2021 AMC 12B Fall Problem 8

Below is the professionally curated solution for Problem 8 of the 2021 AMC 12B Fall, from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2021 AMC 12B Fall solutions, or check the answer key.

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Concepts:isosceles triangle triangle area trigonometry

Difficulty rating: 1530

The product of the lengths of the two congruent sides of an obtuse isosceles triangle is equal to the product of the base and twice the triangle's height to the base. What is the measure, in degrees, of the vertex angle of this triangle?

$105$

$120$

$135$

$150$

$165$

Solution:

Let the congruent sides have length $s,$ the base be $b,$ and the height to the base be $h.$ The given condition is $s^2 = 2bh.$

The area equals $\tfrac12 bh$ and also $\tfrac12 s^2 \sin\theta,$ where $\theta$ is the vertex angle. So $bh = s^2 \sin\theta.$

Substituting $s^2 = 2bh$ gives $bh = 2bh\sin\theta,$ so $\sin\theta = \tfrac12.$ Since the triangle is obtuse, $\theta = 150^\circ.$

Thus, the correct answer is D.

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