AoPSWiki

Looking for a challenging geometry text? Preparing for MATHCOUNTS or the AMC exams? Check out Art of Problem Solving's Introduction to Geometry by Richard Rusczyk.

Personal tools

Search

Toolbox

2005 AIME II Problems/Problem 12

From AoPSWiki

Problem

Square $ABCD$ has center $O,\ AB=900,\ E$ and $F$ are on $AB$ with $AE<BF$ and $E$ between $A$ and $F, m\angle EOF =45^\circ,$ and $EF=400.$ Given that $BF=p+q\sqrt{r},$ where $p,q,$ and $r$ are positive integers and $r$ is not divisible by the square of any prime, find $p+q+r.$

Contents

1 Problem
2 Solution
- 2.1 Solution 1
- 2.2 Solution 2
3 See also

Solution

Solution 1

defaultpen(linewidth(0.7)+fontsize(10)); pair A=(0,9), B=(9,9), C=(9,0), D=(0,0), E=(2.5-0.5*sqrt(7),9), F=(6.5-0.5*sqrt(7),9...

Let $G$ be the foot of the perpendicular from $O$ to $AB$ . Denote $x = EG$ and $y = FG$ , and $x > y$ (since $AE < BF$ and $AG = BG$ ). Then $\tan \angle EOG = \frac{x}{450}$ , and $\tan \angle FOG = \frac{y}{450}$ .

By the tangent addition rule $\left( \tan (a + b) = \frac{\tan a + \tan b}{1 - \tan a \tan b} \right)$ , we see that $\tan 45 = \tan (EOG + FOG) = \frac{\frac{x}{450} + \frac{y}{450}}{1 - \frac{x}{450} \cdot \frac{y}{450}}.$ Since $\tan 45 = 1$ , this simplifies to $1 - \frac{xy}{450^2} = \frac{x + y}{450}$ . We know that $x + y = 400$ , so we can substitute this to find that $1 - \frac{xy}{450^2} = \frac 89 \Longrightarrow xy = 150^2$ .

Substituting $x = 400 - y$ again, we know have $xy = (400 - y)y = 150^2$ . This is a quadratic with roots $200 \pm 50\sqrt{7}$ . Since $y < x$ , use the smaller root, $200 - 50\sqrt{7}$ .

Now, $BF = BG - FG = 450 - (200 - 50\sqrt{7}) = 250 + 50\sqrt{7}$ . The answer is $250 + 50 + 7 = \boxed{307}$ .

Solution 2

defaultpen(linewidth(0.7)+fontsize(10)); pair A=(0,9), B=(9,9), C=(9,0), D=(0,0), E=(2.5-0.5*sqrt(7),9), F=(6.5-0.5*sqrt(7),9...

Label $BF=x$ , so $EA =$ $500 - x$ . Rotate $\triangle{OEF}$ about $O$ until $EF$ lies on $BC$ . Now we know that $\angle{EOF}=45^\circ$ therefore $\angle BOF+\angle AOE=45^\circ$ also since $O$ is the center of the square. Label the new triangle that we created $\triangle OGJ$ . Now we know that rotation preserves angles and side lengths, so $BG=500-x$ and $JC=x$ . Draw $GF$ and $OB$ . Notice that $\angle BOG =\angle OAE$ since rotations preserve the same angles so $\angle{FOG}=45^\circ$ too and by SAS we know that $\triangle FOE\cong \triangle FOG$ so $FG=400$ . Now we have a right $\triangle BFG$ with legs $x$ and $500-x$ and hypotenuse $400$ . Then by the Pythagorean Theorem,

$\begin{align*}(500-x)^2+x^2&=400^2 \\250000-1000x+2x^2&=16000 \\90000-1000x+2x^2&=0 \end{align*}$

and applying the quadratic formula we get that $x=250\pm 50\sqrt{7}$ . Since $BF > AE$ we take the positive sign because and so our answer is $p+q+r = 250 + 50 + 7 = 307$ .

See also

2005 AIME II (Problems • Resources)
Preceded by Problem 11	Followed by Problem 13
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15

Retrieved from "http://www.artofproblemsolving.com/Wiki/index.php/2005_AIME_II_Problems/Problem_12"

Category: Intermediate Geometry Problems

Looking for a challenging geometry text? Preparing for MATHCOUNTS or the AMC exams? Check out Art of Problem Solving's Introduction to Geometry by Richard Rusczyk.

© Copyright 2008 AoPS Incorporated. All Rights Reserved. • Foundation • Privacy • Contact Us