# 1971 AHSME Problems/Problem 9

## Problem

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

$[asy] import geometry; point C = origin; point A = (0,-4); point D = (24,0); point B = (24,-14); point E; // Defining point E pair[] e = intersectionpoints(perpendicular(A,line(B,D)),B--D); E = e[0]; // Circles draw(circle(A,length(segment(A,C)))); draw(circle(B,length(segment(B,D)))); // Segments draw(A--B); draw(A--C); draw(B--D); draw(C--D); label("24",midpoint(C--D),N); draw(A--E); // Labelling Points dot(A); label("A",A,SW); dot(B); label("B",B,SE); dot(C); label("C",C,NW); dot(D); label("D",D,NE); dot(E); label("E",E,(1,0)); // Right Angle Marks markscalefactor = 0.135; draw(rightanglemark(A,C,D)); draw(rightanglemark(C,D,E)); draw(rightanglemark(C,A,E)); draw(rightanglemark(A,E,D)); draw(rightanglemark(A,E,B)); [/asy]$

Let the center of the smaller circle be $A$ and the center of the larger circle be $B$. Draw $\overline{AC}$ and $\overline{BD}$, as in the diagram. We know that $\overline{AC}$ and $\overline{BD}$ are perpendicular to $\overline{CD}$ because $\overline{CD}$ is tangent to both circles. From the problem, we know $CD=24$.

Draw $\overline{AE} \perp \overline{BD}$ with $E$ on $\overline{BD}.$ $ABDC$ is a rectangle, so, knowing the radii of the circles, we see that $BE = BD - ED = BD - AC = 14 - 4 = 10.$

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

Thus, our answer is $\boxed{\textbf{(D) }26}$.

-edited by coolmath34