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Difference between revisions of "2005 AIME II Problems/Problem 12"

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and applying the [[quadratic formula]] we get that  
 
and applying the [[quadratic formula]] we get that  
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<math>x=250\pm 50\sqrt{7}</math>.  We take the positive sign because (WHY?) and so our answer is <math>\displaystyle p+q+r = 250 + 50 + 7 = 307</math>.
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<math>x=250\pm 50\sqrt{7}</math>.  We take the positive sign because (WHY?-BECAUSE WE ARE GIVEN THAT BF<AE THUS ONE WILL BE POSITIVE THE OTHER NEGATIVE) and so our answer is <math>\displaystyle p+q+r = 250 + 50 + 7 = 307</math>.
  
 
== See also ==
 
== See also ==

Revision as of 19:30, 26 July 2007

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 $250 + 50 + 7 = 307$.

Solution 2

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 $\angle BOE+\angle AOE=45^\circ$ 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 $\displaystyle FG=400$. Now we have a right $\triangle BFG$ with legs $x$ and $500-x$ and hypotenuse 400. Then by the Pythagorean Theorem],

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

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


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

and applying the quadratic formula we get that $x=250\pm 50\sqrt{7}$. We take the positive sign because (WHY?-BECAUSE WE ARE GIVEN THAT BF<AE THUS ONE WILL BE POSITIVE THE OTHER NEGATIVE) and so our answer is $\displaystyle p+q+r = 250 + 50 + 7 = 307$.

See also

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
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