Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

Difference between revisions of "2021 AMC 12A Problems/Problem 11"

m (→‎Solution 3 (Slopes and Parallelogram))
m (→‎Solution 4 (Answer Choices and Educated Guesses))
Line 52: Line 52:
 
Define points <math>A,B,C,</math> and <math>D</math> as Solution 2 does.
 
Define points <math>A,B,C,</math> and <math>D</math> as Solution 2 does.
  
−
Since choices <math>\textbf{(B)}, \textbf{(C)},</math> and <math>\textbf{(D)}</math> all involve <math>\sqrt2,</math> we suspect that one of them is the correct answer. We take a guess in faith that <math>\overleftrightarrow{AB}</math> and <math>\overleftrightarrow{BC}</math> form <math>45^\circ</math> angles with the coordinate axes, then we get that <math>B=(0,2)</math> and <math>C=(2,0).</math> This result verifies our guess. Following the penultimate paragraph of Solution 3 gives the answer <math>\boxed{\textbf{(C) }10\sqrt2}.</math>
+
Since choices <math>\textbf{(B)}, \textbf{(C)},</math> and <math>\textbf{(D)}</math> all involve <math>\sqrt2,</math> we suspect that one of them is the correct answer. We take a guess in faith that <math>\overleftrightarrow{AB}</math> and <math>\overleftrightarrow{BC}</math> form <math>45^\circ</math> angles with the coordinate axes, from which <math>B=(0,2)</math> and <math>C=(2,0).</math> This result verifies our guess. Following the penultimate paragraph of Solution 3 gives the answer <math>\boxed{\textbf{(C) }10\sqrt2}.</math>
  
 
~MRENTHUSIASM
 
~MRENTHUSIASM

Revision as of 19:56, 22 March 2021

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 (1) LaTeX Revised.png

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

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) LaTeX.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/lxjt0ewbou

~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. Geometrically, the rays coincide when reflected about the line perpendicular to that coordinate axis, creating a line symmetry. 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 $\overline{BE}\parallel\overline{AD},$ thus $ABED$ is a parallelogram, as shown below.

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

Let $B=(0,b).$ By the property of slopes, we get $E=(4,b).$ By symmetry, we obtain $C=(2,0).$

Applying the slope formula on $\overline{AB}$ and $\overline{DC}$ gives \[m=\frac{5-b}{3-0}=\frac{5-0}{7-2}.\] Equating the last two expressions gives $b=2.$

By the Distance Formula, $AB=3\sqrt2,BC=2\sqrt2,$ and $CD=5\sqrt2.$ The total distance that the beam will travel is \[AB+BC+CD=\boxed{\textbf{(C) }10\sqrt2}.\]

Graph in Desmos: https://www.desmos.com/calculator/lgfiiqgqc2

~MRENTHUSIASM

Solution 4 (Answer Choices and Educated Guesses)

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

Since choices $\textbf{(B)}, \textbf{(C)},$ and $\textbf{(D)}$ all involve $\sqrt2,$ we suspect that one of them is the correct answer. We take a guess in faith that $\overleftrightarrow{AB}$ and $\overleftrightarrow{BC}$ form $45^\circ$ angles with the coordinate axes, from which $B=(0,2)$ and $C=(2,0).$ This result verifies our guess. Following the penultimate paragraph of Solution 3 gives the answer $\boxed{\textbf{(C) }10\sqrt2}.$

~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

Video Solution by TheBeautyofMath

https://youtu.be/ySWSHyY9TwI

~IceMatrix

See also

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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2021_AMC_12A_Problems/Problem_11&oldid=149852"