[Rectangle](https://artofproblemsolving.com/wiki/index.php/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](https://artofproblemsolving.com/wiki/index.php/Inscribed_circle) of [triangle](https://artofproblemsolving.com/wiki/index.php/Triangle) $BEF$ is [tangent](https://artofproblemsolving.com/wiki/index.php/Tangent) to $EF$ at point $P,$ and the inscribed circle of triangle $DEF$ is tangent to $EF$ at [point](https://artofproblemsolving.com/wiki/index.php/Point) $Q.$ Find $PQ.$
[2007 AIME II-9.png](https://artofproblemsolving.com/wiki/index.php/File:2007_AIME_II-9.png)

Solution 1
Several [Pythagorean triples](https://artofproblemsolving.com/wiki/index.php/Pythagorean_triple) 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](https://artofproblemsolving.com/wiki/index.php/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](https://artofproblemsolving.com/wiki/index.php/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](https://artofproblemsolving.com/wiki/index.php/Pythagorean_Theorem) to find $PQ$.