Difference between revisions of "2012 AIME I Problems/Problem 13"
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− | Reinterpret the problem in the following manner. Equilateral triangle ABC has a point X on the interior such that AX = 5, BX = 4, CX = 3. A 60 degree clockwise (or counter clockwise) rotation about vertex A maps | + | Reinterpret the problem in the following manner. Equilateral triangle ABC has a point X on the interior such that AX = 5, BX = 4, CX = 3. A 60 degree clockwise (or counter clockwise) rotation about vertex A maps X to X' and C to C'. Note that angle XAX' is 60 and XA = X'A = 5 which tells us that triangle XAX' is equilateral and that XX' = 5. We now notice that XB = 3, X'B = 4 which tells us that angle XBX' is 90 because there is a 3,4,5 pythagorean triple. Now note that angle ABC + angle ACB = 120, angle XCA + angle XBA = 90, angle XCB+angle XBC = 30, angle BXC = 150. Applying the law of cosines on triangle BXC yields |
<math>BX^2+CX^2-2*BX*CX*cos(150) = BC^2 = 3^2+4^2-24*cos(150) = 25+12\sqrt{3}</math> and since we are looking for the area we have <math>BC^2\frac{\sqrt{3}}{4} = 25\frac{\sqrt{3}}{4}+9</math> | <math>BX^2+CX^2-2*BX*CX*cos(150) = BC^2 = 3^2+4^2-24*cos(150) = 25+12\sqrt{3}</math> and since we are looking for the area we have <math>BC^2\frac{\sqrt{3}}{4} = 25\frac{\sqrt{3}}{4}+9</math> |
Revision as of 12:56, 17 March 2012
Problem 13
Three concentric circles have radii and An equilateral triangle with one vertex on each circle has side length The largest possible area of the triangle can be written as where and are positive integers, and are relatively prime, and is not divisible by the square of any prime. Find
Solution
Reinterpret the problem in the following manner. Equilateral triangle ABC has a point X on the interior such that AX = 5, BX = 4, CX = 3. A 60 degree clockwise (or counter clockwise) rotation about vertex A maps X to X' and C to C'. Note that angle XAX' is 60 and XA = X'A = 5 which tells us that triangle XAX' is equilateral and that XX' = 5. We now notice that XB = 3, X'B = 4 which tells us that angle XBX' is 90 because there is a 3,4,5 pythagorean triple. Now note that angle ABC + angle ACB = 120, angle XCA + angle XBA = 90, angle XCB+angle XBC = 30, angle BXC = 150. Applying the law of cosines on triangle BXC yields
and since we are looking for the area we have
so our final answer is 3+4+25+9 = .
See also
2012 AIME I (Problems • Answer Key • Resources) | ||
Preceded by Problem 12 |
Followed by Problem 14 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |