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fabee2c450d3e0053aeaa4fb8f8acae1 | https://artofproblemsolving.com/wiki/index.php/2020_AIME_II_Problems/Problem_6 | Define a sequence recursively by $t_1 = 20$ $t_2 = 21$ , and \[t_n = \frac{5t_{n-1}+1}{25t_{n-2}}\] for all $n \ge 3$ . Then $t_{2020}$ can be expressed as $\frac{p}{q}$ , where $p$ and $q$ are relatively prime positive integers. Find $p+q$ | Let us examine the first few terms of this sequence and see if we can find a pattern. We are obviously given $t_1 = 20$ and $t_2 = 21$ , so now we are able to determine the numerical value of $t_3$ using this information: \[t_3 = \frac{5t_{3-1}+1}{25t_{3-2}} = \frac{5t_{2}+1}{25t_{1}} = \frac{5(21) + 1}{25(20)} = \frac{105 + 1}{500}t_3 = \frac{106}{500} = \frac{53}{250}\] \[t_4 = \frac{5t_{4-1}+1}{25t_{4-2}} = \frac{5t_{3}+1}{25t_{2}} = \frac{5(\frac{53}{250}) + 1}{25(21)} = \frac{\frac{53}{50} + 1}{525} = \frac{\frac{103}{50}}{525} = \frac{103}{26250}\] \[t_5 = \frac{5t_{5-1}+1}{25t_{5-2}} = \frac{5t_{4}+1}{25t_{3}} = \frac{5(\frac{103}{26250}) + 1}{25(\frac{53}{250})} = \frac{\frac{103}{5250} + 1}{\frac{53}{10}} = \frac{\frac{5353}{5250}}{\frac{53}{10}} = \frac{101}{525}\] \[t_6 = \frac{5t_{6-1}+1}{25t_{6-2}} = \frac{5t_{5}+1}{25t_{4}} = \frac{5(\frac{101}{525}) + 1}{25(\frac{103}{26250})} = \frac{\frac{101}{105} + 1}{\frac{103}{1050}} = \frac{\frac{206}{105}}{\frac{103}{1050}} \implies t_6 = 20\]
Alas, we have figured this sequence is period 5! But since $2020 \equiv 5 \pmod 5$ , we can state that $t_{2020} = t_5 = \frac{101}{525}$ . According to the original problem statement, our answer is $\boxed{626}$ . ~ nikenissan | null | 626 |
ad92d5f7970b850f78bc8590180ba2de | https://artofproblemsolving.com/wiki/index.php/2020_AIME_II_Problems/Problem_7 | Two congruent right circular cones each with base radius $3$ and height $8$ have the axes of symmetry that intersect at right angles at a point in the interior of the cones a distance $3$ from the base of each cone. A sphere with radius $r$ lies within both cones. The maximum possible value of $r^2$ is $\frac{m}{n}$ , where $m$ and $n$ are relatively prime positive integers. Find $m+n$ | Using the diagram above, we notice that the desired length is simply the distance between the point $C$ and $\overline{AB}$ . We can mark $C$ as $(3,3)$ since it is $3$ units away from each of the bases. Point $B$ is $(8,3)$ . Thus, line $\overline{AB}$ is $y = \frac{3}{8}x \Rightarrow 3x + 8y = 0$ . We can use the distance from point to line formula $\frac{|Ax_0 + By_0 + C|}{\sqrt{A^2 + B^2}}$ , where $x_0$ and $y_0$ are the coordinates of the point, and A, B, and C are the coefficients of the line in form $Ax + By + C = 0$ . Plugging everything in, we get \[\frac{|3(3) - 8(3)|}{\sqrt{8^2 + 3^2}} = \frac{15}{\sqrt{73}} \Rightarrow \frac{225}{73} \Rightarrow \boxed{298}\] | null | 298 |
9ae22e2574a0c350dd0e3b0e7f5f5121 | https://artofproblemsolving.com/wiki/index.php/2020_AIME_II_Problems/Problem_8 | Define a sequence recursively by $f_1(x)=|x-1|$ and $f_n(x)=f_{n-1}(|x-n|)$ for integers $n>1$ . Find the least value of $n$ such that the sum of the zeros of $f_n$ exceeds $500,000$ | First it will be shown by induction that the zeros of $f_n$ are the integers $a, {a+2,} {a+4,} \dots, {a + n(n-1)}$ , where $a = n - \frac{n(n-1)}2.$
This is certainly true for $n=1$ . Suppose that it is true for $n = m-1 \ge 1$ , and note that the zeros of $f_m$ are the solutions of $|x - m| = k$ , where $k$ is a nonnegative zero of $f_{m-1}$ . Because the zeros of $f_{m-1}$ form an arithmetic sequence with common difference $2,$ so do the zeros of $f_m$ . The greatest zero of $f_{m-1}$ is \[m-1+\frac{(m-1)(m-2)}2 =\frac{m(m-1)}2,\] so the greatest zero of $f_m$ is $m+\frac{m(m-1)}2$ and the least is $m-\frac{m(m-1)}2$
It follows that the number of zeros of $f_n$ is $\frac{n(n-1)}2+1=\frac{n^2-n+2}2$ , and their average value is $n$ . The sum of the zeros of $f_n$ is \[\frac{n^3-n^2+2n}2.\] Let $S(n)=n^3-n^2+2n$ , so the sum of the zeros exceeds $500000$ if and only if $S(n) > 1000000 = 100^3\!.$ Because $S(n)$ is increasing for $n > 2$ , the values $S(100) = 1000000 - 10000 + 200 = 990200$ and $S(101)=1030301 - 10201 + 202 = 1020302$ show that the requested value of $n$ is $\boxed{101}$ | null | 101 |
9ae22e2574a0c350dd0e3b0e7f5f5121 | https://artofproblemsolving.com/wiki/index.php/2020_AIME_II_Problems/Problem_8 | Define a sequence recursively by $f_1(x)=|x-1|$ and $f_n(x)=f_{n-1}(|x-n|)$ for integers $n>1$ . Find the least value of $n$ such that the sum of the zeros of $f_n$ exceeds $500,000$ | Starting from $f_1(x)=|x-1|$ , we can track the solutions, the number of solutions, and their sum.
\[\begin{array}{c|c|c|c} n&Solutions&number&sum\\ 1&1&1&1\\ 2&1,3&2&4\\ 3&0,2,4,6&4&12\\ 4&-2,0,2...10&7&28\\ 5&-5,-3,-1...15&11&55\\ \end{array}\]
It is clear that the solutions form arithmetic sequences with a difference of 2, and the negative solutions cancel out all but $n$ of the $1+\frac{n(n-1)}{2}$ solutions. Thus, the sum of the solutions is $n \cdot [1+\frac{n(n-1)}{2}]$ , which is a cubic function.
$n \cdot [1+\frac{n(n-1)}{2}]>500,000$
Multiplying both sides by $2$
$n \cdot [2+n(n-1)]>1,000,000$
1 million is $10^6=100^3$ , so the solution should be close to $100$
100 is slightly too small, so $\boxed{101}$ works. | null | 101 |
13dabca62e108a05c44c512a87032cc1 | https://artofproblemsolving.com/wiki/index.php/2020_AIME_II_Problems/Problem_9 | While watching a show, Ayako, Billy, Carlos, Dahlia, Ehuang, and Frank sat in that order in a row of six chairs. During the break, they went to the kitchen for a snack. When they came back, they sat on those six chairs in such a way that if two of them sat next to each other before the break, then they did not sit next to each other after the break. Find the number of possible seating orders they could have chosen after the break. | We proceed with casework based on the person who sits first after the break.
$\textbf{Case 1:}$ A is first. Then the first three people in the row can be ACE, ACF, ADB, ADF, AEB, AEC, AFB, AFC, or AFD, which yield 2, 1, 2, 2, 1, 2, 0, 1, and 1 possible configurations, respectively, implying 2 + 1 + 2 + 2 + 1 + 2 + 0 + 1 + 1 = 12 possible configurations in this case.
$\textbf{Case 2:}$ B is first. Then the first three people in the row can be BDA, BDF, BEA, BEC, BFA, BFC, or BFD, which yield 2, 4, 2, 4, 0, 1, and 2 possible configurations, respectively, implying 2 + 4 + 2 + 4 + 0 + 1 + 2 = 15 possible configurations in this case.
$\textbf{Case 3:}$ C is first. Then the first three people in the row can be CAD, CAE, CAF, CEA, CEB, CFA, CFB, or CFD, which yield 1, 2, 1, 4, 4, 2, 2, and 2 possible configurations, respectively, implying 1 + 2 + 1 + 4 + 4 + 2 + 2 + 2 = 18 possible configurations in this case.
Finally, the number of valid configurations for A and F, B and E, and C and D are equal by symmetry, so our final answer is 2(12 + 15 + 18), which computes to be $\boxed{090}.$ ~peace09 | null | 090 |
13dabca62e108a05c44c512a87032cc1 | https://artofproblemsolving.com/wiki/index.php/2020_AIME_II_Problems/Problem_9 | While watching a show, Ayako, Billy, Carlos, Dahlia, Ehuang, and Frank sat in that order in a row of six chairs. During the break, they went to the kitchen for a snack. When they came back, they sat on those six chairs in such a way that if two of them sat next to each other before the break, then they did not sit next to each other after the break. Find the number of possible seating orders they could have chosen after the break. | We determine the order of A, B, C, relative to each other. Then, we will insert D, E, F into the alignment and calculate the total number of possibilities. This solution can be visualised as standing in lines rather than sitting on chairs.
There are 6 possible alignments for A, B, and C, but we only evaluate 3 because the other 3 cases can be mirrored to overlap these 3 cases.
$\textbf{Case 1: A B C}$ In this case, there must be a person standing between A and B and also between B and C. Also, D cannot be adjacent to C. There are 9 possibilities.
$\textbf{Case 2: A C B}$ In this case, there must be a person standing between B and C. Also, D cannot be adjacent to C. There are 12 possibilities.
$\textbf{Case 3: C A B}$ In this case, there must be a person standing between A and B. Also, D cannot be adjacent to C. There are 24 possibilities.
So the total number of cases is 2(9+12+24)= $\boxed{090}$ | null | 090 |
13dabca62e108a05c44c512a87032cc1 | https://artofproblemsolving.com/wiki/index.php/2020_AIME_II_Problems/Problem_9 | While watching a show, Ayako, Billy, Carlos, Dahlia, Ehuang, and Frank sat in that order in a row of six chairs. During the break, they went to the kitchen for a snack. When they came back, they sat on those six chairs in such a way that if two of them sat next to each other before the break, then they did not sit next to each other after the break. Find the number of possible seating orders they could have chosen after the break. | Consider arrangements of numbers $1, 2, 3, ..., n$
Let $f(n,k)$ be the number of arrangements in which $1$ and $2$ $2$ and $3$ $3$ and $4$ , ..., $k-1$ and $k$ aren't together. We need to find $f(6,6)$
Let $d(n,k)$ be the number of arrangements in which $1$ and $2$ $2$ and $3$ $3$ and $4$ , ..., $k-2$ and $k-1$ aren't together, but $(k-1)$ and $k$ are together.
Then,
\[f(n,k+1) = f(n,k) - d(n,k+1) ......(1)\]
\[d(n,k+1) = 2f(n-1,k) + d(n-1,k) ......(2)\]
Hence, \[f(n,k+1)=f(n,k)-f(n-1,k)-f(n-1,k-1)\]
And because $f(n,0) = f(n,1) = n!$ , it's easy to get $f(6,6) = \boxed{090}$ | null | 090 |
62d95c1bcbbe3077e143667ccf157b98 | https://artofproblemsolving.com/wiki/index.php/2020_AIME_II_Problems/Problem_10 | Find the sum of all positive integers $n$ such that when $1^3+2^3+3^3+\cdots +n^3$ is divided by $n+5$ , the remainder is $17$ | The formula for the sum of cubes, also known as Nicomachus's Theorem, is as follows: \[1^3+2^3+3^3+\dots+k^3=(1+2+3+\dots+k)^2=\left(\frac{k(k+1)}{2}\right)^2\] for any positive integer $k$
So let's apply this to this problem.
Let $m=n+5$ . Then we have \begin{align*} 1^3+2^3+3^3+\dots+(m-5)^3&\equiv 17 \mod m \\ \left(\frac{(m-5)(m-4)}{2}\right)^2&\equiv 17 \mod m \\ \left(\dfrac{m(m-9)+20}2\right)^2&\equiv 17\mod m \\ \left(\dfrac{20}2\right)^2&\equiv 17\mod m \\ \frac{400}{4}&\equiv 17 \mod m \\ 332 &\equiv 0 \mod m \\ \end{align*} So, $m\in\{83,166,332\}$ . Testing the cases, only $332$ fails. This leaves $78+161=\boxed{239}$ | null | 239 |
62d95c1bcbbe3077e143667ccf157b98 | https://artofproblemsolving.com/wiki/index.php/2020_AIME_II_Problems/Problem_10 | Find the sum of all positive integers $n$ such that when $1^3+2^3+3^3+\cdots +n^3$ is divided by $n+5$ , the remainder is $17$ | Using the formula for $\sum_{k=1}^n k^3$ \[1^3 + 2^3 + 3^3 + ... + n^3 = \frac{n^2(n+1)^2}{4}\] Since $1^3 + 2^3 + 3^3 + ... + n^3$ divided by $n + 5$ has a remainder of $17$ \[\frac{n^2(n+1)^2}{4} \equiv 17\pmod {n + 5}\] Using the rules of modular arithmetic, \[n^2(n+1)^2 \equiv 68\pmod {n + 5}\] \[n^2(n+1)^2 - 68\equiv 0\pmod {n + 5}\] Expanding the left hand side, \[n^4 + 2 n^3 + n^2 - 68\equiv 0\pmod {n + 5}\] This means that $n^4 + 2 n^3 + n^2 - 68$ is divisible by ${n + 5}$
\[(n + 5) | (n^4 + 2 n^3 + n^2 - 68)\] Dividing polynomials, \[\frac{n^4 + 2 n^3 + n^2 - 68}{n + 5}\] \[= n^3 - 3 n^2 + 16n - 80 + \frac{332}{(n + 5)}\] $(n + 5)$ $|$ $(n^4 + 2 n^3 + n^2 - 68)$ $\iff$ $\frac{332}{(n + 5)}$ $\in$ $\mathbb{Z}$ $\frac{332}{(n + 5)}$ $\in$ $\mathbb{Z}$ $\iff$ $(n + 5) = \pm 1, \pm 2, \pm 4, \pm 83, \pm 166, \pm 332$ Note that $n$ $\in$ $\mathbb{N}$ and $n + 5 > 17$ (because the remainder when dividing by $n + 5$ is $17$ , so $n + 5$ must be greater than $17$ ), so all options $\leq 17$ can be eliminated. \[(n + 5) = 83, 166, 332\] \[n = 78, 161, 327\] Checking all 3 cases, $n = 78$ and $n = 161$ work; $n = 327$ fails. Therefore, the answer is $78 + 161 = \boxed{239} ~ | null | 239 |
62d95c1bcbbe3077e143667ccf157b98 | https://artofproblemsolving.com/wiki/index.php/2020_AIME_II_Problems/Problem_10 | Find the sum of all positive integers $n$ such that when $1^3+2^3+3^3+\cdots +n^3$ is divided by $n+5$ , the remainder is $17$ | As before, we note that $(5+a)^3 + (n-a)^3 \equiv (5+a)^3 - (n+5 - (n-a))^3 \equiv 0 \pmod {n+5}.$ Thus, we can pair up the terms from $5^3$ to $n^3$ and cancel them. We have to deal with two cases:
If $n$ is even, then $5^3+6^3 + \cdots + n^3 \equiv 0 \pmod {n+5},$ as there are an even number of terms and they pair and cancel. We thus get $1^2+2^3+3^3+4^3 = 100 \equiv 17 \pmod {n+5},$ or $(n+5) | 83,$ which yields $n=78.$
If $n$ is odd, then $1^3+2^3+\cdots + n^3 \equiv 1^3+2^3+3^3+4^3+\left( \frac{n+5}{2} \right)^3 \equiv 17 \pmod {n+5}.$ Letting $k = \frac{n+5}{2}$ yields $k^2 + 83 \equiv 0 \pmod {2k}.$ However, this means that $83$ is divisible by $k,$ so $k=1,83.$ Plugging this back into $n$ yields $n=2(83)-5 = 161$ in the latter case.
Thus, the sum of all possible $n$ is just $78+161 = \boxed{239}.$ | null | 239 |
dfa5b61747b7b585e477651a47272ce2 | https://artofproblemsolving.com/wiki/index.php/2020_AIME_II_Problems/Problem_11 | Let $P(x) = x^2 - 3x - 7$ , and let $Q(x)$ and $R(x)$ be two quadratic polynomials also with the coefficient of $x^2$ equal to $1$ . David computes each of the three sums $P + Q$ $P + R$ , and $Q + R$ and is surprised to find that each pair of these sums has a common root, and these three common roots are distinct. If $Q(0) = 2$ , then $R(0) = \frac{m}{n}$ , where $m$ and $n$ are relatively prime positive integers. Find $m + n$ | Let $Q(x) = x^2 + ax + 2$ and $R(x) = x^2 + bx + c$ . We can write the following: \[P + Q = 2x^2 + (a - 3)x - 5\] \[P + R = 2x^2 + (b - 3)x + (c - 7)\] \[Q + R = 2x^2 + (a + b)x + (c + 2)\] Let the common root of $P+Q,P+R$ be $r$ $P+R,Q+R$ be $s$ ; and $P+Q,Q+R$ be $t$ . We then have that the roots of $P+Q$ are $r,t$ , the roots of $P + R$ are $r, s$ , and the roots of $Q + R$ are $s,t$
By Vieta's, we have: \[r + t = \dfrac{3 - a}{2}\tag{1}\] \[r + s = \dfrac{3 - b}{2}\tag{2}\] \[s + t = \dfrac{-a - b}{2}\tag{3}\] \[rt = \dfrac{-5}{2}\tag{4}\] \[rs = \dfrac{c - 7}{2}\tag{5}\] \[st = \dfrac{c + 2}{2}\tag{6}\]
Subtracting $(3)$ from $(1)$ , we get $r - s = \dfrac{3 + b}{2}$ . Adding this to $(2)$ , we get $2r = 3 \implies r = \dfrac{3}{2}$ . This gives us that $t = \dfrac{-5}{3}$ from $(4)$ . Substituting these values into $(5)$ and $(6)$ , we get $s = \dfrac{c-7}{3}$ and $s = \dfrac{-3c - 6}{10}$ . Equating these values, we get $\dfrac{c-7}{3} = \dfrac{-3c-6}{10} \implies c = \dfrac{52}{19} = R(0)$ . Thus, our answer is $52 + 19 = \boxed{071}$ . ~ TopNotchMath | null | 071 |
dfa5b61747b7b585e477651a47272ce2 | https://artofproblemsolving.com/wiki/index.php/2020_AIME_II_Problems/Problem_11 | Let $P(x) = x^2 - 3x - 7$ , and let $Q(x)$ and $R(x)$ be two quadratic polynomials also with the coefficient of $x^2$ equal to $1$ . David computes each of the three sums $P + Q$ $P + R$ , and $Q + R$ and is surprised to find that each pair of these sums has a common root, and these three common roots are distinct. If $Q(0) = 2$ , then $R(0) = \frac{m}{n}$ , where $m$ and $n$ are relatively prime positive integers. Find $m + n$ | We know that $P(x)=x^2-3x-7$
Since $Q(0)=2$ , the constant term in $Q(x)$ is $2$ . Let $Q(x)=x^2+ax+2$
Finally, let $R(x)=x^2+bx+c$
$P(x)+Q(x)=2x^2+(a-3)x-5$ . Let its roots be $p$ and $q$
$P(x)+R(x)=2x^2+(b-3)x+(c-7)$ Let its roots be $p$ and $r$
$Q(x)+R(x)=2x^2+(a+b)x+(c+2)$ . Let its roots be $q$ and $r$
By vietas, $p+q=\frac{3-a}{2}, p+r=\frac{3-b}{2}, q+r=\frac{-(a+b)}{2}$
We could work out the system of equations, but it's pretty easy to see that $p=\frac32, q=-\frac{a}{2}, r=-\frac{b}{2}$
$\text{Again, by vietas, }pq=-\frac52\text{, } pr=\frac{c-7}{2}\text{, } qr=\frac{c+2}{2}\text{, } \text{multiplying everything together a}\text{nd taking the sqrt of both sides,}$ \[(pqr)^2=\left(-\frac52\right)\left(\frac{c-7}{2}\right)\left(\frac{c+2}{2}\right)\] \[pqr=\sqrt{\left(-\frac52\right)\left(\frac{c-7}{2}\right)\left(\frac{c+2}{2}\right)}\] $\text{Dividing this }\text{equation by }qr=\frac{c+2}{2}$ \[\frac{pqr}{qr}=\frac{\sqrt{\left(-\frac52\right)\left(\frac{c-7}{2}\right)\left(\frac{c+2}{2}\right)}}{\frac{c+2}{2}}\] \[p = \frac{\sqrt{\left(-\frac52\right)\left(\frac{c-7}{2}\right)}}{\sqrt{\frac{c+2}{2}}}\] $\text{Recall th}\text{at }p=\frac32 \text{ and square both sides}$ \[\frac94=\frac{\left(-\frac52\right)\left(\frac{c-7}{2}\right)}{\frac{c+2}{2}}\] $\text{Solving gives } c=\frac{52}{19}, \text{ so our answer is }\boxed{071}$ | null | 071 |
9cde4ddc3daa7e0ba458c2a7f91f81db | https://artofproblemsolving.com/wiki/index.php/2020_AIME_II_Problems/Problem_12 | Let $m$ and $n$ be odd integers greater than $1.$ An $m\times n$ rectangle is made up of unit squares where the squares in the top row are numbered left to right with the integers $1$ through $n$ , those in the second row are numbered left to right with the integers $n + 1$ through $2n$ , and so on. Square $200$ is in the top row, and square $2000$ is in the bottom row. Find the number of ordered pairs $(m,n)$ of odd integers greater than $1$ with the property that, in the $m\times n$ rectangle, the line through the centers of squares $200$ and $2000$ intersects the interior of square $1099$ | Let us take some cases. Since $m$ and $n$ are odds, and $200$ is in the top row and $2000$ in the bottom, $m$ has to be $3$ $5$ $7$ , or $9$ . Also, taking a look at the diagram, the slope of the line connecting those centers has to have an absolute value of $< 1$ . Therefore, $m < 1800 \mod n < 1800-m$
If $m=3$ $n$ can range from $667$ to $999$ . However, $900$ divides $1800$ , so looking at mods, we can easily eliminate $899$ and $901$ . Now, counting these odd integers, we get $167 - 2 = 165$
Similarly, let $m=5$ . Then $n$ can range from $401$ to $499$ . However, $450|1800$ , so one can remove $449$ and $451$ . Counting odd integers, we get $50 - 2 = 48$
Take $m=7$ . Then, $n$ can range from $287$ to $333$ . However, $300|1800$ , so one can verify and eliminate $299$ and $301$ . Counting odd integers, we get $24 - 2 = 22$
Let $m = 9$ . Then $n$ can vary from $223$ to $249$ . However, $225|1800$ . Checking that value and the values around it, we can eliminate $225$ . Counting odd integers, we get $14 - 1 = 13$
Add all of our cases to get \[165+48+22+13 = \boxed{248}\] | null | 248 |
2ec2795a1609a050dbccc146a12c8a27 | https://artofproblemsolving.com/wiki/index.php/2020_AIME_II_Problems/Problem_13 | Convex pentagon $ABCDE$ has side lengths $AB=5$ $BC=CD=DE=6$ , and $EA=7$ . Moreover, the pentagon has an inscribed circle (a circle tangent to each side of the pentagon). Find the area of $ABCDE$ | Assume the incircle touches $AB$ $BC$ $CD$ $DE$ $EA$ at $P,Q,R,S,T$ respectively. Then let $PB=x=BQ=RD=SD$ $ET=y=ES=CR=CQ$ $AP=AT=z$ . So we have $x+y=6$ $x+z=5$ and $y+z$ =7, solve it we have $x=2$ $z=3$ $y=4$ . Let the center of the incircle be $I$ , by SAS we can proof triangle $BIQ$ is congruent to triangle $DIS$ , and triangle $CIR$ is congruent to triangle $SIE$ . Then we have $\angle AED=\angle BCD$ $\angle ABC=\angle CDE$ . Extend $CD$ , cross ray $AB$ at $M$ , ray $AE$ at $N$ , then by AAS we have triangle $END$ is congruent to triangle $BMC$ . Thus $\angle M=\angle N$ . Let $EN=MC=a$ , then $BM=DN=a+2$ . So by law of cosine in triangle $END$ and triangle $ANM$ we can obtain \[\frac{2a+8}{2(a+7)}=\cos N=\frac{a^2+(a+2)^2-36}{2a(a+2)}\] , solved it gives us $a=8$ , which yield triangle $ANM$ to be a triangle with side length 15, 15, 24, draw a height from $A$ to $NM$ divides it into two triangles with side lengths 9, 12, 15, so the area of triangle $ANM$ is 108. Triangle $END$ is a triangle with side lengths 6, 8, 10, so the area of two of them is 48, so the area of pentagon is $108-48=\boxed{60}$ | null | 60 |
2ec2795a1609a050dbccc146a12c8a27 | https://artofproblemsolving.com/wiki/index.php/2020_AIME_II_Problems/Problem_13 | Convex pentagon $ABCDE$ has side lengths $AB=5$ $BC=CD=DE=6$ , and $EA=7$ . Moreover, the pentagon has an inscribed circle (a circle tangent to each side of the pentagon). Find the area of $ABCDE$ | Suppose that the circle intersects $\overline{AB}$ $\overline{BC}$ $\overline{CD}$ $\overline{DE}$ , and $\overline{EA}$ at $P$ $Q$ $R$ $S$ , and $T$ respectively. Then $AT = AP = a$ $BP = BQ = b$ $CQ = CR = c$ $DR = DS = d$ , and $ES = ET = e$ . So $a + b = 5$ $b + c = 6$ $c + d = 6$ $d + e = 6$ , and $e + a = 7$ . Then $2a + 2b + 2c + 2d + 2e = 30$ , so $a + b + c + d + e= 15$ . Then we can solve for each individually. $a = 3$ $b = 2$ $c = 4$ $d = 2$ , and $e = 4$ . To find the radius, we notice that $4 \arctan(\frac{2}{r}) + 4 \arctan(\frac{4}{r}) + 2 \arctan (\frac{3}{r}) = 360 ^ \circ$ , or $2 \arctan(\frac{2}{r}) + 2 \arctan(\frac{4}{r}) + \arctan (\frac{3}{r}) = 180 ^ \circ$ . Each of these angles in this could be represented by complex numbers. When two complex numbers are multiplied, their angles add up to create the angle of the resulting complex number. Thus, $(r + 2i)^2 \cdot (r + 4i)^2 \cdot (r + 3i)$ is real. Expanding, we get:
\[(r^2 + 4ir - 4)(r^2 + 8ir -16)(r + 3i)\]
\[(r^4 + 12ir^3 - 52r^2 - 96ir + 64)(r + 3i)\]
On the last expanding, we only multiply the reals with the imaginaries and vice versa, because we only care that the imaginary component equals 0.
\[15ir^4 - 252ir^2 + 192i = 0\]
\[5r^4 - 84r^2 + 64 = 0\]
\[(5r^2 - 4)(r^2 - 16) = 0\]
$r$ must equal 4, as r cannot be negative or be approximately equal to 1.
