AMC12 2021 A

AMC12 2021 A · Q21

AMC12 2021 A · Q21. It mainly tests Complex numbers (rare), Geometry misc.

The five solutions to the equation\[(z-1)(z^2+2z+4)(z^2+4z+6)=0\] may be written in the form $x_k+y_ki$ for $1\le k\le 5,$ where $x_k$ and $y_k$ are real. Let $\mathcal E$ be the unique ellipse that passes through the points $(x_1,y_1),(x_2,y_2),(x_3,y_3),(x_4,y_4),$ and $(x_5,y_5)$. The eccentricity of $\mathcal E$ can be written in the form $\sqrt{\frac mn}$, where $m$ and $n$ are relatively prime positive integers. What is $m+n$? (Recall that the eccentricity of an ellipse $\mathcal E$ is the ratio $\frac ca$, where $2a$ is the length of the major axis of $\mathcal E$ and $2c$ is the is the distance between its two foci.)

方程 \[(z-1)(z^2+2z+4)(z^2+4z+6)=0\] 的五个解可以写成形式 $x_k+y_ki$，其中 $1\le k\le 5$，$x_k$ 和 $y_k$ 是实数。令 $\mathcal E$ 是通过点 $(x_1,y_1),(x_2,y_2),(x_3,y_3),(x_4,y_4)$ 和 $(x_5,y_5)$ 的唯一椭圆。$\mathcal E$ 的离心率可以写成 $\sqrt{\frac mn}$ 的形式，其中 $m$ 和 $n$ 是互质的正整数。求 $m+n$？（回忆椭圆 $\mathcal E$ 的离心率是 $\frac ca$，其中 $2a$ 是 $\mathcal E$ 的长轴长度，$2c$ 是其两个焦点间的距离。）

(A) 7 7

(B) 9 9

(D) 13 13

(E) 15 15

Answer

Correct choice: (A)

正确答案：（A）

Solution

The solutions to this equation are $z = 1$, $z = -1 \pm i\sqrt 3$, and $z = -2\pm i\sqrt 2$. Consider the five points $(1,0)$, $\left(-1,\pm\sqrt 3\right)$, and $\left(-2,\pm\sqrt 2\right)$; these are the five points which lie on $\mathcal E$. Note that since these five points are symmetric about the $x$-axis, so must $\mathcal E$. Now let $r=b/a$ denote the ratio of the length of the minor axis of $\mathcal E$ to the length of its major axis. Remark that if we perform a transformation of the plane which scales every $x$-coordinate by a factor of $r$, $\mathcal E$ is sent to a circle $\mathcal E'$. Thus, the problem is equivalent to finding the value of $r$ such that $(r,0)$, $\left(-r,\pm\sqrt 3\right)$, and $\left(-2r,\pm\sqrt 2\right)$ all lie on a common circle; equivalently, it suffices to determine the value of $r$ such that the circumcenter of the triangle formed by the points $P_1 = (r,0)$, $P_2 = \left(-r,\sqrt 3\right)$, and $P_3 = \left(-2r,\sqrt 2\right)$ lies on the $x$-axis. Recall that the circumcenter of a triangle $ABC$ is the intersection point of the perpendicular bisectors of its three sides. The equations of the perpendicular bisectors of the segments $\overline{P_1P_2}$ and $\overline{P_1P_3}$ are\[y = \tfrac{\sqrt 3}2 + \tfrac{2r}{\sqrt 3}x\qquad\text{and}\qquad y = \tfrac{\sqrt 2}2 + \tfrac{3r}{\sqrt 2}\left(x + \tfrac r2\right)\]respectively. These two lines have different slopes for $r\neq 0$, so indeed they will intersect at some point $(x_0,y_0)$; we want $y_0 = 0$. Plugging $y = 0$ into the first equation yields $x = -\tfrac{3}{4r}$, and so plugging $y=0$ into the second equation and simplifying yields\[-\tfrac{1}{3r} = x + \tfrac r2 = -\tfrac{3}{4r} + \tfrac{r}{2}.\]Solving yields $r=\sqrt{\tfrac 56}$. Finally, recall that the lengths $a$, $b$, and $c$ (where $c$ is the distance between the foci of $\mathcal E$) satisfy $c = \sqrt{a^2 - b^2}$. Thus the eccentricity of $\mathcal E$ is $\tfrac ca = \sqrt{1 - \left(\tfrac ba\right)^2} = \sqrt{\tfrac 16}$ and the requested answer is $\boxed{\textbf{(A) } 7}$.

该方程的解为 $z = 1$，$z = -1 \pm i\sqrt 3$，和 $z = -2\pm i\sqrt 2$。考虑五个点 $(1,0)$，$\left(-1,\pm\sqrt 3\right)$，和 $\left(-2,\pm\sqrt 2\right)$；这些是躺在 $\mathcal E$ 上的五个点。注意到由于这些五个点关于 $x$ 轴对称，因此 $\mathcal E$ 也必须如此。现在令 $r=b/a$ 表示 $\mathcal E$ 的短轴长度与其长轴长度的比率。注意如果我们对平面进行变换，将每个 $x$ 坐标按因子 $r$ 缩放，则 $\mathcal E$ 被映射到圆 $\mathcal E'$。因此，问题等价于找到 $r$ 的值，使得 $(r,0)$，$\left(-r,\pm\sqrt 3\right)$，和 $\left(-2r,\pm\sqrt 2\right)$ 都躺在同一个圆上；等价地，足以确定 $r$ 的值，使得由点 $P_1 = (r,0)$，$P_2 = \left(-r,\sqrt 3\right)$，和 $P_3 = \left(-2r,\sqrt 2\right)$ 形成的三角形的圆心躺在 $x$ 轴上。回忆三角形 $ABC$ 的圆心是其三边垂直平分线的交点。线段 $\overline{P_1P_2}$ 和 $\overline{P_1P_3}$ 的垂直平分线的方程分别为 \[y = \tfrac{\sqrt 3}2 + \tfrac{2r}{\sqrt 3}x\qquad\text{和}\qquad y = \tfrac{\sqrt 2}2 + \tfrac{3r}{\sqrt 2}\left(x + \tfrac r2\right)\] 这两条直线对于 $r\neq 0$ 有不同的斜率，因此它们将在某点 $(x_0,y_0)$ 相交；我们希望 $y_0 = 0$。将 $y = 0$ 代入第一条方程得到 $x = -\tfrac{3}{4r}$，因此将 $y=0$ 代入第二条方程并化简得到 \[\-\tfrac{1}{3r} = x + \tfrac r2 = -\tfrac{3}{4r} + \tfrac{r}{2}.\] 解得 $r=\sqrt{\tfrac 56}$。最后，回忆长度 $a$，$b$，和 $c$（其中 $c$ 是 $\mathcal E$ 的两个焦点间的距离）满足 $c = \sqrt{a^2 - b^2}$。因此 $\mathcal E$ 的离心率是 $\tfrac ca = \sqrt{1 - \left(\tfrac ba\right)^2} = \sqrt{\tfrac 16}$，所求答案是 $\boxed{\textbf{(A) } 7}$。

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