus minimum). The smallest allocation of $2, 61, 61$ and $2, 11, 3, 3, 11$ fails; the next best is $2, 61, 11, 3, 3$ and $2, 61, 11$ , in which $a=6710$ and $n=5368$ . That is our best solution, upon which we see that $m=3$ , thus $\boxed{371}$ | null | 371 |
0a9ea29e6eb9370f04daf38943debb7d | https://artofproblemsolving.com/wiki/index.php/2013_AIME_I_Problems/Problem_9 | A paper equilateral triangle $ABC$ has side length $12$ . The paper triangle is folded so that vertex $A$ touches a point on side $\overline{BC}$ a distance $9$ from point $B$ . The length of the line segment along which the triangle is folded can be written as $\frac{m\sqrt{p}}{n}$ , where $m$ $n$ , and $p$ are positive integers, $m$ and $n$ are relatively prime, and $p$ is not divisible by the square of any prime. Find $m+n+p$
[asy] import cse5; size(12cm); pen tpen = defaultpen + 1.337; real a = 39/5.0; real b = 39/7.0; pair B = MP("B", (0,0), dir(200)); pair A = MP("A", (9,0), dir(-80)); pair C = MP("C", (12,0), dir(-20)); pair K = (6,10.392); pair M = (a*B+(12-a)*K) / 12; pair N = (b*C+(12-b)*K) / 12; draw(B--M--N--C--cycle, tpen); draw(M--A--N--cycle); fill(M--A--N--cycle, mediumgrey); pair shift = (-20.13, 0); pair B1 = MP("B", B+shift, dir(200)); pair A1 = MP("A", K+shift, dir(90)); pair C1 = MP("C", C+shift, dir(-20)); draw(A1--B1--C1--cycle, tpen);[/asy] | Let $M$ and $N$ be the points on $\overline{AB}$ and $\overline{AC}$ , respectively, where the paper is folded. Let $D$ be the point on $\overline{BC}$ where the folded $A$ touches it. [asy] import cse5; size(8cm); pen tpen = defaultpen + 1.337; real a = 39/5.0; real b = 39/7.0; pair B = MP("B", (0,0), dir(200)); pair A = MP("A", 12*dir(60), dir(90)); pair C = MP("C", (12,0), dir(-20)); pair D = MP("D", (9,0), dir(-80)); pair Y = MP("Y", midpoint(A--D), dir(-50)); pair M = MP("M", extension(A,B,Y,Y+(dir(90)*(D-A))), dir(180)); pair N = MP("N", extension(A,C,M,Y), dir(20)); pair F = MP("F", foot(A,B,C), dir(-90)); pair X = MP("X", extension(A,F,M,N), dir(-120)); draw(B--A--C--cycle, tpen); draw(M--N^^F--A--D); draw(rightanglemark(D,F,A,15)); draw(rightanglemark(A,Y,M,15)); MA("\theta",F,A,D,1.8); [/asy] We have $AF=6\sqrt{3}$ and $FD=3$ , so $AD=3\sqrt{13}$ . Denote $\angle DAF = \theta$ ; we get $\cos\theta = 2\sqrt{3}/\sqrt{13}$
In triangle $AXY$ $AY=\tfrac 12 AD = \tfrac 32 \sqrt{13}$ , and $AX=AY\sec\theta =\tfrac{13}{4}\sqrt{3}$
In triangle $AMX$ , we get $\angle AMX=60^\circ-\theta$ and then use sine-law to get $MX=\tfrac 12 AX\csc(60^\circ-\theta)$ ; similarly, from triangle $ANX$ we get $NX=\tfrac 12 AX\csc(60^\circ+\theta)$ . Thus \[MN=\tfrac 12 AX(\csc(60^\circ-\theta) +\csc(60^\circ+\theta)).\] Since $\sin(60^\circ\pm \theta) = \tfrac 12 (\sqrt{3}\cos\theta \pm \sin\theta)$ , we get \begin{align*} \csc(60^\circ-\theta) +\csc(60^\circ+\theta) &= \frac{\sqrt{3}\cos\theta}{\cos^2\theta - \tfrac 14} = \frac{24 \cdot \sqrt{13}}{35} \end{align*} Then \[MN = \frac 12 AX \cdot \frac{24 \cdot \sqrt{13}}{35} = \frac{39\sqrt{39}}{35}\]
The answer is $39 + 39 + 35 = \boxed{113}$ | null | 113 |
0a9ea29e6eb9370f04daf38943debb7d | https://artofproblemsolving.com/wiki/index.php/2013_AIME_I_Problems/Problem_9 | A paper equilateral triangle $ABC$ has side length $12$ . The paper triangle is folded so that vertex $A$ touches a point on side $\overline{BC}$ a distance $9$ from point $B$ . The length of the line segment along which the triangle is folded can be written as $\frac{m\sqrt{p}}{n}$ , where $m$ $n$ , and $p$ are positive integers, $m$ and $n$ are relatively prime, and $p$ is not divisible by the square of any prime. Find $m+n+p$
[asy] import cse5; size(12cm); pen tpen = defaultpen + 1.337; real a = 39/5.0; real b = 39/7.0; pair B = MP("B", (0,0), dir(200)); pair A = MP("A", (9,0), dir(-80)); pair C = MP("C", (12,0), dir(-20)); pair K = (6,10.392); pair M = (a*B+(12-a)*K) / 12; pair N = (b*C+(12-b)*K) / 12; draw(B--M--N--C--cycle, tpen); draw(M--A--N--cycle); fill(M--A--N--cycle, mediumgrey); pair shift = (-20.13, 0); pair B1 = MP("B", B+shift, dir(200)); pair A1 = MP("A", K+shift, dir(90)); pair C1 = MP("C", C+shift, dir(-20)); draw(A1--B1--C1--cycle, tpen);[/asy] | Let $P$ and $Q$ be the points on $\overline{AB}$ and $\overline{AC}$ , respectively, where the paper is folded.
Let $D$ be the point on $\overline{BC}$ where the folded $A$ touches it.
Let $a$ $b$ , and $x$ be the lengths $AP$ $AQ$ , and $PQ$ , respectively.
We have $PD = a$ $QD = b$ $BP = 12 - a$ $CQ = 12 - b$ $BD = 9$ , and $CD = 3$
Using the Law of Cosines on $BPD$
$a^{2} = (12 - a)^{2} + 9^{2} - 2 \times (12 - a) \times 9 \times \cos{60}$
$a^{2} = 144 - 24a + a^{2} + 81 - 108 + 9a$
$a = \frac{39}{5}$
Using the Law of Cosines on $CQD$
$b^{2} = (12 - b)^{2} +3^{2} - 2 \times (12 - b) \times 3 \times \cos{60}$
$b^{2} = 144 - 24b + b^{2} + 9 - 36 + 3b$
$b = \frac{39}{7}$
Using the Law of Cosines on $DPQ$
$x^{2} = a^{2} + b^{2} - 2ab \cos{60}$
$x^{2} = (\frac{39}{5})^2 + (\frac{39}{7})^2 - (\frac{39}{5} \times \frac{39}{7})$
$x = \frac{39 \sqrt{39}}{35}$
The solution is $39 + 39 + 35 = \boxed{113}$ | null | 113 |
0a9ea29e6eb9370f04daf38943debb7d | https://artofproblemsolving.com/wiki/index.php/2013_AIME_I_Problems/Problem_9 | A paper equilateral triangle $ABC$ has side length $12$ . The paper triangle is folded so that vertex $A$ touches a point on side $\overline{BC}$ a distance $9$ from point $B$ . The length of the line segment along which the triangle is folded can be written as $\frac{m\sqrt{p}}{n}$ , where $m$ $n$ , and $p$ are positive integers, $m$ and $n$ are relatively prime, and $p$ is not divisible by the square of any prime. Find $m+n+p$
[asy] import cse5; size(12cm); pen tpen = defaultpen + 1.337; real a = 39/5.0; real b = 39/7.0; pair B = MP("B", (0,0), dir(200)); pair A = MP("A", (9,0), dir(-80)); pair C = MP("C", (12,0), dir(-20)); pair K = (6,10.392); pair M = (a*B+(12-a)*K) / 12; pair N = (b*C+(12-b)*K) / 12; draw(B--M--N--C--cycle, tpen); draw(M--A--N--cycle); fill(M--A--N--cycle, mediumgrey); pair shift = (-20.13, 0); pair B1 = MP("B", B+shift, dir(200)); pair A1 = MP("A", K+shift, dir(90)); pair C1 = MP("C", C+shift, dir(-20)); draw(A1--B1--C1--cycle, tpen);[/asy] | Proceed with the same labeling as in Solution 1.
$\angle B = \angle C = \angle A = \angle PDQ = 60^\circ$
$\angle PDB + \angle PDQ + \angle QDC = \angle QDC + \angle CQD + \angle C = 180^\circ$
Therefore, $\angle PDB = \angle CQD$
Similarly, $\angle BPD = \angle QDC$
Now, $\bigtriangleup BPD$ and $\bigtriangleup CDQ$ are similar triangles, so
$\frac{3}{12-a} = \frac{12-b}{9} = \frac{b}{a}$
Solving this system of equations yields $a = \frac{39}{5}$ and $b = \frac{39}{7}$
Using the Law of Cosines on $APQ$
$x^{2} = a^{2} + b^{2} - 2ab \cos{60}$
$x^{2} = (\frac{39}{5})^2 + (\frac{39}{7})^2 - (\frac{39}{5} \times \frac{39}{7})$
$x = \frac{39 \sqrt{39}}{35}$
The solution is $39 + 39 + 35 = \boxed{113}$ | null | 113 |
0a9ea29e6eb9370f04daf38943debb7d | https://artofproblemsolving.com/wiki/index.php/2013_AIME_I_Problems/Problem_9 | A paper equilateral triangle $ABC$ has side length $12$ . The paper triangle is folded so that vertex $A$ touches a point on side $\overline{BC}$ a distance $9$ from point $B$ . The length of the line segment along which the triangle is folded can be written as $\frac{m\sqrt{p}}{n}$ , where $m$ $n$ , and $p$ are positive integers, $m$ and $n$ are relatively prime, and $p$ is not divisible by the square of any prime. Find $m+n+p$
[asy] import cse5; size(12cm); pen tpen = defaultpen + 1.337; real a = 39/5.0; real b = 39/7.0; pair B = MP("B", (0,0), dir(200)); pair A = MP("A", (9,0), dir(-80)); pair C = MP("C", (12,0), dir(-20)); pair K = (6,10.392); pair M = (a*B+(12-a)*K) / 12; pair N = (b*C+(12-b)*K) / 12; draw(B--M--N--C--cycle, tpen); draw(M--A--N--cycle); fill(M--A--N--cycle, mediumgrey); pair shift = (-20.13, 0); pair B1 = MP("B", B+shift, dir(200)); pair A1 = MP("A", K+shift, dir(90)); pair C1 = MP("C", C+shift, dir(-20)); draw(A1--B1--C1--cycle, tpen);[/asy] | We let the original position of $A$ be $A$ , and the position of $A$ after folding be $D$ . Also, we put the triangle on the coordinate plane such that $A=(0,0)$ $B=(-6,-6\sqrt3)$ $C=(6,-6\sqrt3)$ , and $D=(3,-6\sqrt3)$
[asy] size(10cm); pen tpen = defaultpen + 1.337; real a = 39/5.0; real b = 39/7.0; pair B = MP("B", (0,0), dir(200)); pair A = (9,0); pair C = MP("C", (12,0), dir(-20)); pair K = (6,10.392); pair M = (a*B+(12-a)*K) / 12; pair N = (b*C+(12-b)*K) / 12; draw(B--M--N--C--cycle); draw(M--A--N--cycle); label("$D$", A, S); pair X = (6,6*sqrt(3)); draw(B--X--C); label("$A$",X,dir(90)); draw(A--X); [/asy]
Note that since $A$ is reflected over the fold line to $D$ , the fold line is the perpendicular bisector of $AD$ . We know $A=(0,0)$ and $D=(3,-6\sqrt3)$ . The midpoint of $AD$ (which is a point on the fold line) is $(\tfrac32, -3\sqrt3)$ . Also, the slope of $AD$ is $\frac{-6\sqrt3}{3}=-2\sqrt3$ , so the slope of the fold line (which is perpendicular), is the negative of the reciprocal of the slope of $AD$ , or $\frac{1}{2\sqrt3}=\frac{\sqrt3}{6}$ . Then, using point slope form, the equation of the fold line is \[y+3\sqrt3=\frac{\sqrt3}{6}\left(x-\frac32\right)\] \[y=\frac{\sqrt3}{6}x-\frac{13\sqrt3}{4}\] Note that the equations of lines $AB$ and $AC$ are $y=\sqrt3x$ and $y=-\sqrt3x$ , respectively. We will first find the intersection of $AB$ and the fold line by substituting for $y$ \[\sqrt3 x=\frac{\sqrt3}{6}x-\frac{13\sqrt3}{4}\] \[\frac{5\sqrt3}{6}x=-\frac{13\sqrt3}{4} \implies x=-\frac{39}{10}\] Therefore, the point of intersection is $\left(-\tfrac{39}{10},-\tfrac{39\sqrt3}{10}\right)$ . Now, lets find the intersection with $AC$ . Substituting for $y$ yields \[-\sqrt3 x=\frac{\sqrt3}{6}x-\frac{13\sqrt3}{4}\] \[\frac{-7\sqrt3}{6}x=-\frac{13\sqrt3}{4} \implies x=\frac{39}{14}\] Therefore, the point of intersection is $\left(\tfrac{39}{14},-\tfrac{39\sqrt3}{14}\right)$ . Now, we just need to use the distance formula to find the distance between $\left(-\tfrac{39}{10},-\tfrac{39\sqrt3}{10}\right)$ and $\left(\tfrac{39}{14},-\tfrac{39\sqrt3}{14}\right)$ \[\sqrt{\left(\frac{39}{14}+\frac{39}{10}\right)^2+\left(-\frac{39\sqrt3}{14}+\frac{39\sqrt3}{10}\right)^2}\] The number 39 is in all of the terms, so let's factor it out: \[39\sqrt{\left(\frac{1}{14}+\frac{1}{10}\right)^2+\left(-\frac{\sqrt3}{14}+\frac{\sqrt3}{10}\right)^2}=39\sqrt{\left(\frac{6}{35}\right)^2+\left(\frac{\sqrt3}{35}\right)^2}\] \[\frac{39}{35}\sqrt{6^2+\sqrt3^2}=\frac{39\sqrt{39}}{35}\] Therefore, our answer is $39+39+35=\boxed{113}$ , and we are done. | null | 113 |
f7f2d214fc2a7032dc7ea572aa528d69 | https://artofproblemsolving.com/wiki/index.php/2013_AIME_I_Problems/Problem_10 | There are nonzero integers $a$ $b$ $r$ , and $s$ such that the complex number $r+si$ is a zero of the polynomial $P(x)={x}^{3}-a{x}^{2}+bx-65$ . For each possible combination of $a$ and $b$ , let ${p}_{a,b}$ be the sum of the zeros of $P(x)$ . Find the sum of the ${p}_{a,b}$ 's for all possible combinations of $a$ and $b$ | Since $r+si$ is a root, by the Complex Conjugate Root Theorem, $r-si$ must be the other imaginary root. Using $q$ to represent the real root, we have
$(x-q)(x-r-si)(x-r+si) = x^3 -ax^2 + bx -65$
Applying difference of squares, and regrouping, we have
$(x-q)(x^2 - 2rx + (r^2 + s^2)) = x^3 -ax^2 + bx -65$
So matching coefficients, we obtain
$q(r^2 + s^2) = 65$
$b = r^2 + s^2 + 2rq$
$a = q + 2r$
By Vieta's each ${p}_{a,b} = a$ so we just need to find the values of $a$ in each pair.
We proceed by determining possible values for $q$ $r$ , and $s$ and using these to determine $a$ and $b$
If $q = 1$ $r^2 + s^2 = 65$ so (r, s) = $(\pm1, \pm 8), (\pm8, \pm 1), (\pm4, \pm 7), (\pm7, \pm 4)$
Similarly, for $q = 5$ $r^2 + s^2 = 13$ so the pairs $(r,s)$ are $(\pm2, \pm 3), (\pm3, \pm 2)$
For $q = 13$ $r^2 + s^2 = 5$ so the pairs $(r,s)$ are $(\pm2, \pm 1), (\pm1, \pm 2)$
Now we can disregard the plus minus signs for s because those cases are included as complex conjugates of the counted cases.
The positive and negative values of r will cancel, so the sum of the ${p}_{a,b} = a$ for $q = 1$ is $q$ times the number of distinct $r$ values (as each value of $r$ generates a pair $(a,b)$ ).
Our answer is then $(1)(8) + (5)(4) + (13)(4) = \boxed{080}$ | null | 080 |
99d2ad6137c794f7a6b0998d758dafdc | https://artofproblemsolving.com/wiki/index.php/2013_AIME_I_Problems/Problem_11 | Ms. Math's kindergarten class has $16$ registered students. The classroom has a very large number, $N$ , of play blocks which satisfies the conditions:
(a) If $16$ $15$ , or $14$ students are present in the class, then in each case all the blocks can be distributed in equal numbers to each student, and
(b) There are three integers $0 < x < y < z < 14$ such that when $x$ $y$ , or $z$ students are present and the blocks are distributed in equal numbers to each student, there are exactly three blocks left over.
Find the sum of the distinct prime divisors of the least possible value of $N$ satisfying the above conditions. | $N$ must be some multiple of $\text{lcm}(14, 15, 16)= 2^{4}\cdot 3\cdot 5\cdot 7$ ; this $lcm$ is hereby denoted $k$ and $N = qk$
$1$ $2$ $3$ $4$ $5$ $6$ $7$ $8$ $10$ , and $12$ all divide $k$ , so $x, y, z = 9, 11, 13$
We have the following three modulo equations:
$nk\equiv 3\pmod{9}$
$nk\equiv 3\pmod{11}$
$nk\equiv 3\pmod{13}$
To solve the equations, you can notice the answer must be of the form $9\cdot 11\cdot 13\cdot m + 3$ where $m$ is an integer.
This must be divisible by $lcm$ $(14, 15, 16)$ , which is $560\cdot 3$
Therefore, $\frac{9\cdot 11\cdot 13\cdot m + 3}{560\cdot 3} = q$ , which is an integer. Factor out $3$ and divide to get $\frac{429m+1}{560} = q$ .
Therefore, $429m+1=560q$ . We can use Bezout's Identity or a Euclidean algorithm bash to solve for the least of $m$ and $q$
We find that the least $m$ is $171$ and the least $q$ is $131$
Since we want to factor $1680\cdot q$ , don't multiply: we already know that the prime factors of $1680$ are $2$ $3$ $5$ , and $7$ , and since $131$ is prime, we have $2 + 3 + 5 + 7 + 131 = \boxed{148}$ | null | 148 |
99d2ad6137c794f7a6b0998d758dafdc | https://artofproblemsolving.com/wiki/index.php/2013_AIME_I_Problems/Problem_11 | Ms. Math's kindergarten class has $16$ registered students. The classroom has a very large number, $N$ , of play blocks which satisfies the conditions:
(a) If $16$ $15$ , or $14$ students are present in the class, then in each case all the blocks can be distributed in equal numbers to each student, and
(b) There are three integers $0 < x < y < z < 14$ such that when $x$ $y$ , or $z$ students are present and the blocks are distributed in equal numbers to each student, there are exactly three blocks left over.
