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

Revision as of 21:46, 12 March 2010

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 $PQ = 259$ .

Solution 2

By the Two Tangent theorem, we have that $FY = PQ + QF$ . Solve for $PQ = FY - QF$ . Also, $QF = EP = EX$ , so $PQ = FY - EX$ . Since $BX = BY$ , this can become $PQ = FY - EX + (BY - BX)$ $= \left(FY + BY\right) - \left(EX + BX\right) = FB - EB$ . Substituting in their values, the answer is $364 - 105 = 259$ .

Solution 3

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 $PQ$ .

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 === 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 + EY\right) = FB - EB</math>. Substituting in their values, the answer is <math>364 - 105 = 259</math>.
+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===

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