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

(Created page with "== Problem == Let <math>A_1A_2\dots A_{11}</math> be a non-convex <math>11</math>-gon such that • The area of <math>A_iA_1A_{i+1}</math> is <math>1</math> for each <math>2...")
 
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If <math>A_1A_2+A_1A_{11}</math> can be expressed as <math>\frac{m\sqrt{n}-p}{q}</math> for positive integers <math>m,n,p,q</math> with <math>n</math> squarefree and <math>\gcd(m,p,q)=1</math>, find <math>m+n+p+q</math>.
 
If <math>A_1A_2+A_1A_{11}</math> can be expressed as <math>\frac{m\sqrt{n}-p}{q}</math> for positive integers <math>m,n,p,q</math> with <math>n</math> squarefree and <math>\gcd(m,p,q)=1</math>, find <math>m+n+p+q</math>.
  
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== Solution ==
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== Solution 1==
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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>
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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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~Bluesoul

Revision as of 00:00, 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

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