Difference between revisions of "1954 AHSME Problems/Problem 26"
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| + | ==Problem 26== | ||
https://artofproblemsolving.com/community/c4h256249s1_lines_and_circles | 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\equal{}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: | ||
| + | |||
| + | <math> \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}</math> | ||
| + | |||
| + | ==Solution== | ||
==See Also== | ==See Also== | ||
Revision as of 17:03, 22 April 2020
Problem 26
https://artofproblemsolving.com/community/c4h256249s1_lines_and_circles
The straight line 
is divided at 
so that $AC\equal{}3CB$ (Error compiling LaTeX. Unknown error_msg). Circles are described on 
and 
as diameters and a common tangent meets 
produced at 
. Then 
equals:
Solution
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. 
