2025 AMC 12B Problem 22

Question

What is the greatest possible area of the triangle in the complex plane with vertices

2z,

(1+i)z,

and

(1-i)z,

where

z

is a complex number satisfying

|4z - 2| = 1?

\dfrac{1}{4}

\dfrac{1}{2}

\dfrac{9}{16}

\dfrac{3}{4}

1

The vertices are

z

times the fixed points

2, 1+i, 1-i,

so the triangle is that fixed triangle scaled by

|z|

Its area is

|z|^2

times a fixed area; the condition

|4z - 2| = 1

is a circle, so maximize

|z|

on it

The vertices are

z \cdot 2,

z(1+i),

and

z(1-i),

so the triangle is the fixed triangle with vertices

2, 1+i, 1-i

— which has area

1

— scaled by

|z|,

giving area

|z|^2.

The condition

|4z - 2| = 1

is the circle

\left|z - \tfrac{1}{2}\right| = \tfrac{1}{4},

on which

|z|

is at most

\tfrac{1}{2} + \tfrac{1}{4} = \tfrac{3}{4}.

So the greatest area is

\left(\tfrac{3}{4}\right)^2 = \tfrac{9}{16}.

Thus, the correct answer is C.

Problem 22 in other years