AMC12 2007 A

AMC12 2007 A · Q23

AMC12 2007 A · Q23. It mainly tests Logarithms (rare), Coordinate geometry.

Square $ABCD$ has area $36,$ and $\overline{AB}$ is parallel to the x-axis. Vertices $A,$ $B$, and $C$ are on the graphs of $y = \log_{a}x,$ $y = 2\log_{a}x,$ and $y = 3\log_{a}x,$ respectively. What is $a?$

正方形 $ABCD$ 的面积为 $36,$ 且 $\overline{AB}$ 平行于 $x$ 轴。顶点 $A,$ $B$, 和 $C$ 分别在 $y = \log_{a}x,$ $y = 2\log_{a}x,$ 和 $y = 3\log_{a}x,$ 的图像上。求 $a$ 的值。

(A) $\sqrt[6]{3}$ $\sqrt[6]{3}$

(B) $\sqrt{3}$ $\sqrt{3}$

(D) $\sqrt{6}$ $\sqrt{6}$

(E) 6 6

Answer

Correct choice: (A)

正确答案：（A）

Solution

Let $x$ be the x-coordinate of $B$ and $C$, and $x_2$ be the x-coordinate of $A$ and $y$ be the y-coordinate of $A$ and $B$. Then $2\log_ax= y \Longrightarrow a^{y/2} = x$ and $\log_ax_2 = y \Longrightarrow x_2 = a^y = \left(a^{y/2}\right)^2 = x^2$. Since the distance between $A$ and $B$ is $6$, we have $x^2 - x - 6 = 0$, yielding $x = -2, 3$. However, we can discard the negative root (all three logarithmic equations are underneath the line $y = 3$ and above $y = 0$ when $x$ is negative, hence we can't squeeze in a square of side 6). Thus $x = 3$. Substituting back, $3\log_{a}x - 2\log_{a}x = 6 \Longrightarrow a^6 = x$, so $a = \sqrt[6]{3}\ \ \mathrm{(A)}$.

设 $B$ 与 $C$ 的横坐标为 $x$，$A$ 的横坐标为 $x_2$，且 $A$ 与 $B$ 的纵坐标为 $y$。则 $2\log_ax= y \Longrightarrow a^{y/2} = x$，且 $\log_ax_2 = y \Longrightarrow x_2 = a^y = \left(a^{y/2}\right)^2 = x^2$。由于 $A$ 与 $B$ 的距离为 $6$，有 $x^2 - x - 6 = 0$，解得 $x = -2, 3$。但可舍去负根（当 $x$ 为负时，这三条对数曲线都在直线 $y = 3$ 下方且在 $y = 0$ 上方，无法容纳边长为 6 的正方形），故 $x = 3$。代回得 $3\log_{a}x - 2\log_{a}x = 6 \Longrightarrow a^6 = x$，所以 $a = \sqrt[6]{3}\ \ \mathrm{(A)}$。

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