Difference between revisions of "1954 AHSME Problems/Problem 26"
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==Solution== | ==Solution== | ||
| + | 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>. | ||
==See Also== | ==See Also== | ||
Revision as of 17:04, 22 April 2020
Problem 26
https://artofproblemsolving.com/community/c4h256249s1_lines_and_circles
The straight line 
is divided at 
so that 
. Circles are described on 
and 
as diameters and a common tangent meets 
produced at 
. Then 
equals:
Solution
Let 
and let 
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, 
, or 
.
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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
