# Difference between revisions of "2011 AIME I Problems/Problem 2"

## Problem

In rectangle $ABCD$, $AB=12$ and $BC=10$. Points $E$ and $F$ lie inside rectangle $ABCD$ so that $BE=9$,$DF=8$,$\overline{BE}||\overline{DF}$,$\overline{EF}||\overline{AB}$, 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 $m+n+p$.

## Solution 1

Let us call the point where $\overline{EF}$ intersects $\overline{AD}$ point $G$, and the point where $\overline{EF}$ intersects $\overline{BC}$ point $H$. Since angles $FHB$ and $EGA$ are both right angles, and angles $BEF$ and $DFE$ are congruent due to parallelism, right triangles $BHE$ and $DGF$ are similar. This implies that $\frac{BH}{GD} = \frac{9}{8}$. Since $BC=10$, $BH+GD=BH+HC=BC=10$. ($HC$ is the same as $GD$ because they are opposite sides of a rectangle.) Now, we have a system:

$\frac{BH}{GD}=\frac{9}8$

$BH+GD=10$

Solving this system (easiest by substitution), we get that:

$BH=\frac{90}{17}$

$GD=\frac{80}{17}$

Using the Pythagorean Theorem, we can solve for the remaining sides of the two right triangles:

$\sqrt{9^2-(\frac{90}{17})^2}$ and $\sqrt{8^2-(\frac{80}{17})^2}$

Notice that adding these two sides would give us twelve plus the overlap $EF$. This means that:

$EF= \sqrt{9^2-(\frac{90}{17})^2}+\sqrt{8^2-(\frac{80}{17})^2}-12=3\sqrt{21}-12$

Since $21$ isn't divisible by any perfect square, our answer is:

$3+21+12=\boxed{36}$

## Solution 2

Extend lines $BE$ and $CD$ to meet at point $G$. Draw the altitude $GH$ from point $G$ to line $BA$ extended.

$GE=CF=8$ $GB=17$

In right $\bigtriangleup GHB$, $GH=10$, $GB=17$, thus by Pythagoras Theorem we have: $HB=\sqrt{17^2-10^2}=3\sqrt{21}$

$HA=EF=3\sqrt{21}-12$

Thus our answer is: $3+21+12=\boxed{36}$

## See also

 2011 AIME I (Problems • Answer Key • Resources) Preceded byProblem 1 Followed byProblem 3 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 All AIME Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.

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