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 $\displaystyle BX = 105 - EX = 105 - \left[\frac{287 - PQ}{2}\right]$ . Also, $\displaystyle BX = BY$ . $FY = 364 - BY = 364 - \left[105 - \left[\frac{287 - PQ}{2}\right]\right]$ .

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 $\displaystyle PQ = 259$ .

Solution 2

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

2007 AIME II (Problems • Answer Key • Resources)
Preceded by Problem 8	Followed by Problem 10
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2007 AIME II Problems/Problem 9

Contents

Problem

Solution

Solution 1

Solution 2

See also