Find the sum of the distinct prime divisors of the least possible value of $N$ satisfying the above conditions. | Note that the number of play blocks is a multiple of the LCM of $16$ $15$ , and $14$ . The value of this can be found to be $(16)(15)(7) = 1680$ . This number is also divisible by $1$ $2$ $3$ $4$ $5$ $6$ $7$ $8$ $10$ , and $12$ , thus, the three numbers $x, y, z$ are $9, 11, 13$
Thus, $1680k\equiv 3$ when taken mod $9$ $11$ $13$ . Since $1680$ is congruent to $6$ mod $9$ and $3$ mod $13$ , and congruent to $8$ mod $11$ , the number $k$ must be a number that is congruent to $1$ mod $13$ $2$ mod $3$ (because $6$ is a multiple of $3$ , which is a factor of $9$ that can be divided out) and cause $8$ to become $3$ when multiplied under modulo $11$
Looking at the last condition shows that $k\equiv 10$ mod $11$ (after a bit of bashing) and is congruent to $1$ mod $13$ and $2$ mod $3$ as previously noted. Listing out the numbers congruent to $10$ mod $11$ and $1$ mod $13$ yield the following lists:
$10$ mod $11$ $21$ $32$ $43$ $54$ $65$ $76$ $87$ $98$ $109$ $120$ $131$ ...
$1$ mod $13$ $14$ $27$ $40$ $53$ $66$ $79$ $92$ $105$ $118$ $131$ $144$ $157$ $170$ ...
Both lists contain $x$ elements where $x$ is the modulo being taken, thus, there must be a solution in these lists as adding $11(13)$ to this solution yields the next smallest solution. In this case, $131$ is the solution for $k$ and thus the answer is $1680(131)$ . Since $131$ is prime, the sum of the prime factors is $2 + 3 + 5 + 7 + 131 = \boxed{148}$ | null | 148 |
99d2ad6137c794f7a6b0998d758dafdc | https://artofproblemsolving.com/wiki/index.php/2013_AIME_I_Problems/Problem_11 | Ms. Math's kindergarten class has $16$ registered students. The classroom has a very large number, $N$ , of play blocks which satisfies the conditions:
(a) If $16$ $15$ , or $14$ students are present in the class, then in each case all the blocks can be distributed in equal numbers to each student, and
(b) There are three integers $0 < x < y < z < 14$ such that when $x$ $y$ , or $z$ students are present and the blocks are distributed in equal numbers to each student, there are exactly three blocks left over.
Find the sum of the distinct prime divisors of the least possible value of $N$ satisfying the above conditions. | It is obvious that $N=a\cdot 2^4 \cdot 3\cdot 5\cdot 7$ and so the only mod $3$ number of students are $9, 11, 13$ . Therefore, $N=1287\cdot k+3$ . Try some approaches and you will see that this one is one of the few successful ones:
Start by setting the two $N$ equations together, then we get $1680a=1287k+3$ . Divide by $3$ . Note that since the RHS is $1\pmod{3}$ , and since $560$ is $2\pmod{3}$ , then $a=3b+2$ , where $b$ is some nonnegative integer, because $a$ must be $2\pmod{3}$
This reduces to $560 \cdot 3b + 1119 = 429k$ . Now, take out the $11!$ With the same procedure, $b=11c-1$ , where $c$ is some nonnegative integer.
You also get $c=13d+4$ , at which point $k=171+560d$ $d$ cannot be equal to $0$ . Therefore, $c=4, b=43, a=131$ , and we know the prime factors of $N$ are $2, 3, 5, 7, 131$ so the answer is $\boxed{148}$ | null | 148 |
99d2ad6137c794f7a6b0998d758dafdc | https://artofproblemsolving.com/wiki/index.php/2013_AIME_I_Problems/Problem_11 | Ms. Math's kindergarten class has $16$ registered students. The classroom has a very large number, $N$ , of play blocks which satisfies the conditions:
(a) If $16$ $15$ , or $14$ students are present in the class, then in each case all the blocks can be distributed in equal numbers to each student, and
(b) There are three integers $0 < x < y < z < 14$ such that when $x$ $y$ , or $z$ students are present and the blocks are distributed in equal numbers to each student, there are exactly three blocks left over.
Find the sum of the distinct prime divisors of the least possible value of $N$ satisfying the above conditions. | We start by noticing that $N = a\textbf{lcm}(14, 15, 16) = 1680a$ for some integer $a$ in order to satisfy the first condition.
Next, we satisfy the second condition. Since $x<y<z < 14$ must leave a remainder when dividing $1680a$ , they are not divisors of $1680x$ . Thus, we can eliminate all $y \le 14$ s.t. $\gcd(y, 1680x) \ne 1$ which leaves $(x, y, z) = (9, 11, 13)$ . Thus, $N = 1680a \equiv 3 \pmod 9 \equiv 3 \pmod {11} \equiv 3 \pmod {13}$ . Now, we seek to find the least $a$ which satisfies this set of congruences.
By Chinese Remainder Theorem on the first two congruences, we find that $a \equiv 32 \pmod {33}$ (we divide by three before proceeding in the first congruence to ensure the minimal solution). Finally, by CRT again on $a \equiv 32 \pmod {33}$ and $1680a \equiv 3 \pmod {13}$ we find that $a \equiv 131 \pmod {429}$
Thus, the minimal value of $N$ is possible at $a = 131$ . The prime factorization of this minimum value is $2^4 \cdot 3 \cdot 5 \cdot 7 \cdot 131$ and so the answer is $2 + 3 + 5 + 7 + 131 = \boxed{148}$ | null | 148 |
99d2ad6137c794f7a6b0998d758dafdc | https://artofproblemsolving.com/wiki/index.php/2013_AIME_I_Problems/Problem_11 | Ms. Math's kindergarten class has $16$ registered students. The classroom has a very large number, $N$ , of play blocks which satisfies the conditions:
(a) If $16$ $15$ , or $14$ students are present in the class, then in each case all the blocks can be distributed in equal numbers to each student, and
(b) There are three integers $0 < x < y < z < 14$ such that when $x$ $y$ , or $z$ students are present and the blocks are distributed in equal numbers to each student, there are exactly three blocks left over.
Find the sum of the distinct prime divisors of the least possible value of $N$ satisfying the above conditions. | As the problem stated, the number of boxes is definitely a multiple of $lcm(14,15,16)=1680$ , so we assume total number of boxes is $1680k$
Then, according to $(b)$ statement, we get $1680k \equiv 3 \pmod x \equiv 3 \pmod {y} \equiv 3 \pmod {z}$ . So we have $lcm(x,y,z)+3+m\cdot lcm(x,y,z)=1680k$ , we just write it to be $(1+n)lcm(x,y,z)=1680k-3$ Which tells that $x,y,z$ must be all odd number. Moreover, we can see $(1+m)lcm(x,y,z)$ can't be a multiple of $3,5,7$ (as $1680$ is a multiple of $5,7$ ) which means that $lcm(x,y,z)=lcm(9,11,13)=1287$ We let $n=1+m$
Now, we write $1287n+3=1680k, 429n+1=560k$ It is true that $n\equiv 1 \pmod{10}$ , let $n=10p+1$ , it has $429p+43=56k$ Then, $p$ must be odd, let $p=2q+1$ , it indicates $429q+236=28k$ Now, $q$ must be even, $q=2s$ tells $429s+118=14k$ Eventually, $s$ must be even, $s=2y$ $858y+59=7k$ $y=1$ is the smallest. This time, $k=131$
So the number of balls is $1680\cdot 131=2^4\cdot 3\cdot 5\cdot 7 \cdot 131$ , the desired value is $2+3+5+7+131=\boxed{148}$ | null | 148 |
99d2ad6137c794f7a6b0998d758dafdc | https://artofproblemsolving.com/wiki/index.php/2013_AIME_I_Problems/Problem_11 | Ms. Math's kindergarten class has $16$ registered students. The classroom has a very large number, $N$ , of play blocks which satisfies the conditions:
(a) If $16$ $15$ , or $14$ students are present in the class, then in each case all the blocks can be distributed in equal numbers to each student, and
(b) There are three integers $0 < x < y < z < 14$ such that when $x$ $y$ , or $z$ students are present and the blocks are distributed in equal numbers to each student, there are exactly three blocks left over.
Find the sum of the distinct prime divisors of the least possible value of $N$ satisfying the above conditions. | From part (a), we know that $2^4\cdot3\cdot5\cdot7 | N$ . From part (b), we know that $N \equiv 3 \pmod {1287}$ . We can expand on part (a) by saying that $N = 1680k$ for some $k$ . Rather than taking the three modulos together, we take them individually. \[1680k \equiv 6k \equiv 3 \pmod 9\] \[k \equiv 2^{-1} \pmod 9\] The inverse of 2 mod 9 is easily seen to be $5$ \[k \equiv 5 \pmod 9\]
Now moving to the second modulo which we leave as follows, \[1680k \equiv 8k \equiv 3 \pmod {11}\] Now the last modulo, \[1680k \equiv 3k \equiv 3 \pmod {13}\] \[k \equiv 1 \pmod{13}\] CRT on the first and the third one results in $k \equiv 4 \pmod {117}$ . Now doing the second one and the one we just made, $k \equiv 131 \pmod{1287}$ . Thus, the smallest value that works for $k = 131$ . Thus $N = 2^4\cdot3\cdot5\cdot7\cdot131$ $2+3+5+7+131 = \boxed{148}$ | null | 148 |
0ffc58fdf7ce9f49c5aadc2be759793a | https://artofproblemsolving.com/wiki/index.php/2013_AIME_I_Problems/Problem_12 | Let $\bigtriangleup PQR$ be a triangle with $\angle P = 75^\circ$ and $\angle Q = 60^\circ$ . A regular hexagon $ABCDEF$ with side length 1 is drawn inside $\triangle PQR$ so that side $\overline{AB}$ lies on $\overline{PQ}$ , side $\overline{CD}$ lies on $\overline{QR}$ , and one of the remaining vertices lies on $\overline{RP}$ . There are positive integers $a, b, c,$ and $d$ such that the area of $\triangle PQR$ can be expressed in the form $\frac{a+b\sqrt{c}}{d}$ , where $a$ and $d$ are relatively prime, and c is not divisible by the square of any prime. Find $a+b+c+d$ | First, find that $\angle R = 45^\circ$ .
Draw $ABCDEF$ . Now draw $\bigtriangleup PQR$ around $ABCDEF$ such that $Q$ is adjacent to $C$ and $D$ . The height of $ABCDEF$ is $\sqrt{3}$ , so the length of base $QR$ is $2+\sqrt{3}$ . Let the equation of $\overline{RP}$ be $y = x$ . Then, the equation of $\overline{PQ}$ is $y = -\sqrt{3} (x - (2+\sqrt{3})) \to y = -x\sqrt{3} + 2\sqrt{3} + 3$ . Solving the two equations gives $y = x = \frac{\sqrt{3} + 3}{2}$ . The area of $\bigtriangleup PQR$ is $\frac{1}{2} * (2 + \sqrt{3}) * \frac{\sqrt{3} + 3}{2} = \frac{5\sqrt{3} + 9}{4}$ $a + b + c + d = 9 + 5 + 3 + 4 = \boxed{021}$ | null | 021 |
0ffc58fdf7ce9f49c5aadc2be759793a | https://artofproblemsolving.com/wiki/index.php/2013_AIME_I_Problems/Problem_12 | Let $\bigtriangleup PQR$ be a triangle with $\angle P = 75^\circ$ and $\angle Q = 60^\circ$ . A regular hexagon $ABCDEF$ with side length 1 is drawn inside $\triangle PQR$ so that side $\overline{AB}$ lies on $\overline{PQ}$ , side $\overline{CD}$ lies on $\overline{QR}$ , and one of the remaining vertices lies on $\overline{RP}$ . There are positive integers $a, b, c,$ and $d$ such that the area of $\triangle PQR$ can be expressed in the form $\frac{a+b\sqrt{c}}{d}$ , where $a$ and $d$ are relatively prime, and c is not divisible by the square of any prime. Find $a+b+c+d$ | Use coordinates. Call $Q$ the origin and $QP$ be on the x-axis. It is easy to see that $F$ is the vertex on $RP$ . After labeling coordinates (noting additionally that $QBC$ is an equilateral triangle), we see that the area is $QP$ times $0.5$ times the coordinate of $R$ . Draw a perpendicular of $F$ , call it $H$ , and note that $QP = 1 + \sqrt{3}$ after using the trig functions for $75$ degrees.
Now, get the lines for $QR$ and $RP$ $y=\sqrt{3}x$ and $y=-(2+\sqrt{3})x + (5+\sqrt{3})$ , whereupon we get the ordinate of $R$ to be $\frac{3+2\sqrt{3}}{2}$ , and the area is $\frac{5\sqrt{3} + 9}{4}$ , so our answer is $\boxed{021}$ | null | 021 |
0ffc58fdf7ce9f49c5aadc2be759793a | https://artofproblemsolving.com/wiki/index.php/2013_AIME_I_Problems/Problem_12 | Let $\bigtriangleup PQR$ be a triangle with $\angle P = 75^\circ$ and $\angle Q = 60^\circ$ . A regular hexagon $ABCDEF$ with side length 1 is drawn inside $\triangle PQR$ so that side $\overline{AB}$ lies on $\overline{PQ}$ , side $\overline{CD}$ lies on $\overline{QR}$ , and one of the remaining vertices lies on $\overline{RP}$ . There are positive integers $a, b, c,$ and $d$ such that the area of $\triangle PQR$ can be expressed in the form $\frac{a+b\sqrt{c}}{d}$ , where $a$ and $d$ are relatively prime, and c is not divisible by the square of any prime. Find $a+b+c+d$ | Angle chasing yields that both triangles $PAF$ and $PQR$ are $75$ $60$ $45$ triangles. First look at triangle $PAF$ . Using Law of Sines, we find:
$\frac{\frac{\sqrt{6} + \sqrt{2}}{4}}{1} = \frac{\frac{\sqrt{2}}{2}}{PA}$
Simplifying, we find $PA = \sqrt{3} - 1$ .
Since $\angle{Q} = 60^\circ$ , WLOG assume triangle $BQC$ is equilateral, so $BQ = 1$ . So $PQ = \sqrt{3} + 1$
Apply Law of Sines again,
$\frac{\frac{\sqrt{2}}{2}}{\sqrt{3} + 1} = \frac{\frac{\sqrt{3}}{2}}{PR}$
Simplifying, we find $PR = \frac{\sqrt{6}}{2} \cdot (1 + \sqrt{3})$
$[PQR] = \frac{1}{2} \cdot PQ \cdot PR \cdot \sin 75^\circ$
Evaluating and reducing, we get $\frac{9 + 5\sqrt{3}}{4},$ thus the answer is $\boxed{021}$ | null | 021 |
94f5c2dc98b595704a85ea8f3d73d20e | https://artofproblemsolving.com/wiki/index.php/2013_AIME_I_Problems/Problem_13 | Triangle $AB_0C_0$ has side lengths $AB_0 = 12$ $B_0C_0 = 17$ , and $C_0A = 25$ . For each positive integer $n$ , points $B_n$ and $C_n$ are located on $\overline{AB_{n-1}}$ and $\overline{AC_{n-1}}$ , respectively, creating three similar triangles $\triangle AB_nC_n \sim \triangle B_{n-1}C_nC_{n-1} \sim \triangle AB_{n-1}C_{n-1}$ . The area of the union of all triangles $B_{n-1}C_nB_n$ for $n\geq1$ can be expressed as $\tfrac pq$ , where $p$ and $q$ are relatively prime positive integers. Find $q$ | Note that every $B_nC_n$ is parallel to each other for any nonnegative $n$ . Also, the area we seek is simply the ratio $k=\frac{[B_0B_1C_1]}{[B_0B_1C_1]+[C_1C_0B_0]}$ , because it repeats in smaller and smaller units. Note that the area of the triangle, by Heron's formula, is 90.
For ease, all ratios I will use to solve this problem are with respect to the area of $[AB_0C_0]$ . For example, if I say some area has ratio $\frac{1}{2}$ , that means its area is 45.
Now note that $k=$ 1 minus ratio of $[B_1C_1A]$ minus ratio $[B_0C_0C_1]$ . We see by similar triangles given that ratio $[B_0C_0C_1]$ is $\frac{17^2}{25^2}$ . Ratio $[B_1C_1A]$ is $(\frac{336}{625})^2$ , after seeing that $C_1C_0 = \frac{289}{625}$ , . Now it suffices to find 90 times ratio $[B_0B_1C_1]$ , which is given by 1 minus the two aforementioned ratios. Substituting these ratios to find $k$ and clearing out the $5^8$ , we see that the answer is $90\cdot \frac{5^8-336^2-17^2\cdot 5^4}{5^8-336^2}$ , which gives $q= \boxed{961}$ | null | 961 |
94f5c2dc98b595704a85ea8f3d73d20e | https://artofproblemsolving.com/wiki/index.php/2013_AIME_I_Problems/Problem_13 | Triangle $AB_0C_0$ has side lengths $AB_0 = 12$ $B_0C_0 = 17$ , and $C_0A = 25$ . For each positive integer $n$ , points $B_n$ and $C_n$ are located on $\overline{AB_{n-1}}$ and $\overline{AC_{n-1}}$ , respectively, creating three similar triangles $\triangle AB_nC_n \sim \triangle B_{n-1}C_nC_{n-1} \sim \triangle AB_{n-1}C_{n-1}$ . The area of the union of all triangles $B_{n-1}C_nB_n$ for $n\geq1$ can be expressed as $\tfrac pq$ , where $p$ and $q$ are relatively prime positive integers. Find $q$ | Let $k$ be the coefficient of the similarity of triangles \[\triangle B_0 C_1 C_0 \sim \triangle AB_0 C_0 \implies k = \frac {B_0 C_0}{AC_0} = \frac {17}{25}.\] Then area $\frac {[B_0 C_1 C_0]}{[AB_0 C_0 ]} = k^2 \implies \frac {[AB_0 C_1]}{[AB_0 C_0]} = 1 – k^2.$
The height of triangles $\triangle B_0C_1A$ and $\triangle AB_0C_0$ from $B_0$ is the same $\implies \frac {AC_1}{AC_0} = 1 – k^2.$
The coefficient of the similarity of triangles $\triangle AB_1C_1 \sim \triangle AB_0C_0$ is $\frac {AC_1}{AC_0} = 1 – k^2 \implies \frac {[B_1C_1C_2 ]}{[AB_0C_0 ]} = k^2 (1 – k^2)^2.$
Analogically the coefficient of the similarity of triangles $\triangle AB_2C_2 \sim \triangle AB_0C_0$ is $(1 – k^2)^2 \implies \frac {[B_2C_2C_3]}{[AB_0C_0 ]} = k^2 (1 – k^2)^4$ and so on.
