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Difference between revisions of "2021 Fall AMC 10B Problems/Problem 17"

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<math>(\textbf{A})\: 5x+2y=0\qquad(\textbf{B}) \: 3x+2y=0\qquad(\textbf{C}) \: x-3y=0\qquad(\textbf{D}) \: 2x-3y=0\qquad(\textbf{E}) \: 5x-3y=0</math>
 
<math>(\textbf{A})\: 5x+2y=0\qquad(\textbf{B}) \: 3x+2y=0\qquad(\textbf{C}) \: x-3y=0\qquad(\textbf{D}) \: 2x-3y=0\qquad(\textbf{E}) \: 5x-3y=0</math>
  
−
==Solutions==
+
==Solution 1==
 +
Denote <math>O</math> as the origin.
  
−
===Solution 1===
+
Even though the problem is phrased as a coordinate bash, that looks disgusting. Instead, let's try to phrase this problem in terms of Euclidean geometry, using the observation that <math>\angle POP'' = 90^{\circ}</math>, and that both <math>\ell</math> and <math>m</math> must pass through <math>O</math> in order to preserve the distance from <math>P</math> to the origin.
 +
<asy>
 +
unitsize(1.4cm);
 +
draw((0,3)--(0,0)--(3,0), dashed);
 +
dot((0,3));
 +
dot((3,0));
 +
label("$P$", (0,3), W);
 +
label("$P''$", (3,0), S);
 +
 
 +
draw((0,0)--(1.5,4.5));
 +
label("$\ell$", (1.5,4.5), N);
 +
draw((0,0)--(4,2));
 +
label("$m$", (4,2), E);
 +
 
 +
dot((1.8,2.4));
 +
label("$P'$", (1.8,2.4), N);
 +
label("$O$",(0,0));
 +
 
 +
dot((1,3)); dot((2.5,1.25));
 +
label("$A$", (1,3), E); label("$B$", (2.5,1.25), N);
 +
</asy>
 +
(<math>A</math> and <math>B</math> are just defined as points on lines <math>\ell</math> and <math>m</math>.)
 +
Because of how reflections work, we have that <math>\angle AOP' = \angle POA</math> and <math>\angle P'OB = \angle BOP''</math>; adding these two equations together and using angle addition, we have that <math>\angle AOB = \angle POA + \angle BOP''</math>. Since the sum of both sides combined must be <math>90^{\circ}</math> by angle addition,
 +
<cmath>\angle AOB = 45^{\circ}.</cmath>
 +
This is helpful! We can now return to using coordinates, with this piece of information in mind:
 +
<asy>
 +
unitsize(0.2cm);
 +
markscalefactor = 0.08;
 +
import graph;
 +
Label f;
 +
f.p=fontsize(9);
 +
xaxis(-2,6,Ticks(f, 2.0));
 +
yaxis(-1,6,Ticks(f, 2.0));
 +
dot((-1,4));
 +
label("$P$", (-1,4), W);
 +
dot((4,1));
 +
label("$P''$", (4,1), W);
 +
 
 +
draw((0,0)--(1.2,6));
 +
label("$\ell$", (1.2,6), N);
 +
dot((0.5,2.5));label("$(0.5,2.5)$", (0.5,2.5), E);label("$A$", (0.5,2.5), W);
 +
dot((3,2));label("$B$", (3,2), E);
 +
draw((0.5,2.5)--(3,2), dashed);
 +
 
 +
draw((0,0)--(6,4));
 +
label("$m$", (6,4), E);
 +
 
 +
draw(anglemark((6, 4), (0, 0), (1, 5)));
 +
label("$45^{\circ}$", (0.54,0.75));
 +
</asy>
 +
The <math>45^{\circ}</math> angle is a little bit unwieldy in the coordinate plane, so we should try to make a <math>45-45-90</math> triangle. Let <math>A</math> be a point on <math>\ell</math>; to make <math>A</math> fit nicely in the diagram, let it be <math>(0.5,2.5)</math>. Now, let's draw a perpendicular to <math>\ell</math> through point <math>A</math>, intersecting <math>m</math> at point <math>B</math>. <math>OAB</math> is a <math>45-45-90</math> triangle, so <math>B</math> is a <math>90</math> degree counterclockwise rotation from <math>O</math> about <math>A</math>. Therefore, the coordinates of <math>B</math> are
 +
<cmath>(0.5+2.5,2.5-0.5) = (3,2).</cmath>
 +
So, <math>(3,2)</math> is a point on line <math>m</math>, which we already know passes through the origin; therefore, <math>m</math>'s equation is <math>y=\frac{2x}{3} \implies \boxed{\textbf{(D) } 2x-3y = 0}.</math>
 +
 
