Difference between revisions of "2009 AIME II Problems/Problem 5"
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<math>(6+m)^2</math> = <math>m^2</math> + <math>64</math> - <math>2(8)(m) cos 60</math>. | <math>(6+m)^2</math> = <math>m^2</math> + <math>64</math> - <math>2(8)(m) cos 60</math>. | ||
− | The <math>2 and the < | + | The <math>2</math> and the <math>cos 60</math> cancel out: |
− | < | + | <math>m^2</math> + <math>12m</math> + <math>36</math> = <math>m^2</math> + <math>64</math> - <math>8m</math> |
− | < | + | <math>12m</math> + <math>36</math> = <math>64</math> - <math>8m</math> |
− | < | + | <math>m</math> = <math>28/20</math> = <math>7/5</math>. The radius of circle <math>E</math> is <math>4</math> + <math>7/5</math> = <math>27/5</math>, so the answer is <math>27</math> + <math>5</math> = <math>\boxed{032}</math>. |
== See Also == | == See Also == | ||
{{AIME box|year=2009|n=II|num-b=4|num-a=6}} | {{AIME box|year=2009|n=II|num-b=4|num-a=6}} |
Revision as of 20:40, 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 be the intersection of the circles with centers
and
, and
be the intersection of the circles with centers
and
. Since the radius of
is
,
=
. Assume
=
. Then
and
are radii of circle
and have length
.
=
, and it can easily be shown that angle
=
degrees. Using the Law of Cosines on triangle
, we obtain
=
+
-
.
The and the
cancel out:
+
+
=
+
-
+
=
-
=
=
. The radius of circle
is
+
=
, so the answer is
+
=
.
See Also
2009 AIME II (Problems • Answer Key • Resources) | ||
Preceded by Problem 4 |
Followed by Problem 6 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |