2007 AIME II Problems/Problem 9
Contents
Problem
Rectangle is given with and Points and lie on and respectively, such that The inscribed circle of triangle is tangent to at point and the inscribed circle of triangle is tangent to at point Find
Solution
Solution 1
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 .
Solution 2
By the Two Tangent Theorem, we have that . Solve for . Also, , so . Since , this can become . Substituting in their values, the answer is .
Solution 3
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 .
Solution 4
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 .
Solution 5
Scaling everything by 7, we have that . Note that if the perpendicular of dropped down to is , then . But and so we have a right triangle with meaning . Now, by symmetry, we know that meaning . If the tangent of the circle inscribed in is tangent to at , then if we have a system of equations. . We can then solve for , and since , the rest follows.
See also
2007 AIME II (Problems • Answer Key • Resources) | ||
Preceded by Problem 8 |
Followed by Problem 10 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |
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