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

2025 AIME II Problems/Problem 12

Revision as of 01:44, 14 February 2025 by Tonyttian (talk | contribs) (→‎Solution 1)

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

$\forall$ 2 $\le$ i $\le$ 10, cos\angle $A_{i}<cmath>A_{1}</cmath>A_{i+1}$=$\frac{12}{13}$\\ So sin\angle $A_{i}<cmath>A_{1}</cmath>A_{i+1}$=$\frac{5}{13}$\\ Since the area of each triangle is 1,\\ $\frac{1}{2}$\times\[A_{1}\]A_{i}$\times$A_{1}$$ (Error compiling LaTeX. Unknown error_msg)A_{i+1}$\times sin$\angle A_{i}\[A_{1}\]A_{i+1}$=1\$ (Error compiling LaTeX. Unknown error_msg)\Rightarrow$$ (Error compiling LaTeX. Unknown error_msg)A_{1}$$ (Error compiling LaTeX. Unknown error_msg)A_{i}$\times$A_{1}$$ (Error compiling LaTeX. Unknown error_msg)A_{i+1}$=$\frac{26}{5} $\\ So$A_{1}$$ (Error compiling LaTeX. Unknown error_msg)A_{i}$\times$A_{1}$$ (Error compiling LaTeX. Unknown error_msg)A_{i+1}$=$A_{1}$$ (Error compiling LaTeX. Unknown error_msg)A_{i+1}$\times$A_{1}$$ (Error compiling LaTeX. Unknown error_msg)A_{i+2}$\\ This means$A_{1}$$ (Error compiling LaTeX. Unknown error_msg)A_{i}$=A_{1}$$A_{i+2}$\\ In $\triangle A_{1}<cmath>A_{i}</cmath>A_{i+1}$ and $\triangle A_{1}<cmath>A_{i+2}</cmath>A_{i+1}$，\\ they share one of the same side and the angles on vertex $A_{1}$ are the same $A_{1}$$A_{i}$= $A_{1}$$A_{i=2}$\\ So they are congruent \\ This means $\forall$ 2 $\le$ i $\le$ 9 $A_{i}$$A_{i+1}$= $A_{i+1}$$A_{i+2}$\\ Perimeter = $A_{1}$$A_{2}$+$\sum_{i=2}^{10}A_{i}$$A_{i+1}$+$A_{11}$$A_{1}$=20\\ Then $A_{1}$$A_{2}$+$A_{11}$$A_{1}$+9$A_{2}$$A_{3}$=20\\ Let us set $A_{1}$$A_{2}$=a $A_{11}$$A_{1}$=b and $A_{2}$$A_{3}$=c\\ Then a+b+9c=20\\ Now, we apply cosine law in $\triangle A_{1}<cmath>A_{2}</cmath>A_{3}$\\ $A_{2}$$A_{3}$^{2}=$A_{1}$$A_{2}$^{2} +$A_{1}$$A_{3}$^{3}-2$A_{1}$$A_{2}$\times$A_{1}$$A_{3}$cos\angle A_{2}\[A_{1}\]A_{3}$\$ (Error compiling LaTeX. Unknown error_msg)\Rightarrow$c^{2}=a^{2}+b^{2}-2ab$\times\frac {12}{13} $\\ Set$A_{1}$$ (Error compiling LaTeX. Unknown error_msg)A_{2}$+$A_{11}$$ (Error compiling LaTeX. Unknown error_msg)A_{1}$=t,\\ then c^{2}=(a+b)^{2}-2ab-2ab$\times\frac {12}{13}=t^{2}-2\times\frac {26}{5}\times (1+\frac {12}{13})\\ $\Rightarrow$ c^{2}=t^{2}- 20 \\ Since 9c=20-a-b=20-t\\ Square both sides, giving 81c^{2}=400+t^{2}-40t\\ $\Rightarrow$ 81t^{2}-20\times81=400+t^{2}-40t\\ $\Rightarrow$ 80t^{2}+40t-20\times101\\ $\Rightarrow$ 80t^{2}+40t-20\times101\\ $\Rightarrow$ 4t^{2}+2t-101\\ Soving it, we get t=\frac{9\sqrt{5}-1 }{4} \\ So m+n+p+q=1+4+5+9=19 is the correct answer

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2025_AIME_II_Problems/Problem_12&oldid=242446"