AMC12 2018 B

AMC12 2018 B · Q16

AMC12 2018 B · Q16. It mainly tests Complex numbers (rare), Geometry misc.

The solutions to the equation $(z+6)^8 = 81$ are connected in the complex plane to form a convex regular polygon, three of whose vertices are labeled $A$, $B$, and $C$. What is the least possible area of $\triangle ABC$?

方程 $(z+6)^8 = 81$ 的解在复平面上连接起来形成一个凸正多边形，其三个顶点标为 $A$、$B$ 和 $C$。$ riangle ABC$ 的最小可能面积是多少？

(A) $\frac{1}{6}\sqrt{6}$ $\frac{1}{6}\sqrt{6}$

(B) $\frac{3}{2}\sqrt{2} - \frac{3}{2}$ $\frac{3}{2}\sqrt{2} - \frac{3}{2}$

(D) $\frac{1}{2}\sqrt{2}$ $\frac{1}{2}\sqrt{2}$

(E) $\sqrt{3} - 1$ $\sqrt{3} - 1$

Answer

Correct choice: (B)

正确答案：（B）

Solution

Answer (B): The answer would be the same if the equation were $z^8=81$, resulting from a horizontal translation of 6 units. The solutions to this equation are the 8 eighth roots of 81, each of which is $\sqrt[8]{3^4}=\sqrt{3}$ units from the origin. These 8 points form a regular octagon. The triangle of minimum area occurs when the vertices of the triangle are consecutive vertices of the octagon, so without loss of generality they have coordinates $A\left(\frac{1}{2}\sqrt{6},\frac{1}{2}\sqrt{6}\right)$, $B\left(\sqrt{3},0\right)$, and $C\left(\frac{1}{2}\sqrt{6},-\frac{1}{2}\sqrt{6}\right)$. This triangle has base $AC=\sqrt{6}$ and height $\sqrt{3}-\frac{1}{2}\sqrt{6}$, so its area is $$ \frac{1}{2}\cdot \sqrt{6}\cdot \left(\sqrt{3}-\frac{1}{2}\sqrt{6}\right)=\frac{3}{2}\sqrt{2}-\frac{3}{2}. $$

答案（B）：如果方程是 $z^8=81$，则答案相同，因为这相当于水平平移 6 个单位得到的。该方程的解是 81 的 8 个八次方根，它们到原点的距离都为 $\sqrt[8]{3^4}=\sqrt{3}$。这 8 个点构成一个正八边形。面积最小的三角形出现在三角形的顶点取为正八边形的相邻三个顶点时，因此不失一般性，可取其坐标为 $A\left(\frac{1}{2}\sqrt{6},\frac{1}{2}\sqrt{6}\right)$，$B\left(\sqrt{3},0\right)$，以及 $C\left(\frac{1}{2}\sqrt{6},-\frac{1}{2}\sqrt{6}\right)$。该三角形的底边 $AC=\sqrt{6}$，高为 $\sqrt{3}-\frac{1}{2}\sqrt{6}$，所以其面积为 $$ \frac{1}{2}\cdot \sqrt{6}\cdot \left(\sqrt{3}-\frac{1}{2}\sqrt{6}\right)=\frac{3}{2}\sqrt{2}-\frac{3}{2}. $$

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