Thus, the area of $ABCDE$ is $4 \cdot (a + b + c + d + e) = 4 \cdot 15 = \boxed{60}$ | null | 60 |
2ec2795a1609a050dbccc146a12c8a27 | https://artofproblemsolving.com/wiki/index.php/2020_AIME_II_Problems/Problem_13 | Convex pentagon $ABCDE$ has side lengths $AB=5$ $BC=CD=DE=6$ , and $EA=7$ . Moreover, the pentagon has an inscribed circle (a circle tangent to each side of the pentagon). Find the area of $ABCDE$ | This pentagon is very close to a regular pentagon with side lengths $6$ . The area of a regular pentagon with side lengths $s$ is $\frac{5s^2}{4\sqrt{5-2\sqrt{5}}}$ $5-2\sqrt{5}$ is slightly greater than $\frac{1}{2}$ given that $2\sqrt{5}$ is slightly less than $\frac{9}{2}$ $4\sqrt{5-2\sqrt{5}}$ is then slightly greater than $2\sqrt{2}$ . We will approximate that to be $2.9$ . The area is now roughly $\frac{180}{2.9}$ , but because the actual pentagon is not regular, but has the same perimeter of the regular one that we are comparing to we can say that this is an overestimate on the area and turn the $2.9$ into $3$ thus turning the area into $\frac{180}{3}$ which is $60$ and since $60$ is a multiple of the semiperimeter $15$ , we can safely say that the answer is most likely $\boxed{60}$ | null | 60 |
2ec2795a1609a050dbccc146a12c8a27 | https://artofproblemsolving.com/wiki/index.php/2020_AIME_II_Problems/Problem_13 | Convex pentagon $ABCDE$ has side lengths $AB=5$ $BC=CD=DE=6$ , and $EA=7$ . Moreover, the pentagon has an inscribed circle (a circle tangent to each side of the pentagon). Find the area of $ABCDE$ | Let $\omega$ be the inscribed circle, $I$ be its center, and $r$ be its radius. The area of $ABCDE$ is equal to its semiperimeter, $15,$ times $r$ , so the problem is reduced to finding $r$ . Let $a$ be the length of the tangent segment from $A$ to $\omega$ , and analogously define $b$ $c$ $d$ , and $e$ . Then $a+b=5$ $b+c= c+d=d+e=6$ , and $e+a=7$ , with a total of $a+b+c+d+e=15$ . Hence $a=3$ $b=d=2$ , and $c=e=4$ . It follows that $\angle B= \angle D$ and $\angle C= \angle E$ . Let $Q$ be the point where $\omega$ is tangent to $\overline{CD}$ . Then $\angle IAE = \angle IAB =\frac{1}{2}\angle A$ . Now we claim that points $A, I, Q$ are collinear, which can be proved if $\angle{AIQ}=\angle{QIA}=180^{\circ}$ . The sum of the internal angles in polygons $ABCQI$ and $AIQDE$ are equal, so $\angle IAE + \angle AIQ + \angle IQD + \angle D + \angle E = \angle IAB + \angle B + \angle C + \angle CQI + \angle QIA$ , which implies that $\angle AIQ$ must be $180^\circ$ . Therefore points $A$ $I$ , and $Q$ are collinear. [asy] defaultpen(fontsize(8pt)); unitsize(0.025cm); pair[] vertices = {(0,0), (5,0), (8.6,4.8), (3.8,8.4), (-1.96, 6.72)}; string[] labels = {"$A$", "$B$", "$C$", "$D$", "$E$"}; pair[] dirs = {SW, SE,E, N, NW}; string[] smallLabels = {"$a$", "$b$", "$c$", "$d$", "$e$"}; pair I = (3,4); real rad = 4; pair Q = foot(I, vertices[2], vertices[3]); pair[] interpoints = {}; for(int i =0; i<vertices.length; ++i){ interpoints.push(foot(I, vertices[i], vertices[(i+1)%vertices.length])); } for(int i = 0; i< vertices.length; ++i){ draw(vertices[i]--vertices[(i+1)%vertices.length]); dot(labels[i],vertices[i],dirs[i]); draw(I--vertices[i]); } draw(Circle(I, rad)); dot("$I$", I, dir(200)); draw(I--Q); dot("$Q$", Q, NE); for(int i = 0; i < vertices.length; ++i){ label(smallLabels[i], vertices[i] --interpoints[i]); //dot(interpoints[i], blue); label(smallLabels[i], interpoints[(i-1)%vertices.length] -- vertices[i]); } [/asy] Because $\overline{AQ} \perp \overline{CD}$ , it follows that \[AC^2-AD^2=CQ^2-DQ^2=c^2-d^2=12.\] Another expression for $AC^2-AD^2$ can be found as follows. Note that $\tan \left(\frac{\angle B}{2}\right) = \frac{r}{2}$ and $\tan \left(\frac{\angle E}{2}\right) = \frac{r}{4}$ , so \[\cos (\angle B) =\frac{1-\tan^2 \left(\frac{\angle B}{2}\right)}{1+\tan^2 \left(\frac{\angle B}{2}\right)} = \frac{4-r^2}{4+r^2}\] and \[\cos (\angle E) = \frac{1-\tan^2 \left(\frac{\angle E}{2}\right)}{1+\tan^2 \left(\frac{\angle E}{2}\right)}= \frac{16-r^2}{16+r^2}.\] Applying the Law of Cosines to $\triangle ABC$ and $\triangle AED$ gives \[AC^2=AB^2+BC^2-2\cdot AB\cdot BC\cdot \cos (\angle B) = 5^2+6^2-2 \cdot 5 \cdot 6 \cdot \frac{4-r^2}{4+r^2}\] and \[AD^2=AE^2+DE^2-2 \cdot AE \cdot DE \cdot \cos(\angle E) = 7^2+6^2-2 \cdot 7 \cdot 6 \cdot \frac{16-r^2}{16+r^2}.\] Hence
\[12=AC^2- AD^2= 5^2-2\cdot 5 \cdot 6\cdot \frac{4-r^2}{4+r^2} -7^2+2\cdot 7 \cdot 6 \cdot \frac{16-r^2}{16+r^2},\] yielding \[2\cdot 7 \cdot 6 \cdot \frac{16-r^2}{16+r^2}- 2\cdot 5 \cdot 6\cdot \frac{4-r^2}{4+r^2}= 36;\] equivalently \[7(16-r^2)(4+r^2)-5(4-r^2)(16+r^2) = 3(16+r^2)(4+r^2).\] Substituting $x=r^2$ gives the quadratic equation $5x^2-84x+64=0$ , with solutions $\frac{42 - 38}{5}=\frac45$ , and $\frac{42 + 38}{5}= 16$ . The solution $r^2=\frac45$ corresponds to a five-pointed star, which is not convex. Indeed, if $r<3$ , then $\tan \left(\frac{\angle A}{2}\right)$ $\tan \left(\frac{\angle C}{2}\right)$ , and $\tan \left(\frac{\angle E}{2}\right)$ are less than $1,$ implying that $\angle A$ $\angle C$ , and $\angle E$ are acute, which cannot happen in a convex pentagon. Thus $r^2=16$ and $r=4$ . The requested area is $15\cdot4 = \boxed{60}$ | null | 60 |
a2db8574305e481f11ae98dcf9222704 | https://artofproblemsolving.com/wiki/index.php/2020_AIME_II_Problems/Problem_14 | For a real number $x$ let $\lfloor x\rfloor$ be the greatest integer less than or equal to $x$ , and define $\{x\} = x - \lfloor x \rfloor$ to be the fractional part of $x$ . For example, $\{3\} = 0$ and $\{4.56\} = 0.56$ . Define $f(x)=x\{x\}$ , and let $N$ be the number of real-valued solutions to the equation $f(f(f(x)))=17$ for $0\leq x\leq 2020$ . Find the remainder when $N$ is divided by $1000$ | To solve $f(f(f(x)))=17$ , we need to solve $f(x) = y$ where $f(f(y))=17$ , and to solve that we need to solve $f(y) = z$ where $f(z) = 17$
It is clear to see for some integer $a \geq 17$ there is exactly one value of $z$ in the interval $[a, a+1)$ where $f(z) = 17$ . To understand this, imagine the graph of $f(z)$ on the interval $[a, a+1)$ The graph starts at $0$ , is continuous and increasing, and approaches $a+1$ . So as long as $a+1 > 17$ , there will be a solution for $z$ in the interval.
Using this logic, we can find the number of solutions to $f(x) = y$ . For every interval $[a, a+1)$ where $a \geq \left \lfloor{y}\right \rfloor$ there will be one solution for $x$ in that interval. However, the question states $0 \leq x \leq 2020$ , but because $x=2020$ doesn't work we can change it to $0 \leq x < 2020$ . Therefore, $\left \lfloor{y}\right \rfloor \leq a \leq 2019$ , and there are $2020 - \left \lfloor{y}\right \rfloor$ solutions to $f(x) = y$
We can solve $f(y) = z$ similarly. $0 \leq y < 2020$ to satisfy the bounds of $x$ , so there are $2020 - \left \lfloor{z}\right \rfloor$ solutions to $f(y) = z$ , and $0 \leq z < 2020$ to satisfy the bounds of $y$
Going back to $f(z) = 17$ , there is a single solution for z in the interval $[a, a+1)$ , where $17 \leq a \leq 2019$ . (We now have an upper bound for $a$ because we know $z < 2020$ .) There are $2003$ solutions for $z$ , and the floors of these solutions create the sequence $17, 18, 19, ..., 2018, 2019$
Lets first look at the solution of $z$ where $\left \lfloor{z}\right \rfloor = 17$ . Then $f(y) = z$ would have $2003$ solutions, and the floors of these solutions would also create the sequence $17, 18, 19, ..., 2018, 2019$
If we used the solution of $y$ where $\left \lfloor{y}\right \rfloor = 17$ , there would be $2003$ solutions for $f(x) = y$ . If we used the solution of $y$ where $\left \lfloor{y}\right \rfloor = 18$ , there would be $2002$ solutions for $x$ , and so on. So for the solution of $z$ where $\left \lfloor{z}\right \rfloor = 17$ , there will be $2003 + 2002 + 2001 + ... + 2 + 1 = \binom{2004}{2}$ solutions for $x$
If we now look at the solution of $z$ where $\left \lfloor{z}\right \rfloor = 18$ , there would be $\binom{2003}{2}$ solutions for $x$ . If we looked at the solution of $z$ where $\left \lfloor{z}\right \rfloor = 19$ , there would be $\binom{2002}{2}$ solutions for $x$ , and so on.
The total number of solutions to $x$ is $\binom{2004}{2} + \binom{2003}{2} + \binom{2002}{2} + ... + \binom{3}{2} + \binom{2}{2}$ . Using the hockey stick theorem, we see this equals $\binom{2005}{3}$ , and when we take the remainder of that number when divided by $1000$ , we get the answer, $\boxed{10}$ | null | 10 |
748b45bb3f22aa9b45ce1e7a7bf1580e | https://artofproblemsolving.com/wiki/index.php/2020_AIME_II_Problems/Problem_15 | Let $\triangle ABC$ be an acute scalene triangle with circumcircle $\omega$ . The tangents to $\omega$ at $B$ and $C$ intersect at $T$ . Let $X$ and $Y$ be the projections of $T$ onto lines $AB$ and $AC$ , respectively. Suppose $BT = CT = 16$ $BC = 22$ , and $TX^2 + TY^2 + XY^2 = 1143$ . Find $XY^2$ | Let $O$ be the circumcenter of $\triangle ABC$ ; say $OT$ intersects $BC$ at $M$ ; draw segments $XM$ , and $YM$ . We have $MT=3\sqrt{15}$
Since $\angle A=\angle CBT=\angle BCT$ , we have $\cos A=\tfrac{11}{16}$ . Notice that $AXTY$ is cyclic, so $\angle XTY=180^{\circ}-A$ , so $\cos XTY=-\cos A$ , and the cosine law in $\triangle TXY$ gives \[1143-2XY^2=-\frac{11}{8}\cdot XT\cdot YT.\]
Since $\triangle BMT \cong \triangle CMT$ , we have $TM\perp BC$ , and therefore quadrilaterals $BXTM$ and $CYTM$ are cyclic. Let $P$ (resp. $Q$ ) be the midpoint of $BT$ (resp. $CT$ ). So $P$ (resp. $Q$ ) is the center of $(BXTM)$ (resp. $CYTM$ ). Then $\theta=\angle ABC=\angle MTX$ and $\phi=\angle ACB=\angle YTM$ . So $\angle XPM=2\theta$ , so \[\frac{\frac{XM}{2}}{XP}=\sin \theta,\] which yields $XM=2XP\sin \theta=BT(=CT)\sin \theta=TY$ . Similarly we have $YM=XT$
Ptolemy's theorem in $BXTM$ gives \[16TY=11TX+3\sqrt{15}BX,\] while Pythagoras' theorem gives $BX^2+XT^2=16^2$ . Similarly, Ptolemy's theorem in $YTMC$ gives \[16TX=11TY+3\sqrt{15}CY\] while Pythagoras' theorem in $\triangle CYT$ gives $CY^2+YT^2=16^2$ . Solve this for $XT$ and $TY$ and substitute into the equation about $\cos XTY$ to obtain the result $XY^2=\boxed{717}$ | null | 717 |
748b45bb3f22aa9b45ce1e7a7bf1580e | https://artofproblemsolving.com/wiki/index.php/2020_AIME_II_Problems/Problem_15 | Let $\triangle ABC$ be an acute scalene triangle with circumcircle $\omega$ . The tangents to $\omega$ at $B$ and $C$ intersect at $T$ . Let $X$ and $Y$ be the projections of $T$ onto lines $AB$ and $AC$ , respectively. Suppose $BT = CT = 16$ $BC = 22$ , and $TX^2 + TY^2 + XY^2 = 1143$ . Find $XY^2$ | Let $M$ denote the midpoint of $\overline{BC}$ . The critical claim is that $M$ is the orthocenter of $\triangle AXY$ , which has the circle with diameter $\overline{AT}$ as its circumcircle. To see this, note that because $\angle BXT = \angle BMT = 90^\circ$ , the quadrilateral $MBXT$ is cyclic, it follows that \[\angle MXA = \angle MXB = \angle MTB = 90^\circ - \angle TBM = 90^\circ - \angle A,\] implying that $\overline{MX} \perp \overline{AC}$ . Similarly, $\overline{MY} \perp \overline{AB}$ . In particular, $MXTY$ is a parallelogram. [asy] defaultpen(fontsize(8pt)); unitsize(0.8cm); pair A = (0,0); pair B = (-1.26,-4.43); pair C = (-1.26+3.89, -4.43); pair M = (B+C)/2; pair O = circumcenter(A,B,C); pair T = (0.68, -6.49); pair X = foot(T,A,B); pair Y = foot(T,A,C); path omega = circumcircle(A,B,C); real rad = circumradius(A,B,C); filldraw(A--B--C--cycle, 0.2*royalblue+white); label("$\omega$", O + rad*dir(45), SW); //filldraw(T--Y--M--X--cycle, rgb(150, 247, 254)); filldraw(T--Y--M--X--cycle, 0.2*heavygreen+white); draw(M--T); draw(X--Y); draw(B--T--C); draw(A--X--Y--cycle); draw(omega); dot("$X$", X, W); dot("$Y$", Y, E); dot("$O$", O, W); dot("$T$", T, S); dot("$A$", A, N); dot("$B$", B, W); dot("$C$", C, E); dot("$M$", M, N); [/asy] Hence, by the Parallelogram Law, \[TM^2 + XY^2 = 2(TX^2 + TY^2) = 2(1143-XY^2).\] But $TM^2 = TB^2 - BM^2 = 16^2 - 11^2 = 135$ . Therefore \[XY^2 = \frac13(2 \cdot 1143-135) = \boxed{717}.\] | null | 717 |
748b45bb3f22aa9b45ce1e7a7bf1580e | https://artofproblemsolving.com/wiki/index.php/2020_AIME_II_Problems/Problem_15 | Let $\triangle ABC$ be an acute scalene triangle with circumcircle $\omega$ . The tangents to $\omega$ at $B$ and $C$ intersect at $T$ . Let $X$ and $Y$ be the projections of $T$ onto lines $AB$ and $AC$ , respectively. Suppose $BT = CT = 16$ $BC = 22$ , and $TX^2 + TY^2 + XY^2 = 1143$ . Find $XY^2$ | Let $H$ be the orthocenter of $\triangle AXY$
Lemma 1: $H$ is the midpoint of $BC$
Proof: Let $H'$ be the midpoint of $BC$ , and observe that $XBH'T$ and $TH'CY$ are cyclical. Define $H'Y \cap BA=E$ and $H'X \cap AC=F$ , then note that: \[\angle H'BT=\angle H'CT=\angle H'XT=\angle H'YT=\angle A.\] That implies that $\angle H'XB=\angle H'YC=90^\circ-\angle A$ $\angle CH'Y=\angle EH'B=90^\circ-\angle B$ , and $\angle BH'Y=\angle FH'C=90^\circ-\angle C$ . Thus $YH'\perp AX$ and $XH' \perp AY$ $H'$ is indeed the same as $H$ , and we have proved lemma 1.
Since $AXTY$ is cyclical, $\angle XTY=\angle XHY$ and this implies that $XHYT$ is a paralelogram.
By the Law of Cosines: \[XY^2=XT^2+TY^2+2(XT)(TY)\cdot \cos(\angle A)\] \[XY^2=XH^2+HY^2+2(XH)(HY) \cdot \cos(\angle A)\] \[HT^2=HX^2+XT^2-2(HX)(XT) \cdot \cos(\angle A)\] \[HT^2=HY^2+YT^2-2(HY)(YT) \cdot \cos(\angle A).\] We add all these equations to get: \[HT^2+XY^2=2(XT^2+TY^2) \qquad (1).\] We have that $BH=HC=11$ and $BT=TC=16$ using our midpoints. Note that $HT \perp BC$ , so by the Pythagorean Theorem, it follows that $HT^2=135$ . We were also given that $XT^2+TY^2=1143-XY^2$ , which we multiply by $2$ to use equation $(1)$ \[2(XT^2+TY^2)=2286-2 \cdot XY^2\] Since $2(XT^2+TY^2)=2(HT^2+TY^2)=HT^2+XY^2$ , we have \[135+XY^2=2286-2 \cdot XY^2\] \[3 \cdot XY^2=2151.\] Therefore, $XY^2=\boxed{717}$ . ~ MathLuis | null | 717 |
748b45bb3f22aa9b45ce1e7a7bf1580e | https://artofproblemsolving.com/wiki/index.php/2020_AIME_II_Problems/Problem_15 | Let $\triangle ABC$ be an acute scalene triangle with circumcircle $\omega$ . The tangents to $\omega$ at $B$ and $C$ intersect at $T$ . Let $X$ and $Y$ be the projections of $T$ onto lines $AB$ and $AC$ , respectively. Suppose $BT = CT = 16$ $BC = 22$ , and $TX^2 + TY^2 + XY^2 = 1143$ . Find $XY^2$ | Using the Claim (below) we get $\triangle ABC \sim \triangle XTM \sim \triangle YMT.$
Corresponding sides of similar $\triangle XTM \sim \triangle YMT$ is $MT,$ so
$\triangle XTM = \triangle YMT \implies MY = XT, MX = TY \implies XMYT$ – parallelogram.
\[4 TD^2 = MT^2 = \sqrt{BT^2 - BM^2} =\sqrt{153}.\] The formula for median $DT$ of triangle $XYT$ is \[2 DT^2 = XT^2 + TY^2 – \frac{XY^2}{2},\] \[3 \cdot XY^2 = 2XT^2 + 2TY^2 + 2XY^2 – 4 DT^2,\] \[3 \cdot XY^2 = 2 \cdot 1143-153 = 2151 \implies XY^2 = \boxed{717}.\] | null | 717 |
fce5e4bed1cc9509c2a1ff4683f8cb8a | https://artofproblemsolving.com/wiki/index.php/2019_AIME_I_Problems/Problem_1 | Consider the integer \[N = 9 + 99 + 999 + 9999 + \cdots + \underbrace{99\ldots 99}_\text{321 digits}.\] Find the sum of the digits of $N$ | Let's express the number in terms of $10^n$ . We can obtain $(10-1)+(10^2-1)+(10^3-1)+\cdots+(10^{321}-1)$ . By the commutative and associative property, we can group it into $(10+10^2+10^3+\cdots+10^{321})-321$ . We know the former will yield $1111....10$ , so we only have to figure out what the last few digits are. There are currently $321$ 1's. We know the last four digits are $1110$ , and that the others will not be affected if we subtract $321$ . If we do so, we get that $1110-321=789$ . This method will remove three $1$ 's, and add a $7$ $8$ and $9$ . Therefore, the sum of the digits is $(321-3)+7+8+9=\boxed{342}$ | null | 342 |
fce5e4bed1cc9509c2a1ff4683f8cb8a | https://artofproblemsolving.com/wiki/index.php/2019_AIME_I_Problems/Problem_1 | Consider the integer \[N = 9 + 99 + 999 + 9999 + \cdots + \underbrace{99\ldots 99}_\text{321 digits}.\] Find the sum of the digits of $N$ | Observe how adding results in the last term but with a $1$ concatenated in front and also a $1$ subtracted ( $09$ $108$ $1107$ $11106$ ). Then for any index of terms, $n$ , the sum is $11...10-n$ , where the first term is of length $n+1$ . Here, that is $\boxed{342}$ | null | 342 |
92ec3ed45a5124c09bd4a43b4d3b8bfc | https://artofproblemsolving.com/wiki/index.php/2019_AIME_I_Problems/Problem_2 | Jenn randomly chooses a number $J$ from $1, 2, 3,\ldots, 19, 20$ . Bela then randomly chooses a number $B$ from $1, 2, 3,\ldots, 19, 20$ distinct from $J$ . The value of $B - J$ is at least $2$ with a probability that can be expressed in the form $\frac{m}{n}$ where $m$ and $n$ are relatively prime positive integers. Find $m+n$ | $B-J \ne 0$ because $B \ne J$ , so the probability that $B-J < 0$ is $\frac{1}{2}$ by symmetry.
