2005 AIME II Problems/Problem 12

Problem

Square $\displaystyle ABCD$ 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 $p+q+r.$

Solution

An image is supposed to go here. You can help us out by creating one and editing it in. Thanks.

Draw the perpendicular from $AB \perp OP$ , with the intersection at $G$ . Denote $x = EG$ and $y = FG$ , and $x > y$ (since $AE < BF$ and $AG = BG$ ). The tangent of $\displaystyle \angle EOG = \frac{x}{450}$ , and of $\tan \angle FOG = \frac{y}{450}$ .

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

A second equation can be set up using $x + y = 400 \displaystyle$ . To solve for $y$ , $x = 400 - y \Longrightarrow (400 - y)y = 150^2$ . 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 $250 + 50 + 7 = 307$ .

Solution 2

Since we are given that EF=400 label BF=x and EA=500-x. Rotate $\triangle{OEF}$ about O until EF lies on BC. Now we know that $\angle{EOF}=45$ therefore $\angle{BOE}$ + $\angle{AOE}=45$ 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 lenghts so $BG=500-x$ and $JC=x$ . Now draw $GF$ and $OB$ . Notice that $\angle{BOG}$ = $\angle{OAE}$ since rotations preserve the same angles so $\angle{FOG}$ = $45$

too and by SAS we know that $\triangle{FOE}=\triangle{FOG}$ so FG=400. Now we have a right $\triangle{BFG}$ with legs x and 500-x and hypotenuse 400. Now by Pythagorean Theorem,

$(500-x)^2+x^2=400^2$

$250000-1000x+2x^2=16000$

$90000-1000x+2x^2=0$

applying the quadratic formula we get that $x$ = $250+/-50\sqrt{7}$ and since we want p+q+r we get that is 307.

--Sonny 19:00, 26 July 2007 (EDT)

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

Page

Toolbox

Search

2005 AIME II Problems/Problem 12

Contents

Problem

Solution

Solution 2

See also