Difference between revisions of "2025 AIME II Problems/Problem 12"

Revision as of 03:07, 14 February 2025

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

Since $[A_1A_iA_{i+1}]$ are the same, we have have $A_1A_{11}=A_1A_{9}=...=A_1A_3=x$ and $A_1A_2=A_1A_4=...=A_1A_{10}=y$ , since $\angle{A_iA_1A_{i+1}}$ is the same for all the $2\leq i\leq 10$ , so $A_iA_{i+1}$ are the same for all $2\leq i\leq 10$ , set them be $d$

Now we have $x+y+9d=20, x^2+y^2-2xy\cdot \frac{12}{13}=d^2, xy\cdot \frac{5}{13}=2$

Solve the system of equations we could get $d=\frac{9-\sqrt{5}}{4}$ , $x+y=20-9d=\frac{9\sqrt{5}-1}{4}\implies \boxed{019}$

~Bluesoul

@@ Line 18: / Line 18: @@
 ~Bluesoul
+==See also==
+{{AIME box|year=2025|num-b=11|num-a=13|n=II}}
+{{MAA Notice}}

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Difference between revisions of "2025 AIME II Problems/Problem 12"

Revision as of 03:07, 14 February 2025

Problem

Solution 1

See also