2025 AIME II Problems/Problem 12

Problem

Let $A_1A_2\dots A_{11}$ be a non-convex $11$ -gon such that

• The area of $A_iA_1A_{i+1}$ is $1$ for each $2 \le i \le 10$ ,

• $\cos(\angle A_iA_1A_{i+1})=\frac{12}{13}$ for each $2 \le i \le 10$ ,

• The perimeter of $A_1A_2\dots A_{11}$ is $20$ .

If $A_1A_2+A_1A_{11}$ can be expressed as $\frac{m\sqrt{n}-p}{q}$ for positive integers $m,n,p,q$ with $n$ squarefree and $\gcd(m,p,q)=1$ , find $m+n+p+q$ .

Solution 1

Set $A_1A_2 = x$ and $A_1A_3 = y$ . By the first condition, we have $\frac{1}{2}xy\sin\theta = 1$ , where $\theta = \angle A_2 A_1 A_3$ . Since $\cos\theta = \frac{12}{13}$ , we have $\sin\theta = \frac{5}{13}$ , so $xy = \frac{26}{5}$ . Repeating this process for $\triangle A_i A_1 A_{i+1}$ , we get $A_1A_2 = A_1A_4 = \ldots A_1A_{10} = x$ and $A_1A_3 = A_1A_5 = \ldots A_1A_{11} = y$ . Since the included angle of these $9$ triangles is $\theta$ , the square of the third side is $x^2 + y^2 - 2xy\cos\theta = x^2 + y^2 - \frac{52}{5}\cdot \frac{12}{13} = x^2 + y^2 - \frac{48}{5} = (x+y)^2 - 20.$ Thus the third side has length $\sqrt{(x+y)^2 - 20}.$ The perimeter is constructed from $9$ of these lengths, plus $A_{11}A_1 + A_1A_2 = x + y$ , so $9\sqrt{(x+y)^2 - 20} + x + y = 20$ . We seek the value of $x + y,$ so let $x + y = a$ so $\begin{align*} 9\sqrt{a^2 - 20} + a &= 20\\ 81(a^2 - 20) &= 400 - 40a + a^2\\ 4a^2 + 2a - 101 &= 0 \\ a &= \frac{-2 \pm \sqrt{1620}}{8} = \frac{-1 \pm \sqrt{405}}{4} = \frac{-1 \pm 9\sqrt{5}}{4}. \end{align*}$ Taking the positive solution gives $m + n + p + q = 1 + 9 + 5 + 4 = \boxed{\textbf{(019)}}.$

-Benedict T (countmath1)

2025 AIME II (Problems • Answer Key • Resources)
Preceded by Problem 11	Followed by Problem 13
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2025 AIME II Problems/Problem 12

Problem

Solution 1

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