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

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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 $250 + 50 + 7 = 307$ .

2005 AIME II (Problems • Answer Key • Resources)
Preceded by Problem 11	Followed by Problem 13
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2005 AIME II Problems/Problem 12

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