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

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 $\textbf{(D)}\ 60^\circ\qquad$ .

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1981 AHSME Problems/Problem 21

Problem 21

Solution