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~[https://artofproblemsolving.com/community/user/1201585 kafuu_chino]
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==Video Solution by Innovative Minds==
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https://youtu.be/9PMdtwkKTlU
  
 
==See also==
 
==See also==
 
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Revision as of 01:23, 14 November 2024

Problem

The measures of the smallest angles of three different right triangles sum to $90^\circ$. All three triangles have side lengths that are primitive Pythagorean triples. Two of them are $3-4-5$ and $5-12-13$. What is the perimeter of the third triangle?

$\textbf{(A) } 40 \qquad\textbf{(B) } 126 \qquad\textbf{(C) } 154 \qquad\textbf{(D) } 176 \qquad\textbf{(E) } 208$

Solution 1

Let $\alpha$ and $\beta$ be the smallest angles of the $3-4-5$ and $5-12-13$ triangles respectively. We have \[\tan(\alpha)=\frac{3}{4} \text{ and } \tan(\beta)=\frac{5}{12}\] Then \[\tan(\alpha+\beta)=\frac{\frac{3}{4}+\frac{5}{12}}{1-\frac{3}{4}\cdot\frac{5}{12}}=\frac{56}{33}\] Let $\theta$ be the smallest angle of the third triangle. Consider \[\tan{90^\circ}=\tan((\alpha+\beta)+\theta)=\frac{\frac{56}{33}+\tan{\theta}}{1-\frac{56}{33}\cdot\tan{\theta}}\] In order for this to be undefined, we need \[1-\frac{56}{33}\cdot\tan{\theta}=0\] so \[\tan{\theta}=\frac{33}{56}\] Hence the base side lengths of the third triangle are $33$ and $56$. By the Pythagorean Theorem, the hypotenuse of the third triangle is $65$, so the perimeter is $33+56+65=\boxed{\textbf{(C) }154}$.

~kafuu_chino

Video Solution by Innovative Minds

https://youtu.be/9PMdtwkKTlU

See also

2024 AMC 12B (Problems • Answer Key • Resources)
Preceded by
Problem 20 		Followed by
Problem 22
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All AMC 12 Problems and Solutions

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