2018 AMC 10B Problem 24

Problem 24 of 25HarderGeometry

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?

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Solution

The triangle

XYZ

is equilateral with side

\frac{3}{2},

so its area is

\dfrac{\sqrt3}{4}\left(\dfrac32\right)^2=\dfrac{9\sqrt3}{16}.

The triangles

ACE

and

XYZ

are concentric and rotated

30^\circ

from each other. At each vertex of

XYZ,

the sides of

ACE

cut off a

30

60

90

triangle whose hypotenuse is the half-side segment

AX=\frac{1}{2}.

Its legs are

\frac{1}{4}

and

\frac{\sqrt3}{4},

so each corner has area

\frac{\sqrt3}{32}.

Removing the three corners gives

\dfrac{9\sqrt3}{16}-3\cdot\dfrac{\sqrt3}{32}=\dfrac{15\sqrt3}{32}.

Therefore, the answer is C.

Tagged: regular polygon · equilateral triangle · area decomposition