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

Revision as of 20:00, 23 November 2014

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

Solution 1

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 $BX = 105 - EX = 105 - \left[\frac{287 - PQ}{2}\right]$ . Also, $BX = BY$ . $FY = 364 - BY = 364 - \left[105 - \left[\frac{287 - PQ}{2}\right]\right]$ .

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 $PQ = \boxed{259}$ .

Solution 2

Call the incenter of $\triangle BEF$ $O_1$ and the incenter of $\triangle DFE$ $O_2$ . Draw triangles $\triangle O_1PQ,\triangle PQO_2$ .

Drawing $BE$ , We find that $BE = \sqrt {63^2 + 84^2} = 105$ . Applying the same thing for $F$ , we find that $FD = 105$ as well. Draw a line through $E,F$ parallel to the sides of the rectangle, to intersect the opposite side at $E_1,F_1$ respectively. Drawing $\triangle EE_1F$ and $FF_1E$ , we can find that $EF = \sqrt {63^2 + 280^2} = 287$ . We then use Heron's formula to get:

$[BEF] = [DEF] = 11 466$ .

So the inradius of the triangle-type things is $\frac {637}{21}$ .

Now, we just have to find $O_1Q = O_2P$ , which can be done with simple subtraction, and then we can use the Pythagorean Theorem to find $PQ$ .

Template:Incomplete

@@ Line 8: / Line 8: @@
 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>. By the Two Tangent theorem, note that <math>EP = EX = \frac{287 - PQ}{2}</math>, making <math>BX = 105 - EX = 105 - \left[\frac{287 - PQ}{2}\right]</math>. Also, <math>BX = BY</math>. <math>FY = 364 - BY = 364 - \left[105 - \left[\frac{287 - PQ}{2}\right]\right]</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>. By the Two Tangent theorem, note that <math>EP = EX = \frac{287 - PQ}{2}</math>, making <math>BX = 105 - EX = 105 - \left[\frac{287 - PQ}{2}\right]</math>. Also, <math>BX = BY</math>. <math>FY = 364 - BY = 364 - \left[105 - \left[\frac{287 - PQ}{2}\right]\right]</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>PQ = \boxed{259}</math>.

2007 AIME II (Problems • Answer Key • Resources)
Preceded by Problem 8	Followed by Problem 10
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