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</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 | 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 | ||
− | <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</math> and the <math>cos 60</math> cancel out: | + | 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>m^2</math> + <math>12m</math> + <math>36</math> = <math>m^2</math> + <math>64</math> - <math>8m</math> |
Revision as of 19:41, 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 |