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] | ||
+ | |||
+ | ==Solution 2 (Complex Number)== | ||
+ | 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>. | ||
+ | |||
+ | 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. | ||
+ | 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. | ||
+ | |||
+ | Solving the equation, we get <cmath>56k-33=0\,,\quad 33k+56=n\,.</cmath> Hence <math>k=\frac{33}{56}</math> | ||
+ | |||
+ | 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>. | ||
+ | |||
+ | ~Prof. Joker | ||
+ | |||
+ | ==Solution 3 (Another Trig)== | ||
+ | |||
+ | 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 | ||
+ | <cmath>\alpha + \beta + \theta = 90</cmath> | ||
+ | <cmath>\alpha + \beta = 90 - \theta</cmath> | ||
+ | <cmath>\sin{(\alpha + \beta)} = \sin{(90 - \theta)} = \cos{\theta}</cmath> | ||
+ | <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> | ||
+ | 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> | ||
+ | |||
+ | Giving us a final answer of <math>65 + 56 + 33 = \boxed{\textbf{(C) }154}</math>. | ||
+ | |||
+ | ~tkl | ||
+ | ==Solution 4 (Similarity)== | ||
+ | [[File:Pithagor triangles 13 5 65.png|300px|right]] | ||
+ | Let's arrange the triangles <math>BCD (5-12-13), BCE (9-12-15)</math> and <math>ABE</math> as shown in the diagram. | ||
+ | <cmath>F = AE \cap BC.</cmath> | ||
+ | <cmath>AE \perp AB, DB \perp AB \implies \triangle BCD \sim \triangle FCE \sim \triangle FAB \implies</cmath> | ||
+ | <cmath>EF = \frac{9 \cdot 13}{5}, CF = \frac{9 \cdot 12}{5}, BF = BC + CF = \frac{12 \cdot 14}{5},</cmath> | ||
+ | <cmath>\frac {AB}{CE} = \frac {BF}{EF} \implies AB = \frac{12 \cdot 14}{13},</cmath> | ||
+ | <cmath>AE = AF - EF = BF \cdot \frac {12}{13} - EF = \frac {99}{13} \implies</cmath> | ||
+ | <cmath>AE : AB : BE = 99 : 12 \cdot 14 : 15 \cdot 13 = 33 : 56 : 65 \implies 65 + 56 + 33 = \boxed{\textbf{(C) }154}</cmath> | ||
+ | '''vladimir.shelomovskii@gmail.com, vvsss''' | ||
+ | |||
+ | ==Video Solution by Innovative Minds== | ||
+ | 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 16:42, 14 November 2024
Contents
Problem
The measures of the smallest angles of three different right triangles sum to . All three triangles have side lengths that are primitive Pythagorean triples. Two of them are and . What is the perimeter of the third triangle?
Solution 1
Let and be the smallest angles of the and triangles respectively. We have Then Let be the smallest angle of the third triangle. Consider In order for this to be undefined, we need so Hence the base side lengths of the third triangle are and . By the Pythagorean Theorem, the hypotenuse of the third triangle is , so the perimeter is .
Solution 2 (Complex Number)
The smallest angle of triangle can be viewed as the arguement of , and the smallest angle of triangle can be viewed as the arguement of .
Hence, if we assume the ratio of the two shortest length of the last triangle is ( being some rational number), then we can derive the following formula of the sum of their arguement. Since their arguement adds up to , it's the arguement of . Hence, where is some real number.
Solving the equation, we get Hence
Since the sidelength of the theird triangle are co-prime integers, two of its sides are and . And the last side is , hence, the parameter of the third triangle if .
~Prof. Joker
Solution 3 (Another Trig)
Denote the smallest angle of the triangle as , the smallest angle of the triangle as , and the smallest angle of the triangle we are trying to solve for as . We then have Taking the hypotenuse to be and one of the legs to be , we compute the last leg to be
Giving us a final answer of .
~tkl
Solution 4 (Similarity)
Let's arrange the triangles and as shown in the diagram. vladimir.shelomovskii@gmail.com, vvsss
Video Solution by Innovative Minds
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.