# 1971 AHSME Problems/Problem 9

## Problem 9

An uncrossed belt is fitted without slack around two circular pulleys with radii of $14$ inches and $4$ inches. If the distance between the points of contact of the belt with the pulleys is $24$ inches, then the distance between the centers of the pulleys in inches is $\textbf{(A) }24\qquad \textbf{(B) }2\sqrt{119}\qquad \textbf{(C) }25\qquad \textbf{(D) }26\qquad \textbf{(E) }4\sqrt{35}$

## Solution

(I don't know how to use Asymptote, so if you are a visual learner, I apologize)

Let the center of the smaller circle be $A$ and the center of the larger circle be $B.$ Draw perpendicular radii $AC$ and $BD.$ The length $CD$ should equal 24.

From $A,$ draw a perpendicular line $AE,$ where $E$ is the foot of the perpendicular on $BD.$ Line $BE = 14 - 4 = 10.$

Points $A, E,$ and $B$ form a right triangle with legs $10$ and $24.$ We are looking for distance $AB,$ which is $\sqrt{10^2 + 24^2} = 26$

The answer is $\textbf{(D)} 26.$

-edited by coolmath34