2007 AIME II Problems/Problem 9
Several Pythagorean triples exist amongst the numbers given. . Also, the length of .
Use the Two Tangent theorem on . Since both circles are inscribed in congruent triangles, they are congruent; therefore, . By the Two Tangent theorem, note that , making . Also, . .
Finally, . Also, . Equating, we see that , so .
By the Two Tangent theorem, we have that . Solve for . Also, , so . Since , this can become . Substituting in their values, the answer is .
Call the incenter of and the incenter of . Draw triangles .
Drawing , We find that . Applying the same thing for , we find that as well. Draw a line through parallel to the sides of the rectangle, to intersect the opposite side at respectively. Drawing and , we can find that . We then use Heron's formula to get:
So the inradius of the triangle-type things is .
Now, we just have to find , which can be done with simple subtraction, and then we can use the Pythagorean Theorem to find .
|2007 AIME II (Problems • Answer Key • Resources)|
|1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15|
|All AIME Problems and Solutions|