2024 AMC 12B Problems/Problem 21
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
Note the 3 smallest angles as A, B and C sinA = 3/5 cosA = 4/5 sinB = 5/13 cosB = 12/13 The angle C would be the third triangle's smallest angle in the satisfying condition. cos(A+B) = cosAcosB-sinAsinB
= (4/5)(12/13)-(3/5)(5/13) = 48/65 - 15/65 = 33/65
We can also conclude that sin(A+B) = 56/65 (By sin^2(x)+cos^2(x) = 1) Method 1: Notice that C = 90-(A+B) We could quickly conclude the triangle is 33-56-65 Method 2: Because A + B + C = 90 degrees, therefore cos((A+B)+C) = 0 This leads to cos(A+B)cos(C)-sin(A+B)sin(C) = 0 Which (33/65)cos(C)=(56/65)sin(C) Therefore 33cos(C) = 56sin(C) tan(C) = 33/56 Therefore the third triangle is 33-56-65
Overall, Answer = 33+56+65 = (C)154
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.