2012 AMC 12B Problem 21

Question

Square

AXYZ

is inscribed in equiangular hexagon

ABCDEF

with

X

on

\overline{BC},

Y

on

\overline{DE},

and

Z

on

\overline{EF}.

Suppose that

AB = 40

and

EF = 41(\sqrt{3} - 1).

What is the side-length of the square?

29\sqrt{3}

\dfrac{21}{2}\sqrt{2} + \dfrac{41}{2}\sqrt{3}

20\sqrt{3} + 16

20\sqrt{2} + 13\sqrt{3}

21\sqrt{6}

Answer: A

Concepts:equiangular polygon congruence (geometry)Pythagorean Theorem

Difficulty rating: 2170

Small hint:

Drop the square’s vertex

A

onto a line through

A

perpendicular to the parallel sides

EF

and

CB

Big hint:

Let

u=BX;

the equiangular

60^\circ

angles give congruent corner triangles, so express

EF

in terms of

u

and solve for the side

s^2=(20+u)^2+(20\sqrt3)^2

Solution:

Extend

EF

and

CB

to a line through

A

perpendicular to both, meeting them at

H

and

J.

Since

\angle ABJ=60^\circ,

we have

BJ=20

and

AJ=20\sqrt3.

With

u=BX,

the Pythagorean theorem gives

s^2=(20+u)^2+(20\sqrt3)^2.

The equiangular angles make the four corner triangles congruent, and chasing the equal segments along

EF

yields

\begin{aligned} u+20\sqrt3 &= 41(\sqrt3-1) \\ &\quad {}+\frac{20+u}{\sqrt3}, \end{aligned}

so

u=21\sqrt3-20.

Since

20+u=21\sqrt3,

we get

\begin{aligned} s^2 &= (21\sqrt3)^2+(20\sqrt3)^2 \\ &= 3(441+400)=3\cdot29^2, \end{aligned}

giving

s=29\sqrt3.

Thus, the correct answer is A.