Question

In rectangle $ABCD,$ $AB = 12$ and $BC = 10.$ Points $E$ and $F$ lie inside rectangle $ABCD$ so that $BE = 9,$ $DF = 8,$ $\overline{BE} \parallel \overline{DF},$ $\overline{EF} \parallel \overline{AB},$ and line $BE$ intersects segment $\overline{AD}.$ The length $EF$ can be expressed in the form $m\sqrt{n} - p,$ where $m,$ $n,$ and $p$ are positive integers and $n$ is not divisible by the square of any prime. Find $m + n + p.$

Answer

Answer: 36

Concepts:coordinate geometry vector Pythagorean Theorem

Difficulty rating: 2390

Solution:

Place $D = (0, 0),$ $C = (12, 0),$ $B = (12, 10),$ $A = (0, 10).$ Since $\overline{BE} \parallel \overline{DF},$ there is a unit vector $(u, v)$ with $u, v \gt 0$ such that $E = B - 9(u, v)$ and $F = D + 8(u, v):$ line $BE$ heads down and to the left so that it can cross $\overline{AD},$ while $\overline{DF}$ points up and to the right into the rectangle.

Because $\overline{EF} \parallel \overline{AB}$ is horizontal, $E$ and $F$ have equal heights: $10 - 9v = 8v,$ so $v = \frac{10}{17}$ and $u = \sqrt{1 - v^2} = \frac{3\sqrt{21}}{17}.$

Then $E$ and $F$ have $x$ -coordinates $12 - 9u$ and $8u,$ so $EF = |12 - 17u|$ $= |12 - 3\sqrt{21}|$ $= 3\sqrt{21} - 12,$ since $3\sqrt{21} \gt 12.$ Thus $m + n + p = 3 + 21 + 12 = 36.$

2011 AIME I Problem 2

Answer: 36

Difficulty rating: 2390

Solution:

Problem 2 in Other Years