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

m (→‎Solution 1 (Geometry))
(→‎Solution 4 (System of linear equations))
 
(17 intermediate revisions by 3 users not shown)
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~MRENTHUSIASM
 
~MRENTHUSIASM
  
−
==Solution 1 (Geometry)==
+
==Solution 1 (Reflections)==
 
Let <math>A=(3,5)</math> and <math>D=(7,5).</math> Suppose that the beam hits and bounces off the <math>y</math>-axis at <math>B,</math> then hits and bounces off the <math>x</math>-axis at <math>C.</math>
 
Let <math>A=(3,5)</math> and <math>D=(7,5).</math> Suppose that the beam hits and bounces off the <math>y</math>-axis at <math>B,</math> then hits and bounces off the <math>x</math>-axis at <math>C.</math>
  
−
When the beam hits and bounces off a coordinate axis, the angle of incidence and the angle of reflection are congruent. So, we straighten up the path of the beam by reflections:
+
When the beam hits and bounces off a coordinate axis, the angle of incidence and the angle of reflection are congruent. Therefore, we straighten up the path of the beam by reflections:
 
<ol style="margin-left: 1.5em;">
 
<ol style="margin-left: 1.5em;">
 
   <li>We reflect <math>\overline{BC}</math> about the <math>y</math>-axis to get <math>\overline{BC'}.</math></li><p>
 
   <li>We reflect <math>\overline{BC}</math> about the <math>y</math>-axis to get <math>\overline{BC'}.</math></li><p>
−
   <li>We reflect <math>\overline{CD}</math> about the <math>x</math>-axis to get <math>\overline{C'D'},</math> then reflect <math>\overline{C'D'}</math> about the <math>y</math>-axis to get <math>\overline{C'D''}.</math></li><p>
+
   <li>We reflect <math>\overline{CD}</math> about the <math>x</math>-axis to get <math>\overline{C'D'}</math> with <math>D'=(7,-5),</math> then reflect <math>\overline{C'D'}</math> about the <math>y</math>-axis to get <math>\overline{C'D''}</math> with <math>D''=(-7,-5).</math></li><p>
 
</ol>
 
</ol>
 
We obtain the following diagram:
 
We obtain the following diagram:
Line 74: Line 74:
 
dot("$D(7,5)$",D,(0,2),linewidth(3.5));
 
dot("$D(7,5)$",D,(0,2),linewidth(3.5));
 
dot("$C'$",E,(0,-2),linewidth(3.5));
 
dot("$C'$",E,(0,-2),linewidth(3.5));
−
dot("$D'$",F,(0,-2),linewidth(3.5));
+
dot("$D'(7,-5)$",F,(0,-2),linewidth(3.5));
−
dot("$D''$",G,(0,-2),linewidth(3.5));
+
dot("$D''(-7,-5)$",G,(0,-2),linewidth(3.5));
 
</asy>
 
</asy>
−
It follows that <math>D'=(7,-5)</math> and <math>D''=(-7,-5).</math> The total distance that the beam will travel is  
+
The total distance that the beam will travel is  
 
<cmath>\begin{align*}
 
<cmath>\begin{align*}
 
AB+BC+CD&=AB+BC+CD' \\
 
AB+BC+CD&=AB+BC+CD' \\
Line 89: Line 89:
 
~JHawk0224 (Proposal)
 
~JHawk0224 (Proposal)
  
−
==Solution 2 (Algebra)==
+
==Solution 2 (Parallelogram)==
−
Define points <math>A,B,C,</math> and <math>D</math> as Solution 1 does.
+
Define points <math>A,B,C,</math> and <math>D</math> as Solution 1 does. Moreover, let <math>E</math> be a point on <math>\overline{CD}</math> such that <math>\overline{BE}</math> is perpendicular to the <math>y</math>-axis, and <math>F</math> be a point on <math>\overline{BE}</math> such that <math>\overline{CF}</math> is perpendicular to the <math>x</math>-axis, as shown below.
−
 
 
−
When a straight line 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. Let <math>\ell</math> be the line containing <math>P</math> and perpendicular to that coordinate axis. <i><b>Geometrically, these two rays coincide when reflected about <math>\boldsymbol{\ell}.</math></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> We obtain parallelogram <math>ABED,</math> as shown below.  
 
