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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]] [[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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__TOC__
 
== Solution ==
 
== Solution ==
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=== Solution 1 ===
 
<center>[[Image:AIME_2005II_Solution_12_1.png]]</center>
 
<center>[[Image:AIME_2005II_Solution_12_1.png]]</center>
 
Let <math>G</math> be a point on <math>AB</math> such that <math>AB\perp OG</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>\angle EOG = \frac{x}{450}</math>, and of <math>\tan \angle FOG = \frac{y}{450}</math>.
 
Let <math>G</math> be a point on <math>AB</math> such that <math>AB\perp OG</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>\angle EOG = \frac{x}{450}</math>, and of <math>\tan \angle FOG = \frac{y}{450}</math>.
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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>.
 
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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== Solution 2 ==
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=== Solution 2 ===
 
Label <math>BF=x</math>, so <math>EA =</math> <math>500 - x</math>. Rotate <math>\triangle{OEF}</math> about <math>O</math> until <math>EF</math> lies on <math>BC</math>. 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  
 
Label <math>BF=x</math>, so <math>EA =</math> <math>500 - x</math>. Rotate <math>\triangle{OEF}</math> about <math>O</math> until <math>EF</math> lies on <math>BC</math>. 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\cong \triangle FOG</math> so <math>\displaystyle 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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<math>\angle{FOG}=45^\circ</math> too and by SAS we know that <math>\triangle FOE\cong \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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<math>\displaystyle(500-x)^2+x^2=400^2</math>
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<math>(500-x)^2+x^2=400^2</math>
  
 
<math> 250000-1000x+2x^2=16000</math>
 
<math> 250000-1000x+2x^2=16000</math>
  
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<math>90000-1000x+2x^2=0</math>
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<math>\displaystyle 90000-1000x+2x^2=0</math>
 
  
 
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>.  Since <math>BF > AE</math> we take the positive sign because 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>.  Since <math>BF > AE</math> we take the positive sign because and so our answer is <math>p+q+r = 250 + 50 + 7 = 307</math>.
  
 
== See also ==
 
== See also ==

Revision as of 09:44, 10 November 2007

Problem

Square $ABCD$ 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

Solution 1

AIME 2005II Solution 12 1.png

Let $G$ be a point on $AB$ such that $AB\perp OG$. Denote $x = EG$ and $y = FG$, and $x > y$ (since $AE < BF$ and $AG = BG$). The tangent of $\angle EOG = \frac{x}{450}$, 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 $\tan 45 = 1$, $1 - \frac{xy}{450^2} = \frac{x + y}{450}$. We know that $x + y = 400$, 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 $x + y = 400$. 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 $FG=400$. Now we have a right $\triangle BFG$ with legs $x$ and $500-x$ and hypotenuse 400. Then by the Pythagorean Theorem,

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

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

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

and applying the quadratic formula we get that $x=250\pm 50\sqrt{7}$. Since $BF > AE$ we take the positive sign because 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
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All AIME Problems and Solutions
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