2025 AMC 12B 第 22 题

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.