The yellow area $[Y]$ is $\frac {[Y]}{[AB_0C_0 ]} = k^2 + k^2 (1 – k^2)^2 + k^2 (1 – k^2)^4 +.. = \frac {k^2}{1 – (1 – k^2)^2} = \frac{1}{2 – k^2}.$
The required area is $[AB_0C_0 ] – [Y] = [AB_0C_0 ] \cdot (1 – \frac{1}{2 – k^2}) = [AB_0C_0 ] \cdot \frac {1 – k^2}{2 – k^2} = [AB_0C_0 ] \cdot \frac {25^2 – 17^2} {2 \cdot 25^2 – 17^2} = [AB_0C_0 ] \cdot \frac {336}{961}.$
The number $961$ is prime, $[AB_0C_0]$ is integer but not $961,$ therefore the answer is $\boxed{961}$ | null | 961 |
7f53ae63b0a3beefdc2298749f9a6aa5 | https://artofproblemsolving.com/wiki/index.php/2013_AIME_I_Problems/Problem_14 | For $\pi \le \theta < 2\pi$ , let
\begin{align*} P &= \frac12\cos\theta - \frac14\sin 2\theta - \frac18\cos 3\theta + \frac{1}{16}\sin 4\theta + \frac{1}{32} \cos 5\theta - \frac{1}{64} \sin 6\theta - \frac{1}{128} \cos 7\theta + \cdots \end{align*}
and
\begin{align*} Q &= 1 - \frac12\sin\theta -\frac14\cos 2\theta + \frac18 \sin 3\theta + \frac{1}{16}\cos 4\theta - \frac{1}{32}\sin 5\theta - \frac{1}{64}\cos 6\theta +\frac{1}{128}\sin 7\theta + \cdots \end{align*}
so that $\frac{P}{Q} = \frac{2\sqrt2}{7}$ . Then $\sin\theta = -\frac{m}{n}$ where $m$ and $n$ are relatively prime positive integers. Find $m+n$ | Noticing the $\sin$ and $\cos$ in both $P$ and $Q,$ we think of the angle addition identities:
\[\sin(a + b) = \sin a \cos b + \cos a \sin b, \cos(a + b) = \cos a \cos b - \sin a \sin b\]
With this in mind, we multiply $P$ by $\sin \theta$ and $Q$ by $\cos \theta$ to try and use some angle addition identities. Indeed, we get \begin{align*} P \sin \theta + Q \cos \theta &= \cos \theta + \dfrac{1}{2}(\cos \theta \sin \theta - \sin \theta \cos \theta) - \dfrac{1}{4}(\sin{2 \theta} \sin \theta + \cos{2 \theta} \cos{\theta}) - \cdots \\ &= \cos \theta - \dfrac{1}{4} \cos \theta + \dfrac{1}{8} \sin{2 \theta} + \dfrac{1}{16} \cos{3 \theta} + \cdots \\ &= \cos \theta - \dfrac{1}{2}P \end{align*} after adding term-by-term. Similar term-by-term adding yields \[P \cos \theta + Q \sin \theta = -2(Q - 1).\] This is a system of equations; rearrange and rewrite to get \[P(1 + 2 \sin \theta) + 2Q \cos \theta = 2 \cos \theta\] and \[P \cos^2 \theta + Q \cos \theta(2 + \sin \theta) = 2 \cos \theta.\] Subtract the two and rearrange to get \[\dfrac{P}{Q} = \dfrac{\cos \theta}{2 + \sin \theta} = \dfrac{2 \sqrt{2}}{7}.\] Then, square both sides and use Pythagorean Identity to get a quadratic in $\sin \theta.$ Factor that quadratic and solve for $\sin \theta = -17/19, 1/3.$ Since we're given $\pi\leq\theta<2\pi,$ $\sin\theta$ is nonpositive. We therefore use the negative solution, and our desired answer is $17 + 19 = \boxed{036}.$ | null | 036 |
7f53ae63b0a3beefdc2298749f9a6aa5 | https://artofproblemsolving.com/wiki/index.php/2013_AIME_I_Problems/Problem_14 | For $\pi \le \theta < 2\pi$ , let
\begin{align*} P &= \frac12\cos\theta - \frac14\sin 2\theta - \frac18\cos 3\theta + \frac{1}{16}\sin 4\theta + \frac{1}{32} \cos 5\theta - \frac{1}{64} \sin 6\theta - \frac{1}{128} \cos 7\theta + \cdots \end{align*}
and
\begin{align*} Q &= 1 - \frac12\sin\theta -\frac14\cos 2\theta + \frac18 \sin 3\theta + \frac{1}{16}\cos 4\theta - \frac{1}{32}\sin 5\theta - \frac{1}{64}\cos 6\theta +\frac{1}{128}\sin 7\theta + \cdots \end{align*}
so that $\frac{P}{Q} = \frac{2\sqrt2}{7}$ . Then $\sin\theta = -\frac{m}{n}$ where $m$ and $n$ are relatively prime positive integers. Find $m+n$ | Use sum to product formulas to rewrite $P$ and $Q$
\[P \sin\theta\ + Q \cos\theta\ = \cos \theta\ - \frac{1}{4}\cos \theta + \frac{1}{8}\sin 2\theta + \frac{1}{16}\cos 3\theta - \frac{1}{32}\sin 4\theta + ...\]
Therefore, \[P \sin \theta - Q \cos \theta = -2P\]
Using \[\frac{P}{Q} = \frac{2\sqrt2}{7}, Q = \frac{7}{2\sqrt2} P\]
Plug in to the previous equation and cancel out the "P" terms to get: \[\sin\theta - \frac{7}{2\sqrt2} \cos\theta = -2\]
Then use the pythagorean identity to solve for $\sin\theta$ \[\sin\theta = -\frac{17}{19} \implies \boxed{036}\] | null | 036 |
7f53ae63b0a3beefdc2298749f9a6aa5 | https://artofproblemsolving.com/wiki/index.php/2013_AIME_I_Problems/Problem_14 | For $\pi \le \theta < 2\pi$ , let
\begin{align*} P &= \frac12\cos\theta - \frac14\sin 2\theta - \frac18\cos 3\theta + \frac{1}{16}\sin 4\theta + \frac{1}{32} \cos 5\theta - \frac{1}{64} \sin 6\theta - \frac{1}{128} \cos 7\theta + \cdots \end{align*}
and
\begin{align*} Q &= 1 - \frac12\sin\theta -\frac14\cos 2\theta + \frac18 \sin 3\theta + \frac{1}{16}\cos 4\theta - \frac{1}{32}\sin 5\theta - \frac{1}{64}\cos 6\theta +\frac{1}{128}\sin 7\theta + \cdots \end{align*}
so that $\frac{P}{Q} = \frac{2\sqrt2}{7}$ . Then $\sin\theta = -\frac{m}{n}$ where $m$ and $n$ are relatively prime positive integers. Find $m+n$ | Note that \[e^{i\theta}=\cos(\theta)+i\sin(\theta)\]
Thus, the following identities follow immediately: \[ie^{i\theta}=i(\cos(\theta)+i\sin(\theta))=-\sin(\theta)+i\cos(\theta)\] \[i^2 e^{i\theta}=-e^{i\theta}=-\cos(\theta)-i\sin(\theta)\] \[i^3 e^{i\theta}=\sin(\theta)-i\cos(\theta)\]
Consider, now, the sum $Q+iP$ . It follows fairly immediately that:
\[Q+iP=1+\left(\frac{i}{2}\right)^1e^{i\theta}+\left(\frac{i}{2}\right)^2e^{2i\theta}+\ldots=\frac{1}{1-\frac{i}{2}e^{i\theta}}=\frac{2}{2-ie^{i\theta}}\] \[Q+iP=\frac{2}{2-ie^{i\theta}}=\frac{2}{2-(-\sin(\theta)+i\cos(\theta))}=\frac{2}{(2+\sin(\theta))-i\cos(\theta)}\]
This follows straight from the geometric series formula and simple simplification. We can now multiply the denominator by it's complex conjugate to find:
\[Q+iP=\frac{2}{(2+\sin(\theta))-i\cos(\theta)}\left(\frac{(2+\sin(\theta))+i\cos(\theta)}{(2+\sin(\theta))+i\cos(\theta)}\right)\] \[Q+iP=\frac{2((2+\sin(\theta))+i\cos(\theta))}{(2+\sin(\theta))^2+\cos^2(\theta)}\]
Comparing real and imaginary parts, we find: \[\frac{P}{Q}=\frac{\cos(\theta)}{2+\sin(\theta)}=\frac{2\sqrt{2}}{7}\]
Squaring this equation and letting $\sin^2(\theta)=x$
$\frac{P^2}{Q^2}=\frac{\cos^2(\theta)}{4+4\sin(\theta)+\sin^2(\theta)}=\frac{1-x^2}{4+4x+x^2}=\frac{8}{49}$
Clearing denominators and solving for $x$ gives sine as $x=-\frac{17}{19}$
$017+019=\boxed{036}$ | null | 036 |
7f53ae63b0a3beefdc2298749f9a6aa5 | https://artofproblemsolving.com/wiki/index.php/2013_AIME_I_Problems/Problem_14 | For $\pi \le \theta < 2\pi$ , let
\begin{align*} P &= \frac12\cos\theta - \frac14\sin 2\theta - \frac18\cos 3\theta + \frac{1}{16}\sin 4\theta + \frac{1}{32} \cos 5\theta - \frac{1}{64} \sin 6\theta - \frac{1}{128} \cos 7\theta + \cdots \end{align*}
and
\begin{align*} Q &= 1 - \frac12\sin\theta -\frac14\cos 2\theta + \frac18 \sin 3\theta + \frac{1}{16}\cos 4\theta - \frac{1}{32}\sin 5\theta - \frac{1}{64}\cos 6\theta +\frac{1}{128}\sin 7\theta + \cdots \end{align*}
so that $\frac{P}{Q} = \frac{2\sqrt2}{7}$ . Then $\sin\theta = -\frac{m}{n}$ where $m$ and $n$ are relatively prime positive integers. Find $m+n$ | A bit similar to Solution 3. We use $\phi = \theta+90^\circ$ because the progression cycles in $P\in (\sin 0\theta,\cos 1\theta,-\sin 2\theta,-\cos 3\theta\dots)$ . So we could rewrite that as $P\in(\sin 0\phi,\sin 1\phi,\sin 2\phi,\sin 3\phi\dots)$
Similarly, $Q\in (\cos 0\theta,-\sin 1\theta,-\cos 2\theta,\sin 3\theta\dots)\implies Q\in(\cos 0\phi,\cos 1\phi, \cos 2\phi, \cos 3\phi\dots)$
Setting complex $z=q_1+p_1i$ , we get $z=\frac{1}{2}\left(\cos\phi+\sin\phi i\right)$
$(Q,P)=\sum_{n=0}^\infty z^n=\frac{1}{1-z}=\frac{1}{1-\frac{1}{2}\cos\phi-\frac{i}{2}\sin\phi}=\frac{1-0.5\cos\phi+0.5i\sin\phi}{\dots}$
The important part is the ratio of the imaginary part $i$ to the real part. To cancel out the imaginary part from the denominator, we must add $0.5i\sin\phi$ to the numerator to make the denominator a difference (or rather a sum) of squares. The denominator does not matter. Only the numerator, because we are trying to find $\frac{P}{Q}=\tan\text{arg}(\Sigma)$ a PROPORTION of values. So denominators would cancel out.
$\frac{P}{Q}=\frac{\text{Re}\Sigma}{\text{Im}\Sigma}=\frac{0.5\sin\phi}{1-0.5\cos\phi}=\frac{\sin\phi}{2-\cos\phi}=\frac{2\sqrt{2}}{7}$
Setting $\sin\theta=y$ , we obtain \[\frac{\sqrt{1-y^2}}{2+y}=\frac{2\sqrt{2}}{7}\] \[7\sqrt{1-y^2}=2\sqrt{2}(2+y)\] \[49-49y^2=8y^2+32y+32\] \[57y^2+32y-17=0\rightarrow y=\frac{-32\pm\sqrt{1024+4\cdot 969}}{114}\] \[y=\frac{-32\pm\sqrt{4900}}{114}=\frac{-16\pm 35}{57}\]
Since $y<0$ because $\pi<\theta<2\pi$ $y=\sin\theta=-\frac{51}{57}=-\frac{17}{19}$ . Adding up, $17+19=\boxed{036}$ | null | 036 |
7f53ae63b0a3beefdc2298749f9a6aa5 | https://artofproblemsolving.com/wiki/index.php/2013_AIME_I_Problems/Problem_14 | For $\pi \le \theta < 2\pi$ , let
\begin{align*} P &= \frac12\cos\theta - \frac14\sin 2\theta - \frac18\cos 3\theta + \frac{1}{16}\sin 4\theta + \frac{1}{32} \cos 5\theta - \frac{1}{64} \sin 6\theta - \frac{1}{128} \cos 7\theta + \cdots \end{align*}
and
\begin{align*} Q &= 1 - \frac12\sin\theta -\frac14\cos 2\theta + \frac18 \sin 3\theta + \frac{1}{16}\cos 4\theta - \frac{1}{32}\sin 5\theta - \frac{1}{64}\cos 6\theta +\frac{1}{128}\sin 7\theta + \cdots \end{align*}
so that $\frac{P}{Q} = \frac{2\sqrt2}{7}$ . Then $\sin\theta = -\frac{m}{n}$ where $m$ and $n$ are relatively prime positive integers. Find $m+n$ | We notice $\sin\theta=-\frac{i}{2}(e^{i\theta}-e^{-i\theta})$ and $\cos\theta=\frac{1}{2}(e^{i\theta}+e^{-i\theta})$
We observe that both $P$ and $Q$ can be split into $2$ parts, namely the terms which contain the $\cos$ and the terms which contain the $\sin .$
The $\cos$ part of $P$ can be expressed as: \begin{align*}\frac12\cos\theta-\frac18\cos3\theta+\cdots&=\frac14\left(e^{i\theta}\left(1-\frac{e^{i2\theta}}{4}+\cdots\right)+e^{-i\theta}\left(1-\frac{e^{-i2\theta}}{4}+\cdots\right)\right) \\ &= \frac{1}{4}\left(\frac{e^{i\theta}}{1+\frac{1}{4}e^{i2\theta}}+\frac{e^{-i\theta}}{1+\frac{1}{4}e^{-i2\theta}}\right)\\ &= \frac{5(e^{i\theta}+e^{-i\theta})}{17+4e^{i2\theta}+4e^{-i2\theta}}.\end{align*}
Repeating the above process, we find that the $\sin$ part of $P$ is \[\frac{2i(e^{i2\theta}-e^{-i2\theta})}{17+4e^{i2\theta}+4e^{-i2\theta}},\] the $\cos$ part of $Q$ is \[\frac{16+2(e^{i2\theta}+e^{-i2\theta})}{17+4e^{i2\theta}+4e^{-i2\theta}},\] and finally, the $\sin$ part of $Q$ is \[\frac{3i(e^{i\theta}-e^{-i\theta})}{17+4e^{i2\theta}+4e^{-i2\theta}}.\]
Converting back to trigonometric form, we have \begin{align*}\frac{2\sqrt{2}}{7}&=\frac{10\cos{\theta}-4\sin{2\theta}}{16+4\cos{2\theta}-6\sin{\theta}}\\ &=\frac{5\cos{\theta}-2\sin{2\theta}}{8+2\cos{2\theta}-3\sin{\theta}}.\end{align*} Using the $\sin$ double identity and simplifying, we have \[\frac{2\sqrt2}{7}=\frac{\cos{\theta}(5-4\sin{\theta})}{10-4\sin^2{\theta}-3\sin{\theta}}.\] Factoring the denominator, we have \[10-4\sin^2{\theta}-3\sin{\theta}=(5-4\sin\theta)(2+\sin\theta).\] Simplifying \begin{align*}\frac{2\sqrt2}{7}&= \frac{\cos{\theta}(5-4\sin{\theta})}{(5-4\sin\theta)(2+\sin\theta)}\\ &=\frac{\cos\theta}{2+\sin\theta}.\end{align*} We set $\sin\theta$ as $x$ , and by the Pythagorean Identity, we have $57x^2+32x-17=0$ . This factors into $(19x+17)(3x-1)=0$ , which yields the 2 solutions $x=-\frac{17}{19}, x=\frac{1}{3}$ . As $\pi\leq\theta<2\pi$ , the latter root is erroneous, and we are left with $\sin\theta=-\frac{17}{19}$ . Thus, our final answer is $17+19=\boxed{036}$ | null | 036 |
7f53ae63b0a3beefdc2298749f9a6aa5 | https://artofproblemsolving.com/wiki/index.php/2013_AIME_I_Problems/Problem_14 | For $\pi \le \theta < 2\pi$ , let
\begin{align*} P &= \frac12\cos\theta - \frac14\sin 2\theta - \frac18\cos 3\theta + \frac{1}{16}\sin 4\theta + \frac{1}{32} \cos 5\theta - \frac{1}{64} \sin 6\theta - \frac{1}{128} \cos 7\theta + \cdots \end{align*}
and
\begin{align*} Q &= 1 - \frac12\sin\theta -\frac14\cos 2\theta + \frac18 \sin 3\theta + \frac{1}{16}\cos 4\theta - \frac{1}{32}\sin 5\theta - \frac{1}{64}\cos 6\theta +\frac{1}{128}\sin 7\theta + \cdots \end{align*}
so that $\frac{P}{Q} = \frac{2\sqrt2}{7}$ . Then $\sin\theta = -\frac{m}{n}$ where $m$ and $n$ are relatively prime positive integers. Find $m+n$ | Follow solution 3, up to the point of using the geometric series formula \[Q+iP=\frac{1}{1+\frac{\sin(\theta)}{2}-\frac{Qi\cos(\theta)}{2}}\]
Moving everything to the other side, and considering only the imaginary part, we get \[Pi+\frac{Pi}{2}\sin\theta-\frac{Qi}{2}\cos\theta = 0\]
We can then write $P = 2 \sqrt{2} k$ , and $Q = 7k$ , ( $k \neq 0$ ). Thus, we can substitute and divide out by k. \[2\sqrt{2}+\sqrt{2}\sin\theta-\frac{7}{2}\cos\theta\ =\ 0\] \[2\sqrt{2}+\sqrt{2}\sin\theta-\frac{7}{2}\sqrt{1-\sin^{2}\theta}=\ 0\] \[2\sqrt{2}+\sqrt{2}\sin\theta\ =\frac{7}{2}\left(\sqrt{1-\sin^{2}\theta}\right)\] \[8+8\sin\theta+2\sin^{2}\theta=\frac{49}{4}-\frac{49}{7}\sin^{2}\theta\] \[\frac{57}{4}\sin^{2}\theta+8\sin\theta-\frac{17}{4} = 0\] \[57\sin^{2}\theta+32\sin\theta-17 = 0\] \[\left(3\sin\theta-1\right)\left(19\sin\theta+17\right) = 0\]
Since $\pi \le \theta < 2\pi$ , we get $\sin \theta < 0$ , and thus, $\sin\theta = \frac{-17}{19} \implies \boxed{036}$ | null | 036 |
2185cc45929fc395aa4c845f81e639ff | https://artofproblemsolving.com/wiki/index.php/2013_AIME_I_Problems/Problem_15 | Let $N$ be the number of ordered triples $(A,B,C)$ of integers satisfying the conditions
(a) $0\le A<B<C\le99$ ,
(b) there exist integers $a$ $b$ , and $c$ , and prime $p$ where $0\le b<a<c<p$ ,
(c) $p$ divides $A-a$ $B-b$ , and $C-c$ , and
(d) each ordered triple $(A,B,C)$ and each ordered triple $(b,a,c)$ form arithmetic sequences. Find $N$ | From condition (d), we have $(A,B,C)=(B-D,B,B+D)$ and $(b,a,c)=(a-d,a,a+d)$ . Condition $\text{(c)}$ states that $p\mid B-D-a$ $p | B-a+d$ , and $p\mid B+D-a-d$ . We subtract the first two to get $p\mid-d-D$ , and we do the same for the last two to get $p\mid 2d-D$ . We subtract these two to get $p\mid 3d$ . So $p\mid 3$ or $p\mid d$ . The second case is clearly impossible, because that would make $c=a+d>p$ , violating condition $\text{(b)}$ . So we have $p\mid 3$ , meaning $p=3$ . Condition $\text{(b)}$ implies that $(b,a,c)=(0,1,2)$ or $(a,b,c)\in (1,0,2)\rightarrow (-2,0,2)\text{ }(D\equiv 2\text{ mod 3})$ . Now we return to condition $\text{(c)}$ , which now implies that $(A,B,C)\equiv(-2,0,2)\pmod{3}$ . Now, we set $B=3k$ for increasing positive integer values of $k$ $B=0$ yields no solutions. $B=3$ gives $(A,B,C)=(1,3,5)$ , giving us $1$ solution. If $B=6$ , we get $2$ solutions, $(4,6,8)$ and $(1,6,11)$ . Proceeding in the manner, we see that if $B=48$ , we get 16 solutions. However, $B=51$ still gives $16$ solutions because $C_\text{max}=2B-1=101>100$ . Likewise, $B=54$ gives $15$ solutions. This continues until $B=96$ gives one solution. $B=99$ gives no solution. Thus, $N=1+2+\cdots+16+16+15+\cdots+1=2\cdot\frac{16(17)}{2}=16\cdot 17=\boxed{272}$ | null | 272 |
2185cc45929fc395aa4c845f81e639ff | https://artofproblemsolving.com/wiki/index.php/2013_AIME_I_Problems/Problem_15 | Let $N$ be the number of ordered triples $(A,B,C)$ of integers satisfying the conditions
(a) $0\le A<B<C\le99$ ,
(b) there exist integers $a$ $b$ , and $c$ , and prime $p$ where $0\le b<a<c<p$ ,
(c) $p$ divides $A-a$ $B-b$ , and $C-c$ , and
(d) each ordered triple $(A,B,C)$ and each ordered triple $(b,a,c)$ form arithmetic sequences. Find $N$ | Let $(A, B, C)$ $(B-x, B, B+x)$ and $(b, a, c) = (a-y, a, a+y)$ . Now the 3 differences would be \begin{align} \label{1} &A-a = B-x-a \\ \label{2} &B - b = B-a+y \\ \label{3} &C - c = B+x-a-y \end{align}
Adding equations $(1)$ and $(3)$ would give $2B - 2a - y$ . Then doubling equation $(2)$ would give $2B - 2a + 2y$ . The difference between them would be $3y$ . Since $p|\{(1), (2), (3)\}$ , then $p|3y$ . Since $p$ is prime, $p|3$ or $p|y$ . However, since $p > y$ , we must have $p|3$ , which means $p=3$
If $p=3$ , the only possible values of $(b, a, c)$ are $(0, 1, 2)$ . Plugging this into our differences, we get \begin{align*} &A-a = B-x-1 \hspace{4cm}(4)\\ &B - b = B \hspace{5.35cm}(5)\\ &C - c = B+x-2 \hspace{4cm}(6) \end{align*} The difference between $(4)$ and $(5)$ is $x+1$ , which should be divisible by 3. So $x \equiv 2 \mod 3$ . Also note that since $3|(5)$ $3|B$ . Now we can try different values of $x$ and $B$
When $x=2$ $B=3, 6, ..., 96 \Rightarrow 17$ triples.
