Difference between revisions of "2007 AIME II Problems/Problem 9"
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[[Image:2007 AIME II-9.png]] | [[Image:2007 AIME II-9.png]] | ||
+ | === Solution 1 === | ||
Several [[Pythagorean triple]]s exist amongst the numbers given. <math>BE = DF = \sqrt{63^2 + 84^2} = 21\sqrt{3^2 + 4^2} = 105</math>. Also, the length of <math>EF = \sqrt{63^2 + (448 - 2\cdot84)^2} = 7\sqrt{9^2 + 40^2} = 287</math>. | Several [[Pythagorean triple]]s exist amongst the numbers given. <math>BE = DF = \sqrt{63^2 + 84^2} = 21\sqrt{3^2 + 4^2} = 105</math>. Also, the length of <math>EF = \sqrt{63^2 + (448 - 2\cdot84)^2} = 7\sqrt{9^2 + 40^2} = 287</math>. | ||
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Finally, <math>FP = FY = 364 - \left[105 - \left[\frac{287 - PQ}{2}\right]\right] = \frac{805 - PQ}{2}</math>. Also, <math>FP = FQ + PQ = \frac{287 - PQ}{2} + PQ</math>. Equating, we see that <math>\frac{805 - PQ}{2} = \frac{287 + PQ}{2}</math>, so <math>\displaystyle PQ = 259</math>. | Finally, <math>FP = FY = 364 - \left[105 - \left[\frac{287 - PQ}{2}\right]\right] = \frac{805 - PQ}{2}</math>. Also, <math>FP = FQ + PQ = \frac{287 - PQ}{2} + PQ</math>. Equating, we see that <math>\frac{805 - PQ}{2} = \frac{287 + PQ}{2}</math>, so <math>\displaystyle PQ = 259</math>. | ||
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+ | === Solution 2 === | ||
+ | By the [[Two Tangent theorem]], we have that <math>\displaystyle FY = PQ + QF</math>. Solve for <math>\displaystyle PQ = FY - QF</math>. Also, <math>\displaystyle QF = EP = EX</math>, so <math>\displaystyle PQ = FY - EX</math>. Since <math>\displaystyle BX = BY</math>, this can become <math>\displastyle PQ = FY - EX + (BY - BX)</math><math> = \left(FY + BY\right) - \left(EX + EY\right) = FB - EB</math>. Substituting in their values, the answer is <math>364 - 105 = 259</math>. | ||
== See also == | == See also == |
Revision as of 09:01, 5 April 2007
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 $\displastyle PQ = FY - EX + (BY - BX)$ (Error compiling LaTeX. Unknown error_msg). Substituting in their values, the answer is .
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 |