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

Revision as of 11:07, 9 July 2021

1 Problem
2 Diagram
3 Solution 1 (Condensed Explanation of Reflections)
4 Solution 2 (Detailed Explanation of Reflections)
5 Solution 3 (Slopes and Parallelogram)
6 Solution 4 (Answer Choices and Educated Guesses)
7 Video Solution by OmegaLearn (Using Reflections and Distance Formula)
8 Video Solution by Hawk Math
9 Video Solution by TheBeautyofMath
10 See also

Problem

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?

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

Diagram

File:2021 AMC 12A Problem 11 (1) LaTeX Revised.png

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

~MRENTHUSIASM

Solution 1 (Condensed Explanation of Reflections)

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 $\sqrt{10^2+10^2}=\boxed{\textbf{(C) }10\sqrt2}$ .

~JHawk0224

Solution 2 (Detailed Explanation of Reflections)

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.

First, reflecting $\overline{BC}$ about the $y$ -axis gives $\overline{BC'}.$ Then, reflecting $\overline{CD}$ about 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

Graph in Desmos: https://www.desmos.com/calculator/lxjt0ewbou

It follows that $D''=(-7,-5).$ The total distance that the beam will travel is $\begin{align*} AB+BC+CD&=AB+BC'+C'D' \\ &=AB+BC'+C'D'' \\ &=AD'' \\ &=\sqrt{((3-(-7))^2+(5-(-5))^2} \\ &=\sqrt{200} \\ &=\boxed{\textbf{(C) }10\sqrt2}. \end{align*}$ ~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, these two rays coincide when reflected about the line containing $\boldsymbol{P}$ and perpendicular to that coordinate axis, creating 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 $\overline{BE}\parallel\overline{AD},$ from which $ABED$ is a parallelogram, as shown below.

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

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

Let $B=(0,b).$ In parallelogram $ABED,$ we get $E=(4,b).$ By symmetry, we obtain $C=(2,0).$

Applying the slope formula to $\overline{AB}$ 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, we have $AB=3\sqrt2,BC=2\sqrt2,$ and $CD=5\sqrt2.$ The total distance that the beam will travel is $AB+BC+CD=\boxed{\textbf{(C) }10\sqrt2}.$ ~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 $\overline{AB},\overline{BC},$ and $\overline{CD}$ all form $45^\circ$ angles with the coordinate axes, from which $B=(0,2)$ and $C=(2,0).$ The given condition $D=(7,5)$ verifies our guess. Following the last 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

@@ Line 38: / Line 38: @@
 Define points <math>A,B,C,</math> and <math>D</math> as Solution 2 does.
-When a line segment hits and bounces off a coordinate axis at point <math>P,</math> the ray entering <math>P</math> and the ray leaving <math>P</math> have negative slopes. <i><b>Geometrically, these two rays coincide when reflected about the line perpendicular to that coordinate axis, creating line symmetry.</b></i> Let the slope of <math>\overline{AB}</math> be <math>m.</math> It follows that the slope of <math>\overline{BC}</math> is <math>-m,</math> and the slope of <math>\overline{CD}</math> is <math>m.</math> Here, we conclude that <math>\overline{AB}\parallel\overline{CD}.</math>
+When a line segment hits and bounces off a coordinate axis at point <math>P,</math> the ray entering <math>P</math> and the ray leaving <math>P</math> have negative slopes. <i><b>Geometrically, these two rays coincide when reflected about the line containing <math>\boldsymbol{P}</math> and perpendicular to that coordinate axis, creating line symmetry.</b></i> Let the slope of <math>\overline{AB}</math> be <math>m.</math> It follows that the slope of <math>\overline{BC}</math> is <math>-m,</math> and the slope of <math>\overline{CD}</math> is <math>m.</math> Here, we conclude that <math>\overline{AB}\parallel\overline{CD}.</math>
 Next, we locate <math>E</math> on <math>\overline{CD}</math> such that <math>\overline{BE}\parallel\overline{AD},</math> from which <math>ABED</math> is a parallelogram, as shown below.

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

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