The probability that $B-J = 1$ is $\frac{19}{20 \times 19} = \frac{1}{20}$ because there are 19 pairs: $(B,J) = (2,1), \ldots, (20,19)$
The probability that $B-J \ge 2$ is $1-\frac{1}{2}-\frac{1}{20} = \frac{9}{20} \implies \boxed{029}$ | null | 029 |
92ec3ed45a5124c09bd4a43b4d3b8bfc | https://artofproblemsolving.com/wiki/index.php/2019_AIME_I_Problems/Problem_2 | Jenn randomly chooses a number $J$ from $1, 2, 3,\ldots, 19, 20$ . Bela then randomly chooses a number $B$ from $1, 2, 3,\ldots, 19, 20$ distinct from $J$ . The value of $B - J$ is at least $2$ with a probability that can be expressed in the form $\frac{m}{n}$ where $m$ and $n$ are relatively prime positive integers. Find $m+n$ | By symmetry, the desired probability is equal to the probability that $J - B$ is at most $-2$ , which is $\frac{1-P}{2}$ where $P$ is the probability that $B$ and $J$ differ by $1$ (no zero, because the two numbers are distinct). There are $20 \cdot 19 = 380$ total possible combinations of $B$ and $J$ , and $1 + 18 \cdot 2 + 1 = 38$ ones that form $P$ , so $P = \frac{38}{380} = \frac{1}{10}$ . Therefore the answer is $\frac{9}{20} \rightarrow \boxed{029}$ | null | 029 |
92ec3ed45a5124c09bd4a43b4d3b8bfc | https://artofproblemsolving.com/wiki/index.php/2019_AIME_I_Problems/Problem_2 | Jenn randomly chooses a number $J$ from $1, 2, 3,\ldots, 19, 20$ . Bela then randomly chooses a number $B$ from $1, 2, 3,\ldots, 19, 20$ distinct from $J$ . The value of $B - J$ is at least $2$ with a probability that can be expressed in the form $\frac{m}{n}$ where $m$ and $n$ are relatively prime positive integers. Find $m+n$ | This problem is essentially asking how many ways there are to choose $2$ distinct elements from a $20$ element set such that no $2$ elements are adjacent. Using the well-known formula $\dbinom{n-k+1}{k}$ , there are $\dbinom{20-2+1}{2} = \dbinom{19}{2} = 171$ ways. Dividing $171$ by $380$ , our desired probability is $\frac{171}{380} = \frac{9}{20}$ . Thus, our answer is $9+20=\boxed{029}$ .
-Fidgetboss_4000 | null | 029 |
92ec3ed45a5124c09bd4a43b4d3b8bfc | https://artofproblemsolving.com/wiki/index.php/2019_AIME_I_Problems/Problem_2 | Jenn randomly chooses a number $J$ from $1, 2, 3,\ldots, 19, 20$ . Bela then randomly chooses a number $B$ from $1, 2, 3,\ldots, 19, 20$ distinct from $J$ . The value of $B - J$ is at least $2$ with a probability that can be expressed in the form $\frac{m}{n}$ where $m$ and $n$ are relatively prime positive integers. Find $m+n$ | Create a grid using graph paper, with $20$ columns for the values of $J$ from $1$ to $20$ and $20$ rows for the values of $B$ from $1$ to $20$ . Since $B$ cannot equal $J$ , we cross out the diagonal line from the first column of the first row to the twentieth column of the last row. Now, since $B - J$ must be at least $2$ , we can mark the line where $B - J = 2$ . Now we sum the number of squares that are on this line and below it. We get $171$ . Then we find the number of total squares, which is $400 - 20 = 380$ . Finally, we take the ratio $\frac{171}{380}$ , which simplifies to $\frac{9}{20}$ . Our answer is $9+20=\boxed{029}$ | null | 029 |
92ec3ed45a5124c09bd4a43b4d3b8bfc | https://artofproblemsolving.com/wiki/index.php/2019_AIME_I_Problems/Problem_2 | Jenn randomly chooses a number $J$ from $1, 2, 3,\ldots, 19, 20$ . Bela then randomly chooses a number $B$ from $1, 2, 3,\ldots, 19, 20$ distinct from $J$ . The value of $B - J$ is at least $2$ with a probability that can be expressed in the form $\frac{m}{n}$ where $m$ and $n$ are relatively prime positive integers. Find $m+n$ | We can see that if $B$ chooses $20$ $J$ can choose from $1$ through $18$ such that $B-J\geq 2$ . If $B$ chooses $19$ $J$ has choices $1$ $17$ . By continuing this pattern, $B$ will choose $3$ and $J$ will have $1$ option. Summing up the total, we get $18+17+\cdots+1$ as the total number of solutions. The total amount of choices is $20\times19$ (B and J must choose different numbers), so the probability is $\frac{18\cdot19\div2}{20\cdot19}=\frac{9}{20}$ . Therefore, the answer is $9+20=\boxed{029}$ | null | 029 |
92ec3ed45a5124c09bd4a43b4d3b8bfc | https://artofproblemsolving.com/wiki/index.php/2019_AIME_I_Problems/Problem_2 | Jenn randomly chooses a number $J$ from $1, 2, 3,\ldots, 19, 20$ . Bela then randomly chooses a number $B$ from $1, 2, 3,\ldots, 19, 20$ distinct from $J$ . The value of $B - J$ is at least $2$ with a probability that can be expressed in the form $\frac{m}{n}$ where $m$ and $n$ are relatively prime positive integers. Find $m+n$ | Similar to solution 4, we can go through the possible values of $J$ to find all the values of $B$ that makes $B-J\geq 2$ . If $J$ chooses $1$ , then $B$ can choose anything from $3$ to $20$ . If $J$ chooses $2$ , then $B$ can choose anything from $4$ to $20$ . By continuing this pattern, we can see that there is $18+17+\cdots+1$ possible solutions. The amount of solutions is, therefore, $\frac{18\cdot19}{2}=171$ . Now, because $B$ and $J$ must be different, we have $20\times19=380$ possible choices, so the probability is $\frac{171}{380}=\frac{9}{20}$ . Therefore, the final answer is $9+20=\boxed{029}$ | null | 029 |
a228ae39c87eef072e3a5ea1618eb54c | https://artofproblemsolving.com/wiki/index.php/2019_AIME_I_Problems/Problem_3 | In $\triangle PQR$ $PR=15$ $QR=20$ , and $PQ=25$ . Points $A$ and $B$ lie on $\overline{PQ}$ , points $C$ and $D$ lie on $\overline{QR}$ , and points $E$ and $F$ lie on $\overline{PR}$ , with $PA=QB=QC=RD=RE=PF=5$ . Find the area of hexagon $ABCDEF$ | We know the area of the hexagon $ABCDEF$ to be $\triangle PQR- \triangle PAF- \triangle BCQ- \triangle RED$ . Since $PR^2+RQ^2=PQ^2$ , we know that $\triangle PRQ$ is a right triangle. Thus the area of $\triangle PQR$ is $150$ . Another way to compute the area is \[\frac12 \cdot PQ\cdot RQ \sin \angle PQR = \frac12 \cdot 500 \cdot \sin \angle PQR=150 \implies \sin \angle PQR = \frac35.\] Then the area of $\triangle BCQ = \frac12 \cdot BQ \cdot CQ \cdot \sin \angle PQR= \frac{25}{2}\cdot \frac{3}{5}=\frac{15}{2}$ . Preceding in a similar fashion for $\triangle PAF$ , the area of $\triangle PAF$ is $10$ . Since $\angle ERD = 90^{\circ}$ , the area of $\triangle RED=\frac{25}{2}$ . Thus our desired answer is $150-\frac{15}{2}-10-\frac{25}{2}=\boxed{120}$ | null | 120 |
a228ae39c87eef072e3a5ea1618eb54c | https://artofproblemsolving.com/wiki/index.php/2019_AIME_I_Problems/Problem_3 | In $\triangle PQR$ $PR=15$ $QR=20$ , and $PQ=25$ . Points $A$ and $B$ lie on $\overline{PQ}$ , points $C$ and $D$ lie on $\overline{QR}$ , and points $E$ and $F$ lie on $\overline{PR}$ , with $PA=QB=QC=RD=RE=PF=5$ . Find the area of hexagon $ABCDEF$ | Let $R$ be the origin. Noticing that the triangle is a 3-4-5 right triangle, we can see that $A=(4,12), B=(16,3), C=(15,0), D=(5,0), E=(0,5)$ , and $F=(0,10)$ . Using the shoelace theorem, the area is $\boxed{120}$ .
Shoelace theorem:Suppose the polygon $P$ has vertices $(a_1, b_1)$ $(a_2, b_2)$ , ... , $(a_n, b_n)$ , listed in clockwise order. Then the area of $P$ is | null | 120 |
a228ae39c87eef072e3a5ea1618eb54c | https://artofproblemsolving.com/wiki/index.php/2019_AIME_I_Problems/Problem_3 | In $\triangle PQR$ $PR=15$ $QR=20$ , and $PQ=25$ . Points $A$ and $B$ lie on $\overline{PQ}$ , points $C$ and $D$ lie on $\overline{QR}$ , and points $E$ and $F$ lie on $\overline{PR}$ , with $PA=QB=QC=RD=RE=PF=5$ . Find the area of hexagon $ABCDEF$ | Note that $\triangle{PQR}$ has area $150$ and is a $3$ $4$ $5$ right triangle. Then, by similar triangles, the altitude from $B$ to $QC$ has length $3$ and the altitude from $A$ to $FP$ has length $4$ , so $[QBC]+[DRE]+[AFP]=\frac{15}{2}+\frac{25}{2}+\frac{20}{2}=30$ , meaning that $[ABCDEF]=\boxed{120}$ .
-Stormersyle | null | 120 |
a228ae39c87eef072e3a5ea1618eb54c | https://artofproblemsolving.com/wiki/index.php/2019_AIME_I_Problems/Problem_3 | In $\triangle PQR$ $PR=15$ $QR=20$ , and $PQ=25$ . Points $A$ and $B$ lie on $\overline{PQ}$ , points $C$ and $D$ lie on $\overline{QR}$ , and points $E$ and $F$ lie on $\overline{PR}$ , with $PA=QB=QC=RD=RE=PF=5$ . Find the area of hexagon $ABCDEF$ | Knowing that $\triangle{PQR}$ has area $150$ and is a $3$ $4$ $5$ triangle, we can find the area of the smaller triangles $\triangle{DRE}$ $\triangle{APF}$ , and $\triangle{CQB}$ and subtract them from $\triangle{PQR}$ to obtain our answer. First off, we know $\triangle{DRE}$ has area $12.5$ since it is a right triangle. To the find the areas of $\triangle{APF}$ and $\triangle{CQB}$ , we can use Law of Cosines ( $c^2 = a^2 + b^2 - 2ab\cos C$ ) to find the lengths of $AF$ and $CB$ , respectively. Computing gives $AF = \sqrt{20}$ and $CB = \sqrt{10}$ . Now, using Heron's Formula, we find $\triangle{APF} = 10$ and $\triangle{CQB} = 7.5$ . Adding these and subtracting from $\triangle{PQR}$ , we get $150 - (10 + 7.5 + 12.5) = \boxed{120}$ -Starsher | null | 120 |
e1a68337dd7f68ad5b9db554e36e4ad1 | https://artofproblemsolving.com/wiki/index.php/2019_AIME_I_Problems/Problem_4 | A soccer team has $22$ available players. A fixed set of $11$ players starts the game, while the other $11$ are available as substitutes. During the game, the coach may make as many as $3$ substitutions, where any one of the $11$ players in the game is replaced by one of the substitutes. No player removed from the game may reenter the game, although a substitute entering the game may be replaced later. No two substitutions can happen at the same time. The players involved and the order of the substitutions matter. Let $n$ be the number of ways the coach can make substitutions during the game (including the possibility of making no substitutions). Find the remainder when $n$ is divided by $1000$ | There are $0-3$ substitutions. The number of ways to sub any number of times must be multiplied by the previous number. This is defined recursively. The case for $0$ subs is $1$ , and the ways to reorganize after $n$ subs is the product of the number of new subs ( $12-n$ ) and the players that can be ejected ( $11$ ). The formula for $n$ subs is then $a_n=11(12-n)a_{n-1}$ with $a_0=1$
Summing from $0$ to $3$ gives $1+11^2+11^{3}\cdot 10+11^{4}\cdot 10\cdot 9$ . Notice that $10+9\cdot11\cdot10=10+990=1000$ . Then, rearrange it into $1+11^2+11^3\cdot (10+11\cdot10\cdot9)= 1+11^2+11^3\cdot (1000)$ . When taking modulo $1000$ , the last term goes away. What is left is $1+11^2=\boxed{122}$ | null | 122 |
3589e36f9b9d8ce349f616d2ff6e7903 | https://artofproblemsolving.com/wiki/index.php/2019_AIME_I_Problems/Problem_5 | A moving particle starts at the point $(4,4)$ and moves until it hits one of the coordinate axes for the first time. When the particle is at the point $(a,b)$ , it moves at random to one of the points $(a-1,b)$ $(a,b-1)$ , or $(a-1,b-1)$ , each with probability $\frac{1}{3}$ , independently of its previous moves. The probability that it will hit the coordinate axes at $(0,0)$ is $\frac{m}{3^n}$ , where $m$ and $n$ are positive integers such that $m$ is not divisible by $3$ . Find $m + n$ | One could recursively compute the probabilities of reaching $(0,0)$ as the first axes point from any point $(x,y)$ as \[P(x,y) = \frac{1}{3} P(x-1,y) + \frac{1}{3} P(x,y-1) + \frac{1}{3} P(x-1,y-1)\] for $x,y \geq 1,$ and the base cases are $P(0,0) = 1, P(x,0) = P(y,0) = 0$ for any $x,y$ not equal to zero.
We then recursively find $P(4,4) = \frac{245}{2187}$ so the answer is $245 + 7 = \boxed{252}$ | null | 252 |
3589e36f9b9d8ce349f616d2ff6e7903 | https://artofproblemsolving.com/wiki/index.php/2019_AIME_I_Problems/Problem_5 | A moving particle starts at the point $(4,4)$ and moves until it hits one of the coordinate axes for the first time. When the particle is at the point $(a,b)$ , it moves at random to one of the points $(a-1,b)$ $(a,b-1)$ , or $(a-1,b-1)$ , each with probability $\frac{1}{3}$ , independently of its previous moves. The probability that it will hit the coordinate axes at $(0,0)$ is $\frac{m}{3^n}$ , where $m$ and $n$ are positive integers such that $m$ is not divisible by $3$ . Find $m + n$ | Obviously, the only way to reach (0,0) is to get to (1,1) and then have a $\frac{1}{3}$ chance to get to (0,0). Let x denote a move left 1 unit, y denote a move down 1 unit, and z denote a move left and down one unit each. The possible cases for these moves are $(x,y,z)=(0,0,3),(1,1,2),(2,2,1)$ and $(3,3,0)$ . This gives a probability of $1 \cdot \frac{1}{27} + \frac{4!}{2!} \cdot \frac{1}{81} + \frac{5!}{2! \cdot 2!} \cdot \frac{1}{243} +\frac{6!}{3! \cdot 3!} \cdot \frac{1}{729}=\frac{245}{729}$ to get to $(1,1)$ . The probability of reaching $(0,0)$ is $\frac{245}{3^7}$ . This gives $245+7=\boxed{252}$ | null | 252 |
3589e36f9b9d8ce349f616d2ff6e7903 | https://artofproblemsolving.com/wiki/index.php/2019_AIME_I_Problems/Problem_5 | A moving particle starts at the point $(4,4)$ and moves until it hits one of the coordinate axes for the first time. When the particle is at the point $(a,b)$ , it moves at random to one of the points $(a-1,b)$ $(a,b-1)$ , or $(a-1,b-1)$ , each with probability $\frac{1}{3}$ , independently of its previous moves. The probability that it will hit the coordinate axes at $(0,0)$ is $\frac{m}{3^n}$ , where $m$ and $n$ are positive integers such that $m$ is not divisible by $3$ . Find $m + n$ | Since the particle stops at one of the axes, we know that the particle most pass through $(1,1)$ . Thus, it suffices to consider the probability our particle will reach $(1,1)$ . Then the only ways to get to $(1,1)$ from $(4,4)$ are the following:
(1) 3 moves diagonally
(2) 2 moves diagonally, 1 move left, 1 move down
(3) 1 move diagonally, 2 moves left and 2 moves down.
(4) 3 moves left, 3 moves down.
The probability of (1) is $\frac{1}{3^3}$ . The probability of (2) is $\frac{\frac{4!}{2!}}{3^4} = \frac{12}{3^4}$ . The probability of (3) is $\frac{\frac{5!}{2!2!}}{3^5} = \frac{30}{3^5}$ . The probability of (4) is $\frac{\frac{6!}{3!3!}}{3^6} = \frac{20}{3^6}$ . Adding all of these together, we obtain a total probability of $\frac{245}{3^6}$ that our particle will hit $(1,1)$ . Trivially, there is a $\frac{1}{3}$ chance our particle will hit $(0,0)$ from $(1,1)$ . So our final probability will be $\frac{245}{3^6} \cdot \frac{1}{3} = \frac{245}{3^7} \implies m = 245, n = 7 \implies \boxed{252}$ | null | 252 |
be0f72ac3b54da39434f3cad14277c0b | https://artofproblemsolving.com/wiki/index.php/2019_AIME_I_Problems/Problem_6 | In convex quadrilateral $KLMN$ side $\overline{MN}$ is perpendicular to diagonal $\overline{KM}$ , side $\overline{KL}$ is perpendicular to diagonal $\overline{LN}$ $MN = 65$ , and $KL = 28$ . The line through $L$ perpendicular to side $\overline{KN}$ intersects diagonal $\overline{KM}$ at $O$ with $KO = 8$ . Find $MO$ | Let $\angle MKN=\alpha$ and $\angle LNK=\beta$ . Let $P$ be the project of $L$ onto line $NK$ . Note $\angle KLP=\beta$
Then, $KP=28\sin\beta=8\cos\alpha$ .
Furthermore, $KN=\frac{65}{\sin\alpha}=\frac{28}{\sin\beta} \Rightarrow 65\sin\beta=28\sin\alpha$
Dividing the equations gives \[\frac{65}{28}=\frac{28\sin\alpha}{8\cos\alpha}=\frac{7}{2}\tan\alpha\Rightarrow \tan\alpha=\frac{65}{98}\]
Thus, $MK=\frac{MN}{\tan\alpha}=98$ , so $MO=MK-KO=\boxed{090}$ | null | 090 |
be0f72ac3b54da39434f3cad14277c0b | https://artofproblemsolving.com/wiki/index.php/2019_AIME_I_Problems/Problem_6 | In convex quadrilateral $KLMN$ side $\overline{MN}$ is perpendicular to diagonal $\overline{KM}$ , side $\overline{KL}$ is perpendicular to diagonal $\overline{LN}$ $MN = 65$ , and $KL = 28$ . The line through $L$ perpendicular to side $\overline{KN}$ intersects diagonal $\overline{KM}$ at $O$ with $KO = 8$ . Find $MO$ | [asy] size(250); real h = sqrt(98^2+65^2); real l = sqrt(h^2-28^2); pair K = (0,0); pair N = (h, 0); pair M = ((98^2)/h, (98*65)/h); pair L = ((28^2)/h, (28*l)/h); pair P = ((28^2)/h, 0); pair O = ((28^2)/h, (8*65)/h); draw(K--L--N); draw(K--M--N--cycle); draw(L--M); label("K", K, SW); label("L", L, NW); label("M", M, NE); label("N", N, SE); draw(L--P); label("P", P, S); dot(O); label("O", shift((1,1))*O, NNE); label("28", scale(1/2)*L, W); label("65", ((M.x+N.x)/2, (M.y+N.y)/2), NE); [/asy]
Because $\angle KLN = \angle KMN = 90^{\circ}$ $KLMN$ is a cyclic quadrilateral. Hence, by Power of Point, \[KO\cdot KM = KL^2 \implies KM=\dfrac{28^2}{8}=98 \implies MO=98-8=\boxed{090}\] as desired. | null | 090 |
be0f72ac3b54da39434f3cad14277c0b | https://artofproblemsolving.com/wiki/index.php/2019_AIME_I_Problems/Problem_6 | In convex quadrilateral $KLMN$ side $\overline{MN}$ is perpendicular to diagonal $\overline{KM}$ , side $\overline{KL}$ is perpendicular to diagonal $\overline{LN}$ $MN = 65$ , and $KL = 28$ . The line through $L$ perpendicular to side $\overline{KN}$ intersects diagonal $\overline{KM}$ at $O$ with $KO = 8$ . Find $MO$ | [asy] size(250); real h = sqrt(98^2+65^2); real l = sqrt(h^2-28^2); pair K = (0,0); pair N = (h, 0); pair M = ((98^2)/h, (98*65)/h); pair L = ((28^2)/h, (28*l)/h); pair P = ((28^2)/h, 0); pair O = ((28^2)/h, (8*65)/h); draw(K--L--N); draw(K--M--N--cycle); draw(L--M); label("K", K, SW); label("L", L, NW); label("M", M, NE); label("N", N, SE); draw(L--P); label("P", P, S); dot(O); label("O", shift((1,1))*O, NNE); label("28", scale(1/2)*L, W); label("65", ((M.x+N.x)/2, (M.y+N.y)/2), NE); [/asy]
First, let $P$ be the intersection of $LO$ and $KN$ as shown above. Note that $m\angle KPL = 90^{\circ}$ as given in the problem. Since $\angle KPL \cong \angle KLN$ and $\angle PKL \cong \angle LKN$ $\triangle PKL \sim \triangle LKN$ by AA similarity. Similarly, $\triangle KMN \sim \triangle KPO$ . Using these similarities we see that \[\frac{KP}{KL} = \frac{KL}{KN}\] \[KP = \frac{KL^2}{KN} = \frac{28^2}{KN} = \frac{784}{KN}\] and \[\frac{KP}{KO} = \frac{KM}{KN}\] \[KP = \frac{KO \cdot KM}{KN} = \frac{8\cdot KM}{KN}\] Combining the two equations, we get \[\frac{8\cdot KM}{KN} = \frac{784}{KN}\] \[8 \cdot KM = 28^2\] \[KM = 98\] Since $KM = KO + MO$ , we get $MO = 98 -8 = \boxed{090}$ | null | 090 |
be0f72ac3b54da39434f3cad14277c0b | https://artofproblemsolving.com/wiki/index.php/2019_AIME_I_Problems/Problem_6 | In convex quadrilateral $KLMN$ side $\overline{MN}$ is perpendicular to diagonal $\overline{KM}$ , side $\overline{KL}$ is perpendicular to diagonal $\overline{LN}$ $MN = 65$ , and $KL = 28$ . The line through $L$ perpendicular to side $\overline{KN}$ intersects diagonal $\overline{KM}$ at $O$ with $KO = 8$ . Find $MO$ | Extend $KL$ and $NM$ past $L$ and $M$ respectively to meet at $P$ . Let $H$ be the intersection of diagonals $KM$ and $LN$ (this is the orthocenter of $\triangle KNP$ ).
