Difference between revisions of "2007 AIME II Problems/Problem 9"
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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>. | ||
− | Use the [[Two Tangent theorem]] on <math>\triangle BEF</math>. Since both circles are inscribed in congruent triangles, they are congruent; therefore, <math>EP = FQ = \frac{287 - PQ}{2}</math>. | + | Use the [[Two Tangent theorem]] on <math>\triangle BEF</math>. Since both circles are inscribed in congruent triangles, they are congruent; therefore, <math>EP = FQ = \frac{287 - PQ}{2}</math>. Repeatedly apply the Two Tangent theorem to find that <math>FP = 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>. |
== See also == | == See also == |
Revision as of 08:34, 31 March 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
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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, . Repeatedly apply the Two Tangent theorem to find that . Also, . Equating, we see that , so .
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 |