Question

In the figure, $ABCD$ is a square of side length $1.$ The rectangles $JKHG$ and $EBCF$ are congruent. What is $BE?$

$\dfrac12(\sqrt{6} - 2)$

$\dfrac14$

$2 - \sqrt{3}$

$\dfrac{\sqrt{3}}{6}$

$1 - \dfrac{\sqrt{2}}{2}$

Solution:

Let $x = BE = GH = CF$ and $\theta = \angle DHG = \angle AGJ = \angle FKH,$ with $AD = GJ = HK = 1.$ In right triangle $GDH,$ $x \sin\theta = DG = 1 - \cos\theta,$ so $x = \dfrac{1 - \cos\theta}{\sin\theta}.$

Along side $CD,$ $1 = CF + FH + HD = x + \sin\theta + x\cos\theta.$ Substituting for $x$ gives $1 = \dfrac{(1-\cos\theta)(1+\cos\theta)}{\sin\theta} + \sin\theta = \dfrac{\sin^2\theta}{\sin\theta} + \sin\theta = 2\sin\theta.$

Hence $\sin\theta = \tfrac12,$ so $\theta = 30^\circ$ and $x = \dfrac{1 - \frac{\sqrt3}{2}}{\frac12} = 2 - \sqrt3.$

Thus, the correct answer is C.

2014 AMC 12B Problem 21

Solution:

Problem 21 in Other Years