As $\triangle KOL \sim \triangle KHP$ (as $LO \parallel PH$ , using the fact that $H$ is the orthocenter), we may let $OH = 8k$ and $LP = 28k$
Then using similarity with triangles $\triangle KLH$ and $\triangle KMP$ we have
\[\frac{28}{8+8k} = \frac{8+8k+HM}{28+28k}\]
Cross-multiplying and dividing by $4+4k$ gives $2(8+8k+HM) = 28 \cdot 7 = 196$ so $MO = 8k + HM = \frac{196}{2} - 8 = \boxed{090}$ . (Solution by scrabbler94) | null | 090 |
be0f72ac3b54da39434f3cad14277c0b | https://artofproblemsolving.com/wiki/index.php/2019_AIME_I_Problems/Problem_6 | In convex quadrilateral $KLMN$ side $\overline{MN}$ is perpendicular to diagonal $\overline{KM}$ , side $\overline{KL}$ is perpendicular to diagonal $\overline{LN}$ $MN = 65$ , and $KL = 28$ . The line through $L$ perpendicular to side $\overline{KN}$ intersects diagonal $\overline{KM}$ at $O$ with $KO = 8$ . Find $MO$ | First, let $P$ be the intersection of $LO$ and $KN$ . We can use the right triangles in the problem to create equations. Let $a=NP, b=PK, c=NO, d=OM, e=OP, f=PC,$ and $g=NC.$ We are trying to find $d.$ We can find $7$ equations. They are \[4225+d^2=c^2,\] \[4225+d^2+16d+64=a^2+2ab+b^2,\] \[a^2+e^2=c^2,\] \[b^2+e^2=64,\] \[b^2+e^2+2ef+f^2=784,\] \[a^2+e^2+2ef+f^2=g^2,\] and \[g^2+784=a^2+2ab+b^2.\] We can subtract the fifth equation from the sixth equation to get $a^2-b^2=g^2-784.$ We can subtract the fourth equation from the third equation to get $a^2-b^2=c^2-64.$ Combining these equations gives $c^2-64=g^2-784$ so $g^2=c^2+720.$ Substituting this into the seventh equation gives $c^2+1504=a^2+2ab+b^2.$ Substituting this into the second equation gives $4225+d^2+16d+64=c^2+1504$ . Subtracting the first equation from this gives $16d+64=1504.$ Solving this equation, we find that $d=\boxed{090}.$ (Solution by DottedCaculator) | null | 090 |
be0f72ac3b54da39434f3cad14277c0b | https://artofproblemsolving.com/wiki/index.php/2019_AIME_I_Problems/Problem_6 | In convex quadrilateral $KLMN$ side $\overline{MN}$ is perpendicular to diagonal $\overline{KM}$ , side $\overline{KL}$ is perpendicular to diagonal $\overline{LN}$ $MN = 65$ , and $KL = 28$ . The line through $L$ perpendicular to side $\overline{KN}$ intersects diagonal $\overline{KM}$ at $O$ with $KO = 8$ . Find $MO$ | [asy] size(8cm); pair K, L, M, NN, X, O; K=(-sqrt(98^2+65^2)/2, 0); NN=(sqrt(98^2+65^2)/2, 0); L=sqrt(98^2+65^2)/2*dir(180-2*aSin(28/sqrt(98^2+65^2))); M=sqrt(98^2+65^2)/2*dir(2*aSin(65/sqrt(98^2+65^2))); X=foot(L, K, NN); O=extension(L, X, K, M); draw(K -- L -- M -- NN -- K -- M); draw(L -- NN); draw(arc((K+NN)/2, NN, K)); draw(L -- X, dashed); draw(arc((O+NN)/2, NN, X), dashed); draw(rightanglemark(K, L, NN, 100)); draw(rightanglemark(K, M, NN, 100)); draw(rightanglemark(L, X, NN, 100)); dot("$K$", K, SW); dot("$L$", L, unit(L)); dot("$M$", M, unit(M)); dot("$N$", NN, SE); dot("$X$", X, S); [/asy] Notice that $KLMN$ is inscribed in the circle with diameter $\overline{KN}$ and $XOMN$ is inscribed in the circle with diameter $\overline{ON}$ . Furthermore, $(XLN)$ is tangent to $\overline{KL}$ . Then, \[KO\cdot KM=KX\cdot KN=KL^2\implies KM=\frac{28^2}{8}=98,\] and $MO=KM-KO=\boxed{090}$ | null | 090 |
be0f72ac3b54da39434f3cad14277c0b | https://artofproblemsolving.com/wiki/index.php/2019_AIME_I_Problems/Problem_6 | In convex quadrilateral $KLMN$ side $\overline{MN}$ is perpendicular to diagonal $\overline{KM}$ , side $\overline{KL}$ is perpendicular to diagonal $\overline{LN}$ $MN = 65$ , and $KL = 28$ . The line through $L$ perpendicular to side $\overline{KN}$ intersects diagonal $\overline{KM}$ at $O$ with $KO = 8$ . Find $MO$ | [asy] size(250); real h = sqrt(98^2+65^2); real l = sqrt(h^2-28^2); pair K = (0,0); pair N = (h, 0); pair M = ((98^2)/h, (98*65)/h); pair L = ((28^2)/h, (28*l)/h); pair P = ((28^2)/h, 0); pair O = ((28^2)/h, (8*65)/h); draw(K--L--N); draw(K--M--N--cycle); draw(L--M); label("K", K, SW); label("L", L, NW); label("M", M, NE); label("N", N, SE); draw(L--P); label("P", P, S); dot(O); label("O", shift((1,1))*O, NNE); label("28", scale(1/2)*L, W); label("65", ((M.x+N.x)/2, (M.y+N.y)/2), NE); [/asy]
(Diagram by vedadehhc)
Call the base of the altitude from $L$ to $NK$ point $P$ . Let $PO=x$ . Now, we have that $KP=\sqrt{64-x^2}$ by the Pythagorean Theorem. Once again by Pythagorean, $LO=\sqrt{720+x^2}-x$ . Using Power of a Point, we have
\[(KO)(OM)=(LO)(OQ)\] $Q$ is the intersection of $OL$ with the circle $\neq L$
\[8(MO)=(\sqrt{720+x^2}-x)(\sqrt{720+x^2}+x)\]
\[8(MO)=720\]
\[MO=\boxed{090}\] | null | 090 |
be0f72ac3b54da39434f3cad14277c0b | https://artofproblemsolving.com/wiki/index.php/2019_AIME_I_Problems/Problem_6 | In convex quadrilateral $KLMN$ side $\overline{MN}$ is perpendicular to diagonal $\overline{KM}$ , side $\overline{KL}$ is perpendicular to diagonal $\overline{LN}$ $MN = 65$ , and $KL = 28$ . The line through $L$ perpendicular to side $\overline{KN}$ intersects diagonal $\overline{KM}$ at $O$ with $KO = 8$ . Find $MO$ | [asy] size(250); real h = sqrt(98^2+65^2); real l = sqrt(h^2-28^2); pair K = (0,0); pair N = (h, 0); pair M = ((98^2)/h, (98*65)/h); pair L = ((28^2)/h, (28*l)/h); pair P = ((28^2)/h, 0); pair O = ((28^2)/h, (8*65)/h); draw(K--L--N); draw(K--M--N--cycle); draw(L--M); label("K", K, SW); label("L", L, NW); label("M", M, NE); label("N", N, SE); draw(L--P); label("P", P, S); dot(O); label("O", shift((1,1))*O, NNE); label("28", scale(1/2)*L, W); label("65", ((M.x+N.x)/2, (M.y+N.y)/2), NE); [/asy] Note that since $\angle KLN = \angle KMN$ , quadrilateral $KLMN$ is cyclic. Therefore, we have \[\angle LMK = \angle LNK = 90^{\circ} - \angle LKN = \angle KLP,\] so $\triangle KLO \sim \triangle KML$ , giving \[\frac{KM}{28} = \frac{28}{8} \implies KM = 98.\] Therefore, $OM = 98-8 = \boxed{90}$ | null | 90 |
be0f72ac3b54da39434f3cad14277c0b | https://artofproblemsolving.com/wiki/index.php/2019_AIME_I_Problems/Problem_6 | In convex quadrilateral $KLMN$ side $\overline{MN}$ is perpendicular to diagonal $\overline{KM}$ , side $\overline{KL}$ is perpendicular to diagonal $\overline{LN}$ $MN = 65$ , and $KL = 28$ . The line through $L$ perpendicular to side $\overline{KN}$ intersects diagonal $\overline{KM}$ at $O$ with $KO = 8$ . Find $MO$ | [asy] size(250); real h = sqrt(98^2+65^2); real l = sqrt(h^2-28^2); pair K = (0,0); pair N = (h, 0); pair M = ((98^2)/h, (98*65)/h); pair L = ((28^2)/h, (28*l)/h); pair P = ((28^2)/h, 0); pair O = ((28^2)/h, (8*65)/h); draw(K--L--N); draw(K--M--N--cycle); draw(L--M); label("K", K, SW); label("L", L, NW); label("M", M, NE); label("N", N, SE); draw(L--P); label("P", P, S); dot(O); label("O", shift((1,1))*O, NNE); label("28", scale(1/2)*L, W); label("65", ((M.x+N.x)/2, (M.y+N.y)/2), NE); [/asy]
By Pythagorean Theorem, $KM^2+65^2 = KN^2 = 28^2 + LN^2$ . Thus, $LN^2 = KM^2 + 65^2 - 28^2$
By Pythagorean Theorem, $KP^2 + LP^2 = 28^2$ , and $PN^2 + LP^2 = LN^2$
\[PN^2 = (KN - KP)^2 = (\sqrt{KM^2 + 65^2} - KP)^2\]
It follows that \[(\sqrt{KM^2 + 65^2} - KP)^2 + LP^2 = KM^2 + 65^2 - 28^2\]
\[KM^2 + 65^2 - 2\sqrt{KM^2 + 65^2}(KP) + KP^2 + LP^2 = KM^2 + 65^2 - 28^2\]
Since $KP^2 + LP^2 = 28^2$ \[KM^2 + 65^2 - 2\sqrt{KM^2 + 65^2}(KP) + 28^2 = KM^2 + 65^2 - 28^2\]
\[-2\sqrt{KM^2 + 65^2}(KP) = -2 \times 28^2\]
\[KP = \frac{28^2}{\sqrt{KM^2 + 65^2}}\]
$\angle OKP = \angle NKM$ (it's the same angle) and $\angle OPK = \angle KMN = 90^{\circ}$ . Thus, $\triangle KOP \sim \triangle KNM$
Thus,
\[\frac{KO}{KN} = \frac{KP}{KM}\]
\[\frac{8}{\sqrt{KM^2 + 65^2}} = \frac{\frac{28^2}{\sqrt{KM^2 + 65^2}}}{KM}\]
Multiplying both sides by $\sqrt{KM^2 + 65^2}$
\[8 = \frac{28^2}{KM}\]
\[KM = 98\]
Therefore, $OM = 98-8 = \boxed{90}$ | null | 90 |
1f4a52c84d38d609e88b31282c1958e5 | https://artofproblemsolving.com/wiki/index.php/2019_AIME_I_Problems/Problem_7 | There are positive integers $x$ and $y$ that satisfy the system of equations \begin{align*} \log_{10} x + 2 \log_{10} (\text{gcd}(x,y)) &= 60\\ \log_{10} y + 2 \log_{10} (\text{lcm}(x,y)) &= 570. \end{align*} Let $m$ be the number of (not necessarily distinct) prime factors in the prime factorization of $x$ , and let $n$ be the number of (not necessarily distinct) prime factors in the prime factorization of $y$ . Find $3m+2n$ | Add the two equations to get that $\log x+\log y+2(\log(\gcd(x,y))+2(\log(\text{lcm}(x,y)))=630$ .
Then, we use the theorem $\log a+\log b=\log ab$ to get the equation, $\log (xy)+2(\log(\gcd(x,y))+\log(\text{lcm}(x,y)))=630$ .
Using the theorem that $\gcd(x,y) \cdot \text{lcm}(x,y)=x\cdot y$ , along with the previously mentioned theorem, we can get the equation $3\log(xy)=630$ .
This can easily be simplified to $\log(xy)=210$ , or $xy = 10^{210}$
$10^{210}$ can be factored into $2^{210} \cdot 5^{210}$ , and $m+n$ equals to the sum of the exponents of $2$ and $5$ , which is $210+210 = 420$ .
Multiply by two to get $2m +2n$ , which is $840$ .
Then, use the first equation ( $\log x + 2\log(\gcd(x,y)) = 60$ ) to show that $x$ has to have lower degrees of $2$ and $5$ than $y$ (you can also test when $x>y$ , which is a contradiction to the restrains we set before). Therefore, $\gcd(x,y)=x$ . Then, turn the equation into $3\log x = 60$ , which yields $\log x = 20$ , or $x = 10^{20}$ .
Factor this into $2^{20} \cdot 5^{20}$ , and add the two 20's, resulting in $m$ , which is $40$ .
Add $m$ to $2m + 2n$ (which is $840$ ) to get $40+840 = \boxed{880}$ | null | 880 |
1f4a52c84d38d609e88b31282c1958e5 | https://artofproblemsolving.com/wiki/index.php/2019_AIME_I_Problems/Problem_7 | There are positive integers $x$ and $y$ that satisfy the system of equations \begin{align*} \log_{10} x + 2 \log_{10} (\text{gcd}(x,y)) &= 60\\ \log_{10} y + 2 \log_{10} (\text{lcm}(x,y)) &= 570. \end{align*} Let $m$ be the number of (not necessarily distinct) prime factors in the prime factorization of $x$ , and let $n$ be the number of (not necessarily distinct) prime factors in the prime factorization of $y$ . Find $3m+2n$ | First simplifying the first and second equations, we get that
\[\log_{10}(x\cdot\text{gcd}(x,y)^2)=60\] \[\log_{10}(y\cdot\text{lcm}(x,y)^2)=570\]
Thus, when the two equations are added, we have that \[\log_{10}(x\cdot y\cdot\text{gcd}^2\cdot\text{lcm}^2)=630\] When simplified, this equals \[\log_{10}(x^3y^3)=630\] so this means that \[x^3y^3=10^{630}\] so \[xy=10^{210}.\]
Now, the following cannot be done on a proof contest but let's (intuitively) assume that $x<y$ and $x$ and $y$ are both powers of $10$ . This means the first equation would simplify to \[x^3=10^{60}\] and \[y^3=10^{570}.\] Therefore, $x=10^{20}$ and $y=10^{190}$ and if we plug these values back, it works! $10^{20}$ has $20\cdot2=40$ total factors and $10^{190}$ has $190\cdot2=380$ so \[3\cdot 40 + 2\cdot 380 = \boxed{880}.\] | null | 880 |
1f4a52c84d38d609e88b31282c1958e5 | https://artofproblemsolving.com/wiki/index.php/2019_AIME_I_Problems/Problem_7 | There are positive integers $x$ and $y$ that satisfy the system of equations \begin{align*} \log_{10} x + 2 \log_{10} (\text{gcd}(x,y)) &= 60\\ \log_{10} y + 2 \log_{10} (\text{lcm}(x,y)) &= 570. \end{align*} Let $m$ be the number of (not necessarily distinct) prime factors in the prime factorization of $x$ , and let $n$ be the number of (not necessarily distinct) prime factors in the prime factorization of $y$ . Find $3m+2n$ | Let $x=10^a$ and $y=10^b$ and $a<b$ . Then the given equations become $3a=60$ and $3b=570$ . Therefore, $x=10^{20}=2^{20}\cdot5^{20}$ and $y=10^{190}=2^{190}\cdot5^{190}$ . Our answer is $3(20+20)+2(190+190)=\boxed{880}$ | null | 880 |
1f4a52c84d38d609e88b31282c1958e5 | https://artofproblemsolving.com/wiki/index.php/2019_AIME_I_Problems/Problem_7 | There are positive integers $x$ and $y$ that satisfy the system of equations \begin{align*} \log_{10} x + 2 \log_{10} (\text{gcd}(x,y)) &= 60\\ \log_{10} y + 2 \log_{10} (\text{lcm}(x,y)) &= 570. \end{align*} Let $m$ be the number of (not necessarily distinct) prime factors in the prime factorization of $x$ , and let $n$ be the number of (not necessarily distinct) prime factors in the prime factorization of $y$ . Find $3m+2n$ | We will use the notation $(a, b)$ for $\gcd(a, b)$ and $[a, b]$ as $\text{lcm}(a, b)$ .
We can start with a similar way to Solution 1. We have, by logarithm properties, $\log_{10}{x}+\log_{10}{(x, y)^2}=60$ or $x(x, y)^2=10^{60}$ . We can do something similar to the second equation and our two equations become \[x(x, y)^2=10^{60}\] \[y[x, y]^2=10^{570}\] Adding the two equations gives us $xy(x, y)^2[x, y]^2=10^{630}$ . Since we know that $(a, b)\cdot[a, b]=ab$ $x^3y^3=10^{630}$ , or $xy=10^{210}$ . We can express $x$ as $2^a5^b$ and $y$ as $2^c5^d$ . Another way to express $(x, y)$ is now $2^{min(a, c)}5^{min(b, d)}$ , and $[x, y]$ is now $2^{max(a, c)}5^{max(b, d)}$ . We know that $x<y$ , and thus, $a<c$ , and $b<d$ . Our equations for $lcm$ and $gcd$ now become \[2^a5^b(2^a5^a)^2=10^{60}\] or $a=b=20$ . Doing the same for the $lcm$ equation, we have $c=d=190$ , and $190+20=210$ , which satisfies $xy=210$ . Thus, $3m+2n=3(20+20)+2(190+190)=\boxed{880}$ .
~awsomek | null | 880 |
1f4a52c84d38d609e88b31282c1958e5 | https://artofproblemsolving.com/wiki/index.php/2019_AIME_I_Problems/Problem_7 | There are positive integers $x$ and $y$ that satisfy the system of equations \begin{align*} \log_{10} x + 2 \log_{10} (\text{gcd}(x,y)) &= 60\\ \log_{10} y + 2 \log_{10} (\text{lcm}(x,y)) &= 570. \end{align*} Let $m$ be the number of (not necessarily distinct) prime factors in the prime factorization of $x$ , and let $n$ be the number of (not necessarily distinct) prime factors in the prime factorization of $y$ . Find $3m+2n$ | Let $x=d\alpha, y=d\beta, (\alpha, \beta)=1$ . Simplifying, $d^3\alpha=10^{60}, d^3\alpha^2\beta^3=10^{510} \implies \alpha\beta^3 = 10^{510}=2^{510} \cdot 5^{510}$ . Notice that since $\alpha, \beta$ are coprime, and $\alpha < 5^{90}$ (Prove it yourself !) , $\alpha=1, \beta = 10^{170}$ . Hence, $x=10^{20}, y=10^{190}$ giving the answer $\boxed{880}$ | null | 880 |
1f4a52c84d38d609e88b31282c1958e5 | https://artofproblemsolving.com/wiki/index.php/2019_AIME_I_Problems/Problem_7 | There are positive integers $x$ and $y$ that satisfy the system of equations \begin{align*} \log_{10} x + 2 \log_{10} (\text{gcd}(x,y)) &= 60\\ \log_{10} y + 2 \log_{10} (\text{lcm}(x,y)) &= 570. \end{align*} Let $m$ be the number of (not necessarily distinct) prime factors in the prime factorization of $x$ , and let $n$ be the number of (not necessarily distinct) prime factors in the prime factorization of $y$ . Find $3m+2n$ | Add the two equations and use the fact that $\gcd\left(x,y\right)\cdot\mathrm{lcm}\left(x,y\right)=xy$ to find that $xy=10^{210}$ . So let $x=2^a5^b$ and $y=2^{210-a}5^{210-b}$ for $0\leq a,b\leq210$ . If $a\geq105$ then the exponent of $2$ in $x\cdot\gcd\left(x,y\right)^2=10^{60}$ is $a+2\left(210-a\right)=420-a$ , so $a=360$ , contradiction. So $a<105$ . Then the exponent of $2$ in $x\cdot\gcd\left(x,y\right)^2$ is $a+2a=3a$ , so $a=20$ . Similarly, $b=20$ . Then $3m+2n=3\left(a+b\right)+2\left(420-a-b\right)=\boxed{880}$ as desired. | null | 880 |
c63020127c196bd4057f71963a487ca1 | https://artofproblemsolving.com/wiki/index.php/2019_AIME_I_Problems/Problem_8 | Let $x$ be a real number such that $\sin^{10}x+\cos^{10} x = \tfrac{11}{36}$ . Then $\sin^{12}x+\cos^{12} x = \tfrac{m}{n}$ where $m$ and $n$ are relatively prime positive integers. Find $m+n$ | We can substitute $y = \sin^2{x}$ . Since we know that $\cos^2{x}=1-\sin^2{x}$ , we can do some simplification.
This yields $y^5+(1-y)^5=\frac{11}{36}$ . From this, we can substitute again to get some cancellation through binomials. If we let $z=\frac{1}{2}-y$ , we can simplify the equation to: \[\left(\frac{1}{2}+z\right)^5+\left(\frac{1}{2}-z\right)^5=\frac{11}{36}.\] After using binomial theorem, this simplifies to: \[\frac{1}{16}(80z^4+40z^2+1)=\frac{11}{36}.\] If we use the quadratic formula, we obtain $z^2=\frac{1}{12}$ , so $z=\pm\frac{1}{2\sqrt{3}}$ (observe that either choice of $z$ doesn't matter). Substituting $z,$ we get:
\[\sin^{12}{x}+\cos^{12}{x}=\left(\frac{1}{2}-z\right)^6+\left(\frac{1}{2}+z\right)^6=2z^6 + \frac{15z^4}{2} + \frac{15z^2}{8} + \frac{1}{32}=\frac{13}{54}.\]
Therefore, the answer is $13+54=\boxed{067}$ | null | 067 |
c63020127c196bd4057f71963a487ca1 | https://artofproblemsolving.com/wiki/index.php/2019_AIME_I_Problems/Problem_8 | Let $x$ be a real number such that $\sin^{10}x+\cos^{10} x = \tfrac{11}{36}$ . Then $\sin^{12}x+\cos^{12} x = \tfrac{m}{n}$ where $m$ and $n$ are relatively prime positive integers. Find $m+n$ | First, for simplicity, let $a=\sin{x}$ and $b=\cos{x}$ . Note that $a^2+b^2=1$ . We then bash the rest of the problem out. Take the fifth power of this expression and get $a^{10}+b^{10}+5a^2b^2(a^6+b^6)+10a^4b^4(a^2+b^2)=\frac{11}{36}+5a^2b^2(a^6+b^6)+10a^4b^4=1$ . Note that we also have $\frac{11}{36}=a^{10}+b^{10}=(a^{10}+b^{10})(a^2+b^2)=a^{12}+b^{12}+a^2b^2(a^8+b^8)$ . So, it suffices to compute $a^2b^2(a^8+b^8)$ . Let $y=a^2b^2$ . We have from cubing $a^2+b^2=1$ that $a^6+b^6+3a^2b^2(a^2+b^2)=1$ or $a^6+b^6=1-3y$ . Next, using $\frac{11}{36}+5a^2b^2(a^6+b^6)+10a^4b^4=1$ , we get $a^2b^2(a^6+b^6)+2a^4b^4=\frac{5}{36}$ or $y(1-3y)+2y^2=y-y^2=\frac{5}{36}$ . Solving gives $y=\frac{5}{6}$ or $y=\frac{1}{6}$ . Clearly $y=\frac{5}{6}$ is extraneous, so $y=\frac{1}{6}$ . Now note that $a^4+b^4=(a^2+b^2)^2-2a^2b^2=\frac{2}{3}$ , and $a^8+b^8=(a^4+b^4)^2-2a^4b^4=\frac{4}{9}-\frac{1}{18}=\frac{7}{18}$ . Thus we finally get $a^{12}+b^{12}=\frac{11}{36}-\frac{7}{18}\cdot\frac{1}{6}=\frac{13}{54}$ , giving $\boxed{067}$ | null | 067 |
c63020127c196bd4057f71963a487ca1 | https://artofproblemsolving.com/wiki/index.php/2019_AIME_I_Problems/Problem_8 | Let $x$ be a real number such that $\sin^{10}x+\cos^{10} x = \tfrac{11}{36}$ . Then $\sin^{12}x+\cos^{12} x = \tfrac{m}{n}$ where $m$ and $n$ are relatively prime positive integers. Find $m+n$ | Newton sums is basically constructing the powers of the roots of the polynomials instead of deconstructing them which was done in Solution $2$ . Let $\sin^2x$ and $\cos^2x$ be the roots of some polynomial $F(a)$ . Then, by Vieta, $F(a)=a^2-a+b$ for some $b=\sin^2x\cdot\cos^2x$
Let $S_k=\left(\sin^2x\right)^k+\left(\cos^2x\right)^k$ . We want to find $S_6$ . Clearly $S_1=1$ and $S_2=1-2b$ . Newton sums tells us that $S_k-S_{k-1}+bS_{k-2}=0\Rightarrow S_k=S_{k-1}-bS_{k-2}$ where $k\ge 3$ for our polynomial $F(a)$
Bashing, we have \[S_3=S_2-bS_1\Rightarrow S_3=(1-2b)-b(1)=1-3b\] \[S_4=S_3-bS_2\Rightarrow S_4=(1-3b)-b(1-2b)=2b^2-4b+1\] \[S_5=S_4-bS_3\Rightarrow S_5=(2b^2-4b+1)-b(1-3b)=5b^2-5b+1=\frac{11}{36}\]
Thus \[5b^2-5b+1=\frac{11}{36}\Rightarrow 5b^2-5b+\frac{25}{36}=0, 36b^2-36b+5=0, (6b-1)(6b-5)=0\] $b=\frac{1}{6} \text{ or } \frac{5}{6}$ . Clearly, $\sin^2x\cdot\cos^2x\not=\frac{5}{6}$ so $\sin^2x\cdot\cos^2x=b=\frac{1}{6}$
Note $S_4=\frac{7}{18}$ . Solving for $S_6$ , we get $S_6=S_5-\frac{1}{6}S_4=\frac{13}{54}$ . Finally, $13+54=\boxed{067}$ | null | 067 |
c63020127c196bd4057f71963a487ca1 | https://artofproblemsolving.com/wiki/index.php/2019_AIME_I_Problems/Problem_8 | Let $x$ be a real number such that $\sin^{10}x+\cos^{10} x = \tfrac{11}{36}$ . Then $\sin^{12}x+\cos^{12} x = \tfrac{m}{n}$ where $m$ and $n$ are relatively prime positive integers. Find $m+n$ | Factor the first equation. \[\sin^{10}x + \cos^{10}x = (\sin^2x+\cos^2x)(\sin^8x-\sin^6x\cos^2x+\sin^4x\cos^4x-\sin^2x\cos^6x+\cos^8x)\] First of all, $\sin^4x+\cos^4x = 1-2\sin^2x\cos^2x$ because $\sin^4x+\cos^4x=(\sin^2x + \cos^2x)^2 -2\sin^2x\cos^2x = 1 - 2\sin^2x\cos^2x$ We group the first, third, and fifth term and second and fourth term. The first group: \begin{align*} \sin^8+\sin^4x\cos^4x+\cos^8x &= (\sin^4x+\cos^4x)^2-\sin^4x\cos^4x)\\ &= (1 - 2\sin^2x\cos^2x)^2-\sin^4x\cos^4x)\\ &= 1+4\sin^4x\cos^4x-4\sin^2x\cos^2x \end{align*} The second group: \begin{align*} -\sin^6x\cos^2x-\sin^2x\cos^6x &= -\sin^2x\cos^2x(\sin^4x+\cos^4x)\\ &= -\sin^2x\cos^2x(1-2\sin^2x\cos^2x)\\ &= -\sin^2x\cos^2x+2\sin^4x\cos^4x \end{align*} Add the two together to make \[1+4\sin^4x\cos^4x-4\sin^2x\cos^2x-\sin^2x\cos^2x+2\sin^4x\cos^4x = 1 - 5\sin^2x\cos^2x+5\sin^4x\cos^4x\] Because this equals $\frac{11}{36}$ , we have \[5\sin^4x\cos^4x- 5\sin^2x\cos^2x+\frac{25}{36}=0\] Let $\sin^2x\cos^2x = a$ so we get \[5a^2- 5a+\frac{25}{36}=0 \Rightarrow a^2-a+\frac{5}{36}\] Solving the quadratic gives us \[a = \frac{1 \pm \frac{2}{3}}{2}\] Because $\sin^2x\cos^2x \le \frac{1}{4}$ , we finally get $a = \frac{1 - \frac{2}{3}}{2} = \frac{1}{6}$
Now from the second equation, \begin{align*} \sin^{12}x + \cos^{12}x &= (\sin^4x+\cos^4x)(\sin^8x-\sin^4x\cos^4x+\cos^8x)\\ &= (1-2\sin^2x\cos^2x)((\sin^4x+\cos^4x)^2-3\sin^4x\cos^4x)\\ &= (1-2\sin^2x\cos^2x)((1-2\sin^2x\cos^2x)^2-3\sin^4x\cos^4x) \end{align*} Plug in $\sin^2x\cos^2x = \frac{1}{6}$ to get \[(1-2(\frac{1}{6}))(1-2(\frac{1}{6})^2-3(\frac{1}{6})^2) = \frac{13}{54}\] which yields the answer $\boxed{067}$ | null | 067 |
c63020127c196bd4057f71963a487ca1 | https://artofproblemsolving.com/wiki/index.php/2019_AIME_I_Problems/Problem_8 | Let $x$ be a real number such that $\sin^{10}x+\cos^{10} x = \tfrac{11}{36}$ . Then $\sin^{12}x+\cos^{12} x = \tfrac{m}{n}$ where $m$ and $n$ are relatively prime positive integers. Find $m+n$ | Define the recursion $a_n=(\sin^2 x)^n+(\cos^2 x)^n$ We know that the characteristic equation of $a_n$ must have 2 roots, so we can recursively define $a_n$ as $a_n=p*a_{n-1}+q*a_{n-2}$ $p$ is simply the sum of the roots of the characteristic equation, which is $\sin^2 x+\cos^2 x=1$ $q$ is the product of the roots, which is $-(\sin^2 x)(\cos^2 x)$ . This value is not trivial and we have to solve for it.
