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

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• 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>.
  
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== Solution 1==
 
== Solution 1==
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Set <math>A_1A_2 = x</math> and <math>A_1A_3 = y</math>. By the first condition, we have <math>\frac{1}{2}xy\sin\theta = 1</math>, where <math>\theta = \angle A_2 A_1 A_3</math>. Since <math>\cos\theta = \frac{12}{13}</math>, we have <math>\sin\theta = \frac{5}{13}</math>, so <math>xy = \frac{26}{5}</math>. Repeating this process for <math>\triangle A_i A_1 A_{i+1}</math>, we get <math>A_1A_2 = A_1A_4 = \ldots A_1A_{10} = x</math> and <math>A_1A_3 = A_1A_5 = \ldots A_1A_{11} = y</math>. Since the included angle of these <math>9</math> triangles is <math>\theta</math>, the square of the third side is
 +
<cmath>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.</cmath>
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Thus the third side has length <math>\sqrt{(x+y)^2 - 20}.</math> The perimeter is constructed from <math>9</math> of these lengths, plus <math>A_{11}A_1 + A_1A_2 = x + y</math>, so <math>9\sqrt{(x+y)^2 - 20} + x + y = 20</math>. We seek the value of <math>x + y,</math> so let <math>x + y = a</math> so
 +
<cmath>
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\begin{align*}
 +
9\sqrt{a^2 - 20} + a &= 20\\
 +
81(a^2 - 20) &= 400 - 40a + a^2\\
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4a^2 + 2a - 101 &= 0 \\
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a &= \frac{-2 \pm \sqrt{1620}}{8} = \frac{-1 \pm \sqrt{405}}{4} = \frac{-1 \pm 9\sqrt{5}}{4}.
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\end{align*}
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</cmath>
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Taking the positive solution gives <math>m + n + p + q = 1 + 9 + 5 + 4 = \boxed{\textbf{(019)}}.</math>
  
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Since <math>[A_1A_iA_{i+1}]</math> are the same, we have have <math>A_1A_{11}=A_1A_{9}=...=A_1A_3=x</math> and <math>A_1A_2=A_1A_4=...=A_1A_{10}=y</math>, since <math>\angle{A_iA_1A_{i+1}}</math> is the same for all the <math>2\leq i\leq 10</math>, so <math>A_iA_{i+1}</math> are the same for all <math>2\leq i\leq 10</math>, set them be <math>d</math>
+
-Benedict T (countmath1)
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Now we have <math>x+y+9d=20, x^2+y^2-2xy\cdot \frac{12}{13}=d^2, xy\cdot \frac{5}{13}=2</math>
 
  
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Solve the system of equations we could get <math>d=\frac{9-\sqrt{5}}{4}</math>, <math>x+y=20-9d=\frac{9\sqrt{5}-1}{4}\implies \boxed{019}</math>
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==See also==
 +
{{AIME box|year=2025|num-b=11|num-a=13|n=II}}
  
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~Bluesoul
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{{MAA Notice}}

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

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