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Difference between revisions of "1954 AHSME Problems/Problem 26"
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==Problem 26==
 
==Problem 26==
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https://artofproblemsolving.com/community/c4h256249s1_lines_and_circles
 
  
 
The straight line <math> \overline{AB}</math> is divided at <math> C</math> so that <math> AC=3CB</math>.  Circles are described on <math> \overline{AC}</math> and <math> \overline{CB}</math> as diameters and a common tangent meets <math> AB</math> produced at <math> D</math>.  Then <math> BD</math> equals:
 
The straight line <math> \overline{AB}</math> is divided at <math> C</math> so that <math> AC=3CB</math>.  Circles are described on <math> \overline{AC}</math> and <math> \overline{CB}</math> as diameters and a common tangent meets <math> AB</math> produced at <math> D</math>.  Then <math> BD</math> equals:
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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://artofproblemsolving.com/community/c4h256249s1_lines_and_circles
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==Video Solution==
  
 
https://youtu.be/cbU3DxJKB8U
 
https://youtu.be/cbU3DxJKB8U

Latest revision as of 15:39, 7 October 2024

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

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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