1981 AHSME Problems/Problem 11

Problem 11

The three sides of a right triangle have integral lengths which form an arithmetic progression. One of the sides could have length

$\textbf{(A)}\ 22\qquad\textbf{(B)}\ 58\qquad\textbf{(C)}\ 81\qquad\textbf{(D)}\ 91\qquad\textbf{(E)}\ 361$


Let the three sides be $a-d,$ $a,$ and $a+d.$ Because of the Pythagorean Theorem, \[(a-d)^2 + a^2 = (a+d)^2\] This can be simplified to \[a(a-4d) = 0.\]

So, $a$ is a multiple of $4d$ and the triangle has sides $3d, 4d, 5d.$ We check the answer choices for anything divisible by 3, 4, or 5. The only one that works is 81, which is divisible by 3.

The answer is $\textbf{(C)}\ 81.$

-edited by coolmath34