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Difference between revisions of "2007 AIME II Problems/Problem 9"

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== Solution ==
 
== Solution ==
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{{image}}
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[[Image:2007 AIME II-9.png]]
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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>.
  
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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>.
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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>. By the Two Tangent theorem, note that <math>EP = EX = \frac{287 - PQ}{2}</math>, making <math>\displaystyle BX = 105 - EX = 105 - \left[\frac{287 - PQ}{2}\right]</math>. Also, <math>\displaystyle BX = BY</math>. <math>FY = 364 - BY = 364 - \left[105 - \left[\frac{287 - PQ}{2}\right]\right]</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>.
  
 
== See also ==
 
== See also ==

Revision as of 09:50, 5 April 2007

Problem

Rectangle $ABCD$ is given with $AB=63$ and $BC=448.$ Points $E$ and $F$ lie on $AD$ and $BC$ respectively, such that $AE=CF=84.$ The inscribed circle of triangle $BEF$ is tangent to $EF$ at point $P,$ and the inscribed circle of triangle $DEF$ is tangent to $EF$ at point $Q.$ Find $PQ.$

Solution

2007 AIME II-9.png

Several Pythagorean triples exist amongst the numbers given. $BE = DF = \sqrt{63^2 + 84^2} = 21\sqrt{3^2 + 4^2} = 105$. Also, the length of $EF = \sqrt{63^2 + (448 - 2\cdot84)^2} = 7\sqrt{9^2 + 40^2} = 287$.

Use the Two Tangent theorem on $\triangle BEF$. Since both circles are inscribed in congruent triangles, they are congruent; therefore, $EP = FQ = \frac{287 - PQ}{2}$. By the Two Tangent theorem, note that $EP = EX = \frac{287 - PQ}{2}$, making $\displaystyle BX = 105 - EX = 105 - \left[\frac{287 - PQ}{2}\right]$. Also, $\displaystyle BX = BY$. $FY = 364 - BY = 364 - \left[105 - \left[\frac{287 - PQ}{2}\right]\right]$.

Finally, $FP = FY = 364 - \left[105 - \left[\frac{287 - PQ}{2}\right]\right] = \frac{805 - PQ}{2}$. Also, $FP = FQ + PQ = \frac{287 - PQ}{2} + PQ$. Equating, we see that $\frac{805 - PQ}{2} = \frac{287 + PQ}{2}$, so $\displaystyle PQ = 259$.

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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