Difference between revisions of "2009 AIME II Problems/Problem 5"
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− | Let <math>X</math> 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 it can easily be shown that angle CAE = 60 degrees. Using the [[Law of Cosines]] on triangle CAE, we obtain | + | 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</math> = <math>4</math>. Assume <math>AE</math> = <math>m</math>. Then <math>EX</math> and <math>EY</math> are radii of circle <math>E</math> and have length <math>4+m</math>. <math>AC</math> = <math>8</math>, and it can easily be shown that angle <math>CAE</math> = <math>60</math> degrees. Using the [[Law of Cosines]] on triangle <math>CAE</math>, we obtain |
− | (6+m)^2 = m^2 + 64 - 2(8)(m) cos 60. | + | <math>(6+m)^2</math> = <math>m^2</math> + <math>64</math> - <math>2(8)(m) cos 60</math>. |
− | The 2 and the cos 60 cancel out: | + | The <math>2 and the </math>cos 60<math> cancel out: |
− | m^2 + 12m + 36 = m^2 + 64 - 8m | + | </math>m^2<math> + </math>12m<math> + </math>36<math> = </math>m^2<math> + </math>64<math> - </math>8m<math> |
− | 12m + 36 = 64 - 8m | + | </math>12m<math> + </math>36<math> = </math>64<math> - </math>8m<math> |
− | m = 28/20 = 7/5. The radius of circle E is 4 + 7/5 = 27/5, so the answer is 27+5 = <math>\boxed{032} | + | </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}$. |
== 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 cos 60
m^2
12m
36
m^2
64
8m$$ (Error compiling LaTeX. Unknown error_msg)12m
36
64
8m$$ (Error compiling LaTeX. Unknown error_msg)m
28/20
7/5
E
4
7/5
27/5
27
5
\boxed{032}$.
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 |