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 |
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