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

Revision as of 08:55, 11 February 2008

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 $PQ = FY - EX + (BY - BX)$ $= \left(FY + BY\right) - \left(EX + EY\right) = FB - EB$ . Substituting in their values, the answer is $364 - 105 = 259$ .

Solution 3

Call the incenter of $\triangle BEF$ $O_1$ and the incenter of $\triangle FDE$ $O_2$ . 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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@@ Line 17: / Line 17: @@
 ===Solution 3===
-Call the incenter of <math>\triangle BEF</math> <math>O_1</math> and the incenter of that other triangle <math>O_2</math>. Draw triangles <math>\triangle O_1PQ,\triangle PQO_2</math>.
+Call the incenter of <math>\triangle BEF</math> <math>O_1</math> and the incenter of <math>\triangle FDE</math> <math>O_2</math>. Draw triangles <math>\triangle O_1PQ,\triangle PQO_2</math>.
 Drawing <math>BE</math>, We find that <math>BE = \sqrt {63^2 + 84^2} = 105</math>. Applying the same thing for <math>F</math>, we find that <math>FD = 105</math> as well. Draw a line through <math>E,F</math> parallel to the sides of the rectangle, to intersect the opposite side at <math>E_1,F_1</math> respectively. Drawing <math>\triangle EE_1F</math> and <math>FF_1E</math>, we can find that <math>EF = \sqrt {63^2 + 280^2} = 287</math>. We then use Heron's formula to get:

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