 +
~ihatemath123
 +
 
 +
(We never actually had to use the information of the exact coordinates of <math>P</math>; as long as <math>\angle POP'' = 90^{\circ}</math>, when we move <math>P</math> around, this will not affect <math>m</math>'s equation.)
 +
===Supplement===
 +
In case you are confused about the coordinates of B, first transform O, A, and B such that A is the origin A'(0,0) in a new coordinate system. From there it is not too hard to see that O now has the coordinates O' (-0.5,-2.5). Thus point B, from a 90 deg CCW rotation around origin A, will have a coordinate of B' (2.5, -0.5).
 +
 
 +
Now in order to go from this new coord system A'(0,0) to its original point A(0.5,2.5), we have to +0.5,+2.5 to the x and y coordinates respectively. Doing this with B' to B, we have (2.5+0.5,-0.5+2.5)=B (3,2) as desired.
 +
 
 +
~mathboy282
 +
 
 +
==Solution 2==
 
It is well known that the composition of 2 reflections , one after another, about two lines <math>l</math> and <math>m</math>, respectively, that meet at an angle <math>\theta</math> is a rotation by <math>2\theta</math> around the intersection of <math>l</math> and <math>m</math>.  
 
It is well known that the composition of 2 reflections , one after another, about two lines <math>l</math> and <math>m</math>, respectively, that meet at an angle <math>\theta</math> is a rotation by <math>2\theta</math> around the intersection of <math>l</math> and <math>m</math>.  
  
 
Now, we note that <math>(4,1)</math> is a 90 degree rotation clockwise of <math>(-1,4)</math> about the origin, which is also where <math>l</math> and <math>m</math> intersect. So <math>m</math> is a 45 degree rotation of <math>l</math> about the origin clockwise.  
 
Now, we note that <math>(4,1)</math> is a 90 degree rotation clockwise of <math>(-1,4)</math> about the origin, which is also where <math>l</math> and <math>m</math> intersect. So <math>m</math> is a 45 degree rotation of <math>l</math> about the origin clockwise.  
  
−
To rotate <math>l</math> 90 degrees clockwise, we build a square with adjacent vertices <math>(0,0)</math> and <math>(1,5)</math>. The other two vertices are at <math>(5,-1)</math> and <math>(6,4)</math>. The center of the square is at <math>(3,2)</math>, which is the midpoint of <math>(1,5)</math> and <math>(5,-1)</math>. The line <math>m</math> passes through the origin and the center of the square we built, namely at <math>(0,0)</math> and <math>(3,2)</math>. Thus the line is <math>y = \frac{2}{3} x</math>. The answer is '''(D)''' <math>\boxed{3y - 2x = 0}</math>.
+
To rotate <math>l</math> 90 degrees clockwise, we build a square with adjacent vertices <math>(0,0)</math> and <math>(1,5)</math>. The other two vertices are at <math>(5,-1)</math> and <math>(6,4)</math>. The center of the square is at <math>(3,2)</math>, which is the midpoint of <math>(1,5)</math> and <math>(5,-1)</math>. The line <math>m</math> passes through the origin and the center of the square we built, namely at <math>(0,0)</math> and <math>(3,2)</math>. Thus the line is <math>y = \frac{2}{3} x</math>. The answer is <math>\boxed{\textbf{(D) } 2x-3y = 0}</math>.
 +
 
 +
~hurdler
 +
 
 +
~minor edits by nightshade2526
 +
 
 +
==Solution 3==
 +
 
 +
We know that the equation of line <math>\ell</math> is <math>y = 5x</math>. This means that <math>P'</math> is <math>(-1,4)</math> reflected over the line <math>y = 5x</math>. This means that the line with <math>P</math> and <math>P'</math> is perpendicular to <math>\ell</math>, so it has slope <math>-\frac{1}{5}</math>. Then the equation of this perpendicular line is <math>y = -\frac{1}{5}x + c</math>, and plugging in <math>(-1,4)</math> for <math>x</math> and <math>y</math> yields <math>c = \frac{19}{5}</math>.
 +
 
