AMC 10 · 2019 · #24

Grade 11 geometry-2d

Archetype Compound Area by Decomposition / Difference └ Sub-archetype Polygon Triangulation

complex-numbersroots-of-unityminkowski-sumvector-additionarea-regular-hexagon easier-related-problemidentify-subproblemssymmetry-argument ↑ Prerequisites: complex-numbersroots-of-unity

📏 Long solution 💡 4 insights

Problem

A complex number is one of the cube roots of one. Sliding three dials independently over the interval from 0 to 1, the combination of the first dial, the second times that number, and the third times its square sweeps out a region in the complex plane. Find the area of that region.

Pick an answer.

(A)

$\frac{1}{2}\sqrt3$

(B)

$\frac{3}{4}\sqrt3$

(C)

$\frac{3}{2}\sqrt3$

(D)

$\frac{1}{2}\pi\sqrt3$

(E)

$\pi$

How to solve

Strategy Draw a Diagram

Tool #1 (Draw a Diagram): a + bω + cω² is a sum of three arrows in the plane, so the whole question is a picture question — draw the arrows and the answer becomes a shape you can name. Tool #9 (Easier Related Problem): three free dials at once is too much, so first freeze two of them and let only c move; that gives a segment. Tool #7 (Subproblems): then release b, then release a, one dial at a time — each release drags the previous figure along a fixed direction, a move that is easy to picture and easy to justify. Tool #15 (Reorganize): rewriting ω and ω² in x + yi form exposes the identity 1 + ω + ω² = 0, which is what makes the final figure symmetric and regular. Tool #3 (Eliminate Possibilities) is held in reserve for the review: since S is a polygon, no answer containing π can be right.

1STEP 1

Draw the three arrows

They sit one hundred twenty degrees apart.

1 = (1, 0), ω = (-1/2, √(3)/2), ω² = (-1/2, -√(3)/2)

2STEP 2

Spot the cancellation

The three arrows add to zero.

1 + ω + ω² = 0 ⟹ z ∈ S ⇔ -z ∈ S

3STEP 3

Move one dial only

One dial traces a segment.

{ cω² : 0 ≤ c ≤ 1 } = segment from 0 to ω²

4STEP 4

Release the second dial

The segment sweeps a rhombus.

{ bω + cω² } = rhombus with corners 0, ω, -1, ω²

5STEP 5

Release the third dial

The rhombus sweeps a hexagon.

corners: ω, 1 + ω, 1, 1 + ω², ω², -1

6STEP 6

Name the hexagon

It is a regular hexagon.

regular hexagon, circumradius 1, side 1

7STEP 7

Area of one triangle

Compute one equilateral triangle.

h = √(1 - 1/4) = √(3)/2, A_△ = 1/2 · 1 · √(3)/2 = √(3)/4

8STEP 8

Add the six pieces

Six of them give three root three over two.

[S] = 6 · √(3)/4 = 3/2√(3) ≈ 2.598

Answer

3/2√3

Three checks. (1) Independent recount: the same hexagon also splits into exactly 3 unit rhombi — the one from step 4 (take a = 0), its shifted copy (take c = 0, corners 0, 1, 1+ω, ω), and the third (take b = 0, corners 0, 1, 1+ω², ω²). Each rhombus has base 1 and height sin 120° = √(3)/2, so area √(3)/2, and 3 · √(3)/2 = 3/2√(3) — the same total by a different cut. (2) Sanity bounds: S is inscribed in the unit circle, so its area must be under π ≈ 3.14, and it contains one rhombus, so it must be over √(3)/2 ≈ 0.87. The value 3/2√(3) ≈ 2.598 sits comfortably between. (3) Shape check: S was built only by sliding straight segments, so it is a polygon with straight edges — no circular arc ever enters, so an answer containing π is impossible. That kills (D) ≈ 2.72 and (E) ≈ 3.14 on sight, and (A) ≈ 0.87 and (B) ≈ 1.30 are too small since (A) is only one of the three rhombi and (B) is only half the figure. Only (C) survives.

💡Key takeaway

Three dials, three unit arrows 120° apart: turn one dial and you draw a segment, turn the second and it sweeps a rhombus, turn the third and it sweeps a regular hexagon of side 1 — six equilateral triangles, area 6 · √(3)/4 = 3/2√(3).

Put the three arrows on the plane
Spot the cancellation
Freeze two dials, move one
Release b: the segment sweeps a rhombus
Release a: the rhombus sweeps a hexagon
Name the hexagon
Area of one equilateral triangle
Add up the six triangles