AMC12 2021 A

AMC12 2021 A · Q11

AMC12 2021 A · Q11. It mainly tests Coordinate geometry, Transformations.

A laser is placed at the point $(3,5)$. The laser beam travels in a straight line. Larry wants the beam to hit and bounce off the $y$-axis, then hit and bounce off the $x$-axis, then hit the point $(7,5)$. What is the total distance the beam will travel along this path?

激光器放置在点$(3,5)$处。激光束沿直线传播。Larry希望光束先击中$y$轴并反弹，然后击中$x$轴并反弹，然后击中点$(7,5)$。这条路径上光束的总行进距离是多少？

(A) 2\sqrt{10} 2\sqrt{10}

(B) 5\sqrt2 5\sqrt2

(D) 15\sqrt2 15\sqrt2

(E) 10\sqrt5 10\sqrt5

Answer

Correct choice: (C)

正确答案：（C）

Solution

Let $A=(3,5)$ and $D=(7,5).$ Suppose that the beam hits and bounces off the $y$-axis at $B,$ then hits and bounces off the $x$-axis at $C.$ When the beam hits and bounces off a coordinate axis, the angle of incidence and the angle of reflection are congruent. Therefore, we straighten up the path of the beam by reflections: 1. We reflect $\overline{BC}$ about the $y$-axis to get $\overline{BC'}.$ 2. We reflect $\overline{CD}$ about the $x$-axis to get $\overline{CD'}$ with $D'=(7,-5),$ then reflect $\overline{CD'}$ about the $y$-axis to get $\overline{C'D''}$ with $D''=(-7,-5).$ We obtain the following diagram: The total distance that the beam will travel is \begin{align*} AB+BC+CD&=AB+BC+CD' \\ &=AB+BC'+C'D'' \\ &=AD'' \\ &=\sqrt{((3-(-7))^2+(5-(-5))^2} \\ &=\boxed{\textbf{(C) }10\sqrt2}. \end{align*}

设$A=(3,5)$和$D=(7,5)$。假设光束在$B$点击中并反弹$y$轴，然后在$C$点击中并反弹$x$轴。当光束击中并反弹坐标轴时，入射角和反射角相等。因此，我们通过反射来展平光束路径： 1. 将$\overline{BC}$关于$y$轴反射得到$\overline{BC'}$。 2. 将$\overline{CD}$关于$x$轴反射得到$\overline{CD'}$，其中$D'=(7,-5)$，然后将$\overline{CD'}$关于$y$轴反射得到$\overline{C'D''}$，其中$D''=(-7,-5)$。我们得到如下图：光束的总行进距离为 \begin{align*} AB+BC+CD&=AB+BC+CD' \\ &=AB+BC'+C'D'' \\ &=AD'' \\ &=\sqrt{((3-(-7))^2+(5-(-5))^2} \\ &=\boxed{\textbf{(C) }10\sqrt2}. \end{align*}

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