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 <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 | + | 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>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 = 8</math>, and it can easily be shown that angle <math>CAE = 60</math> degrees. Using the [[Law of Cosines]] on triangle <math>CAE</math>, we obtain |
− | <math>(6+m)^2 | + | <math>(6+m)^2 =m^2 + 64 - 2(8)(m) \cos 60</math>. |
− | The <math>2</math> and the | + | The <math>2</math> and the <math>\cos 60</math> terms cancel out: |
− | <math>m^2 | + | <math>m^2 + 12m +36 = m^2 + 64 - 8m</math> |
− | <math>12m | + | <math>12m+ 36 = 64 - 8m</math> |
− | <math>m | + | <math>m =\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>. |
== See Also == | == See Also == |
Revision as of 20:14, 13 March 2015
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 terms 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 |
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