2014 AMC 12B Problem 21

Attempt Problem 21 of the 2014 AMC 12B below, then check your answer against the professionally curated solution from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2014 AMC 12B solutions, or check the answer key.

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21.

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}

Answer: C

Concepts:right triangle trigonometry angle chasing

Difficulty rating: 2350

Small hint:

Let

\theta = \angle DHG;

in right triangle

GDH,

BE = \dfrac{1 - \cos\theta}{\sin\theta}

Big hint:

Along the top side,

CD = CF + FH + HD = 1

simplifies to

2\sin\theta = 1

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,

\begin{gathered} 1 = CF + FH + HD \\ = x + \sin\theta + x\cos\theta. \end{gathered}

Substituting for

x

gives

\begin{gathered} 1 = \dfrac{(1-\cos\theta)(1+\cos\theta)}{\sin\theta} \\ {}+ \sin\theta \\ = \dfrac{\sin^2\theta}{\sin\theta} \\ {}+ \sin\theta \\ = 2\sin\theta. \end{gathered}

Hence

\sin\theta = \tfrac12,

so

\theta = 30^\circ

and

x = \dfrac{1 - \frac{\sqrt3}{2}}{\frac12} = 2 - \sqrt3.

Thus, the correct answer is C.

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