AMC12 2023 A

AMC12 2023 A · Q6

AMC12 2023 A · Q6. It mainly tests Quadratic equations, Logarithms (rare).

Points $A$ and $B$ lie on the graph of $y=\log_{2}x$. The midpoint of $\overline{AB}$ is $(6, 2)$. What is the positive difference between the $x$-coordinates of $A$ and $B$?

点 $A$ 和 $B$ 位于 $y=\log_{2}x$ 的图像上。线段 $\overline{AB}$ 的中点为 $(6, 2)$。$A$ 和 $B$ 的 $x$ 坐标的正差是多少？

(A) 2\sqrt{11} 2\sqrt{11}

(B) 4\sqrt{3} 4\sqrt{3}

(D) 4\sqrt{5} 4\sqrt{5}

(E) 9 9

Answer

Correct choice: (D)

正确答案：（D）

Solution

Let $A(6+m,2+n)$ and $B(6-m,2-n)$, since $(6,2)$ is their midpoint. Thus, we must find $2m$. We find two equations due to $A,B$ both lying on the function $y=\log_{2}x$. The two equations are then $\log_{2}(6+m)=2+n$ and $\log_{2}(6-m)=2-n$. Now add these two equations to obtain $\log_{2}(6+m)+\log_{2}(6-m)=4$. By logarithm rules, we get $\log_{2}((6+m)(6-m))=4$. By raising 2 to the power of both sides, we obtain $(6+m)(6-m)=16$. We then get \[36-m^2=16 \rightarrow m^2=20 \rightarrow m=2\sqrt{5}\]. Since we're looking for $2m$, we obtain $(2)(2\sqrt{5})=\boxed{\textbf{(D) }4\sqrt{5}}$

设 $A(6+m,2+n)$ 和 $B(6-m,2-n)$，因为 $(6,2)$ 是它们的中点。因此，我们需要求 $2m$。由于 $A,B$ 都位于函数 $y=\log_{2}x$ 上，我们得到两个方程：$\log_{2}(6+m)=2+n$ 和 $\log_{2}(6-m)=2-n$。将这两个方程相加，得到 $\log_{2}(6+m)+\log_{2}(6-m)=4$。根据对数法则，得到 $\log_{2}((6+m)(6-m))=4$。两边取 $2$ 的幂，得到 $(6+m)(6-m)=16$。从而 $[36-m^2=16 \rightarrow m^2=20 \rightarrow m=2\sqrt{5}]$。我们需要 $2m$，因此 $(2)(2\sqrt{5})=\boxed{\textbf{(D) }4\sqrt{5}}$

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