Difference between revisions of "2007 AIME II Problems/Problem 9"
(→Solution 3) |
(→Solution 2) |
||
Line 13: | Line 13: | ||
=== Solution 2 === | === Solution 2 === | ||
− | By the [[Two Tangent theorem]], we have that <math>FY = PQ + QF</math>. Solve for <math>PQ = FY - QF</math>. Also, <math>QF = EP = EX</math>, so <math>PQ = FY - EX</math>. Since <math>BX = BY</math>, this can become <math>PQ = FY - EX + (BY - BX)</math><math> = \left(FY + BY\right) - \left(EX + | + | By the [[Two Tangent theorem]], we have that <math>FY = PQ + QF</math>. Solve for <math>PQ = FY - QF</math>. Also, <math>QF = EP = EX</math>, so <math>PQ = FY - EX</math>. Since <math>BX = BY</math>, this can become <math>PQ = FY - EX + (BY - BX)</math><math> = \left(FY + BY\right) - \left(EX + BX\right) = FB - EB</math>. Substituting in their values, the answer is <math>364 - 105 = 259</math>. |
===Solution 3=== | ===Solution 3=== |
Revision as of 21:46, 12 March 2010
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 .
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 |