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

m (Diagram)
m (Solution 3 (Slopes and Parallelogram))
Line 46: Line 46:
 
Let <math>B=(0,b).</math> By the property of slopes, we get <math>E=(4,b).</math> By symmetry, we obtain <math>C=(2,0).</math>
 
Let <math>B=(0,b).</math> By the property of slopes, we get <math>E=(4,b).</math> By symmetry, we obtain <math>C=(2,0).</math>
  
Applying the slope formula on <math>\overline{AB}</math> and <math>\overline{DC}</math> gives <cmath>m=\frac{5-b}{3-0}=\frac{5-0}{7-2}.</cmath> Equating the last two expressions gives <math>b=2.</math>
+
Applying the slope formula to <math>\overline{AB}</math> and <math>\overline{DC}</math> gives <cmath>m=\frac{5-b}{3-0}=\frac{5-0}{7-2}.</cmath> Equating the last two expressions gives <math>b=2.</math>
  
 
By the Distance Formula, <math>AB=3\sqrt2,BC=2\sqrt2,</math> and <math>CD=5\sqrt2.</math> The total distance that the beam will travel is <cmath>AB+BC+CD=\boxed{\textbf{(C) }10\sqrt2}.</cmath>  
 
By the Distance Formula, <math>AB=3\sqrt2,BC=2\sqrt2,</math> and <math>CD=5\sqrt2.</math> The total distance that the beam will travel is <cmath>AB+BC+CD=\boxed{\textbf{(C) }10\sqrt2}.</cmath>  

Revision as of 18:06, 7 May 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

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

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

~MRENTHUSIASM

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

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 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.

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

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

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 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, $AB=3\sqrt2,BC=2\sqrt2,$ 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 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 (ProblemsAnswer KeyResources)
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

Invalid username
Login to AoPS