When $x=5$ $B=6, 9, ..., 93\Rightarrow 15$ triples..
... and so on until
When $x=44$ $B=45\Rightarrow 1$ triple.
So the answer is $17 + 15 + \cdots + 1 = \boxed{272}$ | null | 272 |
2840e1d506ce7060c2be1ea0c1c8c5df | https://artofproblemsolving.com/wiki/index.php/2013_AIME_II_Problems/Problem_1 | Suppose that the measurement of time during the day is converted to the metric system so that each day has $10$ metric hours, and each metric hour has $100$ metric minutes. Digital clocks would then be produced that would read $\text{9:99}$ just before midnight, $\text{0:00}$ at midnight, $\text{1:25}$ at the former $\text{3:00}$ AM, and $\text{7:50}$ at the former $\text{6:00}$ PM. After the conversion, a person who wanted to wake up at the equivalent of the former $\text{6:36}$ AM would set his new digital alarm clock for $\text{A:BC}$ , where $\text{A}$ $\text{B}$ , and $\text{C}$ are digits. Find $100\text{A}+10\text{B}+\text{C}$ | There are $24 \cdot 60=1440$ normal minutes in a day , and $10 \cdot 100=1000$ metric minutes in a day. The ratio of normal to metric minutes in a day is $\frac{1440}{1000}$ , which simplifies to $\frac{36}{25}$ . This means that every time 36 normal minutes pass, 25 metric minutes pass. From midnight to $\text{6:36}$ AM, $6 \cdot 60+36=396$ normal minutes pass. This can be viewed as $\frac{396}{36}=11$ cycles of 36 normal minutes, so 11 cycles of 25 metric minutes pass. Adding $25 \cdot 11=275$ to $\text{0:00}$ gives $\text{2:75}$ , so the answer is $\boxed{275}$ | null | 275 |
2840e1d506ce7060c2be1ea0c1c8c5df | https://artofproblemsolving.com/wiki/index.php/2013_AIME_II_Problems/Problem_1 | Suppose that the measurement of time during the day is converted to the metric system so that each day has $10$ metric hours, and each metric hour has $100$ metric minutes. Digital clocks would then be produced that would read $\text{9:99}$ just before midnight, $\text{0:00}$ at midnight, $\text{1:25}$ at the former $\text{3:00}$ AM, and $\text{7:50}$ at the former $\text{6:00}$ PM. After the conversion, a person who wanted to wake up at the equivalent of the former $\text{6:36}$ AM would set his new digital alarm clock for $\text{A:BC}$ , where $\text{A}$ $\text{B}$ , and $\text{C}$ are digits. Find $100\text{A}+10\text{B}+\text{C}$ | First we want to find out what fraction of a day has passed at 6:36 AM. One hour is $\frac{1}{24}$ of a day, and 36 minutes is $\frac{36}{60}=\frac{3}{5}$ of an hour, so at 6:36 AM, $6 \cdot \frac{1}{24} + \frac{1}{24} \cdot \frac{3}{5}=\frac{1}{4}+\frac{1}{40}=\frac{11}{40}$ of a day has passed. Now the metric timing equivalent of $\frac{11}{40}$ of a day is $\frac{11}{40}\cdot 1000=275$ metric minutes, which is equivalent to 2 metric hours and 75 metric minutes, so our answer is $\boxed{275}$ - mathleticguyyy | null | 275 |
4ef77e1a18dac1904b8888838b1c2cbf | https://artofproblemsolving.com/wiki/index.php/2013_AIME_II_Problems/Problem_2 | Positive integers $a$ and $b$ satisfy the condition \[\log_2(\log_{2^a}(\log_{2^b}(2^{1000}))) = 0.\] Find the sum of all possible values of $a+b$ | To simplify, we write this logarithmic expression as an exponential one. Just looking at the first log, it has a base of 2 and an argument of the expression in parenthesis. Therefore, we can make 2 the base, 0 the exponent, and the argument the result. That means $\log_{2^a}(\log_{2^b}(2^{1000}))=1$ (because $2^0=1$ ). Doing this again, we get $\log_{2^b}(2^{1000})=2^a$ . Doing the process one more time, we finally eliminate all of the logs, getting ${(2^{b})}^{(2^a)}=2^{1000}$ . Using the property that ${(a^x)^{y}}=a^{xy}$ , we simplify to $2^{b\cdot2^{a}}=2^{1000}$ . Eliminating equal bases leaves $b\cdot2^a=1000$ . The largest $a$ such that $2^a$ divides $1000$ is $3$ , so we only need to check $1$ $2$ , and $3$ . When $a=1$ $b=500$ ; when $a=2$ $b=250$ ; when $a=3$ $b=125$ . Summing all the $a$ 's and $b$ 's gives the answer of $\boxed{881}$ | null | 881 |
4ef77e1a18dac1904b8888838b1c2cbf | https://artofproblemsolving.com/wiki/index.php/2013_AIME_II_Problems/Problem_2 | Positive integers $a$ and $b$ satisfy the condition \[\log_2(\log_{2^a}(\log_{2^b}(2^{1000}))) = 0.\] Find the sum of all possible values of $a+b$ | We proceed as in Solution 1, raising $2$ to both sides to achieve $\log_{2^a}(\log_{2^b}(2^{1000})) = 1.$ We raise $2^a$ to both sides to get $\log_{2^b}(2^{1000})=2^a$ , then simplify to get $\dfrac{1000}b=2^a$
At this point, we want both $a$ and $b$ to be integers. Thus, $2^a$ can only be a power of $2$ . To help us see the next step, we factorize $1000$ $\dfrac{2^35^3}b=2^a.$ It should be clear that $a$ must be from $1$ to $3$ ; when $a=1$ $b=500$ ; when $a=2$ $b=250$ ; and finally, when $a=3$ $b=125.$ We sum all the pairs to get $\boxed{881}.$ | null | 881 |
05abafbd2e48c7a5241763b34c7e6eeb | https://artofproblemsolving.com/wiki/index.php/2013_AIME_II_Problems/Problem_3 | A large candle is $119$ centimeters tall. It is designed to burn down more quickly when it is first lit and more slowly as it approaches its bottom. Specifically, the candle takes $10$ seconds to burn down the first centimeter from the top, $20$ seconds to burn down the second centimeter, and $10k$ seconds to burn down the $k$ -th centimeter. Suppose it takes $T$ seconds for the candle to burn down completely. Then $\tfrac{T}{2}$ seconds after it is lit, the candle's height in centimeters will be $h$ . Find $10h$ | We find that $T=10(1+2+\cdots +119)$ . From Gauss's formula, we find that the value of $T$ is $10(7140)=71400$ . The value of $\frac{T}{2}$ is therefore $35700$ . We find that $35700$ is $10(3570)=10\cdot \frac{k(k+1)}{2}$ , so $3570=\frac{k(k+1)}{2}$ . As a result, $7140=k(k+1)$ , which leads to $0=k^2+k-7140$ . We notice that $k=84$ , so the answer is $10(119-84)=\boxed{350}$ | null | 350 |
da8e49c48b00ca60db7e57052fd2964b | https://artofproblemsolving.com/wiki/index.php/2013_AIME_II_Problems/Problem_4 | In the Cartesian plane let $A = (1,0)$ and $B = \left( 2, 2\sqrt{3} \right)$ . Equilateral triangle $ABC$ is constructed so that $C$ lies in the first quadrant. Let $P=(x,y)$ be the center of $\triangle ABC$ . Then $x \cdot y$ can be written as $\tfrac{p\sqrt{q}}{r}$ , where $p$ and $r$ are relatively prime positive integers and $q$ is an integer that is not divisible by the square of any prime. Find $p+q+r$ | The distance from point $A$ to point $B$ is $\sqrt{13}$ . The vector that starts at point A and ends at point B is given by $B - A = (1, 2\sqrt{3})$ . Since the center of an equilateral triangle, $P$ , is also the intersection of the perpendicular bisectors of the sides of the triangle, we need first find the equation for the perpendicular bisector to $\overline{AB}$ . The line perpendicular to $\overline{AB}$ through the midpoint, $M = \left(\dfrac{3}{2},\sqrt{3}\right)$ $\overline{AB}$ can be parameterized by $\left(\dfrac{2\sqrt{3}}{\sqrt{13}}, \dfrac{-1}{\sqrt{13}}\right) t + \left(\dfrac{3}{2},\sqrt{3}\right)$ . At this point, it is useful to note that $\Delta BMP$ is a 30-60-90 triangle with $\overline{MB}$ measuring $\dfrac{\sqrt{13}}{2}$ . This yields the length of $\overline{MP}$ to be $\dfrac{\sqrt{13}}{2\sqrt{3}}$ . Therefore, $P =\left(\dfrac{2\sqrt{3}}{\sqrt{13}},\dfrac{-1}{\sqrt{13}}\right)\left(\dfrac{\sqrt{13}}{2\sqrt{3}}\right) + \left(\dfrac{3}{2},\sqrt{3}\right) = \left(\dfrac{5}{2}, \dfrac{5}{2\sqrt{3}}\right)$ . Therefore $xy = \dfrac{25\sqrt{3}}{12}$ yielding an answer of $p + q + r = 25 + 2 + 12 = \boxed{040}$ | null | 040 |
da8e49c48b00ca60db7e57052fd2964b | https://artofproblemsolving.com/wiki/index.php/2013_AIME_II_Problems/Problem_4 | In the Cartesian plane let $A = (1,0)$ and $B = \left( 2, 2\sqrt{3} \right)$ . Equilateral triangle $ABC$ is constructed so that $C$ lies in the first quadrant. Let $P=(x,y)$ be the center of $\triangle ABC$ . Then $x \cdot y$ can be written as $\tfrac{p\sqrt{q}}{r}$ , where $p$ and $r$ are relatively prime positive integers and $q$ is an integer that is not divisible by the square of any prime. Find $p+q+r$ | Rather than considering the Cartesian plane, we use complex numbers. Thus A is 1 and B is $2 + 2\sqrt{3}i$
Recall that a rotation of $\theta$ radians counterclockwise is equivalent to multiplying a complex number by $e^{i\theta}$ , but here we require a clockwise rotation, so we multiply by $e^{-\frac{i\pi}{3}}$ to obtain C. Upon averaging the coordinates of A, B, and C, we obtain the coordinates of P, viz. $\left(\frac{5}{2}, \frac{5\sqrt{3}}{6}\right)$
Therefore $xy$ is $\frac{25\sqrt{3}}{12}$ and the answer is $25 + 12 + 3 = \boxed{040}$ | null | 040 |
da8e49c48b00ca60db7e57052fd2964b | https://artofproblemsolving.com/wiki/index.php/2013_AIME_II_Problems/Problem_4 | In the Cartesian plane let $A = (1,0)$ and $B = \left( 2, 2\sqrt{3} \right)$ . Equilateral triangle $ABC$ is constructed so that $C$ lies in the first quadrant. Let $P=(x,y)$ be the center of $\triangle ABC$ . Then $x \cdot y$ can be written as $\tfrac{p\sqrt{q}}{r}$ , where $p$ and $r$ are relatively prime positive integers and $q$ is an integer that is not divisible by the square of any prime. Find $p+q+r$ | We can also consider the slopes of the lines. Midpoint $M$ of $AB$ has coordinates $\left(\frac{3}{2},\ \sqrt{3}\right)$ . Because line $AB$ has slope $2\sqrt{3}$ , the slope of line $MP$ is $-\frac{1}{2\sqrt{3}}$ (Because of perpendicular slopes).
Since $\Delta ABC$ is equilateral, and since point $P$ is the centroid, we can quickly calculate that $MP = \frac{\sqrt{39}}{6}$ . Then, define $\Delta x$ and $\Delta y$ to be the differences between points $M$ and $P$ . Because of the slope, it is clear that $\Delta x = 2\sqrt{3} \Delta y$
We can then use the Pythagorean Theorem on line segment $MP$ $MP^2 = \Delta x^2 + \Delta y^2$ yields $\Delta y = -\frac{1}{2\sqrt{3}}$ and $\Delta x = 1$ , after substituting $\Delta x$ . The coordinates of P are thus $\left(\frac{5}{2},\ \frac{5\sqrt{3}}{6}\right)$ . Multiplying these together gives us $\frac{25\sqrt{3}}{12}$ , giving us $\boxed{040}$ as our answer. | null | 040 |
da8e49c48b00ca60db7e57052fd2964b | https://artofproblemsolving.com/wiki/index.php/2013_AIME_II_Problems/Problem_4 | In the Cartesian plane let $A = (1,0)$ and $B = \left( 2, 2\sqrt{3} \right)$ . Equilateral triangle $ABC$ is constructed so that $C$ lies in the first quadrant. Let $P=(x,y)$ be the center of $\triangle ABC$ . Then $x \cdot y$ can be written as $\tfrac{p\sqrt{q}}{r}$ , where $p$ and $r$ are relatively prime positive integers and $q$ is an integer that is not divisible by the square of any prime. Find $p+q+r$ | Since $AC$ will be segment $AB$ rotated clockwise $60^{\circ}$ , we can use a rotation matrix to find $C$ . We first translate the triangle $1$ unit to the left, so $A'$ lies on the origin, and $B' = (1, 2\sqrt{3})$ . Rotating clockwise $60^{\circ}$ is the same as rotating $300^{\circ}$ counter-clockwise, so our rotation matrix is $\begin{bmatrix} \cos{300^{\circ}} & -\sin{300^{\circ}}\\ \sin{300^{\circ}} & \cos{300^{\circ}}\\ \end{bmatrix} = \begin{bmatrix} \frac{1}{2} & \frac{\sqrt{3}}{2}\\ -\frac{\sqrt{3}}{2} & \frac{1}{2}\\ \end{bmatrix}$ . Then $C' = \begin{bmatrix} \frac{1}{2} & \frac{\sqrt{3}}{2}\\ -\frac{\sqrt{3}}{2} & \frac{1}{2}\\ \end{bmatrix} \cdot \begin{bmatrix} 1\\ 2\sqrt{3}\\ \end{bmatrix} = \begin{bmatrix} \frac{7}{2}\\ \frac{\sqrt{3}}{2}\\ \end{bmatrix}$ . Thus, $C = (\frac{9}{2}, \frac{\sqrt{3}}{2})$ . Since the triangle is equilateral, the center of the triangle is the average of the coordinates of the vertices. Then $P = (\frac{1 + 2 + \frac{9}{2}}{3}, \frac{2\sqrt{3} + \frac{\sqrt{3}}{2}}{3}) = (\frac{5}{2}, \frac{5\sqrt{3}}{6})$ . Our answer is $\frac{5}{2} \cdot \frac{5\sqrt{3}}{6} = \frac{25\sqrt{3}}{12}$ $25 + 3 + 12 = \boxed{40}$ | null | 40 |
da8e49c48b00ca60db7e57052fd2964b | https://artofproblemsolving.com/wiki/index.php/2013_AIME_II_Problems/Problem_4 | In the Cartesian plane let $A = (1,0)$ and $B = \left( 2, 2\sqrt{3} \right)$ . Equilateral triangle $ABC$ is constructed so that $C$ lies in the first quadrant. Let $P=(x,y)$ be the center of $\triangle ABC$ . Then $x \cdot y$ can be written as $\tfrac{p\sqrt{q}}{r}$ , where $p$ and $r$ are relatively prime positive integers and $q$ is an integer that is not divisible by the square of any prime. Find $p+q+r$ | We construct point $C$ by drawing two circles with radius $r = AB = \sqrt{13}$ . One circle will be centered at $A$ , while the other is centered at $B$ . The equations of the circles are:
$(x - 1)^2 + y^2 = 13$
$(x - 2)^2 + (y - 2\sqrt{3})^2 = 13$
Setting the LHS of each of these equations equal to each other and solving for $x$ yields after simplification:
$x = \frac{15}{2} - 2\sqrt{3}y$
Plugging that into the first equation gives the following quadratic in $y$ after simplification:
$y^2 - 2\sqrt{3}y + \frac{9}{4} = 0$
The quadratic formula gives $y = \frac{\sqrt{3}}{2}, \frac{3\sqrt{3}}{2}$
Since $x > 0$ and $x = \frac{15}{2} - 2\sqrt{3}y$ , we pick $y = \frac{\sqrt{3}}{2}$ in the hopes that it will give $x > 0$ . Plugging $y$ into the equation for $x$ yields $x = \frac{9}{2}$
Thus, $C(\frac{9}{2}, \frac{\sqrt{3}}{2})$ . Averaging the coordinates of the vertices of equilateral triangle $ABC$ will give the center of mass of the triangle.
