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

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.

2012 AMC 12B Problem 21

Answer: A

Difficulty rating: 2170

Solution:

Problem 21 in Other Years