2019 AMC 12B Problem 24

Question

Let

\omega=-\dfrac12+\dfrac12 i\sqrt3.

Let

S

denote all points in the complex plane of the form

a+b\omega+c\omega^2,

where

0\le a\le1,

0\le b\le1,

and

0\le c\le1.

What is the area of

S?

\dfrac12\sqrt3

\dfrac34\sqrt3

\dfrac32\sqrt3

\dfrac12\pi\sqrt3

\pi

S

is the Minkowski sum of three segments along

1,\ \omega,\ \omega^2

A zonogon’s area equals

\displaystyle\sum_{i\lt j}\bigl|v_i\times v_j\bigr|

over the generating vectors

As

a,b,c

range over

[0,1],

the set

S

is the Minkowski sum of the three unit segments along

v_1=1=(1,0),

\ v_2=\omega=\left(-\dfrac12,\dfrac{\sqrt3}{2}\right),

\ v_3=\omega^2=\left(-\dfrac12,-\dfrac{\sqrt3}{2}\right).

This is a zonogon whose area is the sum of the cross-product magnitudes over pairs. Each pair gives

|v_i\times v_j|=\dfrac{\sqrt3}{2}.

Therefore the area is

3\cdot\dfrac{\sqrt3}{2}=\dfrac{3\sqrt3}{2}.

Thus, C is the correct answer.

Problem 24 in other years