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

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== Solution 2 ==
 
== Solution 2 ==
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Label <math>BF=x</math>, so <math>EA=500-x</math>. Rotate <math>\triangle{OEF}</math> about O until EF lies on BC. Now we know that <math>\angle{EOF}=45^\circ</math> therefore <math>\angle{BOE}+\angle{AOE}=45^\circ</math> also since <math>O</math> is the center of the square. Label the new triangle that we created <math>\triangle{OGJ}</math>. Now we know that rotation preserves angles and side lengths, so <math>BG=500-x</math> and <math>JC=x</math>. Draw <math>GF</math> and <math>OB</math>. Notice that <math>\angle{BOG} =\angle{OAE}</math> since rotations preserve the same angles so  
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<math>\angle{FOG}=45^\circ</math> too and by SAS we know that <math>\triangle{FOE}\isom \triangle{FOG}</math> so <math>FG=400</math>. Now we have a right <math>\triangle{BFG}</math> with legs <math>x</math> and <math>500-x</math> and hypotenuse 400. Then by the [[Pythagorean Theorem]]],  
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Since we are given that EF=400 label BF=x and EA=500-x. Rotate <math>\triangle{OEF}</math> about O until EF lies on BC. Now we know that <math>\angle{EOF}=45</math> therefore <math>\angle{BOE}</math>+<math>\angle{AOE}=45</math> also since <math>O</math> is the center of the square. Label the new triangle that we created <math>\triangle{OGJ}</math>.Now we know that rotation preserves angles and side lenghts so <math>BG=500-x</math> and <math>JC=x</math>. Now draw <math>GF</math> and <math>OB</math>. Notice that <math>\angle{BOG}</math>=<math>\angle{OAE}</math> since rotations preserve the same angles so  
 
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<math>\angle{FOG}</math>=<math>45</math>  
 
−
 
 
−
too and by SAS we know that <math>\triangle{FOE}=\triangle{FOG}</math> so FG=400. Now we have a right <math>\triangle{BFG}</math> with legs x and 500-x and hypotenuse 400. Now by [http://www.artofproblemsolving.com/Wiki/index.php/Pythagorean_theorem Pythagorean Theorem],  
 
  
 
<math>\displaystyle(500-x)^2+x^2=400^2</math>
 
<math>\displaystyle(500-x)^2+x^2=400^2</math>
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<math>\displaystyle 90000-1000x+2x^2=0</math>
 
<math>\displaystyle 90000-1000x+2x^2=0</math>
  
−
applying the quadratic formula we get that  
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and applying the [[quadratic formula]] we get that  
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<math>x</math>=<math>250+/-50\sqrt{7}</math> and since we want p+q+r we get that is 307.
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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>p+q+r = 250 + 50 + 7 = 307</math>.
  
 
== See also ==
 
== See also ==

Revision as of 19:21, 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

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

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 $\triangle{FOE}\isom \triangle{FOG}$ (Error compiling LaTeX. Unknown error_msg) so $FG=400$. 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?) and so our answer is $p+q+r = 250 + 50 + 7 = 307$.

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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