We know that $a_0=2$ $a_1=1$ $a_5=\frac{11}{36}$ .
Solving the rest of the recursion gives
\[a_2=1+2q\] \[a_3=1+3q\] \[a_4=1+4q+2q^2\] \[a_5=1+5q+5q^2=\frac{11}{36}\] \[a_6=1+6q+9q^2+2q^3\]
Solving for $q$ in the expression for $a_5$ gives us $q^2+q+\frac{5}{36}=0$ , so $q=-\frac{5}{6}, -\frac{1}{6}$ . Since $q=-(\sin^2 x)(\cos^2 x)$ , we know that the minimum value it can attain is $-\frac{1}{4}$ by AM-GM, so $q$ cannot be $-\frac{5}{6}$ .
Plugging in the value of $q$ into the expression for $a_6$ , we get $a_6=1-1+\frac{1}{4}-\frac{1}{108}=\frac{26}{108}=\frac{13}{54}$ . Our final answer is then $13+54=\boxed{067}$ | null | 067 |
c63020127c196bd4057f71963a487ca1 | https://artofproblemsolving.com/wiki/index.php/2019_AIME_I_Problems/Problem_8 | Let $x$ be a real number such that $\sin^{10}x+\cos^{10} x = \tfrac{11}{36}$ . Then $\sin^{12}x+\cos^{12} x = \tfrac{m}{n}$ where $m$ and $n$ are relatively prime positive integers. Find $m+n$ | Let $m=\sin^2 x$ and $n=\cos^2 x$ , then $m+n=1$ and $m^5+n^5=\frac{11}{36}$
$m^6+n^6=(m^5+n^5)(m+n)-mn(m^4+n^4)=(m^5+n^5)-mn(m^4+n^4)$
Now factoring $m^5+n^5$ as solution 4 yields $m^5+n^5=(m+n)(m^4-m^3n+m^2n^2-mn^3+n^4)$ $=m^4+n^4-mn(m^2-mn+n^2)=m^4+n^4-mn[(m+n)^2-3mn]=m^4+n^4-mn(1-3mn)$
Since $(m+n)^4=m^4+4m^3n+6m^2n^2+4mn^3+n^4$ $m^4+n^4=(m+n)^4-2mn(2m^2+3mn+2n^2)=1-2mn(2m^2+3mn+2n^2)$
Notice that $2m^2+3mn+2n^2$ can be rewritten as $[\sqrt{2}(a+b)]^2-mn=2-mn$ . Thus, $m^4+n^4=1-2mn(2-mn)$ and $m^5+n^5=1-2mn(2-mn)-mn(1-3mn)=1-5mn+5(mn)^2=\frac{11}{36}$ . As in solution 4, we get $mn=\frac{1}{6}$ and $m^4+n^4=1-2*\frac{1}{6}(2-\frac{1}{6})=\frac{7}{18}$
Substitute $m^4+n^4=\frac{7}{18}$ and $mn=\frac{1}{6}$ , then $m^6+n^6=\frac{11}{36}-\frac{1}{6}*\frac{7}{18}=\frac{13}{54}$ , and the desired answer is $\boxed{067}$ | null | 067 |
c63020127c196bd4057f71963a487ca1 | https://artofproblemsolving.com/wiki/index.php/2019_AIME_I_Problems/Problem_8 | Let $x$ be a real number such that $\sin^{10}x+\cos^{10} x = \tfrac{11}{36}$ . Then $\sin^{12}x+\cos^{12} x = \tfrac{m}{n}$ where $m$ and $n$ are relatively prime positive integers. Find $m+n$ | Let $a_n=\sin^nx+\cos^nx$ for non-negative integers $n$ . Then $a_0=2$ and $a_2=1$ . In addition, \[a_n=\sin^nx+\cos^nx=\left(\sin^{n-2}x+\cos^{n-2}x\right)\left(\sin^2x+\cos^2x\right)-\sin^2x\cos^2x\left(\sin^{n-4}x+\cos^{n-4}x\right)=a_{n-2}-Xa_{n-4},\] where $X=\sin^2x\cos^2x$ . So we can compute
\begin{align*}
a_4&=1-2X\\
a_6&=1-3X\\
a_8&=1-4X+2X^2\\
a_{10}&=1-5X+5X^2=\frac{11}{36}
\end{align*}so $X=\frac{1}{6},\frac{5}{6}$ . But by the sin double angle formula, $\sin^2x\cos^2x=\frac{1}{4}\sin^22x\leq\frac{1}{4}$ , so $X=\frac{1}{6}$ . Then \[a_{12}=a_{10}-Xa_8=\frac{11}{36}-\frac{1}{6}\cdot\frac{7}{18}=\frac{13}{54}\] so the answer is $\boxed{067}$ as desired. | null | 067 |
d3618c86ac60651d0fb2a8cf641443fd | https://artofproblemsolving.com/wiki/index.php/2019_AIME_I_Problems/Problem_9 | Let $\tau(n)$ denote the number of positive integer divisors of $n$ . Find the sum of the six least positive integers $n$ that are solutions to $\tau (n) + \tau (n+1) = 7$ | In order to obtain a sum of $7$ , we must have:
Since both of these cases contain a number with an odd number of divisors, that number must be an even power of a prime. These can come in the form of a square-like $3^2$ with $3$ divisors, or a fourth power like $2^4$ with $5$ divisors. We then find the smallest such values by hand.
Having computed the working possibilities, we take the sum of the corresponding values of $n$ $8+9+16+25+121+361 = \boxed{540}$ . ~Kepy. | null | 540 |
d3618c86ac60651d0fb2a8cf641443fd | https://artofproblemsolving.com/wiki/index.php/2019_AIME_I_Problems/Problem_9 | Let $\tau(n)$ denote the number of positive integer divisors of $n$ . Find the sum of the six least positive integers $n$ that are solutions to $\tau (n) + \tau (n+1) = 7$ | Let the ordered pair $(a,b)$ represent the number of divisors of $n$ and $n+1$ respectively.
We see that to obtain a sum of $7$ , we can have $(2,5), (3,4), (4,3),$ and $(5,2)$
Case 1: When we have $(2,5)$ For $n$ to have 2 divisors, it must be a prime number.
For $n+1$ to have 5 divisors, it must be in the form $a^4$ .
If $n+1$ is in the form $a^4$ , then $n = a^4-1 = (a^2+1)(a-1)(a+1)$ . This means that $n$ , or $a^4-1$ has factors other than 1 and itself; $n$ is not prime.
No cases work in this case
Case 2: When we have $(4,3)$ For $n$ to have 4 divisors, it must be in the form $a^3$ or $ab$ , where $a$ and $b$ are distinct prime numbers .
For $n+1$ to have 3 divisors, it must be a square number.
Let $n+1 = A^2$ $A$ is a prime number). When $n = a^3, a^3+1 = A^2, (A-1)(A+1)=a^3$ .
We see that the only case when it works is when $a=2, A=3$ , so $n=8$ works.
Case 3: When we have $(5,2)$ For $n$ to have 5 divisors, it must be in the form $a^4$ , where $a$ is a prime number.
For $n+1$ to have 2 divisors, it must be a prime number.
Notice that $a$ and $a^4$ have the same parity (even/odd). Since every prime greater than 2 are odd, $n = a^4$ must be even. Since $a^4$ is even, $a$ must be even as well, and the only prime number that is even is 2. When $a=2, n=16$
Case 4: When we have $(3,4)$ For $n$ to have 3 divisors, it must be a square number.
For $n+1$ to have 4 divisors, it must be in the form $a^3$ or $ab$ , where $a$ and $b$ are distinct prime numbers.
Similar to Case 2, let $n = A^2$ $A$ is a prime number).
There are no cases that satisfy this equation.
We test squares of primes to find values of n that work.
Now we add up the values of $n$ to get the answer: $8+16+9+25+121+361 = \boxed{540}$ .
~toastybaker | null | 540 |
271d10430ba3ab1a2954e625c4e2dc09 | https://artofproblemsolving.com/wiki/index.php/2019_AIME_I_Problems/Problem_11 | In $\triangle ABC$ , the sides have integer lengths and $AB=AC$ . Circle $\omega$ has its center at the incenter of $\triangle ABC$ . An excircle of $\triangle ABC$ is a circle in the exterior of $\triangle ABC$ that is tangent to one side of the triangle and tangent to the extensions of the other two sides. Suppose that the excircle tangent to $\overline{BC}$ is internally tangent to $\omega$ , and the other two excircles are both externally tangent to $\omega$ . Find the minimum possible value of the perimeter of $\triangle ABC$ | Let the tangent circle be $\omega$ . Some notation first: let $BC=a$ $AB=b$ $s$ be the semiperimeter, $\theta=\angle ABC$ , and $r$ be the inradius. Intuition tells us that the radius of $\omega$ is $r+\frac{2rs}{s-a}$ (using the exradius formula). However, the sum of the radius of $\omega$ and $\frac{rs}{s-b}$ is equivalent to the distance between the incenter and the the $B/C$ excenter. Denote the B excenter as $I_B$ and the incenter as $I$ .
Lemma: $I_BI=\frac{2b*IB}{a}$ We draw the circumcircle of $\triangle ABC$ . Let the angle bisector of $\angle ABC$ hit the circumcircle at a second point $M$ . By the incenter-excenter lemma, $AM=CM=IM$ . Let this distance be $\alpha$ . Ptolemy's theorem on $ABCM$ gives us \[a\alpha+b\alpha=b(\alpha+IB)\to \alpha=\frac{b*IB}{a}\] Again, by the incenter-excenter lemma, $II_B=2IM$ so $II_b=\frac{2b*IB}{a}$ as desired.
Using this gives us the following equation: \[\frac{2b*IB}{a}=r+\frac{2rs}{s-a}+\frac{rs}{s-b}\] Motivated by the $s-a$ and $s-b$ , we make the following substitution: $x=s-a, y=s-b$ This changes things quite a bit. Here's what we can get from it: \[a=2y, b=x+y, s=x+2y\] It is known (easily proved with Heron's and $a=rs$ ) that \[r=\sqrt{\frac{(s-a)(s-b)(s-b)}{s}}=\sqrt{\frac{xy^2}{x+2y}}\] Using this, we can also find $IB$ : let the midpoint of $BC$ be $N$ . Using Pythagorean's Theorem on $\triangle INB$ \[IB^2=r^2+(\frac{a}{2})^2=\frac{xy^2}{x+2y}+y^2=\frac{2xy^2+2y^3}{x+2y}=\frac{2y^2(x+y)}{x+2y}\] We now look at the RHS of the main equation: \[r+\frac{2rs}{s-a}+\frac{rs}{s-b}=r(1+\frac{2(x+2y)}{x}+\frac{x+2y}{y})=r(\frac{x^2+5xy+4y^2}{xy})=\frac{r(x+4y)(x+y)}{xy}=\frac{2(x+y)IB}{2y}\] Cancelling some terms, we have \[\frac{r(x+4y)}{x}=IB\] Squaring, \[\frac{2y^2(x+y)}{x+2y}=\frac{(x+4y)^2*xy^2}{x^2(x+2y)}\to \frac{(x+4y)^2}{x}=2(x+y)\] Expanding and moving terms around gives \[(x-8y)(x+2y)=0\to x=8y\] Reverse substituting, \[s-a=8s-8b\to b=\frac{9}{2}a\] Clearly the smallest solution is $a=2$ and $b=9$ , so our answer is $2+9+9=\boxed{020}$ -franchester | null | 020 |
271d10430ba3ab1a2954e625c4e2dc09 | https://artofproblemsolving.com/wiki/index.php/2019_AIME_I_Problems/Problem_11 | In $\triangle ABC$ , the sides have integer lengths and $AB=AC$ . Circle $\omega$ has its center at the incenter of $\triangle ABC$ . An excircle of $\triangle ABC$ is a circle in the exterior of $\triangle ABC$ that is tangent to one side of the triangle and tangent to the extensions of the other two sides. Suppose that the excircle tangent to $\overline{BC}$ is internally tangent to $\omega$ , and the other two excircles are both externally tangent to $\omega$ . Find the minimum possible value of the perimeter of $\triangle ABC$ | [asy] unitsize(1cm); var x = 9; pair A = (0,sqrt(x^2-1)); pair B = (-1,0); pair C = (1,0); dot(Label("$A$",A,NE),A); dot(Label("$B$",B,SW),B); dot(Label("$C$",C,SE),C); draw(A--B--C--cycle); var r = sqrt((x-1)/(x+1)); pair I = (0,r); dot(Label("$I$",I,SE),I); draw(circle(I,r)); draw(Label("$r$"),I--I+r*SSW,dashed); pair M = intersectionpoint(A--B,circle(I,r)); pair N = (0,0); pair O = intersectionpoint(A--C,circle(I,r)); dot(Label("$M$",M,W),M); dot(Label("$N$",N,S),N); dot(Label("$O$",O,E),O); var rN = sqrt((x+1)/(x-1)); pair EN = (0,-rN); dot(Label("$E_N$",EN,SE),EN); draw(circle(EN,rN)); draw(Label("$r_N$"),EN--EN+rN*SSW,dashed); pair AB = (-1-2/(x-1),-2rN); pair AC = (1+2/(x-1),-2rN); draw(B--AB,EndArrow); draw(C--AC,EndArrow); pair H = intersectionpoint(B--AB,circle(EN,rN)); dot(Label("$H$",H,W),H); var rM = sqrt(x^2-1); pair EM = (-x,rM); dot(Label("$E_M$",EM,SW),EM); draw(Label("$r_M$"),EM--EM+rM*SSE,dashed); pair CB = (-x-1,0); pair CA = (-2/x,sqrt(x^2-1)+2(sqrt(x^2-1)/x)); draw(B--CB,EndArrow); draw(A--CA,EndArrow); pair J = intersectionpoint(A--B,circle(EM,rM)); pair K = intersectionpoint(B--CB,circle(EM,rM)); dot(Label("$J$",J,W),J); dot(Label("$K$",K,S),K); draw(arc(EM,rM,-100,15),Arrows); [/asy]
First, assume $BC=2$ and $AB=AC=x$ . The triangle can be scaled later if necessary. Let $I$ be the incenter and let $r$ be the inradius. Let the points at which the incircle intersects $AB$ $BC$ , and $CA$ be denoted $M$ $N$ , and $O$ , respectively.
Next, we calculate $r$ in terms of $x$ . Note the right triangle formed by $A$ $I$ , and $M$ . The length $IM$ is equal to $r$ . Using the Pythagorean Theorem, the length $AN$ is $\sqrt{x^2-1}$ , so the length $AI$ is $\sqrt{x^2-1}-r$ . Note that $BN$ is half of $BC=2$ , and by symmetry caused by the incircle, $BN=BM$ and $BM=1$ , so $MA=x-1$ . Applying the Pythagorean Theorem to $AIM$ , we get \[r^2+(x-1)^2=\left(\sqrt{x^2-1}-r\right)^2.\] Expanding yields \[r^2+x^2-2x+1=x^2-1-2r\sqrt{x^2-1}+r^2,\] which can be simplified to \[2r\sqrt{x^2-1}=2x-2.\] Dividing by $2$ and then squaring results in \[r^2(x^2-1)=(x-1)^2,\] and isolating $r^2$ gets us \[r^2=\frac{(x-1)^2}{x^2-1}=\frac{(x-1)^2}{(x+1)(x-1)}=\frac{x-1}{x+1},\] so $r=\sqrt{\frac{x-1}{x+1}}$
We then calculate the radius of the excircle tangent to $BC$ . We denote the center of the excircle $E_N$ and the radius $r_N$
Consider the quadrilateral formed by $M$ $I$ $E_N$ , and the point at which the excircle intersects the extension of $AB$ , which we denote $H$ . By symmetry caused by the excircle, $BN=BH$ , so $BH=1$
Note that triangles $MBI$ and $NBI$ are congruent, and $HBE$ and $NBE$ are also congruent. Denoting the measure of angles $MBI$ and $NBI$ measure $\alpha$ and the measure of angles $HBE$ and $NBE$ measure $\beta$ , straight angle $MBH=2\alpha+2\beta$ , so $\alpha + \beta=90^\circ$ . This means that angle $IBE$ is a right angle, so it forms a right triangle.
Setting the base of the right triangle to $IE$ , the height is $BN=1$ and the base consists of $IN=r$ and $EN=r_N$ . Triangles $INB$ and $BNE$ are similar to $IBE$ , so $\frac{IN}{BN}=\frac{BN}{EN}$ , or $\frac{r}{1}=\frac{1}{r_N}$ . This makes $r_N$ the reciprocal of $r$ , so $r_N=\sqrt{\frac{x+1}{x-1}}$
Circle $\omega$ 's radius can be expressed by the distance from the incenter $I$ to the bottom of the excircle with center $E_N$ . This length is equal to $r+2r_N$ , or $\sqrt{\frac{x-1}{x+1}}+2\sqrt{\frac{x+1}{x-1}}$ . Denote this value $r_\omega$
Finally, we calculate the distance from the incenter $I$ to the closest point on the excircle tangent to $AB$ , which forms another radius of circle $\omega$ and is equal to $r_\omega$ . We denote the center of the excircle $E_M$ and the radius $r_M$ . We also denote the points where the excircle intersects $AB$ and the extension of $BC$ using $J$ and $K$ , respectively. In order to calculate the distance, we must find the distance between $I$ and $E_M$ and subtract off the radius $r_M$
We first must calculate the radius of the excircle. Because the excircle is tangent to both $AB$ and the extension of $AC$ , its center must lie on the angle bisector formed by the two lines, which is parallel to $BC$ . This means that the distance from $E_M$ to $K$ is equal to the length of $AN$ , so the radius is also $\sqrt{x^2-1}$
Next, we find the length of $IE_M$ . We can do this by forming the right triangle $IAE_M$ . The length of leg $AI$ is equal to $AN$ minus $r$ , or $\sqrt{x^2-1}-\sqrt{\frac{x-1}{x+1}}$ . In order to calculate the length of leg $AE_M$ , note that right triangles $AJE_M$ and $BNA$ are congruent, as $JE_M$ and $NA$ share a length of $\sqrt{x^2-1}$ , and angles $E_MAJ$ and $NAB$ add up to the right angle $NAE_M$ . This means that $AE_M=BA=x$
Using Pythagorean Theorem, we get \[IE_M=\sqrt{\left(\sqrt{x^2-1}-\sqrt{\frac{x-1}{x+1}}\right)^2+x^2}.\] Bringing back \[r_\omega=IE_M-r_M\] and substituting in some values, the equation becomes \[r_\omega=\sqrt{\left(\sqrt{x^2-1}-\sqrt{\frac{x-1}{x+1}}\right)^2+x^2}-\sqrt{x^2-1}.\] Rearranging and squaring both sides gets \[\left(r_\omega+\sqrt{x^2-1}\right)^2=\left(\sqrt{x^2-1}-\sqrt{\frac{x-1}{x+1}}\right)^2+x^2.\] Distributing both sides yields \[r_\omega^2+2r_\omega\sqrt{x^2-1}+x^2-1=x^2-1-2\sqrt{x^2-1}\sqrt{\frac{x-1}{x+1}}+\frac{x-1}{x+1}+x^2.\] Canceling terms results in \[r_\omega^2+2r_\omega\sqrt{x^2-1}=-2\sqrt{x^2-1}\sqrt{\frac{x-1}{x+1}}+\frac{x-1}{x+1}+x^2.\] Since \[-2\sqrt{x^2-1}\sqrt{\frac{x-1}{x+1}}=-2\sqrt{(x+1)(x-1)\frac{x-1}{x+1}}=-2(x-1),\] We can further simplify to \[r_\omega^2+2r_\omega\sqrt{x^2-1}=-2(x-1)+\frac{x-1}{x+1}+x^2.\] Substituting out $r_\omega$ gets \[\left(\sqrt{\frac{x-1}{x+1}}+2\sqrt{\frac{x+1}{x-1}}\right)^2+2\left(\sqrt{\frac{x-1}{x+1}}+2\sqrt{\frac{x+1}{x-1}}\right)\sqrt{x^2-1}=-2(x-1)+\frac{x-1}{x+1}+x^2\] which when distributed yields \[\frac{x-1}{x+1}+4+4\left(\frac{x+1}{x-1}\right)+2(x-1+2(x+1))=-2(x-1)+\frac{x-1}{x+1}+x^2.\] After some canceling, distributing, and rearranging, we obtain \[4\left(\frac{x+1}{x-1}\right)=x^2-8x-4.\] Multiplying both sides by $x-1$ results in \[4x+4=x^3-x^2-8x^2+8x-4x+4,\] which can be rearranged into \[x^3-9x^2=0\] and factored into \[x^2(x-9)=0.\] This means that $x$ equals $0$ or $9$ , and since a side length of $0$ cannot exist, $x=9$
As a result, the triangle must have sides in the ratio of $9:2:9$ . Since the triangle must have integer side lengths, and these values share no common factors greater than $1$ , the triangle with the smallest possible perimeter under these restrictions has a perimeter of $9+2+9=\boxed{020}$ . ~ emerald_block | null | 020 |
271d10430ba3ab1a2954e625c4e2dc09 | https://artofproblemsolving.com/wiki/index.php/2019_AIME_I_Problems/Problem_11 | In $\triangle ABC$ , the sides have integer lengths and $AB=AC$ . Circle $\omega$ has its center at the incenter of $\triangle ABC$ . An excircle of $\triangle ABC$ is a circle in the exterior of $\triangle ABC$ that is tangent to one side of the triangle and tangent to the extensions of the other two sides. Suppose that the excircle tangent to $\overline{BC}$ is internally tangent to $\omega$ , and the other two excircles are both externally tangent to $\omega$ . Find the minimum possible value of the perimeter of $\triangle ABC$ | Before we start thinking about the problem, let’s draw it out;
[asy] unitsize(1cm); var x = 9; pair A = (0,sqrt(x^2-1)); pair B = (-1,0); pair C = (1,0); dot(Label("$A$",A,NE),A); dot(Label("$B$",B,SW),B); dot(Label("$C$",C,SE),C); draw(A--B--C--cycle); var r = sqrt((x-1)/(x+1)); pair I = (0,r); dot(Label("$I$",I,SE),I); draw(circle(I,r)); pair G = intersectionpoint(A--B,circle(I,r)); pair D = (0,0); dot(Label("$G$",G,W),G); dot(Label("$D$",D,SSE),D); draw(Label("$r$"),I--G,dashed); var rA = sqrt((x+1)/(x-1)); pair IA = (0,-rA); dot(Label("$I_A$",IA,SE),IA); draw(circle(IA,rA)); pair AB = (-1-2/(x-1),-2rA); pair AC = (1+2/(x-1),-2rA); draw(B--AB,EndArrow); draw(C--AC,EndArrow); pair H = intersectionpoint(B--AB,circle(IA,rA)); dot(Label("$H$",H,W),H); draw(Label("$r_{I_A}$"),IA--H,dashed); var rB = sqrt(x^2-1); pair IB = (x,rB); dot(Label("$I_B$",IB,SE),IB); pair BC = (x+1,0); pair BA = (2/x,sqrt(x^2-1)+2(sqrt(x^2-1)/x)); draw(C--BC,EndArrow); draw(A--BA,EndArrow); pair E = intersectionpoint(A--C,circle(IB,rB)); pair F = intersectionpoint(C--BC,circle(IB,rB)); dot(Label("$E$",E,SE),E); dot(Label("$F$",F,S),F); draw(Label("$r_{I_B}$"),IB--F,dashed); draw(circle(IB,rB)); draw(A--IA); draw(B--IB); pair J = intersectionpoint(B--BA,circle(IB,rB)); dot(Label("$J$",J,W),J); draw(circle(I,r+2rA)); pair W = intersectionpoint(B--F,circle(I,r+2rA)); dot(Label("$\omega$",W,SSE),W); [/asy]
For the sake of space, I've drawn only 2 of the 3 excircles because the third one looks the same as the second large one because the triangle is isosceles. By the incenter-excenter lemma, $AII_A$ and $BII_B$ are collinear, $E$ is the tangent of circle $I_B$ to $AC$ $F$ is the tangent of that circle to the extension of $BC$ , and $J$ is the tangent of the circle to the extension of $BA$ . The interesting part of the diagram is circle $\omega$ , which is internally tangent to circle $I_A$ yet externally tangent to circle $I_B$ . Therefore, perhaps we can relate the radius of this circle to the semiperimeter of triangle $ABC$
We can see that the radius of circle $\omega$ is $2r_{I_A}+r$ using the incenter and A-excenter of our main triangle. This radius is also equal to $BI_B - BI - r_{I_B}$ from the incenter and B-excenter of our triangle. Thus, we can solve for each of these separately in terms of the lengths of the triangle and set them equal to each other to form an equation.
