# Problem #2320

 2320 In the figure, $ABCD$ is a square of side length $1$. The rectangles $JKHG$ and $EBCF$ are congruent. What is $BE$? $[asy] pair A=(1,0), B=(0,0), C=(0,1), D=(1,1), E=(2-sqrt(3),0), F=(2-sqrt(3),1), G=(1,sqrt(3)/2), H=(2.5-sqrt(3),1), J=(.5,0), K=(2-sqrt(3),1-sqrt(3)/2); draw(A--B--C--D--cycle); draw(K--H--G--J--cycle); draw(F--E); label("A",A,SE); label("B",B,SW); label("C",C,NW); label("D",D,NE); label("E",E,S); label("F",F,N); label("G",G,E); label("H",H,N); label("J",J,S); label("K",K,W); [/asy]$ $\textbf{(A) }\frac{1}{2}(\sqrt{6}-2)\qquad\textbf{(B) }\frac{1}{4}\qquad\textbf{(C) }2-\sqrt{3}\qquad\textbf{(D) }\frac{\sqrt{3}}{6}\qquad\textbf{(E) } 1-\frac{\sqrt{2}}{2}$ This problem is copyrighted by the American Mathematics Competitions.
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