Thus, $P(\frac{5}{2}, \frac{5\sqrt{3}}{6})$ , and the product of the coordinates is $\frac{25\sqrt{3}}{12}$ , so the desired quantity is $\boxed{040}$ | null | 040 |
da8e49c48b00ca60db7e57052fd2964b | https://artofproblemsolving.com/wiki/index.php/2013_AIME_II_Problems/Problem_4 | In the Cartesian plane let $A = (1,0)$ and $B = \left( 2, 2\sqrt{3} \right)$ . Equilateral triangle $ABC$ is constructed so that $C$ lies in the first quadrant. Let $P=(x,y)$ be the center of $\triangle ABC$ . Then $x \cdot y$ can be written as $\tfrac{p\sqrt{q}}{r}$ , where $p$ and $r$ are relatively prime positive integers and $q$ is an integer that is not divisible by the square of any prime. Find $p+q+r$ | Labeling our points and sketching a graph we get that $C$ is to the right of $AB$ . Of course, we need to find $C$ . Note that the transformation from $A$ to $B$ is $[1,2\sqrt{3}]$ , and if we imagine a height dropped to $AB$ we see that a transformation from the midpoint $(\frac{3}{2},\sqrt {3})$ to $C$ is basically the first transformation, with $\frac{\sqrt{3}}{2}$ the magnitude and the x and y switched– then multiply the new y by -1. Then, applying this transformation of $[3,\frac{-\sqrt{3}}{2}]$ we get that $C=(\frac{9}{2},\frac{\sqrt{3}}{2})$ which means that $P=(\frac{5}{2},\frac{5\sqrt{3}}{6})$ . Then our answer is $\boxed{40}$ | null | 40 |
7a74a436a0fe981fa352ce32fde73358 | https://artofproblemsolving.com/wiki/index.php/2013_AIME_II_Problems/Problem_5 | In equilateral $\triangle ABC$ let points $D$ and $E$ trisect $\overline{BC}$ . Then $\sin(\angle DAE)$ can be expressed in the form $\frac{a\sqrt{b}}{c}$ , where $a$ and $c$ are relatively prime positive integers, and $b$ is an integer that is not divisible by the square of any prime. Find $a+b+c$ | [asy] pair A = (1, sqrt(3)), B = (0, 0), C= (2, 0); pair M = (1, 0); pair D = (2/3, 0), E = (4/3, 0); draw(A--B--C--cycle); label("$A$", A, N); label("$B$", B, SW); label("$C$", C, SE); label("$D$", D, S); label("$M$", M, S); label("$E$", E, S); draw(A--D); draw(A--M); draw(A--E); [/asy]
Without loss of generality, assume the triangle sides have length 3. Then the trisected side is partitioned into segments of length 1, making your computation easier.
Let $M$ be the midpoint of $\overline{DE}$ . Then $\Delta MCA$ is a 30-60-90 triangle with $MC = \dfrac{3}{2}$ $AC = 3$ and $AM = \dfrac{3\sqrt{3}}{2}$ . Since the triangle $\Delta AME$ is right, then we can find the length of $\overline{AE}$ by pythagorean theorem, $AE = \sqrt{7}$ . Therefore, since $\Delta AME$ is a right triangle, we can easily find $\sin(\angle EAM) = \dfrac{1}{2\sqrt{7}}$ and $\cos(\angle EAM) = \sqrt{1-\sin(\angle EAM)^2}=\dfrac{3\sqrt{3}}{2\sqrt{7}}$ . So we can use the double angle formula for sine, $\sin(\angle EAD) = 2\sin(\angle EAM)\cos(\angle EAM) = \dfrac{3\sqrt{3}}{14}$ . Therefore, $a + b + c = \boxed{020}$ | null | 020 |
7a74a436a0fe981fa352ce32fde73358 | https://artofproblemsolving.com/wiki/index.php/2013_AIME_II_Problems/Problem_5 | In equilateral $\triangle ABC$ let points $D$ and $E$ trisect $\overline{BC}$ . Then $\sin(\angle DAE)$ can be expressed in the form $\frac{a\sqrt{b}}{c}$ , where $a$ and $c$ are relatively prime positive integers, and $b$ is an integer that is not divisible by the square of any prime. Find $a+b+c$ | We find that, as before, $AE = \sqrt{7}$ , and also the area of $\Delta DAE$ is 1/3 the area of $\Delta ABC$ . Thus, using the area formula, $1/2 \cdot 7 \cdot \sin(\angle EAD) = 3\sqrt{3}/4$ , and $\sin(\angle EAD) = \dfrac{3\sqrt{3}}{14}$ . Therefore, $a + b + c = \boxed{020}.$ | null | 020 |
7a74a436a0fe981fa352ce32fde73358 | https://artofproblemsolving.com/wiki/index.php/2013_AIME_II_Problems/Problem_5 | In equilateral $\triangle ABC$ let points $D$ and $E$ trisect $\overline{BC}$ . Then $\sin(\angle DAE)$ can be expressed in the form $\frac{a\sqrt{b}}{c}$ , where $a$ and $c$ are relatively prime positive integers, and $b$ is an integer that is not divisible by the square of any prime. Find $a+b+c$ | Let A be the origin of the complex plane, B be $1+i\sqrt{3}$ , and C be $2$ . Also, WLOG, let D have a greater imaginary part than E. Then, D is $\frac{4}{3}+\frac{2i\sqrt{3}}{3}$ and E is $\frac{5}{3}+\frac{i\sqrt{3}}{3}$ . Then, $\sin(\angle DAE) = Im\left(\dfrac{\frac{4}{3}+\frac{2i\sqrt{3}}{3}}{ \frac{5}{3}+\frac{i\sqrt{3}}{3}}\right) = Im\left(\frac{26+6i\sqrt{3}}{28}\right) = \frac{3\sqrt{3}}{14}$ . Therefore, $a + b + c = \boxed{020}$ | null | 020 |
7a74a436a0fe981fa352ce32fde73358 | https://artofproblemsolving.com/wiki/index.php/2013_AIME_II_Problems/Problem_5 | In equilateral $\triangle ABC$ let points $D$ and $E$ trisect $\overline{BC}$ . Then $\sin(\angle DAE)$ can be expressed in the form $\frac{a\sqrt{b}}{c}$ , where $a$ and $c$ are relatively prime positive integers, and $b$ is an integer that is not divisible by the square of any prime. Find $a+b+c$ | Without loss of generality, say that the side length of triangle ABC is 3. EC is 1 and by the law of cosines, $AE^2=1+3^2-2(1)(3)\cos(\angle DAE)$ or $AE=\sqrt7$ The same goes for AD. DE equals 1 because AD and AE trisect BC. By the law of cosines, $\cos(\angle DAE)=(1-7-7)/-2(\sqrt7)(\sqrt7)=13/14$ Since $\sin^2(\angle DAE)=1-cos^2(\angle DAE)$ Then $\sin^2(\angle DAE)= 1-\frac{169}{196}=\frac{27}{196}$ So $\sin(\angle DAE)=\frac{3\sqrt{3}}{14}$ . Therefore, $a + b + c = \boxed{020}$ | null | 020 |
7a74a436a0fe981fa352ce32fde73358 | https://artofproblemsolving.com/wiki/index.php/2013_AIME_II_Problems/Problem_5 | In equilateral $\triangle ABC$ let points $D$ and $E$ trisect $\overline{BC}$ . Then $\sin(\angle DAE)$ can be expressed in the form $\frac{a\sqrt{b}}{c}$ , where $a$ and $c$ are relatively prime positive integers, and $b$ is an integer that is not divisible by the square of any prime. Find $a+b+c$ | Setting up a convinient coordinate system, we let $A$ be at point $(0, 0)$ $B$ be at point $(3, 3\sqrt3)$ , and $C$ be at point $(6, 0)$ . Then $D$ and $E$ will be at points $(4, 2\sqrt3)$ and $(5, \sqrt3)$ . Then $\cos(\angle DAE) = \frac{\vec{AD}\cdot\vec{AE}}{\|\vec{AD}\| \|\vec{AE}\|} = \frac{4\cdot5 + 2\sqrt{3}\cdot\sqrt{3}}{28}=\frac{13}{14}$ . From here, we see that $\sin(\angle DAE) = \sqrt{1-\cos^2(\angle DAE)} = \frac{3\sqrt3}{14}\Longrightarrow\boxed{020}$ | null | 020 |
74495eb5ac70b29fe931a183ef4f5c26 | https://artofproblemsolving.com/wiki/index.php/2013_AIME_II_Problems/Problem_6 | Find the least positive integer $N$ such that the set of $1000$ consecutive integers beginning with $1000\cdot N$ contains no square of an integer. | The difference between consecutive integral squares must be greater than 1000. $(x+1)^2-x^2\geq1000$ , so $x\geq\frac{999}{2}\implies x\geq500$ $x=500$ does not work, so $x>500$ . Let $n=x-500$ . By inspection, $n^2$ should end in a number close to but less than 1000 such that there exists $1000N$ within the difference of the two squares. Examine when $n^2=1000$ . Then, $n=10\sqrt{10}$ . One example way to estimate $\sqrt{10}$ follows.
Then, $n\approx 31.6$ $n^2<1000$ , so $n$ could be $31$ . Add 500 to get the first square and 501 to get the second. Then, the two integral squares are $531^2$ and $532^2$ . Checking, $531^2=281961$ and $532^2=283024$ $282,000$ straddles the two squares, which have a difference of 1063. The difference has been minimized, so $N$ is minimized $N=282000\implies\boxed{282}$ | null | 282 |
74495eb5ac70b29fe931a183ef4f5c26 | https://artofproblemsolving.com/wiki/index.php/2013_AIME_II_Problems/Problem_6 | Find the least positive integer $N$ such that the set of $1000$ consecutive integers beginning with $1000\cdot N$ contains no square of an integer. | Let us first observe the difference between $x^2$ and $(x+1)^2$ , for any arbitrary $x\ge 0$ $(x+1)^2-x^2=2x+1$ . So that means for every $x\ge 0$ , the difference between that square and the next square have a difference of $2x+1$ . Now, we need to find an $x$ such that $2x+1\ge 1000$ . Solving gives $x\ge \frac{999}{2}$ , so $x\ge 500$ . Now we need to find what range of numbers has to be square-free: $\overline{N000}\rightarrow \overline{N999}$ have to all be square-free.
Let us first plug in a few values of $x$ to see if we can figure anything out. $x=500$ $x^2=250000$ , and $(x+1)^2=251001$ . Notice that this does not fit the criteria, because $250000$ is a square, whereas $\overline{N000}$ cannot be a square. This means, we must find a square, such that the last $3$ digits are close to $1000$ , but not there, such as $961$ or $974$ . Now, the best we can do is to keep on listing squares until we hit one that fits. We do not need to solve for each square: remember that the difference between consecutive squares are $2x+1$ , so all we need to do is addition. After making a list, we find that $531^2=281961$ , while $532^2=283024$ . It skipped $282000$ , so our answer is $\boxed{282}$ | null | 282 |
74495eb5ac70b29fe931a183ef4f5c26 | https://artofproblemsolving.com/wiki/index.php/2013_AIME_II_Problems/Problem_6 | Find the least positive integer $N$ such that the set of $1000$ consecutive integers beginning with $1000\cdot N$ contains no square of an integer. | Let $x$ be the number being squared. Based on the reasoning above, we know that $N$ must be at least $250$ , so $x$ has to be at least $500$ . Let $k$ be $x-500$ . We can write $x^2$ as $(500+k)^2$ , or $250000+1000k+k^2$ . We can disregard $250000$ and $1000k$ , since they won't affect the last three digits, which determines if there are any squares between $\overline{N000}\rightarrow \overline{N999}$ . So we must find a square, $k^2$ , such that it is under $1000$ , but the next square is over $1000$ . We find that $k=31$ gives $k^2=961$ , and so $(k+1)^2=32^2=1024$ . We can be sure that this skips a thousand because the $1000k$ increments it up $1000$ each time. Now we can solve for $x$ $(500+31)^2=281961$ , while $(500+32)^2=283024$ . We skipped $282000$ , so the answer is $\boxed{282}$ | null | 282 |
74495eb5ac70b29fe931a183ef4f5c26 | https://artofproblemsolving.com/wiki/index.php/2013_AIME_II_Problems/Problem_6 | Find the least positive integer $N$ such that the set of $1000$ consecutive integers beginning with $1000\cdot N$ contains no square of an integer. | The goal is to find the least $N \in \mathbb{N}$ such that $\exists m \in \mathbb{N}$ where $m^2 + 1 \leq 1000N, 1000N + 1000 \leq (m+1)^2$
Combining the two inequalities leads to $(m+1)^2 \geq m^2 + 1001, m \geq 500$
Let $m = k + 500$ , where $k \in \mathbb{W}$ , then the inequalities become,
$N \geq \frac{(k+500)^2 + 1}{1000} = \frac{k^2 + 1}{1000} + k + 250$ , and
$N \leq \frac{(k+501)^2}{1000} - 1 = \frac{(k+1)^2}{1000} + k + 250.$
For $k=31$ , one can verify that $N = 282$ is the unique integer satisfying the inequalities.
For $k \leq 30$ $k + 250 < \frac{k^2 + 1}{1000} + k + 250 \leq N$ $\leq \frac{(k+1)^2}{1000} + k + 250 \leq \frac{(30+1)^2}{1000} + k + 250 < k + 251$
i.e., $k + 250 < N < k + 251$ , a contradiction.
Note $k \geq 32$ leads to larger $N$ (s).
Hence, the answer is $\boxed{282}$ | null | 282 |
5ab15aa29563ce91d36cde787d04525c | https://artofproblemsolving.com/wiki/index.php/2013_AIME_II_Problems/Problem_7 | A group of clerks is assigned the task of sorting $1775$ files. Each clerk sorts at a constant rate of $30$ files per hour. At the end of the first hour, some of the clerks are reassigned to another task; at the end of the second hour, the same number of the remaining clerks are also reassigned to another task, and a similar assignment occurs at the end of the third hour. The group finishes the sorting in $3$ hours and $10$ minutes. Find the number of files sorted during the first one and a half hours of sorting. | There are $x$ clerks at the beginning, and $t$ clerks are reassigned to another task at the end of each hour. So, $30x+30(x-t)+30(x-2t)+30\cdot\frac{10}{60} \cdot (x-3t)=1775$ , and simplify that we get $19x-21t=355$ .
Now the problem is to find a reasonable integer solution. Now we know $x= \frac{355+21t}{19}$ , so $19$ divides $355+21t$ , AND as long as $t$ is a integer, $19$ must divide $2t+355$ . Now, we suppose that $19m=2t+355$ , similarly we get $t=\frac{19m-355}{2}$ , and so in order to get a minimum integer solution for $t$ , it is obvious that $m=19$ works. So we get $t=3$ and $x=22$ . One and a half hour's work should be $30x+15(x-t)$ , so the answer is $\boxed{945}$ | null | 945 |
5ab15aa29563ce91d36cde787d04525c | https://artofproblemsolving.com/wiki/index.php/2013_AIME_II_Problems/Problem_7 | A group of clerks is assigned the task of sorting $1775$ files. Each clerk sorts at a constant rate of $30$ files per hour. At the end of the first hour, some of the clerks are reassigned to another task; at the end of the second hour, the same number of the remaining clerks are also reassigned to another task, and a similar assignment occurs at the end of the third hour. The group finishes the sorting in $3$ hours and $10$ minutes. Find the number of files sorted during the first one and a half hours of sorting. | We start with the same approach as solution 1 to get $19x-21t=355$ . Then notice that $21t + 355 \equiv 0 \pmod{19}$ , or $2t-6 \equiv 0 \pmod{19}$ , giving the smallest solution at $t=3$ . We find that $x=22$ . Then the number of files they sorted will be $30x+15(x-t)=660+285=\boxed{945}.$ | null | 945 |
1ea3c9b7ff7732e4063360616c08c6ff | https://artofproblemsolving.com/wiki/index.php/2013_AIME_II_Problems/Problem_8 | A hexagon that is inscribed in a circle has side lengths $22$ $22$ $20$ $22$ $22$ , and $20$ in that order. The radius of the circle can be written as $p+\sqrt{q}$ , where $p$ and $q$ are positive integers. Find $p+q$ | Let us call the hexagon $ABCDEF$ , where $AB=CD=DE=AF=22$ , and $BC=EF=20$ .
We can just consider one half of the hexagon, $ABCD$ , to make matters simpler.
Draw a line from the center of the circle, $O$ , to the midpoint of $BC$ $X$ . Now, draw a line from $O$ to the midpoint of $AB$ $Y$ . Clearly, $\angle BXO=90^{\circ}$ , because $BO=CO$ , and $\angle BYO=90^{\circ}$ , for similar reasons. Also notice that $\angle AOX=90^{\circ}$ .
Let us call $\angle BOY=\theta$ . Therefore, $\angle AOB=2\theta$ , and so $\angle BOX=90-2\theta$ . Let us label the radius of the circle $r$ . This means \[\sin{\theta}=\frac{BY}{r}=\frac{11}{r}\] \[\sin{(90-2\theta)}=\frac{BX}{r}=\frac{10}{r}\] Now we can use simple trigonometry to solve for $r$ .
Recall that $\sin{(90-\alpha)}=\cos(\alpha)$ : That means $\sin{(90-2\theta)}=\cos{2\theta}=\frac{10}{r}$ .
