2025 AMC 12B Problem 21

Attempt Problem 21 of the 2025 AMC 12B below, then check your answer against the professionally curated solution from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2025 AMC 12B solutions, or check the answer key.

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21.

Two non-congruent triangles have the same area. Each triangle has sides of length 88 and 9,9, and the third side of each triangle has integer length. What is the sum of the lengths of the third sides?

2020

2222

2424

2626

2828

Answer: C
Concepts:law of cosinestriangle areatriangle inequality
Difficulty rating: 2170
Solution:

The area with included angle θ\theta is 36sinθ,36\sin\theta, so two triangles of equal area use angles θ\theta and 180θ,180^\circ - \theta, with cosines ±cosθ.\pm\cos\theta. By the law of cosines the third sides satisfy t2=145144cosθ,t^2 = 145 \mp 144\cos\theta, hence t12+t22=290.t_1^2 + t_2^2 = 290. Since 2902(mod8),290\equiv2\pmod8, both integer sides must be odd. Checking the odd squares in the triangle-inequality range 1<t<171\lt t\lt17 leaves only 112+132=121+169=290.11^2+13^2=121+169=290. Therefore the sum is 24.24.

Thus, the correct answer is C.

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