AMC 10 · 2025 · #22

Grade 11 algebrageometry-2d
complex-numberssimilar-figuresarea-triangles extreme-principleidentify-subproblems ↑ Prerequisites: complex-numbers
📏 Medium solution 💡 4 insights
Problem
In the complex plane, three points 2z, (1+i)z, and (1-i)z form a triangle. The complex number z is not free to be anything: it must satisfy |4z-2|=1. As z moves over all values allowed by that condition, the triangle changes size and turns. Find the largest area the triangle can reach.

Pick an answer.

(A)
$\frac 14$
(B)
$\frac 12$
(C)
$\frac{9}{16}$
(D)
$\frac 34$
(E)
1
How to solve
Strategy Extreme Principle

The question says "greatest possible," so Tool #14 (Extreme Principle) owns the finish: once the area is written as a single increasing function of |z|, the winner is whichever allowed z sits farthest from the origin, and on a circle that point is easy to name. Getting to that clean form is Tool #7 (Identify Subproblems): the common factor z splits the problem into "what shape is the triangle" and "how big is |z|," two questions that do not interfere. Tool #9 (Solve an Easier Related Problem) does the shape half by throwing z away entirely and measuring the fixed triangle 2, 1+i, 1-i, which is just three plotted points. Tool #1 (Draw a Diagram) makes that measurement a one-line base-times-height. Tool #15 (Organize Information in More Ways) handles the constraint: |4z-2|=1 says nothing useful until it is rearranged into center-and-radius form.

1STEP 1

Pull the common factor z out

Pull out the common factor z.

|2z-(1+i)z| = |1-i| |z| = √(2) |z|, |2z-(1-i)z| = |1+i| |z| = √(2) |z|, |(1+i)z-(1-i)z| = |2i| |z| = 2|z|
2STEP 2

Areas scale by the square of |z|

The area scales by the modulus squared.

Area(2z, (1+i)z, (1-i)z) = |z|² · Area(2, 1+i, 1-i)
3STEP 3

Measure the fixed triangle

The fixed triangle has area 1.

Area(2, 1+i, 1-i) = 1/2 · 2 · 1 = 1 ⟹ Area = |z|²
4STEP 4

Read the constraint as a circle

The condition is a circle of radius one quarter about one half.

|4z-2|=1 ⟺ 4|z-1/2|=1 ⟺ |z-1/2|=1/4
5STEP 5

Push z to the far side of the circle

At the far point the modulus is three quarters, giving nine sixteenths.

|z|_max=1/2+1/4=3/4 ⟹ Area_max=(3/4)²=9/16 (C)
Answer
9/16
Test the winning z=3/4 directly, with no formulas. The vertices become 3/2, 3/4+3/4i, and 3/4-3/4i, that is (1.5, 0), (0.75, 0.75), (0.75, -0.75). The vertical base has length 1.5 and the height is 1.5-0.75=0.75, so the area is 1/2 · 1.5 · 0.75=0.5625=9/16. The size also makes sense: |z| ranges over [1/4, 3/4], so the area ranges over [1/16, 9/16] and can never reach (D) 3/4 or (E) 1. Choice (D) is the trap for anyone who maximizes |z| and forgets that area scales by the square.
💡Key takeaway

When one complex number multiplies every vertex, the shape stays put and only the size changes — so the area is a fixed number times |z|², and you just need the point of the circle farthest from the origin.

  • Pull the common factor z out
  • Areas scale by the square of |z|
  • Measure the fixed triangle
  • Read the constraint as a circle
  • Push z to the far side of the circle