Recall that $\cos{2\alpha}=1-2\sin^2{\alpha}$ : That means $\cos{2\theta}=1-2\sin^2{\theta}=\frac{10}{r}$ .
Let $\sin{\theta}=x$ .
Substitute to get $x=\frac{11}{r}$ and $1-2x^2=\frac{10}{r}$ Now substitute the first equation into the second equation: $1-2\left(\frac{11}{r}\right)^2=\frac{10}{r}$ Multiplying both sides by $r^2$ and reordering gives us the quadratic \[r^2-10r-242=0\] Using the quadratic equation to solve, we get that $r=5+\sqrt{267}$ (because $5-\sqrt{267}$ gives a negative value), so the answer is $5+267=\boxed{272}$ | null | 272 |
1ea3c9b7ff7732e4063360616c08c6ff | https://artofproblemsolving.com/wiki/index.php/2013_AIME_II_Problems/Problem_8 | A hexagon that is inscribed in a circle has side lengths $22$ $22$ $20$ $22$ $22$ , and $20$ in that order. The radius of the circle can be written as $p+\sqrt{q}$ , where $p$ and $q$ are positive integers. Find $p+q$ | Using the trapezoid $ABCD$ mentioned above, draw an altitude of the trapezoid passing through point $B$ onto $AD$ at point $J$ . Now, we can use the pythagorean theorem: $(22^2-(r-10)^2)+10^2=r^2$ . Expanding and combining like terms gives us the quadratic \[r^2-10r-242=0\] and solving for $r$ gives $r=5+\sqrt{267}$ . So the solution is $5+267=\boxed{272}$ | null | 272 |
1ea3c9b7ff7732e4063360616c08c6ff | https://artofproblemsolving.com/wiki/index.php/2013_AIME_II_Problems/Problem_8 | A hexagon that is inscribed in a circle has side lengths $22$ $22$ $20$ $22$ $22$ , and $20$ in that order. The radius of the circle can be written as $p+\sqrt{q}$ , where $p$ and $q$ are positive integers. Find $p+q$ | Join the diameter of the circle $AD$ and let the length be $d$ . By Ptolemy's Theorem on trapezoid $ADEF$ $(AD)(EF) + (AF)(DE) = (AE)(DF)$ . Since it is an isosceles trapezoid, both diagonals are equal. Let them be equal to $x$ each. Then
\[20d + 22^2 = x^2\]
Since $\angle AED$ is subtended by the diameter, it is right. Hence by the Pythagorean Theorem with right $\triangle AED$
\[(AE)^2 + (ED)^2 = (AD)^2\] \[x^2 + 22^2 = d^2\]
From the above equations, we have: \[x^2 = d^2 - 22^2 = 20d + 22^2\] \[d^2 - 20d = 2\times22^2\] \[d^2 - 20d + 100 = 968+100 = 1068\] \[(d-10) = \sqrt{1068}\] \[d = \sqrt{1068} + 10 = 2\times(\sqrt{267}+5)\]
Since the radius is half the diameter, it is $\sqrt{267}+5$ , so the answer is $5+267 \Rightarrow \boxed{272}$ | null | 272 |
1ea3c9b7ff7732e4063360616c08c6ff | https://artofproblemsolving.com/wiki/index.php/2013_AIME_II_Problems/Problem_8 | A hexagon that is inscribed in a circle has side lengths $22$ $22$ $20$ $22$ $22$ , and $20$ in that order. The radius of the circle can be written as $p+\sqrt{q}$ , where $p$ and $q$ are positive integers. Find $p+q$ | As we can see this image, it is symmetrical hence the diameter divides the hexagon into two congruent quadrilateral. Now we can apply the Ptolemy's theorem. Denote the radius is r, we can get \[22*2x+440=\sqrt{4x^2-400}\sqrt{4x^2-484}\] , after simple factorization, we can get \[x^4-342x^2-2420x=0\] , it is easy to see that $x=-10, x=0$ are two solutions for the equation, so we can factorize that into \[x(x+10)(x^2-10x-242)\] so we only need to find the solution for \[x^2-10x-242=0\] and we can get $x=(\sqrt{267}+5)$ is the desired answer for the problem, and our answer is $5+267 \Rightarrow \boxed{272}$ .～bluesoul | null | 272 |
1ea3c9b7ff7732e4063360616c08c6ff | https://artofproblemsolving.com/wiki/index.php/2013_AIME_II_Problems/Problem_8 | A hexagon that is inscribed in a circle has side lengths $22$ $22$ $20$ $22$ $22$ , and $20$ in that order. The radius of the circle can be written as $p+\sqrt{q}$ , where $p$ and $q$ are positive integers. Find $p+q$ | Using solution 1's diagram, extend line segments $AB$ and $CD$ upwards until they meet at point $G$ . Let point $O$ be the center of the hexagon. By the $AA$ postulate, $\Delta ADG \sim \Delta BCG \sim \Delta CDO$ . This means $\frac{DC}{AD} = \frac{22}{2r} = \frac{CO}{AG}$ , so $AG = r \times \frac{2r}{22} = \frac{r^2}{11}$ . We then solve for $AB$ \[\frac{AD-BC}{AD} = \frac{2r-20}{2r} = \frac{AB}{AG}\] \[AB = \frac{r^2}{11} \times \frac{2r-20}{2r} = \frac{r^2-10r}{11}\] Remember that $AB=22$ as well, so $22 = \frac{r^2-10r}{11} \Rightarrow r^2-10r-242=0$ . Solving for $r$ gives $r=5+\sqrt{267}$ . So the solution is $5+267=\boxed{272}$ | null | 272 |
1ea3c9b7ff7732e4063360616c08c6ff | https://artofproblemsolving.com/wiki/index.php/2013_AIME_II_Problems/Problem_8 | A hexagon that is inscribed in a circle has side lengths $22$ $22$ $20$ $22$ $22$ , and $20$ in that order. The radius of the circle can be written as $p+\sqrt{q}$ , where $p$ and $q$ are positive integers. Find $p+q$ | Let $\angle{AOB} = \theta$ . So, we have $\sin \dfrac{\theta}{2} = \dfrac{11}{r}$ and $\cos \dfrac{\theta}{2} = \dfrac{\sqrt{r^{2} - 121}}{r}$ . So, $\sin \theta = 2 \sin \dfrac{\theta}{2} \cos \dfrac{\theta}{2} = \dfrac{22 \sqrt{r^{2} - 121}}{r^{2}}$ . Let $H$ be the foot of the perpendicular from $B$ to $\overline{AD}$ . We have $BF = 2 BH = 2 r \sin \theta = \dfrac{44 \sqrt{r^{2} - 121}}{r}$ . Using Pythagorean theorem on $\triangle BCF$ , to get $(\dfrac{44 \sqrt{r^{2} - 121}}{r})^{2} + 20^{2} = (2r)^{2}$ , or $\dfrac{44^{2}r^{2} - 44^{2} \cdot 121}{r^{2}} + 20^{2} = 4r^{4}$ . Multiplying by $r^{2}$ , we get $44^{2} r^{2} - 44^{2} \cdot 121 + 20^{2} r^{2} = 4r^{4}$ . Rearranging and simplifying, we get a quadratic in $r^{2}$ \[r^{4} - 584r^{2} + 242^{2} = 0 \text\] which gives us $r^{2} = 292 \pm 10\sqrt{267}$ . Because $r$ is in the form $p + \sqrt{q}$ , we know to choose the larger option, meaning $r^2 = 292 + 10\sqrt{267}$ , so $p\sqrt{q} = 5\sqrt{267}$ and $p^2 + q = 292$ . By inspection, we get $(p, q) = (5, 267)$ , so our answer is $5 + 267 = \boxed{272}$ | null | 272 |
1ea3c9b7ff7732e4063360616c08c6ff | https://artofproblemsolving.com/wiki/index.php/2013_AIME_II_Problems/Problem_8 | A hexagon that is inscribed in a circle has side lengths $22$ $22$ $20$ $22$ $22$ , and $20$ in that order. The radius of the circle can be written as $p+\sqrt{q}$ , where $p$ and $q$ are positive integers. Find $p+q$ | [asy] import olympiad; import math; real a; a=2*asin(11/(5+sqrt(267))); pair A,B,C,D,E,F; A=expi(pi); B=expi(pi-a); C=expi(a); D=expi(0); E=expi(-a); F=expi(pi+a); draw(A--B--C--D--E--F--A--D); draw(B--D); draw(A--C); dot(A); dot(B); dot(C); dot(D); dot(E); dot(F); draw(unitcircle); label("$A$",A,W);label("$B$",B,NW);label("$C$",C,NE);label("$D$",D,dir(0));label("$E$",E,SE);label("$F$",F,SW); [/asy]
We know that $AD=x$ is a diameter, hence $ABD$ and $ACD$ are right triangles. Let $AB=BC=22$ , and $CD=20.$ Hence, $ABD$ is a right triangle with legs $22,\sqrt{x^2-484},$ and hypotenuse, $x,$ and $ACD$ is a right triangle with legs $20, \sqrt{x^2-400},$ with hypotenuse $x$ . By Ptolemy's we have \[22(x+20)=\sqrt{x^2-400}\sqrt{x^2-484}\] .
We square both sides to get \[484(x+20)^2=(x^2-400)(x^2-484) \implies 484(x+20)=(x-20)(x^2-484) \implies 484x=x^3-20x^2-484x \implies x(x^2-20x-968)=0\]
We solve for $x$ via the Quadratic Formula and receive $x=10+2\sqrt{267}$ , but we must divide by $2$ since we want the radius, and hence $267+5=\boxed{272}.$ ~SirAppel | null | 272 |
64710504ce1d2bd66a997719d3eec6e2 | https://artofproblemsolving.com/wiki/index.php/2013_AIME_II_Problems/Problem_9 | $7\times 1$ board is completely covered by $m\times 1$ tiles without overlap; each tile may cover any number of consecutive squares, and each tile lies completely on the board. Each tile is either red, blue, or green. Let $N$ be the number of tilings of the $7\times 1$ board in which all three colors are used at least once. For example, a $1\times 1$ red tile followed by a $2\times 1$ green tile, a $1\times 1$ green tile, a $2\times 1$ blue tile, and a $1\times 1$ green tile is a valid tiling. Note that if the $2\times 1$ blue tile is replaced by two $1\times 1$ blue tiles, this results in a different tiling. Find the remainder when $N$ is divided by $1000$ | Firstly, we consider how many different ways possible to divide the $7\times 1$ board.
We ignore the cases of 1 or 2 pieces since we need at least one tile of each color.
Secondly, we use Principle of Inclusion-Exclusion to consider how many ways to color them:
Finally, we combine them together: $15\times 6+20\times 36+15\times 150+6\times 540+1\times 1806= 8106$
So the answer is $\boxed{106}$ | null | 106 |
64710504ce1d2bd66a997719d3eec6e2 | https://artofproblemsolving.com/wiki/index.php/2013_AIME_II_Problems/Problem_9 | $7\times 1$ board is completely covered by $m\times 1$ tiles without overlap; each tile may cover any number of consecutive squares, and each tile lies completely on the board. Each tile is either red, blue, or green. Let $N$ be the number of tilings of the $7\times 1$ board in which all three colors are used at least once. For example, a $1\times 1$ red tile followed by a $2\times 1$ green tile, a $1\times 1$ green tile, a $2\times 1$ blue tile, and a $1\times 1$ green tile is a valid tiling. Note that if the $2\times 1$ blue tile is replaced by two $1\times 1$ blue tiles, this results in a different tiling. Find the remainder when $N$ is divided by $1000$ | This solution is basically solution 1 with more things done at once. The game plan:
$\sum_{i=0}^{7} ($ the amount of ways to divide the board into $i$ pieces $) \cdot ($ the amount of ways to color the respective divisions)
The amount of ways to divide the board is determined using stars and bars. The colorings are found using PIE giving $3^i-3\cdot2^i+3$ . Plus, we don't have to worry about the cases where $i=1$ or $I=2$ since they both give no solutions. So our equation becomes:
$\sum_{i=3}^{7} \left(\dbinom{6}{I}\right)\cdot\left(3^i-3\cdot2^i+3\right)$
Writing it all out and keeping the numbers small with mod 1000, we will eventually arrive at the answer of $\boxed{106}$ | null | 106 |
64710504ce1d2bd66a997719d3eec6e2 | https://artofproblemsolving.com/wiki/index.php/2013_AIME_II_Problems/Problem_9 | $7\times 1$ board is completely covered by $m\times 1$ tiles without overlap; each tile may cover any number of consecutive squares, and each tile lies completely on the board. Each tile is either red, blue, or green. Let $N$ be the number of tilings of the $7\times 1$ board in which all three colors are used at least once. For example, a $1\times 1$ red tile followed by a $2\times 1$ green tile, a $1\times 1$ green tile, a $2\times 1$ blue tile, and a $1\times 1$ green tile is a valid tiling. Note that if the $2\times 1$ blue tile is replaced by two $1\times 1$ blue tiles, this results in a different tiling. Find the remainder when $N$ is divided by $1000$ | 3 colors is a lot. How many ways can we tile an $n \times 1$ board with one color? It's going to be $2^{n-1}$ because if you draw out the $n$ spaces, you can decide whether each of the borders between the tiles are either there or not there. There are $n-1$ borders so there are $2^{n-1}$ tilings. Define a one-tiling of an mx1 as $f_1(m)$
Now let's look at two colors. Let's see if we can get a two-tiling of an $(n+1) \times 1$ based off a $n \times 1$ . There are 2 cases we should consider:
1) The $n \times 1$ was a one-coloring and the block we are going to add consists of the second color. The number of ways we can do this is $2f_1(n)$
2) The $n \times 1$ was a two-color tiling so now we've got 3 cases to form the $(n+1) \times 1$ : We can either add a block of the first color, the second color, or we can adjoin a block to the last block in the $n \times 1$
This gives us $f_2(n+1)=2f_1(n)+3f_2(n)$
Time to tackle the 3-color tiling. Again, we split into 2 cases:
1) The $n \times 1$ was a two-color tiling, and the block we are adding is of the 3rd color. This gives $f_2(n)$ ways but we have to multiply by $3C2 = 3$ because we have to pick 2 different colors for the two-color tiling.
2) The $n \times 1$ was a 3-color tiling, and we have to consider what we can do with the block that we are adding. It can be any of the 3 colors, or we can adjoin it to whatever was the last block in the $n \times 1$ . This gives $4f_3(n)$
So in total we have $f_3(n+1)=3f_2(n)+4f_3(n)$
Finally, we just sorta bash through the computation to get $f_3(7)=8\boxed{106}$ | null | 106 |
64710504ce1d2bd66a997719d3eec6e2 | https://artofproblemsolving.com/wiki/index.php/2013_AIME_II_Problems/Problem_9 | $7\times 1$ board is completely covered by $m\times 1$ tiles without overlap; each tile may cover any number of consecutive squares, and each tile lies completely on the board. Each tile is either red, blue, or green. Let $N$ be the number of tilings of the $7\times 1$ board in which all three colors are used at least once. For example, a $1\times 1$ red tile followed by a $2\times 1$ green tile, a $1\times 1$ green tile, a $2\times 1$ blue tile, and a $1\times 1$ green tile is a valid tiling. Note that if the $2\times 1$ blue tile is replaced by two $1\times 1$ blue tiles, this results in a different tiling. Find the remainder when $N$ is divided by $1000$ | Let $n$ be the length of the board and $x$ be the number of colors. We will find the number of ways to tile the $n \times 1$ board with no color restrictions (some colors may be unused) and then use PIE.
