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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==
 +
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>
 +
 +
-Benedict T (countmath1)
 +
 +
== Solution 2(Very similar to Solution 1)==
 +
 +
Let <math>A_1 A_2 = a_1, A_1 A_3 = a_2, ... A_1 A_{10} = a_9, A_1 A_11 a_{10}</math>. Now the area of each triangle are build up of these lengths multiplied by the sine of the angle in between. This angle's cosine is given to us to be <math>\frac{12}{13}</math>. We recognize this as a <math>5-12-13</math> triangle hence the sine of the angle will be simply <math>\frac{5}{13}</math>. Hence each area will be <math>\frac{1}{2} \cdot a_n a_{n + 1} \cdot \frac{5}{13} = 1 \implies a_n a_{n+ 1} = \frac{26}{5}</math>. Therefore, we have:
 +
 +
<math>a_1 a_2 = \frac{26}{5}</math>
 +
 +
<math>a_2 a_3 = \frac{26}{5}</math>
 +
.
 +
.
 +
.
 +
 +
<math>a_9 a_{10} = \frac{26}{5}</math>
 +
 +
<math>a_{10} a_1 = \frac{26}{5}</math>
 +
 +
Now note that the perimeter will be build up of all the other lengths. For instance, <math>A_2 A_3</math> will be part of this perimeter which happens to be the only unknown side of <math>\triangle A_1 A_2 A_3</math>. By Law of Cosines
 +
 +
<math>(A_2 A_3)^2 = a_1^{2} + a_2^{2} - 2a_1 a_2 \frac{12}{13}</math> after substituting the cosine value. We know <math>a_1 a_2 = \frac{26}{5}</math> so substituting this in we get
 +
 +
<math>(A_2 A_3)^2 = a_1^{2} + a_2^{2} - \frac{48}{5} \implies A_2 A_3 = \sqrt{a_1^{2} + a_2^{2} - \frac{48}{5}}</math>. By symmetry, we conclude
 +
 +
<math>\sqrt{a_1^{2} + a_2^{2} - \frac{48}{5}} + ... + \sqrt{a_{10}^{2} + a_1^{2} - \frac{48}{5}} + a_1 + a_2 = 20</math> because we need to account for <math>A_1 A_2 + A_1 A_11</math> as they are actually known sides that are included in the perimeter.
 +
 +
Note that from the <math>a_1 a_2 = \frac{26}{5}</math> system of equations, we can see clearly that
 +
 +
<math>a_1 = a_3</math>
 +
 +
<math>a_2 = a_4</math>
 +
.
 +
.
 +
.
 +
 +
<math>a_8 = a_{10}</math>
 +
 +
<math>a_9 = a_1</math>
 +
 +
<math>a_{10} = a_2</math>
 +
 +
So we see that <math>a_{2n + 1} = a_1</math> and <math>a_{2n} = a_2</math> so we can substitute this in to get
 +
 +
<math>\sqrt{a_1^{2} + a_2^{2} - \frac{48}{5}} + ... + \sqrt{a_1^{2} + a_2^{2} - \frac{48}{5}} + a_1 + a_2 = 20</math>
 +
 +
Now the square root terms happen <math>9</math> times as there are <math>11</math> sides and two of them are <math>a_1, a_2</math> given at the end of the <math>LHS</math>. So we conclude:
 +
 +
<math>9\sqrt{a_1^{2} + a_2^{2} - \frac{48}{5}} + a_1 + a_2 = 20</math>
 +
 +
Note that <math>a_1^{2} + a_2^{2} - \frac{48}{5} = (a_1 + a_2)^2 - 2a_1 a_2 - \frac{48}{5} = (a_1 + a_2)^2 - \frac{52 + 48}{5} = (a_1 + a_2)^2 - 20</math>. Note that we want to find <math>A_1 A_2 + A_1 A_11 = a_1 + a_{10} = a_1 + a_2</math>. So we let <math>a_1 + a_2 = x</math> to get
 +
 +
<math>9\sqrt{x^2 - 20} + x = 20 \implies 81(x^2 - 20) = (20 - x)^2 \implies 81(x^2 - 20) = 400 - 40x + x^2 \implies 81x^2 - 1620 = 400 - 40x + x^2 \implies 80x^2 + 40x - 2020 = 0 \implies 4x^2 + 2x - 101 = 0 \implies x = \frac{-1 \pm 9\sqrt{5}}{4}</math>. Now take the positive value to get
  
−
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>
+
<math>x = \frac{9\sqrt{5} - 1}{4} \implies 9 + 5 + 1 + 4 = \boxed{19}</math>.
  
