2021 AMC 12A Problems/Problem 11

Problem

A laser is placed at the point $(3,5)$ . The laser bean 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?

$\textbf{(A) }2\sqrt{10} \qquad \textbf{(B) }5\sqrt2 \qquad \textbf{(C) }10\sqrt2 \qquad \textbf{(D) }15\sqrt2 \qquad \textbf{(E) }10\sqrt5$

Diagram

File:2021 AMC 12A Problem 11.png

~MRENTHUSIASM (by Desmos: https://www.desmos.com/calculator/y6wyqok8gm)

Solution 1

Every time the laser bounces off a wall, instead we can imagine it going straight by reflecting it about the wall. Thus, the laser starts at $(3, 5)$ and ends at $(-7, -5)$ , so the path's length is $\sqrt{10^2+10^2}=\boxed{\textbf{(C)} 10\sqrt{2}}$ ~JHawk0224

Solution 2 (Detailed Explanation of Solution 1)

Let $A=(3,5), D=(7,5), B$ be the point where the beam hits the $y$ -axis, and $C$ be the point where the beam hits the $x$ -axis.

Reflecting $\overline{BC}$ about the $y$ -axis gives $\overline{BC'}.$ Then, reflecting $\overline{CD}$ over the $y$ -axis gives $\overline{C'D'}.$ Finally, reflecting $\overline{C'D'}$ about the $x$ -axis gives $\overline{C'D''},$ as shown below.

File:2021 AMC 12A Problem 11(2).png

It follows that $D''=(-7,-5).$ The total distance that the beam will travel is $\begin{align*} AB+BC+CD&=AB+BC'+C'D' \\ &=AB+BC'+C'D'' \\ &=AD'' \\ &=\sqrt{((3-(-7))^2+(5-(-5))^2} \\ &=\sqrt{200} \\ &=\boxed{\textbf{(C) }10\sqrt2} \end{align*}$ Graph in Desmos: https://www.desmos.com/calculator/lqatf2r9bu

~MRENTHUSIASM

Solution 3 (Slopes and Parallelogram)

Define points $A,B,C,$ and $D$ as Solution 2 does.

When a line segment hits and bounces off a coordinate axis at point $P,$ the ray entering $P$ and the ray leaving $P$ have negative slopes. Let the slope of $\overline{AB}$ be $m.$ It follows that the slope of $\overline{BC}$ is $-m,$ and the slope of $\overline{CD}$ is $m.$ Here, we conclude that $\overline{AB}\parallel\overline{CD}.$

Next, we locate $E$ on $\overline{CD}$ such that $\overline{BE}\parallel\overline{AD}.$ So, $ABED$ is a parallelogram.

File:2021 AMC 12A Problem 11(3).png

~MRENTHUSIASM

Video Solution by OmegaLearn (Using Reflections and Distance Formula)

https://youtu.be/e7tNtd-fgeo

~ pi_is_3.14

Video Solution by Hawk Math

https://www.youtube.com/watch?v=AjQARBvdZ20

2021 AMC 12A (Problems • Answer Key • Resources)
Preceded by Problem 10	Followed by Problem 12
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AMC 12 Problems and Solutions

Page

Toolbox

Search