Question

Rectangle $ABCD$ is given with $AB = 63$ and $BC = 448.$ Points $E$ and $F$ lie on $\overline{AD}$ and $\overline{BC}$ respectively, such that $AE = CF = 84.$ The inscribed circle of triangle $BEF$ is tangent to $\overline{EF}$ at point $P,$ and the inscribed circle of triangle $DEF$ is tangent to $\overline{EF}$ at point $Q.$ Find $PQ.$

Answer

Answer: 259

Concepts:incircle, incenter, and inradius Pythagorean Theorem

Difficulty rating: 2650

Solution:

Place $A = (0, 0),$ $B = (63, 0),$ $C = (63, 448),$ $D = (0, 448),$ so $E = (0, 84)$ and $F = (63, 364).$ Then $BE = DF = \sqrt{63^2 + 84^2}$ $= 21\sqrt{3^2 + 4^2} = 105,$ $BF = DE = 448 - 84 = 364,$ and $EF = \sqrt{63^2 + 280^2}$ $= 7\sqrt{9^2 + 40^2} = 287.$ In particular triangles $BEF$ and $DFE$ are congruent, with common semiperimeter $s = \frac{105 + 364 + 287}{2} = 378.$

In any triangle, the distance from a vertex to the incircle's tangency points on its two sides is the semiperimeter minus the opposite side. In triangle $BEF,$ $EP = s - BF$ $= 378 - 364 = 14;$ in triangle $DEF,$ $FQ = s - DE$ $= 378 - 364 = 14.$

Therefore $PQ = EF - EP - FQ$ $= 287 - 14 - 14 = 259.$

2007 AIME II Problem 9

Answer: 259

Difficulty rating: 2650

Solution:

Problem 9 in Other Years