2007 AIME II Problems/Problem 15
Problem
Four circles and with the same radius are drawn in the interior of triangle such that is tangent to sides and , to and , to and , and is externally tangent to and . If the sides of triangle are and the radius of can be represented in the form , where and are relatively prime positive integers. Find
Solution
Solution 1
First, apply Heron's formula to find that the area is . Also the semiperimeter is . So the inradius is .
Now consider the incenter I. Let the radius of one of the small circles be . Let the centers of the three little circles tangent to the sides of be , , and . Let the centre of the circle tangent to those three circles be P. A homothety centered at takes to with factor . The same homothety takes to the circumcentre of , so , where is the circumradius of . The circumradius of can be easily computed by , so doing that reveals . Then , so the answer is .
Solution 2
Consider a 13-14-15 triangle. [By Heron's Formula or by 5-12-13 and 9-12-15 right triangles.]
The inradius is , where is the semiperimeter. Scale the triangle with the inradius by a linear scale factor,
The circumradius is where and are the side-lengths. Scale the triangle with the circumradius by a linear scale factor, .
Cut and combine the triangles, as shown. Then solve for :
The solution is .
See also
2007 AIME II (Problems • Answer Key • Resources) | ||
Preceded by Problem 14 |
Followed by Last question | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.