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Difference between revisions of "2025 AIME II Problems/Problem 12"
(→‎Solution 1)
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~Bluesoul
 
~Bluesoul
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<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>\\
 +
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>\\
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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>，\\
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they share one of the same  side and the  angles on vertex <math>A_{1}</math> are the same
 +
<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 \\
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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\\
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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\\
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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\\
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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>\\
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</math>\Rightarrow<math> c^{2}=a^{2}+b^{2}-2ab</math>\times\frac {12}{13} <math>\\
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Set </math>A_{1}<math></math>A_{2}<math>+</math>A_{11}<math></math>A_{1}<math>=t,\\
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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})\\
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<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\\
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Soving it, we get t=\frac{9\sqrt{5}-1 }{4} \\
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So  m+n+p+q=1+4+5+9=19 is the correct  answer

Revision as of 01:44, 14 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

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

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