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 $S?$

$\dfrac12\sqrt3$

$\dfrac34\sqrt3$

$\dfrac32\sqrt3$

$\dfrac12\pi\sqrt3$

$\pi$

Solution:

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.

2019 AMC 12B Problem 24

Solution:

Problem 24 in Other Years