1981 AHSME Problems/Problem 21

Problem 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 length $c$ is

$\textbf{(A)}\ 15^\circ\qquad\textbf{(B)}\ 30^\circ\qquad\textbf{(C)}\ 45^\circ\qquad\textbf{(D)}\ 60^\circ\qquad\textbf{(E)}\ 150^\circ$

Solution 1

We will try to solve for a possible value of the variables. First notice that exchanging $a$ for $b$ in the original equation must also work. Therefore, $a=b$ works. Replacing $b$ for $a$ and expanding/simplifying in the original equation yields $4a^2-c^2=3a^2$ , or $a^2=c^2$ . Since $a$ and $c$ are positive, $a=c$ . Therefore, we have an equilateral triangle and the angle opposite $c$ is just $\textbf{(D)}\ 60^\circ\qquad$ .

Solution 2

$(a+b+c)(a+b-c)=3ab$ $a^2+2ab+b^2-c^2=3ab$ $a^2+b^2-c^2=ab$ $c^2=a^2+b^2-ab$ This looks a lot like Law of Cosines, which is $c^2=a^2+b^2-2ab\cos{c}$ . $c^2=a^2+b^2-ab=a^2+b^2-2ab\cos{c}$ $ab=2ab\cos{c}$ $\frac{1}{2}=\cos{c}$ $\cos{c}$ is $\frac{1}{2}$ , so the angle opposite side $c$ is $\boxed{60^\circ}$ .

-aopspandy

1981 AHSME (Problems • Answer Key • Resources)
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1981 AHSME Problems/Problem 21

Contents

Problem 21

Solution 1

Solution 2

See also