1959 AMC 12 Problem 7

Attempt Problem 7 of the 1959 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 1959 AMC 12 solutions, or check the answer key.

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The sides of a right triangle are

a,

a+d,

and

a+2d,

with

a

and

d

both positive. The ratio of

a

d

is:

1{:}3

1{:}4

2{:}1

3{:}1

3{:}4

The longest side

a+2d

must be the hypotenuse

Apply the Pythagorean theorem and factor the resulting quadratic in

\frac{a}{d}

The Pythagorean theorem gives

a^2+(a+d)^2=(a+2d)^2.

Simplifying,

\begin{aligned} a^2-2ad-3d^2 &=0,\\ (a-3d)(a+d) &=0. \end{aligned}

Positivity excludes

a=-d,

a=3d.

Thus the ratio is

3:1,

and the correct answer is D.

Problem 7 in other years