2024 AMC 12B Problem 21

Below is the professionally curated solution for Problem 21 of the 2024 AMC 12B, from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2024 AMC 12B solutions, or check the answer key.

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Concepts:trigonometric identityPythagorean Triple

Difficulty rating: 2130

21.

The measures of the smallest angles of three different right triangles sum to 90.90^\circ. All three triangles have side lengths that are primitive Pythagorean triples. Two of them are 33-44-55 and 55-1212-13.13. What is the perimeter of the third triangle?

4040

126126

154154

176176

208208

Solution:

The smallest angles α,β\alpha, \beta of the 33-44-55 and 55-1212-1313 triangles have tanα=34\tan\alpha = \tfrac34 and tanβ=512.\tan\beta = \tfrac{5}{12}. By the tangent addition formula, tan(α+β)=34+512134512=14123348=5633.\tan(\alpha + \beta) = \frac{\tfrac34 + \tfrac{5}{12}}{1 - \tfrac34\cdot\tfrac{5}{12}} = \frac{\tfrac{14}{12}}{\tfrac{33}{48}} = \frac{56}{33}.

The third smallest angle γ\gamma satisfies γ=90(α+β),\gamma = 90^\circ - (\alpha + \beta), so tanγ=3356.\tan\gamma = \dfrac{33}{56}. The right triangle with legs 3333 and 5656 has hypotenuse 332+562=4225=65,\sqrt{33^2 + 56^2} = \sqrt{4225} = 65, a primitive triple. Its perimeter is 33+56+65=154.33 + 56 + 65 = 154.

Thus, the correct answer is C.

Problem 21 in Other Years