Difference between revisions of "1976 AHSME Problems/Problem 24"
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== Problem == | == Problem == | ||
− | In the adjoining figure, circle <math> | + | In the adjoining figure, circle <math>K</math> has diameter <math>AB</math>; circle <math>L</math> is tangent to circle <math>K</math> and to <math>AB</math> at the center of circle <math>K</math>; and circle <math>M</math> tangent to circle <math>K</math>, to circle <math>L</math> and <math>AB</math>. The ratio of the area of circle <math>K</math> to the area of circle <math>M</math> is |
− | and to <math> | ||
− | to circle <math> | ||
<asy> | <asy> | ||
/* Made by Klaus-Anton, Edited by MRENTHUSIASM */ | /* Made by Klaus-Anton, Edited by MRENTHUSIASM */ | ||
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\textbf{(C) }16\qquad | \textbf{(C) }16\qquad | ||
\textbf{(D) }18\qquad | \textbf{(D) }18\qquad | ||
− | \textbf{(E) }\text{not an integer} | + | \textbf{(E) }\text{not an integer}</math> |
== Solution == | == Solution == |
Revision as of 05:49, 6 September 2021
Problem
In the adjoining figure, circle has diameter
; circle
is tangent to circle
and to
at the center of circle
; and circle
tangent to circle
, to circle
and
. The ratio of the area of circle
to the area of circle
is
Solution
See Also
1976 AHSME (Problems • Answer Key • Resources) | ||
Preceded by Problem 23 |
Followed by Problem 25 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 • 26 • 27 • 28 • 29 • 30 | ||
All AHSME Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.