2004 AIME II Problems/Problem 7

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Problem

$ABCD$ is a rectangular sheet of paper that has been folded so that corner $B$ is matched with point $B'$ on edge $AD.$ The crease is $EF,$ where $E$ is on $AB$ and $F$ is on $CD.$ The dimensions $AE=8, BE=17,$ and $CF=3$ are given. The perimeter of rectangle $ABCD$ is $m/n,$ where $m$ and $n$ are relatively prime positive integers. Find $m+n.$

Solution

Solution 1 (synthetic)

pointpen = black; pathpen = black +linewidth(0.7);pair A=(0,0),B=(0,25),C=(70/3,25),D=(70/3,0),E=(0,8),F=(70/3,22),G=(15,0);D...

Since $EF$ is the perpendicular bisector of $\overline{BB'}$ , it follows that $BE = B'E$ (by SAS). By the Pythagorean Theorem, we have $AB' = 15$ . Similarly, from $BF = B'F$ , we have

$\begin{align*}BC^2 + CF^2 = B'D^2 + DF^2 &\Longrightarrow BC^2 + 9 = (BC - 15)^2 + 484 \\BC &= \frac{70}{3}\end{alig...$

Thus the perimeter of $ABCD$ is $2\left(25 + \frac{70}{3}\right) = \frac{290}{3}$ , and the answer is $m+n=\boxed{293}$ .

Solution 2 (analytic)

Let $A = (0,0), B=(0,25)$ , so $E = (0,8)$ and $F = (l,22)$ , and let $l = AD$ be the length of the rectangle. The slope of $EF$ is $\frac{14}{l}$ and so the equation of $EF$ is $y -8 = \frac{14}{l}x$ . We know that $EF$ is perpendicular to and bisects $BB'$ . The slope of $BB'$ is thus $\frac{-l}{14}$ , and so the equation of $BB'$ is $y -25 = \frac{-l}{14}x$ . Let the point of intersection of $EF, BB'$ be $G$ . Then the y-coordinate of $G$ is $\frac{25}{2}$ , so

$\begin{align*}\frac{14}{l}x &= y-8 = \frac{9}{2}\\\frac{-l}{14}x &= y-25 = -\frac{25}{2}\\\end{align*}$

Dividing the two equations yields

$l^2 = \frac{25 \cdot 14^2}{9} \Longrightarrow l = \frac{70}{3}$

The answer is $\boxed{293}$ as above.

2004 AIME II (Problems • Resources)
Preceded by Problem 6	Followed by Problem 8
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15

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