To find the left hand side of the equation, we have to first find $r$ and $r_{I_A}$ . Let $a = AB = AC, b = BD = DC,$ and $h = AD$ . Then since the perimeter of the triangle is $2a+2b$ , the semiperimeter is $a+b$
Now let's take a look at triangle $BDI$ . Because $BI$ is the angle bisector of $\angle B$ , by the angle bisector theorem, $\frac{AI}{ID} = \frac{BA}{BD} \implies \frac{h-r}{r} = \frac{a}{b}$ . Rearranging, we get $r = \frac{hb}{a+b}$
Take a look at triangle $AGI$ $AG = a - GB = a - BD = a-b$ $AI = h-r = \frac{ha}{a+b}$ (angle bisector theorem), and $GI = r = \frac{hb}{a+b}$ . Now let's analyze triangle $AHI_A$ $AH = AB + BH = AB+ BD = a+b$ $AI_A = h+r_{I_A}$ , and $HI_A = r_{I_A}$ . Since $\angle GAI = \angle HAI_A$ and $\angle IGA = \angle I_AHA = 90^{\circ}$ , triangle $AGI$ and $AHI_A$ are similar by AA. Then $\frac{r_{I_A}}{r} = \frac{h+r_{I_A}}{h-r} \implies r_{I_A} = r \cdot \frac{h+r_{I_A}}{h-r} = \frac{hb}{a+b} \cdot \frac{h+r_{I_A}}{\frac{ha}{a+b}} = \frac{b(h+r_{I_A})}{a}$ . Now, solving yields $r_{I_A} = \frac{hb}{a-b}$
Finally, the left hand side of our equation is \[\frac{2hb}{a-b} + \frac{hb}{a+b}\]
Now let's look at triangle $BFI_B$ . How will we find $BI_B$ ? Let's first try to find $BF$ and $I_BF$ in terms of the lengths of the triangle. We recognize:
$BF = BC + CF = BC + CK$ . We really want to have $CA$ instead of $CK$ , and $AK$ looks very similar in length to $DC$ , so let's try to prove that they are equal.
$BJ = BF$ , so we can try to add these two and see if we get anything interesting. We have: $BJ + BF = BA + AJ + BC + CF = BA + AE + BC + CE = BA + BC + CA$ , which is our perimeter. Thus, $BF = a+b$
Triangle $BDI$ is similar to triangle $BFI_B$ by AA, and we know that $BD = b$ , and $ID = r = \frac{hb}{a+b}$ , so thus $I_BF = BF \cdot \frac{ID}{BD} = (a+b) \cdot \frac{\frac{hb}{a+b}}{b} = \frac{hb}{b} = h$ . Thus, the height of this triangle is $h$ by similarity ratios, the same height as vertex $A$ . By the Pythagorean Theorem, $BI_B = \sqrt{(a+b)^2 + h^2}$ and by similarity ratios, $BI = \frac{b}{a+b} \cdot \sqrt{(a+b)^2 + h^2}$ . Finally, $r_{I_B} = I_BF = h$ , and thus the right hand side of our equation is \[\sqrt{(a+b)^2 + h^2} - \frac{b}{a+b} \cdot \sqrt{(a+b)^2 + h^2} - h = \sqrt{(a+b)^2 + h^2}(1 - \frac{b}{a+b}) - h = \sqrt{(a+b)^2 + h^2} \cdot \frac{a}{a+b} - h\]
Setting the two equal, we have \[\frac{2hb}{a-b} + \frac{hb}{a+b} = \sqrt{(a+b)^2 + h^2} \cdot \frac{a}{a+b} - h\]
Multiplying both sides by $(a+b)(a-b)$ we have $2hb(a+b) + hb(a-b) = \sqrt{(a+b)^2 + h^2} \cdot a(a-b) - h(a^2 - b^2)$
From here, let $b = 1$ arbitrarily; note that we can always scale this value to fit the requirements later. Thus our equation is $2h(a+1) + h(a-1) = \sqrt{(a+1)^2 + h^2} \cdot a(a-1) - h(a^2 - 1)$ . Now since $h = \sqrt{a^2 - b^2}$ , we can plug into our equation:
$2h(a+1) + h(a-1) = \sqrt{a² + 2a + 1 + a^2 - b^2 } \cdot a(a-1) - h(a^2 - 1)$ . Remembering $b = 1$
$\implies 2h(a+1) + h(a-1) = \sqrt{2a^2 + 2a} \cdot a(a-1) - h(a^2 - 1)$
$\implies 3ha + h + ha^2 - h = \sqrt{2a^2 + 2a} \cdot a(a-1)$
$\implies ah(3+a) = a(a-1) \cdot \sqrt{2a^2 + 2a}$
$\implies h^2(3+a)^2 = (a-1)^2 \cdot 2a(a+1)$
$\implies (a^2 - 1) (3 + a)^2 = 2a(a+1)(a - 1)^2$
$\implies (3+ a)^2 = 2a(a-1)$
$\implies 9 + 6a + a^2 = 2a^2 - 2a$
$\implies a^2 - 8a - 9 = 0$
$\implies (a-9)(a+1) = 0$
$\implies a = 9$ because the side lengths have to be positive numbers. Furthermore, because our values for $a$ and $b$ are relatively prime, we don't have to scale down our triangle further, and we are done. Therefore, our answer is $2a + 2b = 18 + 2 = \boxed{020}$ | null | 020 |
271d10430ba3ab1a2954e625c4e2dc09 | https://artofproblemsolving.com/wiki/index.php/2019_AIME_I_Problems/Problem_11 | In $\triangle ABC$ , the sides have integer lengths and $AB=AC$ . Circle $\omega$ has its center at the incenter of $\triangle ABC$ . An excircle of $\triangle ABC$ is a circle in the exterior of $\triangle ABC$ that is tangent to one side of the triangle and tangent to the extensions of the other two sides. Suppose that the excircle tangent to $\overline{BC}$ is internally tangent to $\omega$ , and the other two excircles are both externally tangent to $\omega$ . Find the minimum possible value of the perimeter of $\triangle ABC$ | Notice that the $A$ -excircle would have to be very small to fit the property that it is internally tangent to $\omega$ and the other two excircles are both externally tangent, given that circle $\omega$ 's centre is at the incenter of $\triangle ABC$ . If $BC=2$ , we see that $AB=AC$ must be somewhere in the $6$ to $13$ range. If we test $6$ by construction, we notice the $A$ -excircle is too big for it to be internally tangent to $\omega$ while the other two are externally tangent. This means we should test $8$ or $9$ next. I actually did this and found that $9$ worked, so the answer is $2+9+9=\boxed{20}$ . Note that $BC$ cannot be $1$ because then $AB=AC$ would have to be $4.5$ which is not an integer. | null | 20 |
271d10430ba3ab1a2954e625c4e2dc09 | https://artofproblemsolving.com/wiki/index.php/2019_AIME_I_Problems/Problem_11 | In $\triangle ABC$ , the sides have integer lengths and $AB=AC$ . Circle $\omega$ has its center at the incenter of $\triangle ABC$ . An excircle of $\triangle ABC$ is a circle in the exterior of $\triangle ABC$ that is tangent to one side of the triangle and tangent to the extensions of the other two sides. Suppose that the excircle tangent to $\overline{BC}$ is internally tangent to $\omega$ , and the other two excircles are both externally tangent to $\omega$ . Find the minimum possible value of the perimeter of $\triangle ABC$ | Let $M$ be the midpoint $BC, BM = a, AB= BC = b,$ $s = b+a$ be the semiperimeter, $r$ be the inradius. Let $I_A, I_B$ be excenters, $r_A, r_B$ be exradius, $R$ be radius $\omega.$ Then $R = r + 2 r_A,$ \[r = \sqrt{\frac{(s-2a)(s-b)^2}{s}} = a \sqrt{\frac{b-a}{b+a}},\] \[r \cdot s = r_A \cdot (s-2a) \implies r_A = a \sqrt{\frac{b+a}{b–a}},\] \[r \cdot s = r_B \cdot (s-b) \implies r_B = \sqrt{b^2 – a^2} = AM.\] \[II_B =R + r_B = r + 2r_A + r_B \implies\] \[II_B= a \sqrt{\frac{b-a}{b+ a}}+ 2a \sqrt{\frac{b+a}{b – a}} + \sqrt{b^2-a^2} = b\frac{3a+b}{\sqrt{b^2-a^2}}.\] \[\overline{AI_B}= \frac{\overline{A} \cdot 2a - \overline{B} \cdot b + \overline{C}\cdot b}{2a -b + b}-\overline{A} =b \frac{\overline{C}- \overline{B}}{2a},\] \[AI_B = b, \overline{AI_B}\perp AI \implies II_B = \sqrt{AI_B^2 + (AM – r)^2},\] \[II_B = \sqrt{b^2 +\left(\sqrt{b^2-a^2}- a \sqrt{\frac{b-a}{b+ a}}\right)^2} = b\sqrt{\frac{2b}{b+a}}\] Therefore we get problem’s condition in the form of \[b\frac{b +3a}{\sqrt{b^2-a^2}} = b \sqrt{\frac {2b}{b+a}} \implies b + 3a = \sqrt{2b(b-a)} \implies (b-9a)(b+a) = 0 \implies b = 9a.\]
We use $a = 1$ an get $b = 9, 2s = 18+2 = \boxed{020}$ | null | 020 |
dfb54e82313ea3965008fd026797a476 | https://artofproblemsolving.com/wiki/index.php/2019_AIME_I_Problems/Problem_12 | Given $f(z) = z^2-19z$ , there are complex numbers $z$ with the property that $z$ $f(z)$ , and $f(f(z))$ are the vertices of a right triangle in the complex plane with a right angle at $f(z)$ . There are positive integers $m$ and $n$ such that one such value of $z$ is $m+\sqrt{n}+11i$ . Find $m+n$ | Notice that we must have \[\frac{f(f(z))-f(z)}{f(z)-z}=-\frac{f(f(z))-f(z)}{z-f(z)}\in i\mathbb R .\] However, $f(t)-t=t(t-20)$ , so \begin{align*} \frac{f(f(z))-f(z)}{f(z)-z}&=\frac{(z^2-19z)(z^2-19z-20)}{z(z-20)}\\ &=\frac{z(z-19)(z-20)(z+1)}{z(z-20)}\\ &=(z-19)(z+1)\\ &=(z-9)^2-100. \end{align*} Then, the real part of $(z-9)^2$ is $100$ . Since $\text{Im}(z-9)=\text{Im}(z)=11$ , let $z-9=a+11i$ . Then, \[100=\text{Re}((a+11i)^2)=a^2-121\implies a=\pm\sqrt{221}.\] It follows that $z=9+\sqrt{221}+11i$ , and the requested sum is $9+221=\boxed{230}$ | null | 230 |
dfb54e82313ea3965008fd026797a476 | https://artofproblemsolving.com/wiki/index.php/2019_AIME_I_Problems/Problem_12 | Given $f(z) = z^2-19z$ , there are complex numbers $z$ with the property that $z$ $f(z)$ , and $f(f(z))$ are the vertices of a right triangle in the complex plane with a right angle at $f(z)$ . There are positive integers $m$ and $n$ such that one such value of $z$ is $m+\sqrt{n}+11i$ . Find $m+n$ | We will use the fact that segments $AB$ and $BC$ are perpendicular in the complex plane if and only if $\frac{a-b}{b-c}\in i\mathbb{R}$ . To prove this, note that when dividing two complex numbers you subtract the angle of one from the other. Therefore, if the two complex numbers are perpendicular, the difference between their arguments will be 90 degrees, so subtracting the angles will yield an imaginary number with no real part(an argument of 90 degrees puts a complex number on the imaginary axis).
Now to apply this: \[\frac{f(z)-z}{f(f(z))-f(z)}\in i\mathbb{R}\] \[\frac{z^2-19z-z}{(z^2-19z)^2-19(z^2-19z)-(z^2-19z)}\] \[\frac{z^2-20z}{z^4-38z^3+341z^2+380z}\] \[\frac{z(z-20)}{z(z+1)(z-19)(z-20)}\] \[\frac{1}{(z+1)(z-19)}\in i\mathbb{R}\]
The factorization of the nasty denominator above is made easier with the intuition that $(z-20)$ must be a divisor for the problem to lead anywhere. Now we know $(z+1)(z-19)\in i\mathbb{R}$ so using the fact that the imaginary part of $z$ is $11i$ and calling the real part r,
\[(r+1+11i)(r-19+11i)\in i\mathbb{R}\] \[r^2-18r-140=0\]
solving the above quadratic yields $r=9+\sqrt{221}$ so our answer is $9+221=\boxed{230}$ | null | 230 |
dfb54e82313ea3965008fd026797a476 | https://artofproblemsolving.com/wiki/index.php/2019_AIME_I_Problems/Problem_12 | Given $f(z) = z^2-19z$ , there are complex numbers $z$ with the property that $z$ $f(z)$ , and $f(f(z))$ are the vertices of a right triangle in the complex plane with a right angle at $f(z)$ . There are positive integers $m$ and $n$ such that one such value of $z$ is $m+\sqrt{n}+11i$ . Find $m+n$ | I would like to use a famous method, namely the coni method.
According to the question given, we can assume , $A= f(f(z)),B=f(z),C= z$ respectively.
WLOG, $Z_1= \frac{f(f(z))-f(z)}{z-f(z)}$ .
According to the question $\arg{Z_1}=\frac{\pi}{2}$
So, $\Re (Z_1)=0$
Now, $Z_1=\frac{z(z-19)(z+1)(z-20)}{-z(z-20)}$
$\implies Z_1= -(z^2-18z-19)$ .
WLOG, $z=a+11i$ .where $a=m+\sqrt{n}$
So, $\Re (Z_1)= -(a^2-18a-140)$ .
Solving, $a^2-18a-140=0$ .get ,
$a=9$ $\sqrt{221}$ .
So, possible value of $a=9+\sqrt{221}$
$m+n=\boxed{230}$ .
~ftheftics. | null | 230 |
dfb54e82313ea3965008fd026797a476 | https://artofproblemsolving.com/wiki/index.php/2019_AIME_I_Problems/Problem_12 | Given $f(z) = z^2-19z$ , there are complex numbers $z$ with the property that $z$ $f(z)$ , and $f(f(z))$ are the vertices of a right triangle in the complex plane with a right angle at $f(z)$ . There are positive integers $m$ and $n$ such that one such value of $z$ is $m+\sqrt{n}+11i$ . Find $m+n$ | The arguments of the two complex numbers differ by $90^\circ$ if the ratio of the numbers is a pure imaginary number. Thus three distinct complex numbers $A,\,B,$ and $C$ form a right triangle at $B$ if and only if $\tfrac{C-B}{B-A}$ has real part equal to $0.$ Hence \begin{align*} \frac{f(f(z))-f(z)}{f(z)-z}&=\frac{(z^2-19z)^2-19(z^2-19z)-(z^2-19z)}{(z^2-19z)-z}\\ &=\frac{(z^2-19z)(z^2-19z-19-1)}{z^2-20z}\\ &=\frac{z(z-19)(z+1)(z-20)}{z(z-20)} \\ &=z^2-18z-19 \end{align*} must have real part equal to $0.$ If $z=x+11i,$ the real part of $z^2-18z-19$ is $x^2-11^2-18x-19,$ which is $0$ when $x=9\pm\sqrt{221}.$ The requested sum is $9+221=\boxed{230}.$ | null | 230 |
dfb54e82313ea3965008fd026797a476 | https://artofproblemsolving.com/wiki/index.php/2019_AIME_I_Problems/Problem_12 | Given $f(z) = z^2-19z$ , there are complex numbers $z$ with the property that $z$ $f(z)$ , and $f(f(z))$ are the vertices of a right triangle in the complex plane with a right angle at $f(z)$ . There are positive integers $m$ and $n$ such that one such value of $z$ is $m+\sqrt{n}+11i$ . Find $m+n$ | Firstly, the angle between the three complex numbers is equivalent to the angle between $f(z)-z,0,$ and $f(f(z))-f(z)$
Using $f(z)=z(z-19)$ to help expand,
$z-f(z)=20z-z^2$ and $f(f(z))-f(z)=(z^2-19z)(z^2-19z-19)-(z^2-19z)$
The second equation can be rewritten as $(z^2-19z)(z^2-19z-20)=z(z-19)(z-20)(z+1)$
Note that the angle between $(a+bi), 0,$ and $(a+bi)(c+di)$ is the same as the angle between $1, 0,$ and $c+di$ .
The proof of this is as follows:
Treating the complex numbers like vectors (e.g. $a+bi$ turns into $\left[\begin{array}{c} a \\ b \end{array}\right]$ ) we have $cos \theta = \frac{a \cdot (ac-bd) + b \cdot (ad+bc)}{2\sqrt{a^2+b^2}\sqrt{a^2+b^2}\sqrt{c^2+d^2}}$
$cos \theta = \frac{a^2 c - abd + abd + b^2 c}{2(a^2+b^2)\sqrt{c^2+d^2}}$
$cos \theta = \frac{(a^2+b^2)c}{2(a^2+b^2)\sqrt{c^2+d^2}}=\frac{c}{2\sqrt{c^2+d^2}}$
Using the dot product formula for the cosine of the angle between the other two vectors (1, c + di) we get the same result. Thus, it is proven.
Now, we see that the three resulting complex numbers have the same angle as $-1,0,$ and $(z-19)(z+1)$ as a factor of $z(z-20)$ can be taken out of both expressions. Note that the value of $z$ clearly is not 0, nor is it 20 as it has a value of $11i$
Expanding $z^2-18z-19$ with $z=a+11i$ yields
$a^2-121+22ai-18a-198i-19=a^2-18a-140+22ai-198i$ . Turning these into vectors again, the resulting relation is
$cos(90^\circ)=0=\frac{-(a^2-18a-140)}{2\sqrt{(a^2-18a-140)^2 + (22a-198)^2} \cdot 1}$
In order for this thing to be 0, the numerator must be 0, so setting it equal to 0 yields
$a^2-18a-140=0 \Rightarrow a=9\pm \sqrt{81+140}$ . Neither of these make the denominator in the $cos(90^\circ)$ expression 0, so they are valid. The requested result is the positive root, so the answer is $\boxed{230}$ | null | 230 |
d94e833e8150c532227be47ecfc644c4 | https://artofproblemsolving.com/wiki/index.php/2019_AIME_I_Problems/Problem_13 | Triangle $ABC$ has side lengths $AB=4$ $BC=5$ , and $CA=6$ . Points $D$ and $E$ are on ray $AB$ with $AB<AD<AE$ . The point $F \neq C$ is a point of intersection of the circumcircles of $\triangle ACD$ and $\triangle EBC$ satisfying $DF=2$ and $EF=7$ . Then $BE$ can be expressed as $\tfrac{a+b\sqrt{c}}{d}$ , where $a$ $b$ $c$ , and $d$ are positive integers such that $a$ and $d$ are relatively prime, and $c$ is not divisible by the square of any prime. Find $a+b+c+d$ | [asy] unitsize(20); pair A, B, C, D, E, F, X, O1, O2; A = (0, 0); B = (4, 0); C = intersectionpoints(circle(A, 6), circle(B, 5))[0]; D = B + (5/4 * (1 + sqrt(2)), 0); E = D + (4 * sqrt(2), 0); F = intersectionpoints(circle(D, 2), circle(E, 7))[1]; X = extension(A, E, C, F); O1 = circumcenter(C, A, D); O2 = circumcenter(C, B, E); filldraw(A--B--C--cycle, lightcyan, deepcyan); filldraw(D--E--F--cycle, lightmagenta, deepmagenta); draw(B--D, gray(0.6)); draw(C--F, gray(0.6)); draw(circumcircle(C, A, D), dashed); draw(circumcircle(C, B, E), dashed); dot("$A$", A, dir(A-O1)); dot("$B$", B, dir(240)); dot("$C$", C, dir(120)); dot("$D$", D, dir(40)); dot("$E$", E, dir(E-O2)); dot("$F$", F, dir(270)); dot("$X$", X, dir(140)); label("$6$", (C+A)/2, dir(C-A)*I, deepcyan); label("$5$", (C+B)/2, dir(B-C)*I, deepcyan); label("$4$", (A+B)/2, dir(A-B)*I, deepcyan); label("$7$", (F+E)/2, dir(F-E)*I, deepmagenta); label("$2$", (F+D)/2, dir(D-F)*I, deepmagenta); label("$4\sqrt{2}$", (D+E)/2, dir(E-D)*I, deepmagenta); label("$a$", (B+X)/2, dir(B-X)*I, gray(0.3)); label("$a\sqrt{2}$", (D+X)/2, dir(D-X)*I, gray(0.3)); [/asy]
Notice that \[\angle DFE=\angle CFE-\angle CFD=\angle CBE-\angle CAD=180-B-A=C.\] By the Law of Cosines, \[\cos C=\frac{AC^2+BC^2-AB^2}{2\cdot AC\cdot BC}=\frac34.\] Then, \[DE^2=DF^2+EF^2-2\cdot DF\cdot EF\cos C=32\implies DE=4\sqrt2.\] Let $X=\overline{AB}\cap\overline{CF}$ $a=XB$ , and $b=XD$ . Then, \[XA\cdot XD=XC\cdot XF=XB\cdot XE\implies b(a+4)=a(b+4\sqrt2)\implies b=a\sqrt2.\] However, since $\triangle XFD\sim\triangle XAC$ $XF=\tfrac{4+a}3$ , but since $\triangle XFE\sim\triangle XBC$ \[\frac75=\frac{4+a}{3a}\implies a=\frac54\implies BE=a+a\sqrt2+4\sqrt2=\frac{5+21\sqrt2}4,\] and the requested sum is $5+21+2+4=\boxed{032}$ | null | 032 |
d94e833e8150c532227be47ecfc644c4 | https://artofproblemsolving.com/wiki/index.php/2019_AIME_I_Problems/Problem_13 | Triangle $ABC$ has side lengths $AB=4$ $BC=5$ , and $CA=6$ . Points $D$ and $E$ are on ray $AB$ with $AB<AD<AE$ . The point $F \neq C$ is a point of intersection of the circumcircles of $\triangle ACD$ and $\triangle EBC$ satisfying $DF=2$ and $EF=7$ . Then $BE$ can be expressed as $\tfrac{a+b\sqrt{c}}{d}$ , where $a$ $b$ $c$ , and $d$ are positive integers such that $a$ and $d$ are relatively prime, and $c$ is not divisible by the square of any prime. Find $a+b+c+d$ | Define $\omega_1$ to be the circumcircle of $\triangle ACD$ and $\omega_2$ to be the circumcircle of $\triangle EBC$
Because of exterior angles,
$\angle ACB = \angle CBE - \angle CAD$
But $\angle CBE = \angle CFE$ because $CBFE$ is cyclic. In addition, $\angle CAD = \angle CFD$ because $CAFD$ is cyclic. Therefore, $\angle ACB = \angle CFE - \angle CFD$ . But $\angle CFE - \angle CFD = \angle DFE$ , so $\angle ACB = \angle DFE$ . Using Law of Cosines on $\triangle ABC$ , we can figure out that $\cos(\angle ACB) = \frac{3}{4}$ . Since $\angle ACB = \angle DFE$ $\cos(\angle DFE) = \frac{3}{4}$ . We are given that $DF = 2$ and $FE = 7$ , so we can use Law of Cosines on $\triangle DEF$ to find that $DE = 4\sqrt{2}$
Let $G$ be the intersection of segment $\overline{AE}$ and $\overline{CF}$ . Using Power of a Point with respect to $G$ within $\omega_1$ , we find that $AG \cdot GD = CG \cdot GF$ . We can also apply Power of a Point with respect to $G$ within $\omega_2$ to find that $CG \cdot GF = BG \cdot GE$ . Therefore, $AG \cdot GD = BG \cdot GE$
$AG \cdot GD = BG \cdot GE$
$(AB + BG) \cdot GD = BG \cdot (GD + DE)$
$AB \cdot GD + BG \cdot GD = BG \cdot GD + BG \cdot DE$
$AB \cdot GD = BG \cdot DE$
$4 \cdot GD = BG \cdot 4\sqrt{2}$
$GD = BG \cdot \sqrt{2}$
Note that $\triangle GAC$ is similar to $\triangle GFD$ $GF = \frac{BG + 4}{3}$ . Also note that $\triangle GBC$ is similar to $\triangle GFE$ , which gives us $GF = \frac{7 \cdot BG}{5}$ . Solving this system of linear equations, we get $BG = \frac{5}{4}$ . Now, we can solve for $BE$ , which is equal to $BG(\sqrt{2} + 1) + 4\sqrt{2}$ . This simplifies to $\frac{5 + 21\sqrt{2}}{4}$ , which means our answer is $\boxed{032}$ | null | 032 |
d94e833e8150c532227be47ecfc644c4 | https://artofproblemsolving.com/wiki/index.php/2019_AIME_I_Problems/Problem_13 | Triangle $ABC$ has side lengths $AB=4$ $BC=5$ , and $CA=6$ . Points $D$ and $E$ are on ray $AB$ with $AB<AD<AE$ . The point $F \neq C$ is a point of intersection of the circumcircles of $\triangle ACD$ and $\triangle EBC$ satisfying $DF=2$ and $EF=7$ . Then $BE$ can be expressed as $\tfrac{a+b\sqrt{c}}{d}$ , where $a$ $b$ $c$ , and $d$ are positive integers such that $a$ and $d$ are relatively prime, and $c$ is not divisible by the square of any prime. Find $a+b+c+d$ | Construct $FC$ and let $FC\cap AE=K$ . Let $FK=x$ . Using $\triangle FKE\sim \triangle BKC$ \[BK=\frac{5}{7}x\] Using $\triangle FDK\sim ACK$ , it can be found that \[3x=AK=4+\frac{5}{7}x\to x=\frac{7}{4}\] This also means that $BK=\frac{21}{4}-4=\frac{5}{4}$ . It suffices to find $KE$ . It is easy to see the following: \[180-\angle ABC=\angle KBC=\angle KFE\] Using reverse Law of Cosines on $\triangle ABC$ $\cos{\angle ABC}=\frac{1}{8}\to \cos{180-\angle ABC}=\frac{-1}{8}$ . Using Law of Cosines on $\triangle EFK$ gives $KE=\frac{21\sqrt 2}{4}$ , so $BE=\frac{5+21\sqrt 2}{4}\to \boxed{032}$ .