−
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>
+
~ilikemath24736
 +
~Sylesh (Latex errors)
  
−
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>
+
==See also==
 +
{{AIME box|year=2025|num-b=11|num-a=13|n=II}}
  
−
~Bluesoul
+
{{MAA Notice}}

Latest revision as of 21:45, 1 November 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)

Solution 2(Very similar to Solution 1)

Let $A_1 A_2 = a_1, A_1 A_3 = a_2, ... A_1 A_{10} = a_9, A_1 A_11 a_{10}$. Now the area of each triangle are build up of these lengths multiplied by the sine of the angle in between. This angle's cosine is given to us to be $\frac{12}{13}$. We recognize this as a $5-12-13$ triangle hence the sine of the angle will be simply $\frac{5}{13}$. Hence each area will be $\frac{1}{2} \cdot a_n a_{n + 1} \cdot \frac{5}{13} = 1 \implies a_n a_{n+ 1} = \frac{26}{5}$. Therefore, we have:

$a_1 a_2 = \frac{26}{5}$

$a_2 a_3 = \frac{26}{5}$ . . .

$a_9 a_{10} = \frac{26}{5}$

$a_{10} a_1 = \frac{26}{5}$

Now note that the perimeter will be build up of all the other lengths. For instance, $A_2 A_3$ will be part of this perimeter which happens to be the only unknown side of $\triangle A_1 A_2 A_3$. By Law of Cosines

$(A_2 A_3)^2 = a_1^{2} + a_2^{2} - 2a_1 a_2 \frac{12}{13}$ after substituting the cosine value. We know $a_1 a_2 = \frac{26}{5}$ so substituting this in we get

$(A_2 A_3)^2 = a_1^{2} + a_2^{2} - \frac{48}{5} \implies A_2 A_3 = \sqrt{a_1^{2} + a_2^{2} - \frac{48}{5}}$. By symmetry, we conclude

$\sqrt{a_1^{2} + a_2^{2} - \frac{48}{5}} + ... + \sqrt{a_{10}^{2} + a_1^{2} - \frac{48}{5}} + a_1 + a_2 = 20$ because we need to account for $A_1 A_2 + A_1 A_11$ as they are actually known sides that are included in the perimeter.

Note that from the $a_1 a_2 = \frac{26}{5}$ system of equations, we can see clearly that

$a_1 = a_3$

$a_2 = a_4$ . . .

$a_8 = a_{10}$

$a_9 = a_1$

$a_{10} = a_2$

So we see that $a_{2n + 1} = a_1$ and $a_{2n} = a_2$ so we can substitute this in to get

$\sqrt{a_1^{2} + a_2^{2} - \frac{48}{5}} + ... + \sqrt{a_1^{2} + a_2^{2} - \frac{48}{5}} + a_1 + a_2 = 20$

Now the square root terms happen $9$ times as there are $11$ sides and two of them are $a_1, a_2$ given at the end of the $LHS$. So we conclude:

$9\sqrt{a_1^{2} + a_2^{2} - \frac{48}{5}} + a_1 + a_2 = 20$

Note that $a_1^{2} + a_2^{2} - \frac{48}{5} = (a_1 + a_2)^2 - 2a_1 a_2 - \frac{48}{5} = (a_1 + a_2)^2 - \frac{52 + 48}{5} = (a_1 + a_2)^2 - 20$. Note that we want to find $A_1 A_2 + A_1 A_11 = a_1 + a_{10} = a_1 + a_2$. So we let $a_1 + a_2 = x$ to get

$9\sqrt{x^2 - 20} + x = 20 \implies 81(x^2 - 20) = (20 - x)^2 \implies 81(x^2 - 20) = 400 - 40x + x^2 \implies 81x^2 - 1620 = 400 - 40x + x^2 \implies 80x^2 + 40x - 2020 = 0 \implies 4x^2 + 2x - 101 = 0 \implies x = \frac{-1 \pm 9\sqrt{5}}{4}$. Now take the positive value to get

$x = \frac{9\sqrt{5} - 1}{4} \implies 9 + 5 + 1 + 4 = \boxed{19}$.

~ilikemath24736 ~Sylesh (Latex errors)

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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