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

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==Solution 1 (Geometry)==
 
==Solution 1 (Geometry)==
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Let <math>A=(3,5), D=(7,5), B</math> be the point where the beam hits and bounces off the <math>y</math>-axis, and <math>C</math> be the point where the beam hits and bounces off the <math>x</math>-axis.
+
Let <math>A=(3,5), D=(7,5), B</math> be the point where the beam hits and bounces off the <math>y</math>-axis, and <math>C</math> be the point where the beam hits and bounces off the <math>x</math>-axis. Note that <math>AB+BC+CD</math> is the minimum distance from <math>A</math> first to a point on the <math>y</math>-axis, then to a point on the <math>x</math>-axis, and finally to <math>D.</math>
  
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First, we reflect <math>\overline{BC}</math> about the <math>y</math>-axis to get <math>\overline{BC'}.</math> Then, we reflect <math>\overline{CD}</math> about the <math>y</math>-axis to get <math>\overline{C'D'}.</math> Finally, we reflect <math>\overline{C'D'}</math> about the <math>x</math>-axis to get <math>\overline{C'D''}.</math> Note that <math>AB+BC+CD=AB+BC'+C'D'=AB+BC'+C'D''</math> is the minimum distance from <math>A</math> first to a point on the <math>y</math>-axis, then to a point on the <math>x</math>-axis, and finally to <math>D.</math> It follows that <math>A,B,C'</math> and <math>D''</math> are collinear, as shown below.
+
First, we reflect <math>\overline{BC}</math> about the <math>y</math>-axis to get <math>\overline{BC'}.</math> Then, we reflect <math>\overline{CD}</math> about the <math>y</math>-axis to get <math>\overline{C'D'}.</math> Finally, we reflect <math>\overline{C'D'}</math> about the <math>x</math>-axis to get <math>\overline{C'D''}.</math> Since <math>AB+BC+CD=AB+BC'+C'D'=AB+BC'+C'D'',</math> it follows that <math>A,B,C'</math> and <math>D''</math> are collinear, as shown below.
 
<asy>
 
<asy>
 
/* Made by MRENTHUSIASM */
 
/* Made by MRENTHUSIASM */

Revision as of 09:11, 25 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(200); int xMin = -3; int xMax = 9; int yMin = -3; int yMax = 7; draw((xMin,0)--(xMax,0),black+linewidth(1.5),EndArrow(5)); draw((0,yMin)--(0,yMax),black+linewidth(1.5),EndArrow(5)); label("$x$",(xMax,0),(2,0)); label("$y$",(0,yMax),(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)); label("$(3,5)$",A,(0,2)); label("$(7,5)$",D,(0,2)); [/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. Note that $AB+BC+CD$ is the minimum distance from $A$ first to a point on the $y$-axis, then to a point on the $x$-axis, and finally to $D.$

First, we reflect $\overline{BC}$ about the $y$-axis to get $\overline{BC'}.$ Then, we reflect $\overline{CD}$ about the $y$-axis to get $\overline{C'D'}.$ Finally, we reflect $\overline{C'D'}$ about the $x$-axis to get $\overline{C'D''}.$ Since $AB+BC+CD=AB+BC'+C'D'=AB+BC'+C'D'',$ it follows that $A,B,C'$ and $D''$ are collinear, as shown below. [asy] /* Made by MRENTHUSIASM */ size(225); int xMin = -9; int xMax = 9; int yMin = -7; int yMax = 7; draw((xMin,0)--(xMax,0),black+linewidth(1.5),EndArrow(5)); draw((0,yMin)--(0,yMax),black+linewidth(1.5),EndArrow(5)); label("$x$",(xMax,0),(2,0)); label("$y$",(0,yMax),(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)); label("$A(3,5)$",A,(0,2)); label("$B$",B,(-2,0)); label("$C$",C,(0,-2)); label("$D(7,5)$",D,(0,2)); label("$C'$",E,(0,-2)); label("$D'$",F,(0,2)); label("$D''$",G,(0,-2)); [/asy] We have $D'=(-7,5)$ and $D''=(-7,-5).$ Therefore, the total distance that the beam will travel is \[AB+BC+CD=AD''=\sqrt{((3-(-7))^2+(5-(-5))^2}=\boxed{\textbf{(C) }10\sqrt2}.\] ~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 $\overline{BE}\parallel\overline{AD}.$ We obtain parallelogram $ABED,$ as shown below. [asy] /* Made by MRENTHUSIASM */ size(200); int xMin = -3; int xMax = 9; int yMin = -3; int yMax = 7; draw((xMin,0)--(xMax,0),black+linewidth(1.5),EndArrow(5)); draw((0,yMin)--(0,yMax),black+linewidth(1.5),EndArrow(5)); label("$x$",(xMax,0),(2,0)); label("$y$",(0,yMax),(0,2)); pair A = (3,5); pair B = (0,2); pair C = (2,0); pair D = (7,5); pair E = (4,2); draw(A--B--C--D,red); draw(A--D,heavygreen+dashed); draw(B--E,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)); label("$A(3,5)$",A,(0,2)); label("$B$",B,(-2,0)); label("$C$",C,(0,-2)); label("$D(7,5)$",D,(0,2)); label("$E$",E,(2,0)); [/asy] Let $B=(0,b).$ In parallelogram $ABED,$ we get $E=(4,b).$ By symmetry, we get $C=(2,0).$

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, as shown below. [asy] /* Made by MRENTHUSIASM */ size(200); int xMin = -3; int xMax = 9; int yMin = -3; int yMax = 7; //Draws the horizontal gridlines void horizontalLines() { for (int i = yMin+1; i < yMax; ++i) { draw((xMin,i)--(xMax,i), mediumgray+linewidth(0.4)); } } //Draws the vertical gridlines void verticalLines() { for (int i = xMin+1; i < xMax; ++i) { draw((i,yMin)--(i,yMax), mediumgray+linewidth(0.4)); } } //Draws the horizontal ticks void horizontalTicks() { for (int i = yMin+1; i < yMax; ++i) { draw((-3/16,i)--(3/16,i), black+linewidth(1)); } } //Draws the vertical ticks void verticalTicks() { for (int i = xMin+1; i < xMax; ++i) { draw((i,-3/16)--(i,3/16), black+linewidth(1)); } } horizontalLines(); verticalLines(); horizontalTicks(); verticalTicks(); draw((xMin,0)--(xMax,0),black+linewidth(1.5),EndArrow(5)); draw((0,yMin)--(0,yMax),black+linewidth(1.5),EndArrow(5)); label("$x$",(xMax,0),(2,0)); label("$y$",(0,yMax),(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)); label("$A(3,5)$",A,(0,2),UnFill); label("$B$",B,(-2,0),UnFill); label("$C$",C,(0,-2),UnFill); label("$D(7,5)$",D,(0,2),UnFill); [/asy] 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

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