-franchester | null | 032 |
d94e833e8150c532227be47ecfc644c4 | https://artofproblemsolving.com/wiki/index.php/2019_AIME_I_Problems/Problem_13 | Triangle $ABC$ has side lengths $AB=4$ $BC=5$ , and $CA=6$ . Points $D$ and $E$ are on ray $AB$ with $AB<AD<AE$ . The point $F \neq C$ is a point of intersection of the circumcircles of $\triangle ACD$ and $\triangle EBC$ satisfying $DF=2$ and $EF=7$ . Then $BE$ can be expressed as $\tfrac{a+b\sqrt{c}}{d}$ , where $a$ $b$ $c$ , and $d$ are positive integers such that $a$ and $d$ are relatively prime, and $c$ is not divisible by the square of any prime. Find $a+b+c+d$ | Let $P=AE\cap CF$ . Let $CP=5x$ and $BP=5y$ ; from $\triangle{CBP}\sim\triangle{EFP}$ we have $EP=7x$ and $FP=7y$ . From $\triangle{CAP}\sim\triangle{DFP}$ we have $\frac{6}{4+5y}=\frac{2}{7y}$ giving $y=\frac{1}{4}$ . So $BP=\frac{5}{4}$ and $FP=\frac{7}{4}$ . These similar triangles also gives us $DP=\frac{5}{3}x$ so $DE=\frac{16}{3}x$ . Now, Stewart's Theorem on $\triangle{FEP}$ and cevian $FD$ tells us that \[\frac{560}{9}x^3+28x=\frac{49}{3}x+\frac{245}{3}x,\] so $x=\frac{3\sqrt{2}}{4}$ . Then $BE=\frac{5}{4}+7x=\frac{5+21\sqrt{2}}{4}$ so the answer is $\boxed{032}$ as desired. (Solution by Trumpeter, but not added to the Wiki by Trumpeter) | null | 032 |
d94e833e8150c532227be47ecfc644c4 | https://artofproblemsolving.com/wiki/index.php/2019_AIME_I_Problems/Problem_13 | Triangle $ABC$ has side lengths $AB=4$ $BC=5$ , and $CA=6$ . Points $D$ and $E$ are on ray $AB$ with $AB<AD<AE$ . The point $F \neq C$ is a point of intersection of the circumcircles of $\triangle ACD$ and $\triangle EBC$ satisfying $DF=2$ and $EF=7$ . Then $BE$ can be expressed as $\tfrac{a+b\sqrt{c}}{d}$ , where $a$ $b$ $c$ , and $d$ are positive integers such that $a$ and $d$ are relatively prime, and $c$ is not divisible by the square of any prime. Find $a+b+c+d$ | Connect $CF$ meeting $AE$ at $J$ . We can observe that $\triangle{ACJ}\sim \triangle{FJD}$ Getting that $\frac{AJ}{FJ}=\frac{AC}{FD}=3$ . We can also observe that $\triangle{CBJ}\sim \triangle{EFJ}$ , getting that $\frac{CB}{EF}=\frac{BJ}{FJ}=\frac{CJ}{EJ}=\frac{5}{7}$
Assume that $BJ=5x;FJ=7x$ , since $\frac{AJ}{FJ}=3$ , we can get that $\frac{AJ}{FJ}=\frac{AB+BJ}{FJ}=\frac{4+5x}{7x}=3$ , getting that $x=\frac{1}{4}; BJ=\frac{5}{4}; FJ=\frac{7}{4}$
Using Power of Point, we can get that $BJ * EJ=CJ*FJ; DJ * AJ=CJ * FJ$ Assume that $DJ=5k,CJ=15k$ , getting that $JE=21k, DE=16k$
Now applying Law of Cosine on two triangles, $\triangle{ACJ};\triangle{FJE}$ separately, we can get two equations
$(1): (15k)^2+(\frac{21}{4})^2-2*15k *\frac{21}{4} * cos\angle{CJA}=36$
$(2):(21k)^2+(\frac{7}{4})^2-2*\frac{7}{4} * 21k*cos\angle{FJE}=49$
Since $\angle{CJA}=\angle{FJE}$ , we can use $15(2)-7(1)$ to eliminate the $cos$ term
Then we can get that $5040k^2=630$ , getting $k=\frac{\sqrt{2}}{4}$
$BE=21k=\frac{21\sqrt{2}}{4}; BJ=\frac{5}{4}$ , so the desired answer is $\frac{21\sqrt{2}+5}{4}$ , which leads to the answer $\boxed{032}$ | null | 032 |
d94e833e8150c532227be47ecfc644c4 | https://artofproblemsolving.com/wiki/index.php/2019_AIME_I_Problems/Problem_13 | Triangle $ABC$ has side lengths $AB=4$ $BC=5$ , and $CA=6$ . Points $D$ and $E$ are on ray $AB$ with $AB<AD<AE$ . The point $F \neq C$ is a point of intersection of the circumcircles of $\triangle ACD$ and $\triangle EBC$ satisfying $DF=2$ and $EF=7$ . Then $BE$ can be expressed as $\tfrac{a+b\sqrt{c}}{d}$ , where $a$ $b$ $c$ , and $d$ are positive integers such that $a$ and $d$ are relatively prime, and $c$ is not divisible by the square of any prime. Find $a+b+c+d$ | Nice problem!
First, let $AE$ and $CF$ intersect at $X$ . Our motivation here is to introduce cyclic quadrilaterals and find useful relationships in terms of angles. Observe that \[\angle DFE = \angle XFE - \angle XFD = \angle CBE - \angle CAB = 180 - \angle ABC - \angle CAB = \angle BAC\] By the so-called "Reverse Law of Cosines" on $\triangle ABC$ we have \[\cos(\angle BAC) = \frac{4^2 - 5^2 - 6^2}{-2 \cdot 5 \cdot 6} = \frac{3}{4}\] Applying on $\triangle DFE$ gives \[DE^2 = 2^2 + 7^2 - 2 \cdot 2 \cdot 7 \cos(\angle DFE)\] \[= 2^2 + 7^2 - 2 \cdot 2 \cdot 7 \cdot \frac{3}{4}\] \[=32\] So $DE = 4 \sqrt{2}$ , now by our cyclic quadrilaterals again, we are motivated by the multiple appearances of similar triangles throughout the figure. We want some that are related to $BX$ and $XD$ , which are crucial lengths in the problem. Suppose $BX = r, XD = s$ for simplicity. We have:
$\bullet~~~~\triangle AXC \sim \triangle FXD$ $\bullet~~~~\triangle BXC \sim \triangle FXE$
So \[\frac{AX}{FX} = \frac{XC}{XD} = \frac{AC}{FD} \implies \frac{4 + r}{FX} = \frac{XC}{s} = 3\] \[\frac{BX}{FX} = \frac{XC}{XE} = \frac{BC}{FE} \implies \frac{r}{FX} = \frac{XC}{s + 4 \sqrt{2}} = \frac{5}{7}\] \[\implies \frac{4 + r}{r} = \frac{s + 4 \sqrt{2}}{s} = \frac{21}{5}\] \[\implies r = \frac{5}{4}, s = \frac{5 \sqrt{2}}{4}\] So $BE = r + s + 4 \sqrt{2} = \frac{5 + 21 \sqrt{2}}{4}$ . The requested sum is $5 + 21 + 2 + 4 = \boxed{032}$ | null | 032 |
409c3242de66d15ac0fdb0c8b1a01b40 | https://artofproblemsolving.com/wiki/index.php/2019_AIME_I_Problems/Problem_15 | Let $\overline{AB}$ be a chord of a circle $\omega$ , and let $P$ be a point on the chord $\overline{AB}$ . Circle $\omega_1$ passes through $A$ and $P$ and is internally tangent to $\omega$ . Circle $\omega_2$ passes through $B$ and $P$ and is internally tangent to $\omega$ . Circles $\omega_1$ and $\omega_2$ intersect at points $P$ and $Q$ . Line $PQ$ intersects $\omega$ at $X$ and $Y$ . Assume that $AP=5$ $PB=3$ $XY=11$ , and $PQ^2 = \frac{m}{n}$ , where $m$ and $n$ are relatively prime positive integers. Find $m+n$ | [asy] size(8cm); pair O, A, B, P, O1, O2, Q, X, Y; O=(0, 0); A=dir(140); B=dir(40); P=(3A+5B)/8; O1=extension((A+P)/2, (A+P)/2+(0, 1), A, O); O2=extension((B+P)/2, (B+P)/2+(0, 1), B, O); Q=intersectionpoints(circle(O1, length(A-O1)), circle(O2, length(B-O2)))[1]; X=intersectionpoint(P -- (P+(P-Q)*100), circle(O, 1)); Y=intersectionpoint(Q -- (Q+(Q-P)*100), circle(O, 1)); draw(circle(O, 1)); draw(circle(O1, length(A-O1))); draw(circle(O2, length(B-O2))); draw(A -- B); draw(X -- Y); draw(A -- O -- B); draw(O1 -- P -- O2); dot("$O$", O, S); dot("$A$", A, A); dot("$B$", B, B); dot("$P$", P, dir(70)); dot("$Q$", Q, dir(200)); dot("$O_1$", O1, SW); dot("$O_2$", O2, SE); dot("$X$", X, X); dot("$Y$", Y, Y); [/asy] Let $O_1$ and $O_2$ be the centers of $\omega_1$ and $\omega_2$ , respectively. There is a homothety at $A$ sending $\omega$ to $\omega_1$ that sends $B$ to $P$ and $O$ to $O_1$ , so $\overline{OO_2}\parallel\overline{O_1P}$ . Similarly, $\overline{OO_1}\parallel\overline{O_2P}$ , so $OO_1PO_2$ is a parallelogram. Moreover, \[\angle O_1QO_2=\angle O_1PO_2=\angle O_1OO_2,\] whence $OO_1O_2Q$ is cyclic. However, \[OO_1=O_2P=O_2Q,\] so $OO_1O_2Q$ is an isosceles trapezoid. Since $\overline{O_1O_2}\perp\overline{XY}$ $\overline{OQ}\perp\overline{XY}$ , so $Q$ is the midpoint of $\overline{XY}$
By Power of a Point, $PX\cdot PY=PA\cdot PB=15$ . Since $PX+PY=XY=11$ and $XQ=11/2$ \[XP=\frac{11-\sqrt{61}}2\implies PQ=XQ-XP=\frac{\sqrt{61}}2\implies PQ^2=\frac{61}4,\] and the requested sum is $61+4=\boxed{065}$ | null | 065 |
409c3242de66d15ac0fdb0c8b1a01b40 | https://artofproblemsolving.com/wiki/index.php/2019_AIME_I_Problems/Problem_15 | Let $\overline{AB}$ be a chord of a circle $\omega$ , and let $P$ be a point on the chord $\overline{AB}$ . Circle $\omega_1$ passes through $A$ and $P$ and is internally tangent to $\omega$ . Circle $\omega_2$ passes through $B$ and $P$ and is internally tangent to $\omega$ . Circles $\omega_1$ and $\omega_2$ intersect at points $P$ and $Q$ . Line $PQ$ intersects $\omega$ at $X$ and $Y$ . Assume that $AP=5$ $PB=3$ $XY=11$ , and $PQ^2 = \frac{m}{n}$ , where $m$ and $n$ are relatively prime positive integers. Find $m+n$ | One may solve for $PX$ first using PoAP, $PX = \frac{11}{2} - \frac{\sqrt{61}}{2}$ . Then, notice that $PQ^2$ is rational but $PX^2$ is not, also $PX = \frac{XY}{2} - \frac{\sqrt{61}}{2}$ . The most likely explanation for this is that $Q$ is the midpoint of $XY$ , so that $XQ = \frac{11}{2}$ and $PQ=\frac{\sqrt{61}}{2}$ . Then our answer is $m+n=61+4=\boxed{065}$ . One can rigorously prove this using the methods above | null | 065 |
409c3242de66d15ac0fdb0c8b1a01b40 | https://artofproblemsolving.com/wiki/index.php/2019_AIME_I_Problems/Problem_15 | Let $\overline{AB}$ be a chord of a circle $\omega$ , and let $P$ be a point on the chord $\overline{AB}$ . Circle $\omega_1$ passes through $A$ and $P$ and is internally tangent to $\omega$ . Circle $\omega_2$ passes through $B$ and $P$ and is internally tangent to $\omega$ . Circles $\omega_1$ and $\omega_2$ intersect at points $P$ and $Q$ . Line $PQ$ intersects $\omega$ at $X$ and $Y$ . Assume that $AP=5$ $PB=3$ $XY=11$ , and $PQ^2 = \frac{m}{n}$ , where $m$ and $n$ are relatively prime positive integers. Find $m+n$ | Let the tangents to $\omega$ at $A$ and $B$ intersect at $R$ . Then, since $RA^2=RB^2$ $R$ lies on the radical axis of $\omega_1$ and $\omega_2$ , which is $\overline{PQ}$ . It follows that \[-1=(A,B;X,Y)\stackrel{A}{=}(R,P;X,Y).\] Let $Q'$ denote the midpoint of $\overline{XY}$ . By the Midpoint of Harmonic Bundles Lemma(EGMO 9.17), \[RP\cdot RQ'=RX\cdot RY=RA^2=RP\cdot RQ,\] whence $Q=Q'$ . Like above, $XP=\tfrac{11-\sqrt{61}}2$ . Since $XQ=\tfrac{11}2$ , we establish that $PQ=\tfrac{\sqrt{61}}2$ , from which $PQ^2=\tfrac{61}4$ , and the requested sum is $61+4=\boxed{065}$ | null | 065 |
409c3242de66d15ac0fdb0c8b1a01b40 | https://artofproblemsolving.com/wiki/index.php/2019_AIME_I_Problems/Problem_15 | Let $\overline{AB}$ be a chord of a circle $\omega$ , and let $P$ be a point on the chord $\overline{AB}$ . Circle $\omega_1$ passes through $A$ and $P$ and is internally tangent to $\omega$ . Circle $\omega_2$ passes through $B$ and $P$ and is internally tangent to $\omega$ . Circles $\omega_1$ and $\omega_2$ intersect at points $P$ and $Q$ . Line $PQ$ intersects $\omega$ at $X$ and $Y$ . Assume that $AP=5$ $PB=3$ $XY=11$ , and $PQ^2 = \frac{m}{n}$ , where $m$ and $n$ are relatively prime positive integers. Find $m+n$ | Firstly we need to notice that $Q$ is the middle point of $XY$ . Assume the center of circle $w, w_1, w_2$ are $O, O_1, O_2$ , respectively. Then $A, O_1, O$ are collinear and $O, O_2, B$ are collinear. Link $O_1P, O_2P, O_1Q, O_2Q$ . Notice that, $\angle B=\angle A=\angle APO_1=\angle BPO_2$ . As a result, $PO_1\parallel O_2O$ and $OO_1\parallel O_2P$ . So we have parallelogram $PO_2OO_1$ . So $\angle O_2PO_1=\angle O$ Notice that, $O_1O_2\bot PQ$ and $O_1O_2$ divides $PQ$ into two equal length pieces, So we have $\angle O_2QO_1=\angle O_2PO_1=\angle O$ . As a result, $O_2, Q, O, O_1,$ lie on one circle. So $\angle OQO_1=\angle OO_2O_1=\angle O_2O_1P$ . Notice that since $\angle O_1PQ+\angle O_2O_1P=90^{\circ}$ , we have $\angle OQP=\angle OQO_1 + \angle O_1QP = \angle O_2O_1P + O_1PQ=90^{\circ}$ . As a result, $OQ\bot PQ$ . So $Q$ is the middle point of $XY$
Back to our problem. Assume $XP=x$ $PY=y$ and $x<y$ . Then we have $AP\cdot PB=XP\cdot PY$ , that is, $xy=15$ . Also, $XP+PY=x+y=XY=11$ . Solve these above, we have $x=\frac{11-\sqrt{61}}{2}=XP$ . As a result, we have $PQ=XQ-XP=\frac{11}{2}-\frac{11-\sqrt{61}}{2}=\frac{\sqrt{61}}{2}$ . So, we have $PQ^2=\frac{61}{4}$ . As a result, our answer is $m+n=61+4=\boxed{065}$ | null | 065 |
409c3242de66d15ac0fdb0c8b1a01b40 | https://artofproblemsolving.com/wiki/index.php/2019_AIME_I_Problems/Problem_15 | Let $\overline{AB}$ be a chord of a circle $\omega$ , and let $P$ be a point on the chord $\overline{AB}$ . Circle $\omega_1$ passes through $A$ and $P$ and is internally tangent to $\omega$ . Circle $\omega_2$ passes through $B$ and $P$ and is internally tangent to $\omega$ . Circles $\omega_1$ and $\omega_2$ intersect at points $P$ and $Q$ . Line $PQ$ intersects $\omega$ at $X$ and $Y$ . Assume that $AP=5$ $PB=3$ $XY=11$ , and $PQ^2 = \frac{m}{n}$ , where $m$ and $n$ are relatively prime positive integers. Find $m+n$ | Note that the tangents to the circles at $A$ and $B$ intersect at a point $Z$ on $XY$ by radical axis theorem. Since $\angle ZAB = \angle ZQA$ and $\angle ZBA = \angle ZQB$ , we have \[\angle AZB + \angle AQB = \angle AZB + \angle ZAB + \angle ZBA = 180^{\circ},\] so $ZAQB$ is cyclic.