 
<asy>
 
<asy>
 
/* Made by MRENTHUSIASM */
 
/* Made by MRENTHUSIASM */
Line 116: Line 110:
 
pair D = (7,5);
 
pair D = (7,5);
 
pair E = (4,2);
 
pair E = (4,2);
 +
pair F = (2,2);
  
 
draw(A--B--C--D,red);
 
draw(A--B--C--D,red);
−
draw(A--D,heavygreen+dashed);
+
draw(A--D^^B--E^^C--F,heavygreen+dashed);
−
draw(B--E,heavygreen+dashed);
+
dot("$A(3,5)$",A,(0,2),linewidth(3.5));
−
dot(A,linewidth(3.5));
+
dot("$B$",B,(-2,0),linewidth(3.5));
−
dot(B,linewidth(3.5));
+
dot("$C$",C,(0,-2),linewidth(3.5));
−
dot(C,linewidth(3.5));
+
dot("$D(7,5)$",D,(0,2),linewidth(3.5));
−
dot(D,linewidth(3.5));
+
dot("$E$",E,(2,0),linewidth(3.5));
−
dot(E,linewidth(3.5));
+
dot("$F$",F,(0,1.5),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>
 
</asy>
−
Let <math>B=(0,b).</math> In parallelogram <math>ABED,</math> we get <math>E=(4,b).</math> By symmetry, we get <math>C=(2,0).</math>
+
When the beam hits and bounces off a coordinate axis, the angle of incidence and the angle of reflection are congruent, from which <math>\angle ABF=\angle CBF</math> and <math>\angle BCF=\angle ECF.</math> We conclude that <math>\triangle BCF\cong\triangle ECF</math> by ASA, so <math>\angle CBF=\angle CEF.</math> It follows that <math>\angle ABF=\angle CEF</math> by transitive, so <math>\overline{AB}\parallel\overline{CD}</math> by the Converse of the Alternate Interior Angles Theorem.
  
−
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>
+
Note that <math>\overline{AD}\parallel\overline{BE}.</math> Since the opposite sides are parallel, quadrilateral <math>ABED</math> is a parallelogram. From <math>\triangle BCF\cong\triangle ECF,</math> we get <math>BF=EF=2,</math> so <math>C=(2,0).</math>
 +
 
 +
Let <math>B=(0,b).</math> We equate the slopes of <math>\overline{AB}</math> and <math>\overline{DC}:</math> <cmath>\frac{5-b}{3-0}=\frac{5-0}{7-2},</cmath> from which <math>b=2,</math> or <math>B=(0,2).</math>
  
 
By the Distance Formula, we have <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, we have <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>
 +
 +
<u><b>Remark</b></u>
 +
 +
When a straight line 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> always have negative slopes. In this problem, <math>\overline{AB}</math> and <math>\overline{BC}</math> have negative slopes; <math>\overline{BC}</math> and <math>\overline{CD}</math> have negative slopes. So, <math>\overline{AB}</math> and <math>\overline{CD}</math> have the same slope, or <math>\overline{AB}\parallel\overline{CD}.</math>
 +
 
~MRENTHUSIASM
 
~MRENTHUSIASM
  
−
==Solution 3 (Answer Choices and Educated Guesses)==
+
==Solution 3 (Educated Guess)==
 
Define points <math>A,B,C,</math> and <math>D</math> as Solution 1 does.
 
Define points <math>A,B,C,</math> and <math>D</math> as Solution 1 does.
  
