Difference between revisions of "2009 AIME II Problems/Problem 5"
Aimesolver (talk | contribs) (New page: == Problem 5 == Equilateral triangle <math>T</math> is inscribed in circle <math>A</math>, which has radius <math>10</math>. Circle <math>B</math> with radius <math>3</math> is internally ...) |
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Let X 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 angle CAE = 60 degrees. Using the [[Law of Cosines]] on triangle CAE, we obtain | Let X 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 angle CAE = 60 degrees. Using the [[Law of Cosines]] on triangle CAE, we obtain |
Revision as of 18:18, 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 X 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 angle CAE = 60 degrees. Using the Law of Cosines on triangle CAE, we obtain
(6+m)^2 = m^2 + 64 - 2(8)(m) cos 60.
The 2 and the cos 60 cancel out:
m^2 + 12m + 36 = m^2 + 64 - 8m
12m + 36 = 64 - 8m
m = 28/20 = 7/5. The radius of circle E is 4 + 7/5 = 27/5, so the answer is 27+5 = 032.