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Difference between revisions of "2014 AMC 12B Problems/Problem 21"
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==Problem==
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==Problem 21==
  
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In the figure, <math>ABCD</math> is a square of side length 1. The rectangles <math>JKHG</math> and <math>EBCF</math> are congruent. What is <math>BE</math> ?
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In the figure, <math> ABCD </math> is a square of side length <math> 1 </math>. The rectangles <math> JKHG </math> and <math> EBCF </math> are congruent. What is <math> BE </math>?
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<asy>
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pair A=(1,0), B=(0,0), C=(0,1), D=(1,1), E=(2-sqrt(3),0), F=(2-sqrt(3),1), G=(1,sqrt(3)/2), H=(2.5-sqrt(3),1), J=(.5,0), K=(2-sqrt(3),1-sqrt(3)/2);
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draw(A--B--C--D--cycle);
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draw(K--H--G--J--cycle);
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draw(F--E);
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label("$A$",A,SE); label("$B$",B,SW); label("$C$",C,NW); label("$D$",D,NE); label("$E$",E,S); label("$F$",F,N);
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label("$G$",G,E); label("$H$",H,N); label("$J$",J,S); label("$K$",K,W);
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</asy>
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<math> \textbf{(A) }\frac{1}{2}(\sqrt{6}-2)\qquad\textbf{(B) }\frac{1}{4}\qquad\textbf{(C) }2-\sqrt{3}\qquad\textbf{(D) }\frac{\sqrt{3}}{6}\qquad\textbf{(E) } 1-\frac{\sqrt{2}}{2}</math>
  
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-Insert Diagram because I don't know how-
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[[2014 AMC 12B Problems/Problem 21|Solution]]
  
 
==Solution==
 
==Solution==

Revision as of 22:36, 20 February 2014

Problem 21

In the figure, $ABCD$ is a square of side length $1$. The rectangles $JKHG$ and $EBCF$ are congruent. What is $BE$? [asy] pair A=(1,0), B=(0,0), C=(0,1), D=(1,1), E=(2-sqrt(3),0), F=(2-sqrt(3),1), G=(1,sqrt(3)/2), H=(2.5-sqrt(3),1), J=(.5,0), K=(2-sqrt(3),1-sqrt(3)/2); draw(A--B--C--D--cycle); draw(K--H--G--J--cycle); draw(F--E); label("$A$",A,SE); label("$B$",B,SW); label("$C$",C,NW); label("$D$",D,NE); label("$E$",E,S); label("$F$",F,N); label("$G$",G,E); label("$H$",H,N); label("$J$",J,S); label("$K$",K,W); [/asy] $\textbf{(A) }\frac{1}{2}(\sqrt{6}-2)\qquad\textbf{(B) }\frac{1}{4}\qquad\textbf{(C) }2-\sqrt{3}\qquad\textbf{(D) }\frac{\sqrt{3}}{6}\qquad\textbf{(E) } 1-\frac{\sqrt{2}}{2}$


Solution

Solution

Let $BE = x$. Let $JA = y$. Because $\angle FKH = \angle EJK = \angle AGJ = \angle DHG$ and $\angle FHK = \angle EKJ = \angle AJG = \angle DGH$, $\triangle KEJ, \triangle JAG, \triangle GDH, \triangle HFK$ are all similar. Using proportions and the pythagorean theorem, we find \[EK = xy\] \[FK = \sqrt{1-y^2}\] \[EJ = x\sqrt{1-y^2}\] Because we know that $BE+EJ+AJ = EK + FK = 1$, we can set up a systems of equations \[x + x\sqrt{1-y^2} + y = 1\] \[xy + \sqrt{1-y^2} = 1\] Solving for $x$ in the second equation, we get \[x= \frac{1-\sqrt{1-y^2}}{y}\] Plugging this into the first equation, we get \[\frac{1-\sqrt{1-y^2}}{y} + (\sqrt{1-y^2})\frac{1-\sqrt{1-y^2}}{y} + y = 1 \implies \frac{2y^2}{y}=1 \implies y=\frac{1}{2}\] Plugging into the previous equation with $x$, we get \[x= 2\left(1-\sqrt{1-\frac{1}{4}}\right) = 2\left(\frac{2-\sqrt{3}}{2} \right) = \boxed{\textbf{(C)}\ 2-\sqrt{3}}\]

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