Line 214: Line 211:
  
 
~MRENTHUSIASM
 
~MRENTHUSIASM
 +
 +
==Solution 4 (System of Linear Equations)==
 +
Denote <math>(3, 5)</math> as point <math>P</math> and <math>(7, 5)</math> as point <math>S</math>. Define the point <math>Q</math> on the <math>y</math>-axis that the laser hits as <math>(0, y_1)</math> and the point <math>R</math> on the <math>x</math>-axis that the laser hits as <math>(x_1,0)</math>. The laser will bounce off of the two axes at right angles, so we have that line <math>PQ \parallel RS</math>, meaning they have the same slope which we will denote as <math>k</math>. Line <math>QR</math> will be perpendicular to both <math>PQ</math> and <math>RS</math>, meaning it will have slope <math>-\frac{1}{k}</math>.
 +
 +
We can write the equations for lines <math>PQ</math> and <math>RS</math> in point-slope form, and then plug in points <math>Q</math> and <math>R</math> into those respective equations to get equations in terms of our three variables. We can also write the equation for line <math>QR</math> in slope-intercept form and plug in point <math>R</math>. Doing this yields
 +
<cmath>\begin{align*}
 +
    y_1-5&=k(0-3)\\
 +
    0-5&=k(x_1-7)\\
 +
    0&=-\frac{1}{k}x_1+y_1.
 +
\end{align*}</cmath>
 +
We can simplify this into the following cubic equation in terms of <math>k</math>:
 +
<cmath>-3k^3+5k^2-7k+5=0.</cmath>
 +
Luckily, we notice that the coefficients sum up to zero, so <math>k=1</math>. Now we can solve for the other two, yielding <math>x_1=y_1=2</math>. Now we simply have three <math>45^{\circ}</math>-<math>45^{\circ}</math>-<math>90^{\circ}</math> triangles with leg lengths of <math>3, 2,</math> and <math>5</math>, so our final distance is <cmath>3\sqrt{2}+2\sqrt{2}+5\sqrt{2}=\boxed{\textbf{(C) }10\sqrt2}.</cmath>
 +
~Mooshiros
  
 
== Video Solution by OmegaLearn (Using Reflections and Distance Formula) ==
 
== Video Solution by OmegaLearn (Using Reflections and Distance Formula) ==
Line 227: Line 238:
  
 
~IceMatrix
 
~IceMatrix
 +
 +
==Video Solution (Quick and Easy)==
 +
https://youtu.be/Hw-mYGUvflQ
 +
 +
~Education, the Study of Everything
  
 
==See also==
 
==See also==
 
{{AMC12 box|year=2021|ab=A|num-b=10|num-a=12}}
 
{{AMC12 box|year=2021|ab=A|num-b=10|num-a=12}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Latest revision as of 00:28, 1 November 2023

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 (Reflections)

Let $A=(3,5)$ and $D=(7,5).$ Suppose that the beam hits and bounces off the $y$-axis at $B,$ then hits and bounces off the $x$-axis at $C.$

When the beam hits and bounces off a coordinate axis, the angle of incidence and the angle of reflection are congruent. Therefore, we straighten up the path of the beam by reflections:

  1. We reflect $\overline{BC}$ about the $y$-axis to get $\overline{BC'}.$

  2. We reflect $\overline{CD}$ about the $x$-axis to get $\overline{C'D'}$ with $D'=(7,-5),$ then reflect $\overline{C'D'}$ about the $y$-axis to get $\overline{C'D''}$ with $D''=(-7,-5).$

We obtain the following diagram: [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(C--F,heavygreen+dashed); draw(E--G,heavygreen+dashed); dot("$A(3,5)$",A,(0,2),linewidth(3.5)); dot("$B$",B,(-2,0),linewidth(3.5)); dot("$C$",C,(0,-2),linewidth(3.5)); dot("$D(7,5)$",D,(0,2),linewidth(3.5)); dot("$C'$",E,(0,-2),linewidth(3.5)); dot("$D'(7,-5)$",F,(0,-2),linewidth(3.5)); dot("$D''(-7,-5)$",G,(0,-2),linewidth(3.5)); [/asy] The total distance that the beam will travel is \begin{align*} AB+BC+CD&=AB+BC+CD' \\ &=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 (Parallelogram)

