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

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

Solution

Let the three sides be $a-d$ , $a$ , and $a+d$ . By the Pythagorean Theorem, $(a-d)^2 + a^2 = (a+d)^2$ This can be simplified to $a(a-4d) = 0$ which has solutions $a = 0, a = 4d$

Since $a=0$ is invalid, $a=4d$ and the triangle has sides $3d$ , $4d$ , $5d$ . Therefore, the correct answer must be divisible by $3$ , $4$ , or $5$ . The only valid answer choice is $\boxed{\textbf{(C)}\ 81}$ , since it is divisible by $3$ .

-edited by coolmath34

1981 AHSME (Problems • Answer Key • Resources)
Preceded by Problem 10	Followed by Problem 12
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1981 AHSME Problems/Problem 11

Problem 11

Solution

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