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Difference between revisions of "2005 AIME II Problems/Problem 12"
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== Problem ==
 
== Problem ==
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[[Square]] <math>\displaystyle ABCD </math> has [[center]] <math> O,\ AB=900,\ E </math> and <math> F </math> are on <math> AB </math> with <math> AE<BF </math> and <math> E </math> between <math> A </math> and <math> F, m\angle EOF =45^\circ, </math> and <math> EF=400. </math> Given that <math> BF=p+q\sqrt{r}, </math> where <math> p,q, </math> and <math> r </math> are [[positive]] [[integer]]s and <math> r </math> is not divisible by the [[square]] of any [[prime]], find <math> p+q+r. </math>
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Square <math> ABCD </math> has center <math> O, AB=900, E </math> and <math> F </math> are on <math> AB </math> with <math> AE<BF </math> and <math> E </math> between <math> A </math> and <math> F, m\angle EOF =45^\circ, </math> and <math> EF=400. </math> Given that <math> BF=p+q\sqrt{r}, </math> where <math> p,q, </math> and <math> r </math> are positive integers and <math> r </math> is not divisible by the square of any prime, find <math> p+q+r. </math>
 
  
 
== Solution ==
 
== Solution ==
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{{image}}
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Draw the perpendicular from <math>AB \perp OP</math>, with the intersection at <math>G</math>. Denote <math>x = EG</math> and <math>y = FG</math>, and <math>x > y</math> (since <math>AE < BF</math> and <math>AG = BG</math>). The tangent of <math>\displaystyle \angle EOG = \frac{x}{450}</math>, and of <math>\tan \angle FOG = \frac{y}{450}</math>.
  
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{{solution}}
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By the [[trigonometric identity|tangent addition rule]] <math>\left( \tan (a + b) = \frac{\tan a + \tan b}{1 - \tan a \tan b} \right)</math>, we see that <math>\tan 45 = \tan (EOG + FOG) = \frac{\frac{x}{450} + \frac{y}{450}}{1 - \frac{x}{450} \frac{y}{450}}</math>. Since <math>\displaystyle \tan 45 = 1</math>, <math>1 - \frac{xy}{450^2} = \frac{x + y}{450}</math>. We know that <math>x + y = 400 \displaystyle</math>, so we can substitute this to find that <math>1 - \frac{xy}{450^2} = \frac 89 \Longrightarrow xy = 150^2</math>.
  
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== See also ==
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A second equation can be set up using <math>x + y = 400 \displaystyle</math>. To solve for <math>y</math>, <math>x = 400 - y \Longrightarrow (400 - y)y = 150^2</math>. This is a quadratic with roots <math>200 \pm 50\sqrt{7}</math>. Since <math>y < x</math>, use the smaller root, <math>200 - 50\sqrt{7}</math>.
  
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* [[2005 AIME II Problems/Problem 11| Previous problem]]
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Now, <math>BF = BG - FG = 450 - (200 - 50\sqrt{7}) = 250 + 50\sqrt{7}</math>. The answer is <math>250 + 50 + 7 = 307</math>.
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* [[2005 AIME II Problems/Problem 13| Next problem]]
 
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* [[2005 AIME II Problems]]
 
  
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== See also ==
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{{AIME box|year=2005|n=II|num-b=11|num-a=13}}
  
 
[[Category:Intermediate Geometry Problems]]
 
[[Category:Intermediate Geometry Problems]]

Revision as of 17:20, 22 March 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 $250 + 50 + 7 = 307$.

See also

2005 AIME II (Problems • Answer Key • Resources)
Preceded by
Problem 11 		Followed by
Problem 13
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All AIME Problems and Solutions
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