Define points $A,B,C,$ and $D$ as Solution 1 does. Moreover, let $E$ be a point on $\overline{CD}$ such that $\overline{BE}$ is perpendicular to the $y$-axis, and $F$ be a point on $\overline{BE}$ such that $\overline{CF}$ is perpendicular to the $x$-axis, 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); pair F = (2,2); draw(A--B--C--D,red); draw(A--D^^B--E^^C--F,heavygreen+dashed); dot("$A(3,5)$",A,(0,2),linewidth(3.5)); dot("$B$",B,(-2,0),linewidth(3.5)); dot("$C$",C,(0,-2),linewidth(3.5)); dot("$D(7,5)$",D,(0,2),linewidth(3.5)); dot("$E$",E,(2,0),linewidth(3.5)); dot("$F$",F,(0,1.5),linewidth(3.5)); [/asy] When the beam hits and bounces off a coordinate axis, the angle of incidence and the angle of reflection are congruent, from which $\angle ABF=\angle CBF$ and $\angle BCF=\angle ECF.$ We conclude that $\triangle BCF\cong\triangle ECF$ by ASA, so $\angle CBF=\angle CEF.$ It follows that $\angle ABF=\angle CEF$ by transitive, so $\overline{AB}\parallel\overline{CD}$ by the Converse of the Alternate Interior Angles Theorem.

Note that $\overline{AD}\parallel\overline{BE}.$ Since the opposite sides are parallel, quadrilateral $ABED$ is a parallelogram. From $\triangle BCF\cong\triangle ECF,$ we get $BF=EF=2,$ so $C=(2,0).$

Let $B=(0,b).$ We equate the slopes of $\overline{AB}$ and $\overline{DC}:$ \[\frac{5-b}{3-0}=\frac{5-0}{7-2},\] from which $b=2,$ or $B=(0,2).$

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 \[AB+BC+CD=\boxed{\textbf{(C) }10\sqrt2}.\]

Remark

When a straight line hits and bounces off a coordinate axis at point $P,$ the ray entering $P$ and the ray leaving $P$ always have negative slopes. In this problem, $\overline{AB}$ and $\overline{BC}$ have negative slopes; $\overline{BC}$ and $\overline{CD}$ have negative slopes. So, $\overline{AB}$ and $\overline{CD}$ have the same slope, or $\overline{AB}\parallel\overline{CD}.$

~MRENTHUSIASM

Solution 3 (Educated Guess)

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

Solution 4 (System of Linear Equations)

Denote $(3, 5)$ as point $P$ and $(7, 5)$ as point $S$. Define the point $Q$ on the $y$-axis that the laser hits as $(0, y_1)$ and the point $R$ on the $x$-axis that the laser hits as $(x_1,0)$. The laser will bounce off of the two axes at right angles, so we have that line $PQ \parallel RS$, meaning they have the same slope which we will denote as $k$. Line $QR$ will be perpendicular to both $PQ$ and $RS$, meaning it will have slope $-\frac{1}{k}$.

We can write the equations for lines $PQ$ and $RS$ in point-slope form, and then plug in points $Q$ and $R$ into those respective equations to get equations in terms of our three variables. We can also write the equation for line $QR$ in slope-intercept form and plug in point $R$. Doing this yields \begin{align*} y_1-5&=k(0-3)\\ 0-5&=k(x_1-7)\\ 0&=-\frac{1}{k}x_1+y_1. \end{align*} We can simplify this into the following cubic equation in terms of $k$: \[-3k^3+5k^2-7k+5=0.\] Luckily, we notice that the coefficients sum up to zero, so $k=1$. Now we can solve for the other two, yielding $x_1=y_1=2$. Now we simply have three $45^{\circ}$-$45^{\circ}$-$90^{\circ}$ triangles with leg lengths of $3, 2,$ and $5$, so our final distance is \[3\sqrt{2}+2\sqrt{2}+5\sqrt{2}=\boxed{\textbf{(C) }10\sqrt2}.\] ~Mooshiros

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

Video Solution (Quick and Easy)

https://youtu.be/Hw-mYGUvflQ

~Education, the Study of Everything

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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