1954 AHSME Problems/Problem 26

Problem 26

The straight line $\overline{AB}$ is divided at $C$ so that $AC=3CB$ . Circles are described on $\overline{AC}$ and $\overline{CB}$ as diameters and a common tangent meets $AB$ produced at $D$ . Then $BD$ equals:

$\textbf{(A)}\ \text{diameter of the smaller circle} \\ \textbf{(B)}\ \text{radius of the smaller circle} \\ \textbf{(C)}\ \text{radius of the larger circle} \\ \textbf{(D)}\ CB\sqrt{3}\\ \textbf{(E)}\ \text{the difference of the two radii}$

Solution

Let $x=\overline{BD}$ and let $r$ be the radius of the small circle. Draw the line from the center of each of the circles to the point of contact of the tangent of the circle. By similar triangles, $\frac{x+r}{r}=\frac{x+5r}{3r} \implies x=r$ , or $\boxed{\textbf{(B)}}$ .

https://artofproblemsolving.com/community/c4h256249s1_lines_and_circles

Video Solution

https://youtu.be/cbU3DxJKB8U

~Lucas

1954 AHSC (Problems • Answer Key • Resources)
Preceded by Problem 25	Followed by Problem 27
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1954 AHSME Problems/Problem 26

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Problem 26

Solution

Video Solution

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