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

(→‎Diagram: Used Asy for diagrams instead of Desmos.)
(→‎Solution 1 (Geometry))
Line 32: Line 32:
  
 
First, reflecting <math>\overline{BC}</math> about the <math>y</math>-axis gives <math>\overline{BC'}.</math> Then, reflecting <math>\overline{CD}</math> about the <math>y</math>-axis gives <math>\overline{C'D'}.</math> Finally, reflecting <math>\overline{C'D'}</math> about the <math>x</math>-axis gives <math>\overline{C'D''},</math> as shown below.
 
First, reflecting <math>\overline{BC}</math> about the <math>y</math>-axis gives <math>\overline{BC'}.</math> Then, reflecting <math>\overline{CD}</math> about the <math>y</math>-axis gives <math>\overline{C'D'}.</math> Finally, reflecting <math>\overline{C'D'}</math> about the <math>x</math>-axis gives <math>\overline{C'D''},</math> as shown below.
 +
<asy>
 +
/* Made by MRENTHUSIASM */
 +
size(220);
 +
import TrigMacros;
  
−
[[File:2021 AMC 12A Problem 11 (2) LaTeX.png|center]]
+
rr_cartesian_axes(-9,9,-7,7,usegrid=false,complexplane=false);
  
−
Graph in Desmos: https://www.desmos.com/calculator/lxjt0ewbou
+
label("$A(3,5)$",(3,5),(0,2));
 +
label("$B$",(0,2),(-2,0));
 +
label("$C$",(2,0),(0,-2));
 +
label("$D(7,5)$",(7,5),(0,2));
 +
label("$C'$",(-2,0),(0,-2));
 +
label("$D'$",(-7,5),(0,2));
 +
label("$D''$",(-7,-5),(0,-2));
  
 +
pair A = (3,5);
 +
pair B = (0,2);
 +
pair C = (2,0);
 +
pair D = (7,5);
 +
pair E = (-2,0);
 +
pair F = (-7,5);
 +
pair G = (-7,-5);
 +
 +
draw(A--B--C--D,red);
 +
draw(B--E,heavygreen+dashed);
 +
draw(E--F,heavygreen+dashed);
 +
draw(E--G,heavygreen+dashed);
 +
dot(A,linewidth(3.5));
 +
dot(B,linewidth(3.5));
 +
dot(C,linewidth(3.5));
 +
dot(D,linewidth(3.5));
 +
dot(E,linewidth(3.5));
 +
dot(F,linewidth(3.5));
 +
dot(G,linewidth(3.5));
 +
</asy>
 
It follows that <math>D''=(-7,-5).</math> The total distance that the beam will travel is  
 
It follows that <math>D''=(-7,-5).</math> The total distance that the beam will travel is  
 
<cmath>\begin{align*}
 
<cmath>\begin{align*}
Line 43: Line 73:
 
&=AD'' \\
 
&=AD'' \\
 
&=\sqrt{((3-(-7))^2+(5-(-5))^2} \\
 
&=\sqrt{((3-(-7))^2+(5-(-5))^2} \\
−
&=\sqrt{200} \\
 
 
&=\boxed{\textbf{(C) }10\sqrt2}.
 
&=\boxed{\textbf{(C) }10\sqrt2}.
 
\end{align*}</cmath>
 
\end{align*}</cmath>

Revision as of 12:09, 1 September 2021

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

[asy] /* Made by MRENTHUSIASM */ size(220); import TrigMacros; rr_cartesian_axes(-3,9,-3,7,usegrid=false,complexplane=false); label("$(3,5)$",(3,5),(0,2)); label("$(7,5)$",(7,5),(0,2)); pair A = (3,5); pair B = (0,2); pair C = (2,0); pair D = (7,5); draw(A--B--C--D,red); dot(A,linewidth(3.5)); dot(B,linewidth(3.5)); dot(C,linewidth(3.5)); dot(D,linewidth(3.5)); [/asy] ~MRENTHUSIASM

Solution 1 (Geometry)

Let $A=(3,5), D=(7,5), B$ be the point where the beam hits and bounces off the $y$-axis, and $C$ be the point where the beam hits and bounces off 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. [asy] /* Made by MRENTHUSIASM */ size(220); import TrigMacros; rr_cartesian_axes(-9,9,-7,7,usegrid=false,complexplane=false); label("$A(3,5)$",(3,5),(0,2)); label("$B$",(0,2),(-2,0)); label("$C$",(2,0),(0,-2)); label("$D(7,5)$",(7,5),(0,2)); label("$C'$",(-2,0),(0,-2)); label("$D'$",(-7,5),(0,2)); label("$D''$",(-7,-5),(0,-2)); pair A = (3,5); pair B = (0,2); pair C = (2,0); pair D = (7,5); pair E = (-2,0); pair F = (-7,5); pair G = (-7,-5); draw(A--B--C--D,red); draw(B--E,heavygreen+dashed); draw(E--F,heavygreen+dashed); draw(E--G,heavygreen+dashed); dot(A,linewidth(3.5)); dot(B,linewidth(3.5)); dot(C,linewidth(3.5)); dot(D,linewidth(3.5)); dot(E,linewidth(3.5)); dot(F,linewidth(3.5)); dot(G,linewidth(3.5)); [/asy] 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} \\ &=\boxed{\textbf{(C) }10\sqrt2}. \end{align*} ~MRENTHUSIASM (Solution)

~JHawk0224 (Proposal)

Solution 2 (Algebra)

Define points $A,B,C,$ and $D$ as Solution 1 does.

When a straight line hits and bounces off a coordinate axis at point $P,$ the ray entering $P$ and the ray leaving $P$ have negative slopes. Let $\ell$ be the line containing $P$ and perpendicular to that coordinate axis. Geometrically, these two rays coincide when reflected about $\boldsymbol{\ell}.$

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 3 (Answer Choices and Educated Guesses)

Define points $A,B,C,$ and $D$ as Solution 1 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 2 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

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