By stars and bars, the number of ways to divide the board into $k$ pieces is ${n-1 \choose k-1}$ . There are $x^k$ ways to color each of these divisions. Therefore, the total number of ways to divide and color the board is \begin{align*} \chi(n, x) &:= \sum_{k=1}^n {n-1 \choose k-1} x^k \\ &= x\sum_{k=0}^{n-1} {n-1 \choose k} x^k \\ &= x(x+1)^{n-1}. \end{align*}
In the given problem, we have $n=7$ . By PIE, we have \begin{align*} &\quad {3 \choose 3} \chi(7, 3) - {3 \choose 2} \, \chi(7, 2) + {3 \choose 1} \, \chi(7, 1) \\ &= 3 \cdot 4^6 - 3(2 \cdot 3^6) + 3(1 \cdot 2^6) \\ &= 8106 \rightarrow \boxed{106} | null | 106 |
ded6b33a1066af211ee912e8bb536d09 | https://artofproblemsolving.com/wiki/index.php/2013_AIME_II_Problems/Problem_10 | Given a circle of radius $\sqrt{13}$ , let $A$ be a point at a distance $4 + \sqrt{13}$ from the center $O$ of the circle. Let $B$ be the point on the circle nearest to point $A$ . A line passing through the point $A$ intersects the circle at points $K$ and $L$ . The maximum possible area for $\triangle BKL$ can be written in the form $\frac{a - b\sqrt{c}}{d}$ , where $a$ $b$ $c$ , and $d$ are positive integers, $a$ and $d$ are relatively prime, and $c$ is not divisible by the square of any prime. Find $a+b+c+d$ | [asy] import math; import olympiad; import graph; pair A, B, K, L; B = (sqrt(13), 0); A=(4+sqrt(13), 0); dot(B); dot(A); draw(Circle((0,0), sqrt(13))); label("$O$", (0,0), S);label("$B$", B, SW);label("$A$", A, S); dot((0,0)); [/asy]
Now we put the figure in the Cartesian plane, let the center of the circle $O (0,0)$ , then $B (\sqrt{13},0)$ , and $A(4+\sqrt{13},0)$
The equation for Circle O is $x^2+y^2=13$ , and let the slope of the line $AKL$ be $k$ , then the equation for line $AKL$ is $y=k(x-4-\sqrt{13})$
Then we get $(k^2+1)x^2-2k^2(4+\sqrt{13})x+k^2\cdot (4+\sqrt{13})^2-13=0$ . According to Vieta's Formulas , we get
$x_1+x_2=\frac{2k^2(4+\sqrt{13})}{k^2+1}$ , and $x_1x_2=\frac{(4+\sqrt{13})^2\cdot k^2-13}{k^2+1}$
So, $LK=\sqrt{1+k^2}\cdot \sqrt{(x_1+x_2)^2-4x_1x_2}$
Also, the distance between $B$ and $LK$ is $\frac{k\times \sqrt{13}-(4+\sqrt{13})\cdot k}{\sqrt{1+k^2}}=\frac{-4k}{\sqrt{1+k^2}}$
So the area $S=0.5ah=\frac{-4k\sqrt{(16+8\sqrt{13})k^2-13}}{k^2+1}$
Then the maximum value of $S$ is $\frac{104-26\sqrt{13}}{3}$
So the answer is $104+26+13+3=\boxed{146}$ | null | 146 |
ded6b33a1066af211ee912e8bb536d09 | https://artofproblemsolving.com/wiki/index.php/2013_AIME_II_Problems/Problem_10 | Given a circle of radius $\sqrt{13}$ , let $A$ be a point at a distance $4 + \sqrt{13}$ from the center $O$ of the circle. Let $B$ be the point on the circle nearest to point $A$ . A line passing through the point $A$ intersects the circle at points $K$ and $L$ . The maximum possible area for $\triangle BKL$ can be written in the form $\frac{a - b\sqrt{c}}{d}$ , where $a$ $b$ $c$ , and $d$ are positive integers, $a$ and $d$ are relatively prime, and $c$ is not divisible by the square of any prime. Find $a+b+c+d$ | [asy] import math; import olympiad; import graph; pair A, B, K, L; B = (sqrt(13), 0); A=(4+sqrt(13), 0); dot(B); dot(A); draw(Circle((0,0), sqrt(13))); label("$O$", (0,0), S);label("$B$", B, SW);label("$A$", A, S); dot((0,0)); [/asy]
Draw $OC$ perpendicular to $KL$ at $C$ . Draw $BD$ perpendicular to $KL$ at $D$
\[\frac{\triangle OKL}{\triangle BKL} = \frac{OC}{BD} = \frac{AO}{AB} = \frac{4+\sqrt{13}}{4}\]
Therefore, to maximize area of $\triangle BKL$ , we need to maximize area of $\triangle OKL$
\[\triangle OKL = \frac12 r^2 \sin{\angle KOL}\]
So when area of $\triangle OKL$ is maximized, $\angle KOL = \frac{\pi}{2}$
Eventually, we get \[\triangle BKL= \frac12 \cdot (\sqrt{13})^2\cdot(\frac{4}{4+\sqrt{13}})=\frac{104-26\sqrt{13}}{3}\]
So the answer is $104+26+13+3=\boxed{146}$ | null | 146 |
ded6b33a1066af211ee912e8bb536d09 | https://artofproblemsolving.com/wiki/index.php/2013_AIME_II_Problems/Problem_10 | Given a circle of radius $\sqrt{13}$ , let $A$ be a point at a distance $4 + \sqrt{13}$ from the center $O$ of the circle. Let $B$ be the point on the circle nearest to point $A$ . A line passing through the point $A$ intersects the circle at points $K$ and $L$ . The maximum possible area for $\triangle BKL$ can be written in the form $\frac{a - b\sqrt{c}}{d}$ , where $a$ $b$ $c$ , and $d$ are positive integers, $a$ and $d$ are relatively prime, and $c$ is not divisible by the square of any prime. Find $a+b+c+d$ | Let $N,M$ les on $AL$ such that $BM\bot AL, ON\bot AL$ , call $BM=h, ON=k,LN=KN=d$ We call $\angle{LON}=\alpha$ By similar triangle, we have $\frac{h}{k}=\frac{4}{4+\sqrt{13}}, h=\frac{4k}{4+\sqrt{13}}$ . Then, we realize the area is just $dh=d\cdot \frac{4K}{4+\sqrt{13}}$ As $\sin \alpha=\frac{d}{\sqrt{13}}, \cos \alpha=\frac{k}{\sqrt{13}}$ . Now, we have to maximize $\frac{52\sin \alpha \cos \alpha}{4+\sqrt{13}}=\frac{26\sin 2\alpha}{4+\sqrt{13}}$ , which is obviously reached when $\alpha=45^{\circ}$ , the answer is $\frac{104-26\sqrt{13}}{3}$ leads to $\boxed{146}$ | null | 146 |
ded6b33a1066af211ee912e8bb536d09 | https://artofproblemsolving.com/wiki/index.php/2013_AIME_II_Problems/Problem_10 | Given a circle of radius $\sqrt{13}$ , let $A$ be a point at a distance $4 + \sqrt{13}$ from the center $O$ of the circle. Let $B$ be the point on the circle nearest to point $A$ . A line passing through the point $A$ intersects the circle at points $K$ and $L$ . The maximum possible area for $\triangle BKL$ can be written in the form $\frac{a - b\sqrt{c}}{d}$ , where $a$ $b$ $c$ , and $d$ are positive integers, $a$ and $d$ are relatively prime, and $c$ is not divisible by the square of any prime. Find $a+b+c+d$ | Let C and D be the base of perpendiculars dropped from points O and B to AK. Denote BD = h, OC = H. \[\triangle ABD \sim \triangle AOC \implies \frac {h}{H} = \frac {4}{4 + \sqrt{13}}.\] $KL$ is the base of triangles $\triangle OKL$ and $\triangle BKL \implies \frac {[BKL]}{[OKL]} = \frac{h}{H} =$ const $\implies$ The maximum possible area for $\triangle BKL$ and $\triangle OKL$ are at the same position of point $K$
$\triangle OKL$ has sides $OK = OL = \sqrt{13}\implies \max[\triangle OKL] = \frac {OK^2}{2} = \frac {13}{2}$
in the case $\angle KOL = 90^\circ.$ It is possible – if we rotate such triangle, we can find position when $A$ lies on $KL.$ \[\max[\triangle BKL] = \max[\triangle OKL] \cdot \frac {4}{4+\sqrt{13}} = \frac {26}{4+\sqrt{13}} \implies \boxed{146}\] vladimir.shelomovskii@gmail.com, vvsss | null | 146 |
cb5d1e83a3d4b5de259f1b0bbdac27b5 | https://artofproblemsolving.com/wiki/index.php/2013_AIME_II_Problems/Problem_11 | Let $A = \{1, 2, 3, 4, 5, 6, 7\}$ , and let $N$ be the number of functions $f$ from set $A$ to set $A$ such that $f(f(x))$ is a constant function. Find the remainder when $N$ is divided by $1000$ | Any such function can be constructed by distributing the elements of $A$ on three tiers.
The bottom tier contains the constant value, $c=f(f(x))$ for any $x$ . (Obviously $f(c)=c$ .)
The middle tier contains $k$ elements $x\ne c$ such that $f(x)=c$ , where $1\le k\le 6$
The top tier contains $6-k$ elements such that $f(x)$ equals an element on the middle tier.
There are $7$ choices for $c$ . Then for a given $k$ , there are $\tbinom6k$ ways to choose the elements on the middle tier, and then $k^{6-k}$ ways to draw arrows down from elements on the top tier to elements on the middle tier.
Thus $N=7\cdot\sum_{k=1}^6\tbinom6k\cdot k^{6-k}=7399$ , giving the answer $\boxed{399}$ | null | 399 |
cb5d1e83a3d4b5de259f1b0bbdac27b5 | https://artofproblemsolving.com/wiki/index.php/2013_AIME_II_Problems/Problem_11 | Let $A = \{1, 2, 3, 4, 5, 6, 7\}$ , and let $N$ be the number of functions $f$ from set $A$ to set $A$ such that $f(f(x))$ is a constant function. Find the remainder when $N$ is divided by $1000$ | Define the three layers as domain $x$ codomain $f(x)$ , and codomain $f(f(x))$ . Each one of them is contained in the set $A$ . We know that $f(f(x))$ is a constant function , or in other words, can only take on one value. So, we can start off by choosing that value $c$ in $7$ ways. So now, we choose the values that can be $f(x)$ for all those values should satisfy $f(f(x))=c$ . Let's $S$ be that set of values. First things first, we must have $c$ to be part of $S$ , for the $S$ is part of the domain of $x$ . Since the values in $i\in S$ all satisfy $f(i) = c$ , we have $c$ to be a value that $f(x)$ can be. Now, for the elements other than $5$
If we have $k$ elements other than $5$ that can be part of $S$ , we will have $\binom{6}{k}$ ways to choose those values. There will also be $k$ ways for each of the elements in $A$ other than $c$ and those in set $S$ (for when function $f$ is applied on those values, we already know it would be $c$ ). There are $6-k$ elements in $A$ other than $c$ and those in set $S$ . Thus, there should be $6^{6-k}$ ways to match the domain $x$ to the values of $f(x)$ . Summing up all possible values of $k$ $[1,6]$ ), we have
\[\sum_{k=1}^6 k^{6-k} = 6\cdot 1 + 15\cdot 16 + 20\cdot 27 + 15\cdot 16 + 6\cdot 5 + 1) = 1057\]
Multiplying that by the original $7$ for the choice of $c$ , we have $7 \cdot 1057 = 7\boxed{399}.$ | null | 399 |
78507ebfe5db5721a83855c3a4f6b5c3 | https://artofproblemsolving.com/wiki/index.php/2013_AIME_II_Problems/Problem_12 | Let $S$ be the set of all polynomials of the form $z^3 + az^2 + bz + c$ , where $a$ $b$ , and $c$ are integers. Find the number of polynomials in $S$ such that each of its roots $z$ satisfies either $|z| = 20$ or $|z| = 13$ | Every cubic with real coefficients has to have either three real roots or one real and two nonreal roots which are conjugates. This follows from Vieta's formulas
The real root $r$ must be one of $-20$ $20$ $-13$ , or $13$ . By Viète's formulas, $a=-(r+\omega+\omega^*)$ $b=|\omega|^2+r(\omega+\omega^*)$ , and $c=-r|\omega|^2$ . But $\omega+\omega^*=2\Re{(\omega)}$ (i.e., adding the conjugates cancels the imaginary part). Therefore, to make $a$ an integer, $2\Re{(\omega)}$ must be an integer. Conversely, if $\omega+\omega^*=2\Re{(\omega)}$ is an integer, then $a,b,$ and $c$ are clearly integers. Therefore $2\Re{(\omega)}\in \mathbb{Z}$ is equivalent to the desired property. Let $\omega=\alpha+i\beta$
In this case, $\omega$ lies on a circle of radius $20$ in the complex plane. As $\omega$ is nonreal, we see that $\beta\ne 0$ . Hence $-20<\Re{(\omega)}< 20$ , or rather $-40<2\Re{(\omega)}< 40$ . We count $79$ integers in this interval, each of which corresponds to a unique complex number on the circle of radius $20$ with positive imaginary part.
In this case, $\omega$ lies on a circle of radius $13$ in the complex plane. As $\omega$ is nonreal, we see that $\beta\ne 0$ . Hence $-13<\Re{(\omega)}< 13$ , or rather $-26<2\Re{(\omega)}< 26$ . We count $51$ integers in this interval, each of which corresponds to a unique complex number on the circle of radius $13$ with positive imaginary part.
Therefore, there are $79+51=130$ choices for $\omega$ . We also have $4$ choices for $r$ , hence there are $4\cdot 130=520$ total polynomials in this case.
In this case, there are four possible real roots, namely $\pm 13, \pm20$ . Let $p$ be the number of times that $13$ appears among $r_1,r_2,r_3$ , and define $q,r,s$ similarly for $-13,20$ , and $-20$ , respectively. Then $p+q+r+s=3$ because there are three roots. We wish to find the number of ways to choose nonnegative integers $p,q,r,s$ that satisfy that equation. By balls and urns, these can be chosen in $\binom{6}{3}=20$ ways.
Therefore, there are a total of $520+20=\boxed{540}$ polynomials with the desired property. | null | 540 |
78507ebfe5db5721a83855c3a4f6b5c3 | https://artofproblemsolving.com/wiki/index.php/2013_AIME_II_Problems/Problem_12 | Let $S$ be the set of all polynomials of the form $z^3 + az^2 + bz + c$ , where $a$ $b$ , and $c$ are integers. Find the number of polynomials in $S$ such that each of its roots $z$ satisfies either $|z| = 20$ or $|z| = 13$ | There are two cases: either all the roots are real, or one is real and two are imaginary.
Case 1: All roots are real.
Then each of the roots is a member of the set $\{-20, 20, -13, 13\}$ . It splits into three sub-cases: either no two are the same, exactly two are the same, or all three are the same.
Sub-case 1.1: No two are the same.
This is obviously $\dbinom{4}{3}=4$
Sub-case 1.2: Exactly two are the same.
There are four ways to choose the root that will repeat twice, and three ways to choose the remaining root. For this sub-case, $4\cdot 3=12$
Sub-case 1.3: All three are the same.
This is obviously $4$
Thus for case one, we have $4+12+4=20$ polynomials in $S$ . We now have case two, which we state below.
Case 2: Two roots are imaginary and one is real.
Let these roots be $p-qi$ $p+qi$ , and $r$ . Then by Vieta's formulas
Since $a$ $b$ $c$ , and $r$ are integers, we have that $p=\frac{1}{2}k$ for some integer $k$ . Case two splits into two sub-cases now:
Sub-case 2.1: $|p-qi|=|p+qi|=13$ .
Obviously, $|p|<13$ . The $51$ cases in which $p$ is either $0,\pm\frac{1}{2},\pm\frac{2}{2},\pm\frac{3}{2},\ldots,\pm\frac{25}{2}$ are acceptable. Each can pair with one value of $q$ and four values of $r$ , adding $51\cdot 4=204$ polynomials to $S$
Sub-case 2.2: $|p-qi|=|p+qi|=20$ .
Obviously, $|p|<20$ . Here, the $79$ cases in which $p$ is either $0,\pm\frac{1}{2},\pm\frac{2}{2},\pm\frac{3}{2},\ldots,\pm\frac{39}{2}$ are acceptable. Again, each can pair with a single value of $q$ as well as four values of $r$ , adding $79\cdot 4=316$ polynomials to $S$
Thus for case two, $204+316=520$ polynomials are part of $S$
All in all, $20+204+316=\boxed{540}$ polynomials can call $S$ home. | null | 540 |
5facd3d4e3dd0f98887638f3d4cf67a9 | https://artofproblemsolving.com/wiki/index.php/2013_AIME_II_Problems/Problem_14 | For positive integers $n$ and $k$ , let $f(n, k)$ be the remainder when $n$ is divided by $k$ , and for $n > 1$ let $F(n) = \max_{\substack{1\le k\le \frac{n}{2}}} f(n, k)$ . Find the remainder when $\sum\limits_{n=20}^{100} F(n)$ is divided by $1000$ | We can find that
$20\equiv 6 \pmod{7}$
$21\equiv 5 \pmod{8}$
$22\equiv 6 \pmod{8}$
$23\equiv 7 \pmod{8}$
$24\equiv 6 \pmod{9}$
$25\equiv 7 \pmod{9}$
$26\equiv 8 \pmod{9}$
Observing these and we can find that the reminders are in groups of three continuous integers, considering this is true, and we get
$99\equiv 31 \pmod{34}$
$100\equiv 32 \pmod{34}$
So the sum is $6+3\times(6+...+31)+31+32=1512$ ,it is also 17+20+23+...+95, so the answer is $\boxed{512}$ .
By: Kris17 | null | 512 |
5facd3d4e3dd0f98887638f3d4cf67a9 | https://artofproblemsolving.com/wiki/index.php/2013_AIME_II_Problems/Problem_14 | For positive integers $n$ and $k$ , let $f(n, k)$ be the remainder when $n$ is divided by $k$ , and for $n > 1$ let $F(n) = \max_{\substack{1\le k\le \frac{n}{2}}} f(n, k)$ . Find the remainder when $\sum\limits_{n=20}^{100} F(n)$ is divided by $1000$ | $Lemma:$ Highest remainder when $n$ is divided by $1\leq k\leq n/2$ is obtained for $k_0 = (n + (3 - n \pmod{3}))/3$ and the remainder thus obtained is $(n - k_0*2) = [(n - 6)/3 + (2/3)*n \pmod{3}]$
$Note:$ This is the second highest remainder when $n$ is divided by $1\leq k\leq n$ and the highest remainder occurs when $n$ is divided by $k_M$ $(n+1)/2$ for odd $n$ and $k_M$ $(n+2)/2$ for even $n$
Using the lemma above:
$\sum\limits_{n=20}^{100} F(n) = \sum\limits_{n=20}^{100} [(n - 6)/3 + (2/3)*n \pmod{3}]$ $= [(120*81/2)/3 - 2*81 + (2/3)*81] = 1512$
So the answer is $\boxed{512}$ | null | 512 |
578a8bbe9268f4a46d86c205bd8b46ed | https://artofproblemsolving.com/wiki/index.php/2013_AIME_II_Problems/Problem_15 | Let $A,B,C$ be angles of a triangle with \begin{align*} \cos^2 A + \cos^2 B + 2 \sin A \sin B \cos C &= \frac{15}{8} \text{ and} \\ \cos^2 B + \cos^2 C + 2 \sin B \sin C \cos A &= \frac{14}{9} \end{align*} There are positive integers $p$ $q$ $r$ , and $s$ for which \[\cos^2 C + \cos^2 A + 2 \sin C \sin A \cos B = \frac{p-q\sqrt{r}}{s},\] where $p+q$ and $s$ are relatively prime and $r$ is not divisible by the square of any prime. Find $p+q+r+s$ | Let's draw the triangle. Since the problem only deals with angles, we can go ahead and set one of the sides to a convenient value. Let $BC = \sin{A}$
By the Law of Sines, we must have $CA = \sin{B}$ and $AB = \sin{C}$
Now let us analyze the given:
\begin{align*} \cos^2A + \cos^2B + 2\sin A\sin B\cos C &= 1-\sin^2A + 1-\sin^2B + 2\sin A\sin B\cos C \\ &= 2-(\sin^2A + \sin^2B - 2\sin A\sin B\cos C) \end{align*}
Now we can use the Law of Cosines to simplify this:
\[= 2-\sin^2C\]
Therefore: \[\sin C = \sqrt{\dfrac{1}{8}},\cos C = \sqrt{\dfrac{7}{8}}.\] Similarly, \[\sin A = \sqrt{\dfrac{4}{9}},\cos A = \sqrt{\dfrac{5}{9}}.\] Note that the desired value is equivalent to $2-\sin^2B$ , which is $2-\sin^2(A+C)$ . All that remains is to use the sine addition formula and, after a few minor computations, we obtain a result of $\dfrac{111-4\sqrt{35}}{72}$ . Thus, the answer is $111+4+35+72 = \boxed{222}$ | null | 222 |
578a8bbe9268f4a46d86c205bd8b46ed | https://artofproblemsolving.com/wiki/index.php/2013_AIME_II_Problems/Problem_15 | Let $A,B,C$ be angles of a triangle with \begin{align*} \cos^2 A + \cos^2 B + 2 \sin A \sin B \cos C &= \frac{15}{8} \text{ and} \\ \cos^2 B + \cos^2 C + 2 \sin B \sin C \cos A &= \frac{14}{9} \end{align*} There are positive integers $p$ $q$ $r$ , and $s$ for which \[\cos^2 C + \cos^2 A + 2 \sin C \sin A \cos B = \frac{p-q\sqrt{r}}{s},\] where $p+q$ and $s$ are relatively prime and $r$ is not divisible by the square of any prime. Find $p+q+r+s$ | Let \begin{align*} \cos^2 A + \cos^2 B + 2 \sin A \sin B \cos C &= \frac{15}{8} \text{ ------ (1)}\\ \cos^2 B + \cos^2 C + 2 \sin B \sin C \cos A &= \frac{14}{9} \text{ ------ (2)}\\ \cos^2 C + \cos^2 A + 2 \sin C \sin A \cos B &= x \text{ ------ (3)}\\ \end{align*}
Adding (1) and (3) we get: \[2 \cos^2 A + \cos^2 B + \cos^2 C + 2 \sin A( \sin B \cos C + \sin C \cos B) = \frac{15}{8} + x\] or \[2 \cos^2 A + \cos^2 B + \cos^2 C + 2 \sin A \sin (B+C) = \frac{15}{8} + x\] or \[2 \cos^2 A + \cos^2 B + \cos^2 C + 2 \sin ^2 A = \frac{15}{8} + x\] or \[\cos^2 B + \cos^2 C = x - \frac{1}{8} \text{ ------ (4)}\]
Similarly adding (2) and (3) we get: \[\cos^2 A + \cos^2 B = x - \frac{4}{9} \text{ ------ (5)}\] Similarly adding (1) and (2) we get: \[\cos^2 A + \cos^2 C = \frac{14}{9} - \frac{1}{8} \text{ ------ (6)}\]
And (4) - (5) gives: \[\cos^2 C - \cos^2 A = \frac{4}{9} - \frac{1}{8} \text{ ------ (7)}\]
Now (6) - (7) gives: $\cos^2 A = \frac{5}{9}$ or $\cos A = \sqrt{\dfrac{5}{9}}$ and $\sin A = \frac{2}{3}$ so $\cos C$ is $\sqrt{\dfrac{7}{8}}$ and therefore $\sin C$ is $\sqrt{\dfrac{1}{8}}$
Now $\sin B = \sin(A+C)$ can be computed first and then $\cos^2 B$ is easily found.