But if $O$ is the center of $\omega$ , clearly $ZAOB$ is cyclic with diameter $ZO$ , so $\angle ZQO = 90^{\circ}$ implies that $Q$ is the midpoint of $XY$ . Then, by power of point $P$ \[PY \cdot PX = PA \cdot PB = 15,\] whereas it is given that $PY+PX = 11$ . Thus \[PY, PX \in \left\{\tfrac 12 (11 \pm \sqrt{61})\right\}\] so $PQ = \frac{\sqrt{61}}{2}$ , i.e. $PQ^2 = \frac{61}{4}$ and the answer is $61+4 = \boxed{065}$ | null | 065 |
409c3242de66d15ac0fdb0c8b1a01b40 | https://artofproblemsolving.com/wiki/index.php/2019_AIME_I_Problems/Problem_15 | Let $\overline{AB}$ be a chord of a circle $\omega$ , and let $P$ be a point on the chord $\overline{AB}$ . Circle $\omega_1$ passes through $A$ and $P$ and is internally tangent to $\omega$ . Circle $\omega_2$ passes through $B$ and $P$ and is internally tangent to $\omega$ . Circles $\omega_1$ and $\omega_2$ intersect at points $P$ and $Q$ . Line $PQ$ intersects $\omega$ at $X$ and $Y$ . Assume that $AP=5$ $PB=3$ $XY=11$ , and $PQ^2 = \frac{m}{n}$ , where $m$ and $n$ are relatively prime positive integers. Find $m+n$ | Connect $AQ,QB$ , since $\angle{AO_1P}=\angle{AOB}=\angle{BO_2P}$ , so $\angle{AQP}=\frac{\angle{AO_1P}}{2}=\angle{BQP}=\frac{\angle{BO_2P}}{2}, \angle{AQB}=\angle{AOB}$ then, so $A,O,Q,B$ are concyclic
We let $\angle{AO_1P}=\angle{AOB}=\angle{BO_2P}=2\alpha$ , it is clear that $\angle{BQP}=\alpha, \angle{O_1AP}=90^{\circ}-\alpha$ , which leads to the conclusion $OQ\bot XY$ which tells $Q$ is the midpoint of $XY$
Then it is clear, $XP\cdot PY=15, XP=\frac{11-\sqrt{61}}{2}, PQ=\frac{11}{2}-\frac{11-\sqrt{61}}{2}=\frac{\sqrt{61}}{2}$ , the answer is $\boxed{065}$ | null | 065 |
409c3242de66d15ac0fdb0c8b1a01b40 | https://artofproblemsolving.com/wiki/index.php/2019_AIME_I_Problems/Problem_15 | Let $\overline{AB}$ be a chord of a circle $\omega$ , and let $P$ be a point on the chord $\overline{AB}$ . Circle $\omega_1$ passes through $A$ and $P$ and is internally tangent to $\omega$ . Circle $\omega_2$ passes through $B$ and $P$ and is internally tangent to $\omega$ . Circles $\omega_1$ and $\omega_2$ intersect at points $P$ and $Q$ . Line $PQ$ intersects $\omega$ at $X$ and $Y$ . Assume that $AP=5$ $PB=3$ $XY=11$ , and $PQ^2 = \frac{m}{n}$ , where $m$ and $n$ are relatively prime positive integers. Find $m+n$ | [asy] size(8cm); pair O, A, B, P, O1, O2, Q, X, Y; O=(0, 0); A=dir(140); B=dir(40); P=(3A+5B)/8; O1=extension((A+P)/2, (A+P)/2+(0, 1), A, O); O2=extension((B+P)/2, (B+P)/2+(0, 1), B, O); Q=intersectionpoints(circle(O1, length(A-O1)), circle(O2, length(B-O2)))[1]; X=intersectionpoint(P -- (P+(P-Q)*100), circle(O, 1)); Y=intersectionpoint(Q -- (Q+(Q-P)*100), circle(O, 1)); draw(circle(O, 1)); draw(circle(O1, length(A-O1))); draw(circle(O2, length(B-O2))); draw(A -- B,red); draw(X -- Y,green); dot("$A$", A, A); dot("$B$", B, B); dot("$P$", P, dir(70),blue); dot("$Q$", Q, dir(200)); dot("$X$", X, X); dot("$Y$", Y, Y); label("$3$", (A+P)/2, N, red); label("$5$", (B+P)/2, N, red); draw(brace(X,Y)); label("$11$",brace(X,Y),dir(20)); [/asy] $PX \cdot PY=AP \cdot PB=5 \cdot 3=15$ by power of a point. Also, $PX+PY=XY=11$ , so $PX$ and $PY$ are solutions to the quadratic $x^2-11x+15=0$ so $PX$ and $PY$ is $\frac{11\pm\sqrt{61}}{2}$ in some order. Now, because we want $PQ^2$ and it is known to be rational, we can guess that $PQ$ is irrational or the problem would simply ask for $PQ$ . We can also figure out that since $PQ^2$ is rational, $PQ$ is $\sqrt{\text{[something]}}$ $PQ=QX-PX$ , and chances are low that $QX$ is some number with a square root plus or minus $\frac{\sqrt{61}}{2}$ to cancel out the $\frac{\sqrt{61}}{2}$ in $PX$ , so one can see that $PQ^2$ is most likely to be $\left(\frac{\sqrt{61}}{2}\right)^2=\frac{61}{4}$ , and our answer is $61+4=\boxed{065}$ | null | 065 |
409c3242de66d15ac0fdb0c8b1a01b40 | https://artofproblemsolving.com/wiki/index.php/2019_AIME_I_Problems/Problem_15 | Let $\overline{AB}$ be a chord of a circle $\omega$ , and let $P$ be a point on the chord $\overline{AB}$ . Circle $\omega_1$ passes through $A$ and $P$ and is internally tangent to $\omega$ . Circle $\omega_2$ passes through $B$ and $P$ and is internally tangent to $\omega$ . Circles $\omega_1$ and $\omega_2$ intersect at points $P$ and $Q$ . Line $PQ$ intersects $\omega$ at $X$ and $Y$ . Assume that $AP=5$ $PB=3$ $XY=11$ , and $PQ^2 = \frac{m}{n}$ , where $m$ and $n$ are relatively prime positive integers. Find $m+n$ | We will show that $Q$ is the midpoint of $XY.$ To do this, let $Q^{\prime}$ be the altitude from $O$ to $XY$ or, equivalently, to $PQ.$ Notice that $O_{1}00_{2}P$ is a parallelogram. Thus, the height from $O$ to $O_{1}O_{2}$ is equal to the height from $P$ to $O_{1}O_{2}.$ Say that the line through $P$ perpendicular to $O_{1}O_{2}$ intersects $O_{1}O_{2}$ at $H.$ Then, $PQ$ is perpendicular to $O_{1}O_{2},$ so $H$ is on $PQ.$ Now, we have that the altitude from $O$ to $O_{1}O_{2}$ is equal to the altitude from $Q^{\prime}$ to $O_{1}O_{2}$ (since $OQ^{\prime} \parallel O_{1}O_{2}$ ). However, the altitude from $Q^{\prime}$ to $O_{1}O_{2}$ is just $Q^{\prime}H.$ Also, the altitude from $P$ to $O_{1}O_{2}$ is $PH$ , so $PH = Q^{\prime}H.$ Thus, $O_{1}O_{2}$ bisects $PQ^{\prime}.$ However, this is true for $Q,$ too, so $Q = Q^{\prime},$ and we are done. Now, by PoP, we have \[AP \cdot BP = XP \cdot YP = 15.\] Also, we have $XY = XP+YP = 11,$ so $XP = \frac{11 \pm \sqrt{61}}{2}$ . Notice that $XQ = \frac{XY}{2} = \frac{11}{2},$ so $PQ = \frac{\sqrt{61}}{2},$ giving us our answer of $\boxed{065}.$ | null | 065 |
409c3242de66d15ac0fdb0c8b1a01b40 | https://artofproblemsolving.com/wiki/index.php/2019_AIME_I_Problems/Problem_15 | Let $\overline{AB}$ be a chord of a circle $\omega$ , and let $P$ be a point on the chord $\overline{AB}$ . Circle $\omega_1$ passes through $A$ and $P$ and is internally tangent to $\omega$ . Circle $\omega_2$ passes through $B$ and $P$ and is internally tangent to $\omega$ . Circles $\omega_1$ and $\omega_2$ intersect at points $P$ and $Q$ . Line $PQ$ intersects $\omega$ at $X$ and $Y$ . Assume that $AP=5$ $PB=3$ $XY=11$ , and $PQ^2 = \frac{m}{n}$ , where $m$ and $n$ are relatively prime positive integers. Find $m+n$ | Like Solution 7, let $Q'$ be the altitude from $O$ to $XY$ . And, let $M$ be the intersection of $O_1O_2$ and $PQ$ . Construct $P'$ on line $AO$ such that $PP' \parallel O_2O_1$ . First, because of isosceles triangles $OAB$ $O_1AP$ , and $O_2BP$ , we have $\angle{OAP} = \angle{OBA} = \angle{APO_1} = \angle{BPO_2}$ , which means $OO_1PO_2$ is a parallelogram. So, $O_2P = OO_1$ . It is also clear that $PP'O_1O_2$ is a parallelogram by virtue of our definition. Thus, $O_2P = O_1P' = OO_1$ . Since $OQ' \parallel O_1O_2 \parallel P'P$ (because of the right angles), $\frac{Q'M}{MP} = \frac{OO_1}{O_1P'} = 1 \implies Q'M = MP$ . And, because $QM = MP$ $Q = Q'$ . From Power of a Point on $P$ , we have $XP(11-XP) = 15$ , giving us $XP = \frac{11 - \sqrt{61}}{2}$ . Since $OQ$ is perpendicular to $XY$ $Q$ is the midpoint of $XY$ , so $XQ = \frac{11}{2}$ . Thus, $PQ = \frac{11}{2} - \frac{11 - \sqrt{61}}{2} = \frac{\sqrt{61}}{2} \implies {PQ}^2 = \frac{61}{4}$ . Therefore, our answer is $\boxed{65}$ | null | 65 |
e2f39f89371a439a30aab759b8857df9 | https://artofproblemsolving.com/wiki/index.php/2019_AIME_II_Problems/Problem_1 | Two different points, $C$ and $D$ , lie on the same side of line $AB$ so that $\triangle ABC$ and $\triangle BAD$ are congruent with $AB = 9$ $BC=AD=10$ , and $CA=DB=17$ . The intersection of these two triangular regions has area $\tfrac mn$ , where $m$ and $n$ are relatively prime positive integers. Find $m+n$ | [asy] unitsize(10); pair A = (0,0); pair B = (9,0); pair C = (15,8); pair D = (-6,8); pair E = (-6,0); draw(A--B--C--cycle); draw(B--D--A); label("$A$",A,dir(-120)); label("$B$",B,dir(-60)); label("$C$",C,dir(60)); label("$D$",D,dir(120)); label("$E$",E,dir(-135)); label("$9$",(A+B)/2,dir(-90)); label("$10$",(D+A)/2,dir(-150)); label("$10$",(C+B)/2,dir(-30)); label("$17$",(D+B)/2,dir(60)); label("$17$",(A+C)/2,dir(120)); draw(D--E--A,dotted); label("$8$",(D+E)/2,dir(180)); label("$6$",(A+E)/2,dir(-90)); [/asy] - Diagram by Brendanb4321
Extend $AB$ to form a right triangle with legs $6$ and $8$ such that $AD$ is the hypotenuse and connect the points $CD$ so
that you have a rectangle. (We know that $\triangle ADE$ is a $6-8-10$ , since $\triangle DEB$ is an $8-15-17$ .) The base $CD$ of the rectangle will be $9+6+6=21$ . Now, let $O$ be the intersection of $BD$ and $AC$ . This means that $\triangle ABO$ and $\triangle DCO$ are with ratio $\frac{21}{9}=\frac73$ . Set up a proportion, knowing that the two heights add up to 8. We will let $y$ be the height from $O$ to $DC$ , and $x$ be the height of $\triangle ABO$ \[\frac{7}{3}=\frac{y}{x}\] \[\frac{7}{3}=\frac{8-x}{x}\] \[7x=24-3x\] \[10x=24\] \[x=\frac{12}{5}\]
This means that the area is $A=\tfrac{1}{2}(9)(\tfrac{12}{5})=\tfrac{54}{5}$ . This gets us $54+5=\boxed{059}.$ | null | 059 |
e2f39f89371a439a30aab759b8857df9 | https://artofproblemsolving.com/wiki/index.php/2019_AIME_II_Problems/Problem_1 | Two different points, $C$ and $D$ , lie on the same side of line $AB$ so that $\triangle ABC$ and $\triangle BAD$ are congruent with $AB = 9$ $BC=AD=10$ , and $CA=DB=17$ . The intersection of these two triangular regions has area $\tfrac mn$ , where $m$ and $n$ are relatively prime positive integers. Find $m+n$ | Using the diagram in Solution 1, let $E$ be the intersection of $BD$ and $AC$ . We can see that angle $C$ is in both $\triangle BCE$ and $\triangle ABC$ . Since $\triangle BCE$ and $\triangle ADE$ are congruent by AAS, we can then state $AE=BE$ and $DE=CE$ . It follows that $BE=AE$ and $CE=17-BE$ . We can now state that the area of $\triangle ABE$ is the area of $\triangle ABC-$ the area of $\triangle BCE$ . Using Heron's formula, we compute the area of $\triangle ABC=36$ . Using the Law of Cosines on angle $C$ , we obtain
\[9^2=17^2+10^2-2(17)(10)cosC\] \[-308=-340cosC\] \[cosC=\frac{308}{340}\] (For convenience, we're not going to simplify.)
Applying the Law of Cosines on $\triangle BCE$ yields \[BE^2=10^2+(17-BE)^2-2(10)(17-BE)cosC\] \[BE^2=389-34BE+BE^2-20(17-BE)(\frac{308}{340})\] \[0=389-34BE-(340-20BE)(\frac{308}{340})\] \[0=389-34BE+\frac{308BE}{17}\] \[0=81-\frac{270BE}{17}\] \[81=\frac{270BE}{17}\] \[BE=\frac{51}{10}\] This means $CE=17-BE=17-\frac{51}{10}=\frac{119}{10}$ . Next, apply Heron's formula to get the area of $\triangle BCE$ , which equals $\frac{126}{5}$ after simplifying. Subtracting the area of $\triangle BCE$ from the area of $\triangle ABC$ yields the area of $\triangle ABE$ , which is $\frac{54}{5}$ , giving us our answer, which is $54+5=\boxed{059}.$ -Solution by flobszemathguy | null | 059 |
e2f39f89371a439a30aab759b8857df9 | https://artofproblemsolving.com/wiki/index.php/2019_AIME_II_Problems/Problem_1 | Two different points, $C$ and $D$ , lie on the same side of line $AB$ so that $\triangle ABC$ and $\triangle BAD$ are congruent with $AB = 9$ $BC=AD=10$ , and $CA=DB=17$ . The intersection of these two triangular regions has area $\tfrac mn$ , where $m$ and $n$ are relatively prime positive integers. Find $m+n$ | [asy] unitsize(10); pair A = (0,0); pair B = (9,0); pair C = (15,8); pair D = (-6,8); draw(A--B--C--cycle); draw(B--D--A); label("$A$",A,dir(-120)); label("$B$",B,dir(-60)); label("$C$",C,dir(60)); label("$D$",D,dir(120)); label("$9$",(A+B)/2,dir(-90)); label("$10$",(D+A)/2,dir(-150)); label("$10$",(C+B)/2,dir(-30)); label("$17$",(D+B)/2,dir(60)); label("$17$",(A+C)/2,dir(120)); draw(D--(-6,0)--A,dotted); label("$8$",(D+(-6,0))/2,dir(180)); label("$6$",(A+(-6,0))/2,dir(-90)); draw((4.5,0)--(4.5,2.4),dotted); label("$h$", (4.5,1.2), dir(180)); label("$4.5$", (6,0), dir(90)); [/asy] - Diagram by Brendanb4321 extended by Duoquinquagintillion
Begin with the first step of solution 1, seeing $AD$ is the hypotenuse of a $6-8-10$ triangle and calling the intersection of $DB$ and $AC$ point $E$ . Next, notice $DB$ is the hypotenuse of an $8-15-17$ triangle. Drop an altitude from $E$ with length $h$ , so the other leg of the new triangle formed has length $4.5$ . Notice we have formed similar triangles, and we can solve for $h$
\[\frac{h}{4.5} = \frac{8}{15}\] \[h = \frac{36}{15} = \frac{12}{5}\]
So $\triangle ABE$ has area \[\frac{ \frac{12}{5} \cdot 9}{2} = \frac{54}{5}\] And $54+5=\boxed{059}.$ - Solution by Duoquinquagintillion | null | 059 |
e2f39f89371a439a30aab759b8857df9 | https://artofproblemsolving.com/wiki/index.php/2019_AIME_II_Problems/Problem_1 | Two different points, $C$ and $D$ , lie on the same side of line $AB$ so that $\triangle ABC$ and $\triangle BAD$ are congruent with $AB = 9$ $BC=AD=10$ , and $CA=DB=17$ . The intersection of these two triangular regions has area $\tfrac mn$ , where $m$ and $n$ are relatively prime positive integers. Find $m+n$ | Let $a = \angle{CAB}$ . By Law of Cosines, \[\cos a = \frac{17^2+9^2-10^2}{2*9*17} = \frac{15}{17}\] \[\sin a = \sqrt{1-\cos^2 a} = \frac{8}{17}\] \[\tan a = \frac{8}{15}\] \[A = \frac{1}{2}* 9*\frac{9}{2}\tan a = \frac{54}{5}\] And $54+5=\boxed{059}.$ | null | 059 |
e2f39f89371a439a30aab759b8857df9 | https://artofproblemsolving.com/wiki/index.php/2019_AIME_II_Problems/Problem_1 | Two different points, $C$ and $D$ , lie on the same side of line $AB$ so that $\triangle ABC$ and $\triangle BAD$ are congruent with $AB = 9$ $BC=AD=10$ , and $CA=DB=17$ . The intersection of these two triangular regions has area $\tfrac mn$ , where $m$ and $n$ are relatively prime positive integers. Find $m+n$ | Let $A=(0,0), B=(9,0)$ . Now consider $C$ , and if we find the coordinates of $C$ , by symmetry about $x=4.5$ , we can find the coordinates of D.
So let $C=(a,b)$ . So the following equations hold:
$\sqrt{(a-9)^2+(b)^2}=17$
$\sqrt{a^2+b^2}=10$
Solving by squaring both equations and then subtracting one from the other to eliminate $b^2$ , we get $C=(-6,8)$ because $C$ is in the second quadrant.
Now by symmetry, $D=(16, 8)$
So now you can proceed by finding the intersection and then calculating the area directly. We get $\boxed{059}$ | null | 059 |
e2f39f89371a439a30aab759b8857df9 | https://artofproblemsolving.com/wiki/index.php/2019_AIME_II_Problems/Problem_1 | Two different points, $C$ and $D$ , lie on the same side of line $AB$ so that $\triangle ABC$ and $\triangle BAD$ are congruent with $AB = 9$ $BC=AD=10$ , and $CA=DB=17$ . The intersection of these two triangular regions has area $\tfrac mn$ , where $m$ and $n$ are relatively prime positive integers. Find $m+n$ | [asy] unitsize(10); pair A = (0,0); pair B = (9,0); pair C = (15,8); pair D = (-6,8); pair O = (4.5,2.4); pair E = (-6,0); pair K = (1.993,3.737); draw(A--B--C--cycle); draw(B--D--A); label("$A$",A,dir(-120)); label("$B$",B,dir(-60)); label("$C$",C,dir(60)); label("$D$",D,dir(120)); label("$9$",(A+B)/2,dir(-90)); label("$10$",(D+A)/2,dir(-150)); label("$10$",(C+B)/2,dir(-30)); label("$17$",(D+B)/2,dir(60)); label("$17$",(A+C)/2,dir(120)); label("$O$",O,dir(90)); label("$K$",K,dir(-120)); draw(A--K,dotted); draw(D--E--A,dotted); label("$8$",(D+E)/2,dir(180)); label("$6$",(A+E)/2,dir(-90)); label("$E$",E,dir(-135)); [/asy]
Since $\triangle ABC \cong \triangle BAD$ $\angle ADB = \angle BCA$ . Thus, $A$ $B$ $C$ $D$ are concyclic.
By Ptolemy's Theorem on $ABCD$ \[(AD)(BC) + (AB)(DC) = (BD)(AC)\] \[10^2 + 9(DC) = 17^2\] \[DC = 21\]
The altitudes dropped from $C$ and $D$ onto the extension of AB are equal, meaning that $DC \parallel AB$ . Therefore, $\triangle DCO \sim \triangle ABO$ . It follows that \[\frac{OB}{17 - OB} = \frac{AB}{DC} = \frac{9}{21} = \frac{3}{7}\] Solving yields $OB = \frac{51}{10}$
In $\triangle ABD$ , drop an altitude from $A$ to $BD$ . Call the intersection of this altitude and $BD$ $K$
The area of $\triangle ABD$ is $\frac{1}{2}(AB)(DE) = 36$ . Thus, $\frac{1}{2}(AK)(BD) = 36$ , and $AK = \frac{72}{17}$
Therefore, the area of $\triangle AOB$ is $\frac{1}{2}(OB)(AK) = \frac{1}{2}(\frac{51}{10})(\frac{72}{17}) = \frac{54}{5}$
The requested answer is $54 + 5 = \boxed{59}$ | null | 59 |
d034bafe19129106b55d3236d006a395 | https://artofproblemsolving.com/wiki/index.php/2019_AIME_II_Problems/Problem_2 | Lilypads $1,2,3,\ldots$ lie in a row on a pond. A frog makes a sequence of jumps starting on pad $1$ . From any pad $k$ the frog jumps to either pad $k+1$ or pad $k+2$ chosen randomly with probability $\tfrac{1}{2}$ and independently of other jumps. The probability that the frog visits pad $7$ is $\tfrac{p}{q}$ , where $p$ and $q$ are relatively prime positive integers. Find $p+q$ | Let $P_n$ be the probability the frog visits pad $7$ starting from pad $n$ . Then $P_7 = 1$ $P_6 = \frac12$ , and $P_n = \frac12(P_{n + 1} + P_{n + 2})$ for all integers $1 \leq n \leq 5$ . Working our way down, we find \[P_5 = \frac{3}{4}\] \[P_4 = \frac{5}{8}\] \[P_3 = \frac{11}{16}\] \[P_2 = \frac{21}{32}\] \[P_1 = \frac{43}{64}\] $43 + 64 = \boxed{107}$ | null | 107 |
d034bafe19129106b55d3236d006a395 | https://artofproblemsolving.com/wiki/index.php/2019_AIME_II_Problems/Problem_2 | Lilypads $1,2,3,\ldots$ lie in a row on a pond. A frog makes a sequence of jumps starting on pad $1$ . From any pad $k$ the frog jumps to either pad $k+1$ or pad $k+2$ chosen randomly with probability $\tfrac{1}{2}$ and independently of other jumps. The probability that the frog visits pad $7$ is $\tfrac{p}{q}$ , where $p$ and $q$ are relatively prime positive integers. Find $p+q$ | Define a one jump to be a jump from $k$ to $k + 1$ and a two jump to be a jump from $k$ to $k + 2$
Case 1: (6 one jumps) $\left (\frac{1}{2} \right)^6 = \frac{1}{64}$
Case 2: (4 one jumps and 1 two jumps) $\binom{5}{1} \cdot \left(\frac{1}{2}\right)^5 = \frac{5}{32}$
Case 3: (2 one jumps and 2 two jumps) $\binom{4}{2} \cdot \left(\frac{1}{2}\right)^4 = \frac{3}{8}$
Case 4: (3 two jumps) $\left(\frac{1}{2}\right)^3 = \frac{1}{8}$
Summing the probabilities gives us $\frac{43}{64}$ so the answer is $\boxed{107}$ | null | 107 |
d034bafe19129106b55d3236d006a395 | https://artofproblemsolving.com/wiki/index.php/2019_AIME_II_Problems/Problem_2 | Lilypads $1,2,3,\ldots$ lie in a row on a pond. A frog makes a sequence of jumps starting on pad $1$ . From any pad $k$ the frog jumps to either pad $k+1$ or pad $k+2$ chosen randomly with probability $\tfrac{1}{2}$ and independently of other jumps. The probability that the frog visits pad $7$ is $\tfrac{p}{q}$ , where $p$ and $q$ are relatively prime positive integers. Find $p+q$ | Let $P_n$ be the probability that the frog lands on lily pad $n$ . The probability that the frog never lands on pad $n$ is $\frac{1}{2}P_{n-1}$ , so $1-P_n=\frac{1}{2}P_{n-1}$ . This rearranges to $P_n=1-\frac{1}{2}P_{n-1}$ , and we know that $P_1=1$ , so we can compute $P_7$ \begin{align*} P_1&=1\\ P_2&=1-\dfrac{1}{2} \cdot 1=\dfrac{1}{2}\\ P_3&=1-\dfrac{1}{2} \cdot \dfrac{1}{2}=\dfrac{3}{4}\\ P_4&=\dfrac{5}{8}\\ P_5&=\dfrac{11}{16}\\ P_6&=\dfrac{21}{32}\\ P_7&=\dfrac{43}{64}\\ \end{align*} We calculate $P_7$ to be $\frac{43}{64}$ , meaning that our answer is $\boxed{107}$ | null | 107 |
d034bafe19129106b55d3236d006a395 | https://artofproblemsolving.com/wiki/index.php/2019_AIME_II_Problems/Problem_2 | Lilypads $1,2,3,\ldots$ lie in a row on a pond. A frog makes a sequence of jumps starting on pad $1$ . From any pad $k$ the frog jumps to either pad $k+1$ or pad $k+2$ chosen randomly with probability $\tfrac{1}{2}$ and independently of other jumps. The probability that the frog visits pad $7$ is $\tfrac{p}{q}$ , where $p$ and $q$ are relatively prime positive integers. Find $p+q$ | For any point $n$ , let the probability that the frog lands on lily pad $n$ be $P_n$ . The frog can land at lily pad $n$ with either a double jump from lily pad $n-2$ or a single jump from lily pad $n-1$ . Since the probability when the frog is at $n-2$ to make a double jump is $\frac{1}{2}$ and same for when it's at $n-1$ , the recursion is just $P_n = \frac{P_{n-2}+P_{n-1}}{2}$ . Using the fact that $P_1 = 1$ , and $P_2 = \frac{1}{2}$ , we find that $P_7 = \frac{43}{64}$ $43 + 64 = \boxed{107}$ | null | 107 |
2118f7af9560fb57f37c2f6c0b0d15fa | https://artofproblemsolving.com/wiki/index.php/2019_AIME_II_Problems/Problem_3 | Find the number of $7$ -tuples of positive integers $(a,b,c,d,e,f,g)$ that satisfy the following systems of equations: \begin{align*} abc&=70,\\ cde&=71,\\ efg&=72. \end{align*} | As 71 is prime, $c$ $d$ , and $e$ must be 1, 1, and 71 (in some order). However, since $c$ and $e$ are divisors of 70 and 72 respectively, the only possibility is $(c,d,e) = (1,71,1)$ . Now we are left with finding the number of solutions $(a,b,f,g)$ satisfying $ab = 70$ and $fg = 72$ , which separates easily into two subproblems. The number of positive integer solutions to $ab = 70$ simply equals the number of divisors of 70 (as we can choose a divisor for $a$ , which uniquely determines $b$ ). As $70 = 2^1 \cdot 5^1 \cdot 7^1$ , we have $d(70) = (1+1)(1+1)(1+1) = 8$ solutions. Similarly, $72 = 2^3 \cdot 3^2$ , so $d(72) = 4 \times 3 = 12$
Then the answer is simply $8 \times 12 = \boxed{096}$ | null | 096 |
2118f7af9560fb57f37c2f6c0b0d15fa | https://artofproblemsolving.com/wiki/index.php/2019_AIME_II_Problems/Problem_3 | Find the number of $7$ -tuples of positive integers $(a,b,c,d,e,f,g)$ that satisfy the following systems of equations: \begin{align*} abc&=70,\\ cde&=71,\\ efg&=72. \end{align*} | We know that any two consecutive numbers are coprime. Using this, we can figure out that $c=1$ and $e=1$ $d$ then has to be $71$ . Now we have two equations left. $ab=70$ and $fg=72$ . To solve these we just need to figure out all of the factors. Doing the prime factorization of $70$ and $72$ , we find that they have $8$ and $12$ factors, respectively. The answer is $8 \times 12=\boxed{096}$ | null | 096 |
5611776d841d723d0718847e18dbe2f6 | https://artofproblemsolving.com/wiki/index.php/2019_AIME_II_Problems/Problem_4 | A standard six-sided fair die is rolled four times. The probability that the product of all four numbers rolled is a perfect square is $\tfrac{m}{n}$ , where $m$ and $n$ are relatively prime positive integers. Find $m+n$ | Notice that, other than the number 5, the remaining numbers 1, 2, 3, 4, 6 are only divisible by 2 and/or 3. We can do some cases on the number of 5's rolled (note that there are $6^4 = 1296$ outcomes).
Case 1 (easy): Four 5's are rolled. This has probability $\frac{1}{6^4}$ of occurring.
Case 2: Two 5's are rolled.
Case 3: No 5's are rolled.
To find the number of outcomes for the latter two cases, we will use recursion. Consider a 5-sided die with faces numbered 1, 2, 3, 4, 6. For $n \ge 1$ , let $a_n$ equal the number of outcomes after rolling the die $n$ times, with the property that the product is a square. Thus, $a_1 = 2$ as 1 and 4 are the only possibilities.
To find $a_{n+1}$ given $a_n$ (where $n \ge 1$ ), we observe that if the first $n$ rolls multiply to a perfect square, then the last roll must be 1 or 4. This gives $2a_n$ outcomes. Otherwise, the first $n$ rolls do not multiply to a perfect square ( $5^n - a_n$ outcomes). In this case, we claim that the last roll is uniquely determined (either 2, 3, or 6). If the product of the first $n$ rolls is $2^x 3^y$ where $x$ and $y$ are not both even, then we observe that if $x$ and $y$ are both odd, then the last roll must be 6; if only $x$ is odd, the last roll must be 2, and if only $y$ is odd, the last roll must be 3. Thus, we have $5^n - a_n$ outcomes in this case, and $a_{n+1} = 2a_n + (5^n - a_n) = 5^n + a_n$
Computing $a_2$ $a_3$ $a_4$ gives $a_2 = 7$ $a_3 = 32$ , and $a_4 = 157$ . Thus for Case 3, there are 157 outcomes. For case 2, we multiply $a_2$ by $\binom{4}{2} = 6$ to distribute the two 5's among four rolls. Thus the probability is
\[\frac{1 + 6 \cdot 7 + 157}{6^4} = \frac{200}{6^4} = \frac{25}{162} \implies m+n = \boxed{187}\] | null | 187 |