2018 AMC 12B Problem 20

Question

Let

ABCDEF

be a regular hexagon with side length

1.

Denote by

X,

Y,

and

Z

the midpoints of sides

AB,

CD,

and

EF,

respectively. What is the area of the convex hexagon whose interior is the intersection of the interiors of

\triangle ACE

and

\triangle XYZ?

\dfrac{3}{8}\sqrt{3}

\dfrac{7}{16}\sqrt{3}

\dfrac{15}{32}\sqrt{3}

\dfrac{1}{2}\sqrt{3}

\dfrac{9}{16}\sqrt{3}

Put the regular hexagon on the unit circle and write coordinates for

A,C,E

and the three midpoints

Intersect the side lines of the two equilateral triangles to find the six vertices, then use shoelace

Place the regular hexagon on the unit circle with

A=(1,0),

C=(-\tfrac12,\tfrac{\sqrt3}{2}),

and

E=(-\tfrac12,-\tfrac{\sqrt3}{2}).

The three specified midpoints are

X=(\tfrac34,\tfrac{\sqrt3}{4}),

Y=(-\tfrac34,\tfrac{\sqrt3}{4}),

and

Z=(0,-\tfrac{\sqrt3}{2}).

Intersecting the side lines of

\triangle ACE

and

\triangle XYZ

gives the six vertices of their common interior, in cyclic order:

\begin{gathered} (-\tfrac12,0),\quad (-\tfrac18,-\tfrac{3\sqrt3}{8}),\\ (\tfrac14,-\tfrac{\sqrt3}{4}),\quad (\tfrac58,\tfrac{\sqrt3}{8}),\\ (\tfrac14,\tfrac{\sqrt3}{4}),\quad (-\tfrac12,\tfrac{\sqrt3}{4}). \end{gathered}

The shoelace formula applied to these vertices gives area

\dfrac{15\sqrt3}{32}.

Thus, the correct answer is C.

Problem 20 in other years