 +
The midpoint of <math>P'</math> and <math>P</math> lies at the intersection of <math>y = 5x</math> and <math>y = -\frac{1}{5}x + \frac{19}{5}</math>. Solving, we get the x-value of the intersection is <math>\frac{19}{26}</math> and the y-value is <math>\frac{95}{26}</math>. Let the x-value of <math>P'</math> be <math>x'</math> - then by the midpoint formula, <math>\frac{x' - 1}{2} = \frac{19}{26} \implies x' = \frac{32}{13}</math>. We can find the y-value of <math>P'</math> the same way, so <math>P' = (\frac{32}{13},\frac{43}{13})</math>.
 +
 
 +
Now we have to reflect <math>P'</math> over <math>m</math> to get to <math>(4,1)</math>. The midpoint of <math>P'</math> and <math>P''</math> will lie on <math>m</math>, and this midpoint is, by the midpoint formula, <math>(\frac{42}{13},\frac{28}{13})</math>. <math>y = mx</math> must satisfy this point, so <math>m = \frac{\frac{28}{13}}{\frac{42}{13}} = \frac{28}{42} = \frac{2}{3}</math>.
 +
 
 +
Now the equation of line <math>m</math> is <math>y = \frac{2}{3}x \implies 2x-3y = 0 = \boxed{D}</math>
 +
 
 +
 
 +
~KingRavi
 +
== Solution 4 ==
 +
 
 +
First, use Solution 1's method to get <math>\angle POP'' = 90^\circ</math> and that the angle between lines <math>\ell</math> and <math>m</math> is <math>45^\circ</math>. From here, note that the slope of line <math>m</math> is less than that of line <math>\ell</math> as otherwise <math>P''</math> wouldn't even be close to <math>(4, 1)</math>. Thus, line <math>m</math> is a <math>45^\circ</math> clockwise rotation of line <math>\ell</math>. Line <math>\ell</math> makes an angle of <math>\tan^{-1}(5)</math> with the positive x axis. Thus, line <math>m</math> makes an angle of <math>\tan^{-1}(5) - 45^\circ</math> with the positive x axis. Thus, the slope of line <math>m</math> is
 +
<cmath> \tan (\tan^{-1}(5) - 45^\circ) = \frac{5 - 1}{1 + 5\cdot 1} = \frac{2}{3},</cmath>
 +
by the tangent addition formula. Since the slope of line <math>m</math> is <math>\frac{2}{3}</math>, its equation is <math>y = \frac{2}{3}x \implies 2x - 3y = 0</math>, which is choice <math>\boxed{\textbf{D}}</math>.
 +
 
 +
== Solution 5(cheese) ==
 +
When we graph all the lines and points with a ruler, you can see that a slope of <math>\frac{5}{3}</math> is too big while <math>\frac{1}{3}</math> is too small. We also see that the slope cannot be negative, therefore the answer is <math>\boxed{D}.</math>
 +
~ agentdabber
 +
 
 +
==Video Solution==
 +
 
 +
[https://youtu.be/bqCacR8aXmc|Video Solution 2021 Fall 10B #17]
 +
 
 +
~hurdler
 +
==Video Solution 2 (by Interstigation)==
 +
https://www.youtube.com/watch?v=KdrYlPmqqv0
 +
 
 +
~Interstigation
 +
 
 +
==Video Solution by WhyMath==
 +
https://youtu.be/Wwzqihd3cUg
 +
 
 +
~savannahsolver
 +
==Video Solution by TheBeautyofMath==
 +
https://youtu.be/YOpyq7Zu_hA
 +
==Video Solution by SpreadTheMathLove==
 +
https://www.youtube.com/watch?v=PgFX55o6h1g
 +
 
 +
~IceMatrix
  
 
==See Also==
 
==See Also==
 
{{AMC10 box|year=2021 Fall|ab=B|num-a=18|num-b=16}}
 
{{AMC10 box|year=2021 Fall|ab=B|num-a=18|num-b=16}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Latest revision as of 14:09, 29 September 2024

Problem

Distinct lines $\ell$ and $m$ lie in the $xy$-plane. They intersect at the origin. Point $P(-1, 4)$ is reflected about line $\ell$ to point $P'$, and then $P'$ is reflected about line $m$ to point $P''$. The equation of line $\ell$ is $5x - y = 0$, and the coordinates of $P''$ are $(4,1)$. What is the equation of line $m?$

$(\textbf{A})\: 5x+2y=0\qquad(\textbf{B}) \: 3x+2y=0\qquad(\textbf{C}) \: x-3y=0\qquad(\textbf{D}) \: 2x-3y=0\qquad(\textbf{E}) \: 5x-3y=0$

Solution 1

Denote $O$ as the origin.

