# 2007 AIME II Problems/Problem 9

## 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

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 $PQ$.

## Solution 4

Why not first divide everything by its greatest common factor, $7$? Then we're left with much simpler numbers which saves a lot of time. In the end, we will multiply by $7$.

From there, we draw the same diagram as above (with smaller numbers). We soon find that the longest side of both triangles is 52 (64 - 12). That means:

$A = rs$ indicating $26(9)=r(54)$ so $r = 13/3$.

Now, we can start applying the equivalent tangents. Calling them $a$, $b$, and $c$ (with $c$ being the longest and a being the shortest),

$a+b+c$ is the semi perimeter or $54$. And since the longest side (which has $b+c$) is $52$, $a=2$.

Note that the distance $PQ$ we desired to find is just $c - a$. What is $b$ then? $b = 13$. And $c$ is $39$. Therefore the answer is $37$... $NOT.$

Multiply by $7$ back again (I hope you remembered to write this in $huge$ letters on top of the scrap paper!), we actually get $259$.

## Solution 5

Scaling everything by 7, we have that $AE = 12, AB = 9, BF = 52$. Note that if the perpendicular of $F$ dropped down to $ED$ is $X$, then $EX = 52-12 = 40$. But $FX = 9$ and so we have a $9-40-41$ right triangle with $EFX$ meaning $EF = 41$. Now, by symmetry, we know that $EP = QF = a$ meaning $PF = 41-a$. If the tangent of the circle inscribed in $BEF$ is tangent to $BE$ at $Y$, then if $BY = b$ we have a system of equations. $a+b = 15, b+41-a = 52$. We can then solve for $a$, and since $PQ = 41-2a$, the rest follows.

## See also

 2007 AIME II (Problems • Answer Key • Resources) Preceded byProblem 8 Followed byProblem 10 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 All AIME Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.

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