Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

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

m (→‎Solution 1)
m (→‎Problem)
 
Line 4: Line 4:
  
 
• The area of <math>A_iA_1A_{i+1}</math> is <math>1</math> for each <math>2 \le i \le 10</math>,
 
• The area of <math>A_iA_1A_{i+1}</math> is <math>1</math> for each <math>2 \le i \le 10</math>,
 +
 
• <math>\cos(\angle A_iA_1A_{i+1})=\frac{12}{13}</math> for each <math>2 \le i \le 10</math>,
 
• <math>\cos(\angle A_iA_1A_{i+1})=\frac{12}{13}</math> for each <math>2 \le i \le 10</math>,
 +
 
• The perimeter of <math>A_1A_2\dots A_{11}</math> is <math>20</math>.
 
• The perimeter of <math>A_1A_2\dots A_{11}</math> is <math>20</math>.
  

Latest revision as of 22:03, 15 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

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)

See also

2025 AIME II (Problems • Answer Key • Resources)
Preceded by
Problem 11 		Followed by
Problem 13
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15
All AIME Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2025_AIME_II_Problems/Problem_12&oldid=242855"