Question

一个激光器放在点 $(3, 5)$ 。激光束沿直线传播。Larry 希望光束先击中 $y$ -轴并反射，再击中 $x$ -轴并反射，最后击中点 $(7, 5)$ 。光束沿这条路径传播的总距离是多少？

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?

$2\sqrt{10}$

$5\sqrt{2}$

$10\sqrt{2}$

$15\sqrt{2}$

$10\sqrt{5}$

答案：C

知识点：反射（几何）距离公式

难度评级：1600

小提示：

从一条直线反射，可以通过把一个端点关于这条直线反射来把路径拉直

A bounce off a line means reflecting one endpoint across that line straightens the path

大提示：

将起点关于 $y$ -轴反射，并将终点关于 $x$ -轴反射，然后求两个像点之间的直线距离

Reflect the start across the $y$ -axis and the end across the $x$ -axis, then take the straight-line distance between the images

解答：

每次反射都可以把路径展开成一条直线。将起点 $(3, 5)$ 关于 $y$ -轴反射到 $(-3, 5)$ ，并将目标点 $(7, 5)$ 关于 $x$ -轴反射到 $(7, -5)$ 。总传播距离等于这两个像点之间的直线距离： $\begin{aligned} &\sqrt{(-3 - 7)^2 + (5 - (-5))^2} \\ &= \sqrt{100 + 100} = 10\sqrt{2} \end{aligned}\text{。}$

因此，正确答案是 C。

Reflecting the path at each bounce turns it into a single straight segment. Reflect the start $(3, 5)$ across the $y$ -axis to $(-3, 5),$ and reflect the target $(7, 5)$ across the $x$ -axis to $(7, -5).$ The total travel distance equals the straight-line distance between these two images: $\begin{aligned} &\sqrt{(-3 - 7)^2 + (5 - (-5))^2} \\ &= \sqrt{100 + 100} = 10\sqrt{2}. \end{aligned}$

Thus, the correct answer is C.

2021 AMC 12A Spring 第 11 题

答案：C

难度评级：1600

小提示：

大提示：

解答：

其他年份的第 11 题