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 (ProblemsAnswer KeyResources)
Preceded by
Problem 41
Followed by
Problem 43
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