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

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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.

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