1981 AMC 12 Problem 21

Attempt Problem 21 of the 1981 AMC 12 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 1981 AMC 12 solutions, or check the answer key.

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21.

In a triangle with sides of lengths

a,

b,

and

c,

(a+b+c)(a+b-c)=3ab.

The measure of the angle opposite the side of length

c

15^\circ

30^\circ

45^\circ

60^\circ

150^\circ

Expand the left side as a difference of squares

Compare the resulting expression for

c^2

with the law of cosines

The condition gives

(a+b)^2-c^2=3ab,

c^2=a^2+b^2-ab.

\theta

is opposite

c,

the law of cosines says

c^2=a^2+b^2-2ab\cos\theta.

Therefore

2ab\cos\theta=ab,

\cos\theta=\frac{1}{2}

and

\theta=60^\circ.

Therefore, the correct answer is D.

Problem 21 in other years