# 1953 AHSME Problems/Problem 42

## Problem

The centers of two circles are $41$ inches apart. The smaller circle has a radius of $4$ inches and the larger one has a radius of $5$ inches. The length of the common internal tangent is: $\textbf{(A)}\ 41\text{ inches} \qquad \textbf{(B)}\ 39\text{ inches} \qquad \textbf{(C)}\ 39.8\text{ inches} \qquad \textbf{(D)}\ 40.1\text{ inches}\\ \textbf{(E)}\ 40\text{ inches}$

## Solution $[asy] size(400); draw((0,0)--(41,0)); draw((0,0)--(45/41,200/41)--(1645/41,-160/41)); draw((0,0)--(1600/41,-360/41)--(41,0)); draw(circle((0,0),5)); draw(circle((41,0),4)); label("A",(0,0),W); label("B",(41,0),E); label("C",(45/41,200/41),N); label("D",(1645/41,-160/41),SE); label("E",(1600/41,-360/41),E); [/asy]$

Let $A$ be the center of the circle with radius $5$, and $B$ be the center of the circle with radius $4$. Let $\overline{CD}$ be the common internal tangent of circle $A$ and circle $B$. Extend $\overline{BD}$ past $D$ to point $E$ such that $\overline{BE}\perp\overline{AE}$. Since $\overline{AC}\perp\overline{CD}$ and $\overline{BD}\perp\overline{CD}$, $ACDE$ is a rectangle. Therefore, $AC=DE$ and $CD=AE$.

Since the centers of the two circles are $41$ inches apart, $AB=41$. Also, $BE=4+5=9$. Using the Pythagorean Theorem on right triangle $ABE$, $CD=AE=\sqrt{41^2-9^2}=\sqrt{1600}=40$. The length of the common internal tangent is $\boxed{\textbf{(E) } 40\text{ inches}}$

## See Also

 1953 AHSC (Problems • Answer Key • Resources) Preceded byProblem 41 Followed byProblem 43 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 • 26 • 27 • 28 • 29 • 30 • 31 • 32 • 33 • 34 • 35 • 36 • 37 • 38 • 39 • 40 • 41 • 42 • 43 • 44 • 45 • 46 • 47 • 48 • 49 • 50 All AHSME Problems and Solutions

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