Difference between revisions of "2009 AIME II Problems/Problem 5"
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m = 28/20 = 7/5. The radius of circle E is 4 + 7/5 = 27/5, so the answer is 27+5 = 032. | m = 28/20 = 7/5. The radius of circle E is 4 + 7/5 = 27/5, so the answer is 27+5 = 032. | ||
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+ | == See Also == | ||
+ | |||
+ | {{AIME box|year=2009|n=II|num-b=1|num-a=3}} |
Revision as of 19:19, 17 April 2009
Problem 5
Equilateral triangle is inscribed in circle
, which has radius
. Circle
with radius
is internally tangent to circle
at one vertex of
. Circles
and
, both with radius
, are internally tangent to circle
at the other two vertices of
. Circles
,
, and
are all externally tangent to circle
, which has radius
, where
and
are relatively prime positive integers. Find
.
Solution
Let X be the intersection of the circles with centers B and E, and Y be the intersection of the circles with centers C and E. Since the radius of B is 3, AX = 4. Assume AE = m. Then EX and EY are radii of circle E and have length 4+m. AC = 8, and angle CAE = 60 degrees. Using the Law of Cosines on triangle CAE, we obtain
(6+m)^2 = m^2 + 64 - 2(8)(m) cos 60.
The 2 and the cos 60 cancel out:
m^2 + 12m + 36 = m^2 + 64 - 8m
12m + 36 = 64 - 8m
m = 28/20 = 7/5. The radius of circle E is 4 + 7/5 = 27/5, so the answer is 27+5 = 032.
See Also
2009 AIME II (Problems • Answer Key • Resources) | ||
Preceded by Problem 1 |
Followed by Problem 3 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |