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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>.
  
Line 10: Line 12:
  
 
== Solution 1==
 
== Solution 1==
 +
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>
 +
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>
 +
\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*}
 +
</cmath>
 +
Taking the positive solution gives <math>m + n + p + q = 1 + 9 + 5 + 4 = \boxed{\textbf{(019)}}.</math>
  
−
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)
−
 
 
−
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>
 
−
 
 
−
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>
 
  
−
~Bluesoul
+
==See also==
 +
{{AIME box|year=2025|num-b=11|num-a=13|n=II}}
  
−
<math>\forall</math> 2 <math>\le</math> i <math>\le</math> 10,  cos\angle <math>A_{i}<cmath>A_{1}</cmath>A_{i+1}</math>=<math>\frac{12}{13} </math>\\
+
{{MAA Notice}}
−
So sin\angle <math>A_{i}<cmath>A_{1}</cmath>A_{i+1}</math>=<math>\frac{5}{13} </math>\\
 
−
Since the area of each triangle is 1,\\
 
−
<math>\frac{1}{2} </math>\times<cmath>A_{1}</cmath>A_{i}<math>\times</math>A_{1}<math></math>A_{i+1}<math>\times  sin</math>\angle A_{i}<cmath>A_{1}</cmath>A_{i+1}<math>=1\\
 
−
</math>\Rightarrow<math> </math>A_{1}<math></math>A_{i}<math>\times</math>A_{1}<math></math>A_{i+1}<math>= </math>\frac{26}{5} <math>\\
 
−
So </math>A_{1}<math></math>A_{i}<math>\times</math>A_{1}<math></math>A_{i+1}<math>=</math>A_{1}<math></math>A_{i+1}<math>\times</math>A_{1}<math></math>A_{i+2}<math>\\
 
−
This means  </math>A_{1}<math></math>A_{i}<math>=A_{1}</math><math>A_{i+2}</math>\\
 
−
In <math>\triangle A_{1}<cmath>A_{i}</cmath>A_{i+1}</math> and <math>\triangle A_{1}<cmath>A_{i+2}</cmath>A_{i+1}</math>，\\
 
−
they share one of the same  side and the  angles on vertex <math>A_{1}</math> are the same
 
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<math>A_{1}</math><math>A_{i}</math>= <math>A_{1}</math><math>A_{i=2}</math>\\
 
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So they are congruent \\
 
−
This means <math>\forall</math> 2 <math>\le</math> i <math>\le</math> 9 <math>A_{i}</math><math>A_{i+1}</math>= <math>A_{i+1}</math><math>A_{i+2}</math>\\
 
−
Perimeter = <math>A_{1}</math><math>A_{2}</math>+<math>\sum_{i=2}^{10}A_{i}</math><math>A_{i+1}</math>+<math>A_{11}</math><math>A_{1}</math>=20\\
 
−
Then <math>A_{1}</math><math>A_{2}</math>+<math>A_{11}</math><math>A_{1}</math>+9<math>A_{2}</math><math>A_{3}</math>=20\\
 
−
Let us set <math>A_{1}</math><math>A_{2}</math>=a <math>A_{11}</math><math>A_{1}</math>=b and <math>A_{2}</math><math>A_{3}</math>=c\\
 
−
Then a+b+9c=20\\
 
−
Now, we apply cosine law  in <math>\triangle A_{1}<cmath>A_{2}</cmath>A_{3}</math>\\
 
−
<math>A_{2}</math><math>A_{3}</math>^{2}=<math>A_{1}</math><math>A_{2}</math>^{2} +<math>A_{1}</math><math>A_{3}</math>^{3}-2<math>A_{1}</math><math>A_{2}</math>\times<math>A_{1}</math><math>A_{3}</math>cos\angle A_{2}<cmath>A_{1}</cmath>A_{3}<math>\\
 
−
</math>\Rightarrow<math> c^{2}=a^{2}+b^{2}-2ab</math>\times\frac {12}{13} <math>\\
 
−
Set </math>A_{1}<math></math>A_{2}<math>+</math>A_{11}<math></math>A_{1}<math>=t,\\
 
−
then c^{2}=(a+b)^{2}-2ab-2ab</math>\times\frac {12}{13}=t^{2}-2\times\frac {26}{5}\times (1+\frac {12}{13})\\
 
−
<math>\Rightarrow</math> c^{2}=t^{2}- 20      \\
 
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Since 9c=20-a-b=20-t\\
 
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Square both sides, giving 81c^{2}=400+t^{2}-40t\\
 
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<math>\Rightarrow</math> 81t^{2}-20\times81=400+t^{2}-40t\\
 
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<math>\Rightarrow</math> 80t^{2}+40t-20\times101\\
 
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<math>\Rightarrow</math> 80t^{2}+40t-20\times101\\
 
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<math>\Rightarrow</math> 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
 

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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