Even though the problem is phrased as a coordinate bash, that looks disgusting. Instead, let's try to phrase this problem in terms of Euclidean geometry, using the observation that $\angle POP'' = 90^{\circ}$, and that both $\ell$ and $m$ must pass through $O$ in order to preserve the distance from $P$ to the origin. [asy] unitsize(1.4cm); draw((0,3)--(0,0)--(3,0), dashed); dot((0,3)); dot((3,0)); label("$P$", (0,3), W); label("$P''$", (3,0), S); draw((0,0)--(1.5,4.5)); label("$\ell$", (1.5,4.5), N); draw((0,0)--(4,2)); label("$m$", (4,2), E); dot((1.8,2.4)); label("$P'$", (1.8,2.4), N); label("$O$",(0,0)); dot((1,3)); dot((2.5,1.25)); label("$A$", (1,3), E); label("$B$", (2.5,1.25), N); [/asy] ($A$ and $B$ are just defined as points on lines $\ell$ and $m$.) Because of how reflections work, we have that $\angle AOP' = \angle POA$ and $\angle P'OB = \angle BOP''$; adding these two equations together and using angle addition, we have that $\angle AOB = \angle POA + \angle BOP''$. Since the sum of both sides combined must be $90^{\circ}$ by angle addition, \[\angle AOB = 45^{\circ}.\] This is helpful! We can now return to using coordinates, with this piece of information in mind: [asy] unitsize(0.2cm); markscalefactor = 0.08; import graph; Label f; f.p=fontsize(9); xaxis(-2,6,Ticks(f, 2.0)); yaxis(-1,6,Ticks(f, 2.0)); dot((-1,4)); label("$P$", (-1,4), W); dot((4,1)); label("$P''$", (4,1), W); draw((0,0)--(1.2,6)); label("$\ell$", (1.2,6), N); dot((0.5,2.5));label("$(0.5,2.5)$", (0.5,2.5), E);label("$A$", (0.5,2.5), W); dot((3,2));label("$B$", (3,2), E); draw((0.5,2.5)--(3,2), dashed); draw((0,0)--(6,4)); label("$m$", (6,4), E); draw(anglemark((6, 4), (0, 0), (1, 5))); label("$45^{\circ}$", (0.54,0.75)); [/asy] The $45^{\circ}$ angle is a little bit unwieldy in the coordinate plane, so we should try to make a $45-45-90$ triangle. Let $A$ be a point on $\ell$; to make $A$ fit nicely in the diagram, let it be $(0.5,2.5)$. Now, let's draw a perpendicular to $\ell$ through point $A$, intersecting $m$ at point $B$. $OAB$ is a $45-45-90$ triangle, so $B$ is a $90$ degree counterclockwise rotation from $O$ about $A$. Therefore, the coordinates of $B$ are \[(0.5+2.5,2.5-0.5) = (3,2).\] So, $(3,2)$ is a point on line $m$, which we already know passes through the origin; therefore, $m$'s equation is $y=\frac{2x}{3} \implies \boxed{\textbf{(D) } 2x-3y = 0}.$

~ihatemath123

(We never actually had to use the information of the exact coordinates of $P$; as long as $\angle POP'' = 90^{\circ}$, when we move $P$ around, this will not affect $m$'s equation.)

Supplement

In case you are confused about the coordinates of B, first transform O, A, and B such that A is the origin A'(0,0) in a new coordinate system. From there it is not too hard to see that O now has the coordinates O' (-0.5,-2.5). Thus point B, from a 90 deg CCW rotation around origin A, will have a coordinate of B' (2.5, -0.5).

Now in order to go from this new coord system A'(0,0) to its original point A(0.5,2.5), we have to +0.5,+2.5 to the x and y coordinates respectively. Doing this with B' to B, we have (2.5+0.5,-0.5+2.5)=B (3,2) as desired.

~mathboy282

Solution 2

It is well known that the composition of 2 reflections , one after another, about two lines $l$ and $m$, respectively, that meet at an angle $\theta$ is a rotation by $2\theta$ around the intersection of $l$ and $m$.

