2007 AIME II Problems/Problem 15
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
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 .
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 .
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