AMC12 2025 A

AMC12 2025 A · Q17

AMC12 2025 A · Q17. It mainly tests Complex numbers (rare), Algebra misc.

The polynomial $(z + i)(z + 2i)(z + 3i) + 10$ has three roots in the complex plane, where $i = \sqrt{-1}$. What is the area of the triangle formed by these three roots?

多项式 $(z + i)(z + 2i)(z + 3i) + 10$ 在复平面中有三个根，其中 $i = \sqrt{-1}$。这三个根形成的三角形的面积是多少？

(A) 6 6

(B) 8 8

(D) 12 12

(E) 14 14

Answer

Correct choice: (A)

正确答案：（A）

Solution

Noticing the symmetry, we begin with a substitution: $w = z+2i$. We now have the polynomial $(w-i)(w)(w+i)+10$. Expanding, we get \[w^3+w+10.\] Using the Rational Root Theorem, we notice that $-2$ is a root of this polynomial. Upon dividing the polynomial by $w+2$, we get that \[w^3+w+10 = (w+2)(w^2-2w+5).\] Using the Quadratic Formula upon $w^2-2w+5$, we get that the other two roots are $1+2i$ and $1-2i$. From here, notice that the area of the triangle formed by the roots of this polynomial will be equal to that of the original polynomial because the substitution only shifted the graph $2i$ up, not affecting the distances between each root. Graphing the roots onto the complex plane, the vertical side of the triangle has length $4$, with the altitude to that side having length $3$. Therefore, the triangle has area $\frac{4 \cdot 3}{2} = \boxed{6}.$

注意到对称性，进行变量替换：$w = z+2i$。此时多项式为 $(w-i)(w)(w+i)+10$。展开得 \[w^3+w+10.\] 根据有理根定理，注意到 $-2$ 是该多项式的根。将多项式除以 $w+2$，得 \[w^3+w+10 = (w+2)(w^2-2w+5).\] 对 $w^2-2w+5$ 使用二次公式，得其他两个根为 $1+2i$ 和 $1-2i$。由此，注意到该多项式的根形成的三角形面积与原多项式的根形成的三角形面积相等，因为该替换仅将图像向上平移 $2i$，不影响根之间的距离。将根标在复平面上，三角形的垂直边长为 $4$，到该边的垂线高度为 $3$。因此，三角形面积为 $\frac{4 \cdot 3}{2} = \boxed{6}$。

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