Question

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

Answer

Answer: 293

Concepts:paper folding Pythagorean Theorem coordinate geometry

Difficulty rating: 2650

Solution:

Folding reflects $B$ to $B'$ across the crease, so $B'E = BE = 17.$ In right triangle $AEB',$ $AB' = \sqrt{17^2 - 8^2} = 15,$ and $AB = AE + EB = 25.$ Place $A = (0, 0),$ $B = (25, 0),$ $B' = (0, 15).$

Points on the crease are equidistant from $B$ and $B',$ so $\overline{EF}$ is perpendicular to $\overline{BB'}.$ Since $BB'$ has slope $-\frac{3}{5},$ the crease through $E = (8, 0)$ has slope $\frac{5}{3},$ and it meets the line $CD$ (at height $h = BC$ ) at $x = 8 + \frac{3h}{5}.$ The condition $CF = 3$ gives $25 - \left(8 + \frac{3h}{5}\right) = 3,$ so $h = \frac{70}{3}.$

The perimeter is $2\left(25 + \frac{70}{3}\right) = \frac{290}{3},$ so $m + n = 290 + 3 = 293.$

2004 AIME II Problem 7

Answer: 293

Difficulty rating: 2650

Solution:

Problem 7 in Other Years