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 .
Why not first divide everything by its greatest common factor, ? Then we're left with much simpler numbers which saves a lot of time. In the end, we will multiply by .
From there, we draw the same diagram as above (with smaller numbers). We soon find that the longest side of both triangles is 52 (64 - 12). That means:
indicating so .
Now, we can start applying the equivalent tangents. Calling them , , and (with being the longest and a being the shortest),
is the semi perimeter or . And since the longest side (which has ) is , .
Note that the distance we desired to find is just . What is then? . And is . Therefore the answer is ...
Multiply by back again (I hope you remembered to write this in letters on top of the scrap paper!), we actually get .
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