Now, we note that $(4,1)$ is a 90 degree rotation clockwise of $(-1,4)$ about the origin, which is also where $l$ and $m$ intersect. So $m$ is a 45 degree rotation of $l$ about the origin clockwise.

To rotate $l$ 90 degrees clockwise, we build a square with adjacent vertices $(0,0)$ and $(1,5)$. The other two vertices are at $(5,-1)$ and $(6,4)$. The center of the square is at $(3,2)$, which is the midpoint of $(1,5)$ and $(5,-1)$. The line $m$ passes through the origin and the center of the square we built, namely at $(0,0)$ and $(3,2)$. Thus the line is $y = \frac{2}{3} x$. The answer is $\boxed{\textbf{(D) } 2x-3y = 0}$.

~hurdler

~minor edits by nightshade2526

Solution 3

We know that the equation of line $\ell$ is $y = 5x$. This means that $P'$ is $(-1,4)$ reflected over the line $y = 5x$. This means that the line with $P$ and $P'$ is perpendicular to $\ell$, so it has slope $-\frac{1}{5}$. Then the equation of this perpendicular line is $y = -\frac{1}{5}x + c$, and plugging in $(-1,4)$ for $x$ and $y$ yields $c = \frac{19}{5}$.

The midpoint of $P'$ and $P$ lies at the intersection of $y = 5x$ and $y = -\frac{1}{5}x + \frac{19}{5}$. Solving, we get the x-value of the intersection is $\frac{19}{26}$ and the y-value is $\frac{95}{26}$. Let the x-value of $P'$ be $x'$ - then by the midpoint formula, $\frac{x' - 1}{2} = \frac{19}{26} \implies x' = \frac{32}{13}$. We can find the y-value of $P'$ the same way, so $P' = (\frac{32}{13},\frac{43}{13})$.

Now we have to reflect $P'$ over $m$ to get to $(4,1)$. The midpoint of $P'$ and $P''$ will lie on $m$, and this midpoint is, by the midpoint formula, $(\frac{42}{13},\frac{28}{13})$. $y = mx$ must satisfy this point, so $m = \frac{\frac{28}{13}}{\frac{42}{13}} = \frac{28}{42} = \frac{2}{3}$.

Now the equation of line $m$ is $y = \frac{2}{3}x \implies 2x-3y = 0 = \boxed{D}$


~KingRavi

Solution 4

First, use Solution 1's method to get $\angle POP'' = 90^\circ$ and that the angle between lines $\ell$ and $m$ is $45^\circ$. From here, note that the slope of line $m$ is less than that of line $\ell$ as otherwise $P''$ wouldn't even be close to $(4, 1)$. Thus, line $m$ is a $45^\circ$ clockwise rotation of line $\ell$. Line $\ell$ makes an angle of $\tan^{-1}(5)$ with the positive x axis. Thus, line $m$ makes an angle of $\tan^{-1}(5) - 45^\circ$ with the positive x axis. Thus, the slope of line $m$ is \[\tan (\tan^{-1}(5) - 45^\circ) = \frac{5 - 1}{1 + 5\cdot 1} = \frac{2}{3},\] by the tangent addition formula. Since the slope of line $m$ is $\frac{2}{3}$, its equation is $y = \frac{2}{3}x \implies 2x - 3y = 0$, which is choice $\boxed{\textbf{D}}$.

Solution 5(cheese)

When we graph all the lines and points with a ruler, you can see that a slope of $\frac{5}{3}$ is too big while $\frac{1}{3}$ is too small. We also see that the slope cannot be negative, therefore the answer is $\boxed{D}.$ ~ agentdabber

Video Solution

Solution 2021 Fall 10B #17

~hurdler

Video Solution 2 (by Interstigation)

https://www.youtube.com/watch?v=KdrYlPmqqv0

~Interstigation

Video Solution by WhyMath

https://youtu.be/Wwzqihd3cUg

~savannahsolver

Video Solution by TheBeautyofMath

https://youtu.be/YOpyq7Zu_hA

Video Solution by SpreadTheMathLove

https://www.youtube.com/watch?v=PgFX55o6h1g

~IceMatrix

See Also

2021 Fall AMC 10B (Problems • Answer Key • Resources)
Preceded by
Problem 16 		Followed by
Problem 18
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 10 Problems and Solutions

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