AMC12 2025 B

AMC12 2025 B · Q22

AMC12 2025 B · Q22. It mainly tests Complex numbers (rare), Area & perimeter.

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 $|4z - 2| = 1$?

在复平面中，顶点为$2z$、$(1+i)z$和$(1-i)z$的三角形，$z$是满足$|4z - 2| = 1$的复数，该三角形的最大可能面积是多少？

(A) \frac 14 \frac 14

(B) \frac 12 \frac 12

(D) \frac 34 \frac 34

(E) 1 1

Answer

Correct choice: (C)

正确答案：（C）

Solution

Because we can factor $z$ from each vertex, the area of the triangle is equal to $|z|^2$ multiplied by the area of the triangle with vertices $2, 1+i, 1-i$. The vertical side of that triangle has length $2$ and the altitude to that side has length $1$, so the area is $1$. We now just need the maximum possible value of $|z|^2$. The equation $|4z-2| = 1$ can be rewritten as $|z-1/2| = 1/4$. Therefore, it exists as a circle with radius $1/4$ and center at $(1/2, 0)$ on the complex plane. The maximum possible magnitude of a complex number on this circle occurs with $\frac{3}{4}$, with magnitude $\frac{3}{4}$. We can justify this because the line connecting the origin and the complex number $\frac{3}{4}$ goes through the center of the circle. Our answer is then $\left(\frac{3}{4}\right)^2 = \boxed{\frac{9}{16}}.$

因为我们可以从每个顶点提取$z$，三角形的面积等于$|z|^2$乘以顶点为$2$、$1+i$、$1-i$的三角形的面积。该三角形的垂直边长为$2$，到该边的垂线高度为$1$，所以面积为$1$。现在只需找到$|z|^2$的最大可能值。方程$|4z-2| = 1$可以改写为$|z-1/2| = 1/4$。因此，它是复平面中心为$(1/2, 0)$、半径为$1/4$的圆。该圆上复数的最大模发生在$\frac{3}{4}$处，模为$\frac{3}{4}$。这是因为连接原点和复数$\frac{3}{4}$的直线穿过圆心。答案于是$\left(\frac{3}{4}\right)^2 = \boxed{\frac{9}{16}}$。

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