Thus $\cos^2 B$ and $\cos^2 C$ can be plugged into (4) above to give x = $\dfrac{111-4\sqrt{35}}{72}$
Hence the answer is = $111+4+35+72 = \boxed{222}$ | null | 222 |
578a8bbe9268f4a46d86c205bd8b46ed | https://artofproblemsolving.com/wiki/index.php/2013_AIME_II_Problems/Problem_15 | Let $A,B,C$ be angles of a triangle with \begin{align*} \cos^2 A + \cos^2 B + 2 \sin A \sin B \cos C &= \frac{15}{8} \text{ and} \\ \cos^2 B + \cos^2 C + 2 \sin B \sin C \cos A &= \frac{14}{9} \end{align*} There are positive integers $p$ $q$ $r$ , and $s$ for which \[\cos^2 C + \cos^2 A + 2 \sin C \sin A \cos B = \frac{p-q\sqrt{r}}{s},\] where $p+q$ and $s$ are relatively prime and $r$ is not divisible by the square of any prime. Find $p+q+r+s$ | Let's take the first equation $\cos^2 A + \cos^2 B + 2 \sin A \sin B \cos C = \frac{15}{8}$ . Substituting $180 - A - B$ for C, given A, B, and C form a triangle, and that $\cos C = \cos(A + B)$ , gives us:
$\cos^2 A + \cos^2 B + 2 \sin A \sin B \cos (A+B) = \frac{15}{8}$
Expanding out gives us $\cos^2 A + \cos^2 B + 2 \sin^2 A \sin^2 B - 2 \sin A \sin B \cos A \cos B = \frac{15}{8}$
Using the double angle formula $\cos^2 k = \frac{\cos (2k) + 1}{2}$ , we can substitute for each of the squares $\cos^2 A$ and $\cos^2 B$ . Next we can use the Pythagorean identity on the $\sin^2 A$ and $\sin^2 B$ terms. Lastly we can use the sine double angle to simplify.
$\cos^2 A + \cos^2 B + 2(1 - \cos^2 A)(1 - \cos^2 B) - \frac{1}{2} \cdot \sin 2A \sin 2B = \frac{15}{8}$
Expanding and canceling yields, and again using double angle substitution,
$1 + 2 \cdot \frac{\cos (2A) + 1}{2} \cdot \frac{\cos (2B) + 1}{2} - \frac{1}{2} \cdot \sin 2A \sin 2B = \frac{15}{8}$
Further simplifying yields:
$\frac{3}{2} + \frac{\cos 2A \cos 2B - \sin 2A \sin 2B}{2} = \frac{15}{8}$
Using cosine angle addition formula and simplifying further yields, and applying the same logic to Equation $2$ yields:
$\cos (2A + 2B) = \frac{3}{4}$ and $\cos (2B + 2C) = \frac{1}{9}$
Substituting the identity $\cos (2A + 2B) = \cos(2C)$ , we get:
$\cos (2C) = \frac{3}{4}$ and $\cos (2A) = \frac{1}{9}$
Since the third expression simplifies to the expression $\frac{3}{2} + \frac{\cos (2A + 2C)}{2}$ , taking inverse cosine and using the angles in angle addition formula yields the answer, $\frac{111 - 4\sqrt{35}}{72}$ , giving us the answer $\boxed{222}$ | null | 222 |
578a8bbe9268f4a46d86c205bd8b46ed | https://artofproblemsolving.com/wiki/index.php/2013_AIME_II_Problems/Problem_15 | Let $A,B,C$ be angles of a triangle with \begin{align*} \cos^2 A + \cos^2 B + 2 \sin A \sin B \cos C &= \frac{15}{8} \text{ and} \\ \cos^2 B + \cos^2 C + 2 \sin B \sin C \cos A &= \frac{14}{9} \end{align*} There are positive integers $p$ $q$ $r$ , and $s$ for which \[\cos^2 C + \cos^2 A + 2 \sin C \sin A \cos B = \frac{p-q\sqrt{r}}{s},\] where $p+q$ and $s$ are relatively prime and $r$ is not divisible by the square of any prime. Find $p+q+r+s$ | We will use the sum to product formula to simply these equations. Recall \[2\sin{\frac{\alpha+\beta}{2}}\sin{\frac{\alpha-\beta}{2}} = \cos{\alpha}+\cos{\beta}.\] Using this, let's rewrite the first equation: \[\cos^2(A) + \cos^2(B) + 2 \sin(A) \sin(B) \cos(C) = \frac{15}{8}\] \[\cos^2(A) + \cos^2(B) + (\cos(A+B)+\cos(A-B))\cos(C).\] Now, note that $\cos(C)=-\cos(A+B)$ \[\cos^2(A) + \cos^2(B) + (\cos(A+B)+\cos(A-B))(-\cos(A+B))\] \[\cos^2(A) + \cos^2(B) - \cos(A+B)\cos(A-B)+cos^2(A+B)=\frac{15}{8}.\] We apply the sum to product formula again. \[\cos^2(A) + \cos^2(B) - \frac{\cos(2A)+\cos(2B)}{2}+cos^2(A+B)=\frac{15}{8}.\] Now, recall that $\cos(2\alpha)=2\cos^2(\alpha)-1$ . We apply this and simplify our expression to get: \[\cos^2(A+B)=\frac{7}{8}\] \[\cos^2(C)=\frac{7}{8}.\] Analogously, \[\cos^2(A)=\frac{5}{9}.\] \[\cos^2(A+C)=\frac{p-q\sqrt{r}}{s}-1.\] We can find this value easily by angle sum formula. After a few calculations, we get $\frac{111 - 4\sqrt{35}}{72}$ , giving us the answer $\boxed{222}$ .
~superagh | null | 222 |
578a8bbe9268f4a46d86c205bd8b46ed | https://artofproblemsolving.com/wiki/index.php/2013_AIME_II_Problems/Problem_15 | Let $A,B,C$ be angles of a triangle with \begin{align*} \cos^2 A + \cos^2 B + 2 \sin A \sin B \cos C &= \frac{15}{8} \text{ and} \\ \cos^2 B + \cos^2 C + 2 \sin B \sin C \cos A &= \frac{14}{9} \end{align*} There are positive integers $p$ $q$ $r$ , and $s$ for which \[\cos^2 C + \cos^2 A + 2 \sin C \sin A \cos B = \frac{p-q\sqrt{r}}{s},\] where $p+q$ and $s$ are relatively prime and $r$ is not divisible by the square of any prime. Find $p+q+r+s$ | According to LOC $a^2+b^2-2ab\cos{\angle{c}}=c^2$ , we can write it into $\sin^2{\angle{A}}+\sin^2{\angle{B}}-2\sin{\angle{A}}\sin{\angle{B}}\cos{\angle{C}}=\sin^2{\angle{C}}$ $\sin^2{\angle{A}}+\sin^2{\angle{B}}-2\sin{\angle{A}}\sin{\angle{B}}\cos{\angle{C}}+cos^2A+cos^2B+2sinAsinBcosC=\frac{15}{8}+sin^2C$ We can simplify to $2=sin^2C+\frac{15}{8}$ . Similarly, we can generalize $2=sin^2A+\frac{14}{9}$ . After solving, we can get that $sinA=\frac{2}{3}; cosA=\frac{\sqrt{5}}{3}; sinC=\frac{\sqrt{2}}{4}; cosC=\frac{\sqrt{14}}{4}$ Assume the value we are looking for is $x$ , we get $sin^2B+x=2$ , while $sinB=sin(180^{\circ}-A-C)=sin(A+C)$ which is $\frac{2\sqrt{14}+\sqrt{10}}{12}$ , so $x=\frac{111 - 4\sqrt{35}}{72}$ , giving us the answer $\boxed{222}$ .~bluesoul | null | 222 |
8eda77e9d119cc01dece5c4e5b120c14 | https://artofproblemsolving.com/wiki/index.php/2012_AIME_I_Problems/Problem_1 | Find the number of positive integers with three not necessarily distinct digits, $abc$ , with $a \neq 0$ and $c \neq 0$ such that both $abc$ and $cba$ are multiples of $4$ | A positive integer is divisible by $4$ if and only if its last two digits are divisible by $4.$ For any value of $b$ , there are two possible values for $a$ and $c$ , since we find that if $b$ is even, $a$ and $c$ must be either $4$ or $8$ , and if $b$ is odd, $a$ and $c$ must be either $2$ or $6$ . There are thus $2 \cdot 2 = 4$ ways to choose $a$ and $c$ for each $b,$ and $10$ ways to choose $b$ since $b$ can be any digit. The final answer is then $4 \cdot 10 = \boxed{040}$ | null | 040 |
8eda77e9d119cc01dece5c4e5b120c14 | https://artofproblemsolving.com/wiki/index.php/2012_AIME_I_Problems/Problem_1 | Find the number of positive integers with three not necessarily distinct digits, $abc$ , with $a \neq 0$ and $c \neq 0$ such that both $abc$ and $cba$ are multiples of $4$ | A number is divisible by four if its last two digits are divisible by 4. Thus, we require that $10b + a$ and $10b + c$ are both divisible by $4$ . If $b$ is odd, then $a$ and $c$ must both be $2 \pmod 4$ meaning that $a$ and $c$ are $2$ or $6$ . If $b$ is even, then $a$ and $c$ must be $0 \pmod 4$ meaning that $a$ and $c$ are $4$ or $8$ . For each choice of $b$ there are $2$ choices for $a$ and $2$ for $c$ for a total of $10 \cdot 2 \cdot 2 = \boxed{040}$ numbers. | null | 040 |
8eda77e9d119cc01dece5c4e5b120c14 | https://artofproblemsolving.com/wiki/index.php/2012_AIME_I_Problems/Problem_1 | Find the number of positive integers with three not necessarily distinct digits, $abc$ , with $a \neq 0$ and $c \neq 0$ such that both $abc$ and $cba$ are multiples of $4$ | For this number to fit the requirements $bc$ and $ba$ must be divisible by 4. So $bc = 00, 04, 08, 12, 16, ... , 92, 96$ and so must $ba$ for each two digits of $bc$ . There are two possibilities for $ba$ if $b$ is odd and three possibilities if $b$ is even. So there are $2^{2} \cdot 5+3^{2} \cdot 4 = 65$ possibilities but this overcounts when $a$ or $c = 0$ . So when $bc = 00, 20, 40, 60, 80$ and the corresponding $ba$ should be removed, so $65 - 5 \cdot 3 = 50$ . But we are still overcounting when $b$ is even because then $a$ can be 0. So the answer is $50 - 10 \cdot 1 = \boxed{040}$ | null | 040 |
457d3aeabe0d3764843bc55f8ee5cff9 | https://artofproblemsolving.com/wiki/index.php/2012_AIME_I_Problems/Problem_2 | The terms of an arithmetic sequence add to $715$ . The first term of the sequence is increased by $1$ , the second term is increased by $3$ , the third term is increased by $5$ , and in general, the $k$ th term is increased by the $k$ th odd positive integer. The terms of the new sequence add to $836$ . Find the sum of the first, last, and middle terms of the original sequence. | If the sum of the original sequence is $\sum_{i=1}^{n} a_i$ then the sum of the new sequence can be expressed as $\sum_{i=1}^{n} a_i + (2i - 1) = n^2 + \sum_{i=1}^{n} a_i.$ Therefore, $836 = n^2 + 715 \rightarrow n=11.$ Now the middle term of the original sequence is simply the average of all the terms, or $\frac{715}{11} = 65,$ and the first and last terms average to this middle term, so the desired sum is simply three times the middle term, or $\boxed{195}.$ | null | 195 |
457d3aeabe0d3764843bc55f8ee5cff9 | https://artofproblemsolving.com/wiki/index.php/2012_AIME_I_Problems/Problem_2 | The terms of an arithmetic sequence add to $715$ . The first term of the sequence is increased by $1$ , the second term is increased by $3$ , the third term is increased by $5$ , and in general, the $k$ th term is increased by the $k$ th odd positive integer. The terms of the new sequence add to $836$ . Find the sum of the first, last, and middle terms of the original sequence. | After the adding of the odd numbers, the total of the sequence increases by $836 - 715 = 121 = 11^2$ . Since the sum of the first $n$ positive odd numbers is $n^2$ , there must be $11$ terms in the sequence, so the mean of the sequence is $\dfrac{715}{11} = 65$ . Since the first, last, and middle terms are centered around the mean, our final answer is $65 \cdot 3 = \boxed{195}$ | null | 195 |
457d3aeabe0d3764843bc55f8ee5cff9 | https://artofproblemsolving.com/wiki/index.php/2012_AIME_I_Problems/Problem_2 | The terms of an arithmetic sequence add to $715$ . The first term of the sequence is increased by $1$ , the second term is increased by $3$ , the third term is increased by $5$ , and in general, the $k$ th term is increased by the $k$ th odd positive integer. The terms of the new sequence add to $836$ . Find the sum of the first, last, and middle terms of the original sequence. | Proceed as in Solution 2 until it is noted that there are 11 terms in the sequence. Since the sum of the terms of the original arithmetic sequence is 715, we note that $\frac{2a_1 + 10d}{2} \cdot 11 = 715$ or $2a_1 + 10d = 130$ for all sets of first terms and common differences that fit the conditions given in the problem. Assume WLOG that $a_1 = 60$ and $d = 1$ . Then the first term of the corresponding arithmetic sequence will be $60$ , the sixth (middle) term will be $65$ , and the eleventh (largest) term will be $70$ . Thus, our final answer is $60 + 65 + 70 = \boxed{195}$ | null | 195 |
513de467dfe8463f8c969903d7557136 | https://artofproblemsolving.com/wiki/index.php/2012_AIME_I_Problems/Problem_3 | Nine people sit down for dinner where there are three choices of meals. Three people order the beef meal, three order the chicken meal, and three order the fish meal. The waiter serves the nine meals in random order. Find the number of ways in which the waiter could serve the meal types to the nine people so that exactly one person receives the type of meal ordered by that person. | Call a beef meal $B,$ a chicken meal $C,$ and a fish meal $F.$ Now say the nine people order meals $\text{BBBCCCFFF}$ respectively and say that the person who receives the correct meal is the first person. We will solve for this case and then multiply by $9$ to account for the $9$ different ways in which the person to receive the correct meal could be picked. Note, this implies that the dishes are indistinguishable, though the people aren't. For example, two people who order chicken are separate, though if they receive fish, there is only 1 way to order them.
The problem we must solve is to distribute meals $\text{BBCCCFFF}$ to orders $\text{BBCCCFFF}$ with no matches. The two people who ordered $B$ 's can either both get $C$ 's, both get $F$ 's, or get one $C$ and one $F.$ We proceed with casework.
Summing across the cases we see there are $24$ possibilities, so the answer is $9 \cdot 24 = \boxed{216.}$ | null | 216. |
513de467dfe8463f8c969903d7557136 | https://artofproblemsolving.com/wiki/index.php/2012_AIME_I_Problems/Problem_3 | Nine people sit down for dinner where there are three choices of meals. Three people order the beef meal, three order the chicken meal, and three order the fish meal. The waiter serves the nine meals in random order. Find the number of ways in which the waiter could serve the meal types to the nine people so that exactly one person receives the type of meal ordered by that person. | We only need to figure out the number of ways to order the string $BBBCCCFFF$ , where exactly one $B$ is in the first three positions, one $C$ is in the $4^{th}$ to $6^{th}$ positions, and one $F$ is in the last three positions. There are $3^3=27$ ways to place the first $3$ meals. Then for the other two people, there are $2$ ways to serve their meals. Thus, there are $(3\cdot2)^3=\boxed{216}$ ways to serve their meals. | null | 216 |
fbf2d1b099c7f0534786d6d5d8982d53 | https://artofproblemsolving.com/wiki/index.php/2012_AIME_I_Problems/Problem_4 | Butch and Sundance need to get out of Dodge. To travel as quickly as possible, each alternates walking and riding their only horse, Sparky, as follows. Butch begins by walking while Sundance rides. When Sundance reaches the first of the hitching posts that are conveniently located at one-mile intervals along their route, he ties Sparky to the post and begins walking. When Butch reaches Sparky, he rides until he passes Sundance, then leaves Sparky at the next hitching post and resumes walking, and they continue in this manner. Sparky, Butch, and Sundance walk at $6,$ $4,$ and $2.5$ miles per hour, respectively. The first time Butch and Sundance meet at a milepost, they are $n$ miles from Dodge, and they have been traveling for $t$ minutes. Find $n + t$ | When they meet at the milepost, Sparky has been ridden for $n$ miles total. Assume Butch rides Sparky for $a$ miles, and Sundance rides for $n-a$ miles. Thus, we can set up an equation, given that Sparky takes $\frac{1}{6}$ hours per mile, Butch takes $\frac{1}{4}$ hours per mile, and Sundance takes $\frac{2}{5}$ hours per mile:
\[\frac{a}{6} + \frac{n-a}{4} = \frac{n-a}{6} + \frac{2a}{5} \rightarrow a = \frac{5}{19}n.\]
The smallest possible integral value of $n$ is $19$ , so we plug in $n = 19$ and $a = 5$ and get $t = \frac{13}{3}$ hours, or $260$ minutes. So our answer is $19 + 260 = \boxed{279}$ | null | 279 |
91ebe2f4fdadfbef5fe69a4b9ce9ed71 | https://artofproblemsolving.com/wiki/index.php/2012_AIME_I_Problems/Problem_5 | Let $B$ be the set of all binary integers that can be written using exactly $5$ zeros and $8$ ones where leading zeros are allowed. If all possible subtractions are performed in which one element of $B$ is subtracted from another, find the number of times the answer $1$ is obtained. | When $1$ is subtracted from a binary number, the number of digits will remain constant if and only if the original number ended in $10.$ Therefore, every subtraction involving two numbers from $B$ will necessarily involve exactly one number ending in $10.$ To solve the problem, then, we can simply count the instances of such numbers. With the $10$ in place, the seven remaining $1$ 's can be distributed in any of the remaining $11$ spaces, so the answer is ${11 \choose 7} = \boxed{330}$ | null | 330 |