Difference between revisions of "2024 AMC 12B Problems/Problem 21"

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~[https://artofproblemsolving.com/community/user/1201585 kafuu_chino]
 
~[https://artofproblemsolving.com/community/user/1201585 kafuu_chino]
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==Solution 2 (Complex Number)==
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The smallest angle of <math>3-4-5</math> triangle can be viewed as the arguement of <math>4+3i</math>, and the smallest angle of <math>5-12-13</math> triangle can be viewed as the arguement of <math>12+5i</math>.
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Hence, if we assume the ratio of the two shortest length of the last triangle is <math>1:k</math> (<math>k</math> being some rational number), then we can derive the following formula of the sum of their arguement.
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Since their arguement adds up to <math>\frac{\pi}{2}</math>, it's the arguement of <math>i</math>. Hence, <cmath>\left(4+3i\right)\left(5+12i\right)\left(k+i\right)=ni\,,</cmath> where <math>n</math> is some real number.
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Solving the equation, we get <cmath>56k-33=0\,,\quad 33k+56=n\,.</cmath> Hence <math>k=\frac{33}{56}</math>
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Since the sidelength of the theird triangle are co-prime integers, two of its sides are <math>33</math> and <math>56</math>. And the last side is <math>\sqrt{33^2+56^2}=65</math>, hence, the parameter of the third triangle if <math>33+56+65=\boxed{\mathbf{(C)}\,154}</math>.
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~Prof. Joker
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==Video Solution by Innovative Minds==
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https://youtu.be/9PMdtwkKTlU
  
 
==See also==
 
==See also==
 
{{AMC12 box|year=2024|ab=B|num-b=20|num-a=22}}
 
{{AMC12 box|year=2024|ab=B|num-b=20|num-a=22}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Latest revision as of 02:42, 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

Solution 2 (Complex Number)

The smallest angle of $3-4-5$ triangle can be viewed as the arguement of $4+3i$, and the smallest angle of $5-12-13$ triangle can be viewed as the arguement of $12+5i$.

Hence, if we assume the ratio of the two shortest length of the last triangle is $1:k$ ($k$ being some rational number), then we can derive the following formula of the sum of their arguement. Since their arguement adds up to $\frac{\pi}{2}$, it's the arguement of $i$. Hence, \[\left(4+3i\right)\left(5+12i\right)\left(k+i\right)=ni\,,\] where $n$ is some real number.

Solving the equation, we get \[56k-33=0\,,\quad 33k+56=n\,.\] Hence $k=\frac{33}{56}$

Since the sidelength of the theird triangle are co-prime integers, two of its sides are $33$ and $56$. And the last side is $\sqrt{33^2+56^2}=65$, hence, the parameter of the third triangle if $33+56+65=\boxed{\mathbf{(C)}\,154}$.

~Prof. Joker

Video Solution by Innovative Minds

https://youtu.be/9PMdtwkKTlU

See also

2024 AMC 12B (ProblemsAnswer KeyResources)
Preceded by
Problem 20
Followed by
Problem 22
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25
All AMC 12 Problems and Solutions

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