Difference between revisions of "2009 AIME II Problems/Problem 5"
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== Solution == | == Solution == | ||
+ | Let <math>X</math> be the intersection of the circles with centers <math>B</math> and <math>E</math>, and <math>Y</math> be the intersection of the circles with centers <math>C</math> and <math>E</math>. Since the radius of <math>B</math> is <math>3</math>, <math>AX =4</math>. Assume <math>AE</math> = <math>p</math>. Then <math>EX</math> and <math>EY</math> are radii of circle <math>E</math> and have length <math>4+p</math>. <math>AC = 8</math>, and angle <math>CAE = 60</math> degrees because we are given that triangle <math>T</math> is equilateral. Using the [[Law of Cosines]] on triangle <math>CAE</math>, we obtain | ||
− | + | <math>(6+p)^2 =p^2 + 64 - 2(8)(p) \cos 60</math>. | |
− | + | The <math>2</math> and the <math>\cos 60</math> terms cancel out: | |
− | + | <math>p^2 + 12p +36 = p^2 + 64 - 8p</math> | |
− | + | <math>12p+ 36 = 64 - 8p</math> | |
− | + | <math>p =\frac {28}{20} = \frac {7}{5}</math>. The radius of circle <math>E</math> is <math>4 + \frac {7}{5} = \frac {27}{5}</math>, so the answer is <math>27 + 5 = \boxed{032}</math>. | |
− | + | ==Video Solution== | |
+ | https://www.youtube.com/watch?v=KKVxQV4hszo&t=7s | ||
== 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}} | ||
+ | [[Category: Intermediate Geometry Problems]] | ||
+ | {{MAA Notice}} |
Latest revision as of 20:54, 9 August 2020
Contents
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 angle
degrees because we are given that triangle
is equilateral. Using the Law of Cosines on triangle
, we obtain
.
The and the
terms cancel out:
. The radius of circle
is
, so the answer is
.
Video Solution
https://www.youtube.com/watch?v=KKVxQV4hszo&t=7s
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 |
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