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Difference between revisions of "1954 AHSME Problems/Problem 26"
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Let <math> x=\overline{BD}</math> and let <math> r</math> 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, <math> \frac{x+r}{r}=\frac{x+5r}{3r} \implies x=r</math>, or <math> \boxed{\textbf{(B)}}</math>.
 
Let <math> x=\overline{BD}</math> and let <math> r</math> 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, <math> \frac{x+r}{r}=\frac{x+5r}{3r} \implies x=r</math>, or <math> \boxed{\textbf{(B)}}</math>.
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==Video Solution
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https://youtu.be/cbU3DxJKB8U
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~Lucas
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==See Also==
 
==See Also==
  

Revision as of 16:31, 19 September 2022

Problem 26

https://artofproblemsolving.com/community/c4h256249s1_lines_and_circles

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 $\boxed{\textbf{(B)}}$.

==Video Solution

https://youtu.be/cbU3DxJKB8U

~Lucas

See Also

1954 AHSC (Problems • Answer Key • Resources)
Preceded by
Problem 25 		Followed by
Problem 27
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All AHSME Problems and Solutions


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