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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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==Solution 3 (Another Trig)==
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Denote the smallest angle of the <math>3-4-5</math> triangle as <math>\alpha</math>, the smallest angle of the <math>5-12-13</math> triangle as <math>\beta</math>, and the smallest angle of the triangle we are trying to solve for as <math>\theta</math>. We then have
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<cmath>\alpha + \beta + \theta = 90</cmath>
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<cmath>\alpha + \beta = 90 - \theta</cmath>
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<cmath>\sin{(\alpha + \beta)} = \sin{(90 - \theta)} = \cos{\theta}</cmath>
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<cmath>\cos{\theta} = \sin{\alpha}\cos{\beta} + \cos{\alpha}\sin{\beta} = (\frac{3}{5})(\frac{12}{13}) + (\frac{4}{5})(\frac{5}{13}) = \frac{56}{65}</cmath>
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Taking the hypotenuse to be <math>65</math> and one of the legs to be <math>56</math>, we compute the last leg to be <math>\sqrt{65^2 - 56^2} = \sqrt{(65-56)(65+56)} = \sqrt{9*121} = 3*11 = 33</math>
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Giving us a final answer of <math>65 + 56 + 33 = \boxed{\textbf{(C) }154}</math>.
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~tkl
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==Solution 4 (Similarity)==
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[[File:Pithagor triangles 13 5 65.png|300px|right]]
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Let's arrange the triangles <math>BCD (5-12-13), BCE (9-12-15)</math> and <math>ABE</math> as shown in the diagram.
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<cmath>F = AE \cap BC.</cmath>
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<cmath>AE \perp AB, DB \perp AB \implies \triangle BCD \sim \triangle FCE \sim \triangle FAB \implies</cmath>
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<cmath>EF = \frac{9 \cdot 13}{5}, CF = \frac{9 \cdot 12}{5}, BF = BC + CF = \frac{12 \cdot 14}{5},</cmath>
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<cmath>\frac {AB}{CE} = \frac {BF}{EF} \implies AB =  \frac{12 \cdot 14}{13},</cmath>
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<cmath>AE = AF - EF = BF \cdot \frac {12}{13} - EF = \frac {99}{13} \implies</cmath>
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<cmath>AE : AB : BE = 99 : 12 \cdot 14 : 15 \cdot 13 = 33 : 56 : 65 \implies 65 + 56 + 33 = \boxed{\textbf{(C) }154}</cmath>
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'''vladimir.shelomovskii@gmail.com, vvsss'''
  
 
==Video Solution by Innovative Minds==
 
==Video Solution by Innovative Minds==

Latest revision as of 16:42, 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

Solution 3 (Another Trig)

Denote the smallest angle of the $3-4-5$ triangle as $\alpha$, the smallest angle of the $5-12-13$ triangle as $\beta$, and the smallest angle of the triangle we are trying to solve for as $\theta$. We then have \[\alpha + \beta + \theta = 90\] \[\alpha + \beta = 90 - \theta\] \[\sin{(\alpha + \beta)} = \sin{(90 - \theta)} = \cos{\theta}\] \[\cos{\theta} = \sin{\alpha}\cos{\beta} + \cos{\alpha}\sin{\beta} = (\frac{3}{5})(\frac{12}{13}) + (\frac{4}{5})(\frac{5}{13}) = \frac{56}{65}\] Taking the hypotenuse to be $65$ and one of the legs to be $56$, we compute the last leg to be $\sqrt{65^2 - 56^2} = \sqrt{(65-56)(65+56)} = \sqrt{9*121} = 3*11 = 33$

Giving us a final answer of $65 + 56 + 33 = \boxed{\textbf{(C) }154}$.

~tkl

Solution 4 (Similarity)

Pithagor triangles 13 5 65.png

Let's arrange the triangles $BCD (5-12-13), BCE (9-12-15)$ and $ABE$ as shown in the diagram. \[F = AE \cap BC.\] \[AE \perp AB, DB \perp AB \implies \triangle BCD \sim \triangle FCE \sim \triangle FAB \implies\] \[EF = \frac{9 \cdot 13}{5}, CF = \frac{9 \cdot 12}{5}, BF = BC + CF = \frac{12 \cdot 14}{5},\] \[\frac {AB}{CE} = \frac {BF}{EF} \implies AB = \frac{12 \cdot 14}{13},\] \[AE = AF - EF = BF \cdot \frac {12}{13} - EF = \frac {99}{13} \implies\] \[AE : AB : BE = 99 : 12 \cdot 14 : 15 \cdot 13 = 33 : 56 : 65 \implies 65 + 56 + 33 = \boxed{\textbf{(C) }154}\] vladimir.shelomovskii@gmail.com, vvsss

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
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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