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Difference between revisions of "Mock AIME 1 Pre 2005 Problems/Problem 12"

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== Solution ==
 
== Solution ==
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Let <math>\angle{AB'C'} = \theta</math>. By some angle chasing in <math>\triangle{AB'E}</math>, we find that <math>\angle{EAB'} = 90^{\circ} - 2 \theta</math>. Before we apply the law of sines, we're going to want to get everything in terms of <math>\sin \theta</math>, so note that <math>\sin \angle{EAB'} = \sin(90^{\circ} - 2 \theta) = \cos 2 \theta = 1 - 2 \sin^2 \theta</math>. Now, we use law of sines, which gives us the following:
  
 +
<math>\frac{\sin \theta}{5}=\frac{1 - 2 \sin^2 \theta}{23} \implies \sin \theta = \frac{-23 \pm 27}{20}</math>,
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but since <math>\theta < 180^{\circ}</math>, we go with the positive solution. Thus, <math>\sin \theta = \frac15</math>.
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Denote the intersection of <math>B'C'</math> and <math>AE</math> with <math>G</math>. By another application of the law of sines, <math>B'G = \frac{23}{\sqrt{24}}</math> and <math>AE = 10\sqrt{6}</math>. Since <math>\sin \theta = \frac15, GE = \frac{115}{\sqrt{24}}</math>, and <math>AG = AE - GE = 10\sqrt{6} - \frac{115}{\sqrt{24}} = \frac{5}{\sqrt{24}}</math>. Note that <math>\triangle{EB'G} \sim \triangle{C'AG}</math>, so <math>\frac{EG}{B'G}=\frac{C'G}{AG} \implies C'G = \frac{25}{\sqrt{24}}</math>.
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 +
Now we have that <math>AB = AE + EB = 10\sqrt{6} + 23</math>, and <math>B'C' = BC = B'G + C'G = \frac{23}{\sqrt{24}} + \frac{25}{\sqrt{24}} = \frac{48}{\sqrt{24}}=4\sqrt{6}</math>. Thus, the area of <math>ABCD</math> is <math>(10\sqrt{6} + 23)(4\sqrt{6}) = 92\sqrt{6} + 240</math>, and our final answer is <math>92 + 6 + 240 = \boxed{338}</math>.
  
 
== See also ==
 
== See also ==

Latest revision as of 21:41, 26 December 2016

Problem

$ABCD$ is a rectangular sheet of paper. $E$ and $F$ are points on $AB$ and $CD$ respectively such that $BE < CF$. If $BCFE$ is folded over $EF$, $C$ maps to $C'$ on $AD$ and $B$ maps to $B'$ such that $\angle{AB'C'} \cong \angle{B'EA}$. If $AB' = 5$ and $BE = 23$, then the area of $ABCD$ can be expressed as $a + b\sqrt{c}$ square units, where $a, b,$ and $c$ are integers and $c$ is not divisible by the square of any prime. Compute $a + b + c$.

Solution

Let $\angle{AB'C'} = \theta$. By some angle chasing in $\triangle{AB'E}$, we find that $\angle{EAB'} = 90^{\circ} - 2 \theta$. Before we apply the law of sines, we're going to want to get everything in terms of $\sin \theta$, so note that $\sin \angle{EAB'} = \sin(90^{\circ} - 2 \theta) = \cos 2 \theta = 1 - 2 \sin^2 \theta$. Now, we use law of sines, which gives us the following:

$\frac{\sin \theta}{5}=\frac{1 - 2 \sin^2 \theta}{23} \implies \sin \theta = \frac{-23 \pm 27}{20}$, but since $\theta < 180^{\circ}$, we go with the positive solution. Thus, $\sin \theta = \frac15$.

Denote the intersection of $B'C'$ and $AE$ with $G$. By another application of the law of sines, $B'G = \frac{23}{\sqrt{24}}$ and $AE = 10\sqrt{6}$. Since $\sin \theta = \frac15, GE = \frac{115}{\sqrt{24}}$, and $AG = AE - GE = 10\sqrt{6} - \frac{115}{\sqrt{24}} = \frac{5}{\sqrt{24}}$. Note that $\triangle{EB'G} \sim \triangle{C'AG}$, so $\frac{EG}{B'G}=\frac{C'G}{AG} \implies C'G = \frac{25}{\sqrt{24}}$.

Now we have that $AB = AE + EB = 10\sqrt{6} + 23$, and $B'C' = BC = B'G + C'G = \frac{23}{\sqrt{24}} + \frac{25}{\sqrt{24}} = \frac{48}{\sqrt{24}}=4\sqrt{6}$. Thus, the area of $ABCD$ is $(10\sqrt{6} + 23)(4\sqrt{6}) = 92\sqrt{6} + 240$, and our final answer is $92 + 6 + 240 = \boxed{338}$.

See also

Mock AIME 1 Pre 2005 (Problems, Source)
Preceded by
Problem 11 		Followed by
Problem 13
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