AMC 10 · 2024 · #12

Grade 12 algebrageometry-2d
Archetype Compound Area by Decomposition / Difference └ Sub-archetype Polygon Triangulation
complex-polar-formsine-area-formulacomplex-numbersarea-triangles identify-subproblemsconvert-to-algebra ↑ Prerequisites: complex-numbersarea-trianglestrigonometric-ratios
📏 Long solution 💡 4 insights
Problem
A complex number z sits on the circle of radius 2 about the origin, in the part of the plane where the real part is greater than 1 and the imaginary part is positive. The origin together with the first three powers z, z², z³ makes four points, and read in that order they are the corners of a quadrilateral of area 15. Find the imaginary part of z.

Pick an answer.

(A)
$\frac{3}{4}$
(B)
1
(C)
$\frac{4}{3}$
(D)
$\frac{3}{2}$
(E)
$\frac{5}{3}$
How to solve
Strategy Identify Subproblems

There is no area formula for a general quadrilateral, so Tool #7 (Identify Subproblems) owns the whole solution: the diagonal from the origin to z² cuts the region into two triangles, and both of them turn out to have their apex at the origin with the same opening angle, so one sine value measures both. Getting there needs Tool #4 (Introduce a Variable), because |z| = 2 leaves only a direction free, and naming that direction θ reduces the problem to one unknown. Tool #1 (Draw a Diagram) is not decoration here — it is the step that proves the vertices really do fan outward in the listed order, which is exactly what licenses the two-triangle split. Tool #13 (Convert to Algebra) turns the given area into one equation in sinθ. Tool #3 (Eliminate Possibilities) closes it: a sine value names two angles, and the condition Re z ＞ 1 is what throws the obtuse one away.

1STEP 1

Name the angle z turns by

The real-part condition traps the angle between 0 and 60 degrees.

z = 2(cosθ + isinθ), Im z = 2sinθ, Re z = 2cosθ ＞ 1 ⇔ cosθ ＞ 1/2 ⇔ 0° ＜ θ ＜ 60°
2STEP 2

Watch the powers fan out

The sizes grow 2, 4, 8.

|z| = 2, |z²| = 4, |z³| = 8; arg z = θ, arg z² = 2θ, arg z³ = 3θ; 0° ＜ θ ＜ 2θ ＜ 3θ ＜ 180°
3STEP 3

Cut along the diagonal from 0

The two triangles total twenty sine theta.

[0 z z²] = 1/2 · 2 · 4 sinθ = 4sinθ, [0 z² z³] = 1/2 · 4 · 8 sinθ = 16sinθ, Area = 4sinθ + 16sinθ = 20sinθ
4STEP 4

Set the area equal to 15

Sine theta is three quarters.

20sinθ = 15 ⟹ sinθ = 15/20 = 3/4
5STEP 5

Read off the imaginary part

Twice three quarters is three halves.

Im z = 2sinθ = 2 · 3/4 = 3/2, Re z = 2cosθ = 2 · √(7)/4 = √(7)/2 ＞ 1 (D)
Answer
3/2
Build the number and measure the quadrilateral by hand, with no formula from the solution. Take z = √(7)/2 + 3/2i. Multiplying out gives z² = -1/2 + 3√(7)/2i and z³ = -5√(7)/2 + 9/2i, whose moduli are √(1/4 + 63/4) = 4 and √(175/4 + 81/4) = 8, exactly as required. Now run the shoelace formula on (0,0), (√(7)/2, 3/2), (-1/2, 3√(7)/2), (-5√(7)/2, 9/2) in that order. The two edges touching the origin contribute 0; the edge from z to z² contributes 21/4 + 3/4 = 6 and the edge from z² to z³ contributes -9/4 + 105/4 = 24, for a total of 30, so the area is 30/2 = 15 exactly. The size is believable too: θ is confined to (0°, 60°), so sinθ ＜ √(3)/2 and the area can never exceed 10√(3) ≈ 17.3 — an area of 15 fits, but only just, which matches how stretched the fan looks. Finally, the area equals 20sinθ = 10 Im z, so the five choices would give areas 15/2, 10, 40/3, 15, 50/3; only (D) lands on 15.
💡Key takeaway

Multiplying by z turns every point by the same angle and stretches it by the same factor, so the quadrilateral splits into two triangles that share that angle at the origin — and the area comes out to exactly ten times the imaginary part of z.

  • Name the angle z turns by
  • Watch the powers fan out
  • Cut along the diagonal from 0
  • Set the area equal to 15
  • Read off the imaginary part

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