AMC12 2020 A

AMC12 2020 A · Q15

AMC12 2020 A · Q15. It mainly tests Quadratic equations, Complex numbers (rare).

In the complex plane, let $A$ be the set of solutions to $z^3-8=0$ and let $B$ be the set of solutions to $z^3-8z^2-8z+64=0$. What is the greatest distance between a point of $A$ and a point of $B$?

在复平面中，设 $A$ 为 $z^3-8=0$ 的解集，$B$ 为 $z^3-8z^2-8z+64=0$ 的解集。$A$ 中一点与 $B$ 中一点之间的最大距离是多少？

(A) 2\sqrt{3} 2\sqrt{3}

(B) 6 6

(D) 2\sqrt{21} 2\sqrt{21}

(E) 9+\sqrt{3} 9+\sqrt{3}

Answer

Correct choice: (D)

正确答案：（D）

Solution

Answer (D): The solutions to the first equation are 2 times the third roots of unity, namely, 2, $-1 + i\sqrt{3}$, and $-1 - i\sqrt{3}$. The polynomial in the second equation can be rewritten as $(z - 8)(z^2 - 8) = (z - 8)(z - 2\sqrt{2})(z + 2\sqrt{2}),$ so the roots are 8, $2\sqrt{2}$, and $-2\sqrt{2}$. The maximum distance between solutions of the respective equations is the distance between $-1 \pm i\sqrt{3}$ and 8. This is $\sqrt{(-1 - 8)^2 + (\sqrt{3} - 0)^2} = \sqrt{84} = 2\sqrt{21}.$

答案（D）：第一个方程的解是 2 倍的三次单位根，即 2、$-1 + i\sqrt{3}$、以及 $-1 - i\sqrt{3}$。第二个方程中的多项式可以改写为 $(z - 8)(z^2 - 8) = (z - 8)(z - 2\sqrt{2})(z + 2\sqrt{2}),$ 因此其根为 8、$2\sqrt{2}$、以及 $-2\sqrt{2}$。两个方程各自解集之间的最大距离是 $-1 \pm i\sqrt{3}$ 与 8 之间的距离。该距离为 $\sqrt{(-1 - 8)^2 + (\sqrt{3} - 0)^2} = \sqrt{84} = 2\sqrt{21}.$

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