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Difference between revisions of "2023 AMC 10A Problems/Problem 19"

(Created page with "@Announcement Ping @offtopic announce ping @random A lot of people have been asking me whether I am joking or not when I say the rumored leak that was posted on math discords...")
 
m (→‎Solution 3 (Coordinate Geometry))
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@Announcement Ping @offtopic announce ping @random A lot of people have been asking me whether I am joking or not when I say the rumored leak that was posted on math discords and AoPS today was legit. I am not joking, and I have very reasonable evidence to suggest that the leak is legit. The person who has posted this leak has been banned from this server. People have been reporting this individual spamming various math servers and spamming numerous dms. It's pretty clear this individual does not have good intentions. I think this individual is trying to maliciously sabotage the MAA's efforts (wouldn't be the first time something like this has happened).
+
==Problem==
 +
The line segment formed by <math>A(1, 2)</math> and <math>B(3, 3)</math> is rotated to the line segment formed by <math>A'(3, 1)</math> and <math>B'(4, 3)</math> about the point <math>P(r, s)</math>. What is <math>|r-s|</math>?
  
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Theres been two reputable people who've come to me independently, saying that they asked their proctor, and the proctor confirmed that the test matched. How would the proctor know what the test is? They get access to the test in advance for print and scan administration (see screenshot from MAA's official proctor instructions slideshow).
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<math>\textbf{(A) } \frac{1}{4} \qquad \textbf{(B) } \frac{1}{2} \qquad \textbf{(C) } \frac{3}{4}  \qquad \textbf{(D) } \frac{2}{3} \qquad  \textbf{(E) } 1</math>
  
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How would this individual get access to the pdf? I suspect negligence on their proctors part, and somehow they weren't careful with keeping the pdf they received private. I along with other people are trying to contact MAA to resolve this situation. A lot of people still refuse to believe me, and I admit, I first thought this possibility was preposterous. But I have my entire organization at stake with this, and I will take no chances. If you see this individual anywhere, please notify me immediately.
+
==Solution 1==
  
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Will the test actually get fixed? I'm worried it might be too late. AMC questions get leaked on stackexchange and wechat all the time, and the MAA hasn't done much to address this. In the end, math competitions are about a learning experience, not about getting a score.
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Due to rotations preserving an equal distance, we can bash the answer with the distance formula. <math>D(A, P) = D(A', P)</math>, and <math>D(B, P) = D(B',P)</math>.
 +
Thus we will square our equations to yield:
 +
<math>(1-r)^2+(2-s)^2=(3-r)^2+(1-s)^2</math>, and <math>(3-r)^2+(3-s)^2=(4-r)^2+(3-s)^2</math>.
 +
Canceling <math>(3-s)^2</math> from the second equation makes it clear that <math>r</math> equals <math>3.5</math>.
 +
 
 +
Substituting will yield
 +
 
 +
<cmath>\begin{align*}(2.5)^2+(2-s)^2 &= (-0.5)^2+(1-s)^2 \\
 +
6.25+4-4s+s^2 &= 0.25+1-2s+s^2 \\
 +
2s &= 9 \\
 +
s &=4.5 \\
 +
\end{align*}
 +
</cmath>.
 +
 
 +
Now <math>|r-s| = |3.5-4.5| = \boxed{\textbf{(E) } 1}</math>.
 +
 
 +
-Antifreeze5420
 +
 
 +
==Solution 2==
 +
 
 +
Due to rotations preserving distance, we have that <math>BP = B^\prime P</math>, as well as <math>AP = A^\prime P</math>. From here, we can see that P must be on the perpendicular bisector of <math>\overline{BB^\prime}</math> due to the property of perpendicular bisectors keeping the distance to two points constant.
 +
 
 +
From here, we proceed to find the perpendicular bisector of <math>\overline{BB^\prime}</math>. We can see that this is just a horizontal line segment with midpoint at <math>(3.5, 3)</math>. This means that the equation of the perpendicular bisector is <math>x = 3.5</math>.
 +
 
 +
Similarly, we find the perpendicular bisector of <math>\overline{AA^\prime}</math>. We find the slope to be <math>\frac{1-2}{3-1} = -\frac12</math>, so our new slope will be <math>2</math>. The midpoint of <math>A</math> and <math>A^\prime</math> is <math>(2, \frac32)</math>, which we can use with our slope to get another equation of <math>y = 2x - \frac52</math>.
 +
 
 +
Now, point P has to lie on both of these perpendicular bisectors, meaning that it has to satisfy both equations. Plugging in the value of <math>x</math> we found earlier, we find that <math>y=4.5</math>. This means that <math>|r - s| = |3.5 - 4.5| = \boxed{\textbf{(E) } 1}</math>.
 +
 
 +
-DEVSAXENA
 +
 
 +
==Solution 3 (Coordinate Geometry)==
 +
 
 +
To find the center of rotation, we find the intersection point of the perpendicular bisectors of <math>\overline{AA^\prime}</math> and <math>\overline{BB^\prime}</math>.
 +
 
 +
We can find that the equation of the line <math>\overline{AA^\prime}</math> is <math>y = -\frac{1}{2}x + \frac{5}{2}</math>, and that the equation of the line <math>\overline{BB^\prime}</math> is <math>y = 3</math>.
 +
 
 +
When we solve for the perpendicular bisector of  <math>y = -\frac{1}{2}x + \frac{5}{2}</math>, we determine that it has a slope of 2, and it runs through <math>(2, 1.5)</math>. Plugging in <math>1.5 = 2(2)-n</math>, we get than <math>n = \frac{5}{2}</math>. Therefore our perpendicular bisector is <math>y=2x-\frac{5}{2}</math>. Next, we solve for the perpendicular of <math>y = 3</math>. We know that it has an undefined slope, and it runs through <math>(3.5, 3)</math>. We can determine that our second perpendicular bisector is <math>x = 3.5</math>.
 +
 
 +
Setting the equations equal to each other, we get <math>y = \frac{9}{2}</math>. Therefore, <math>|r - s| = |3.5 - 4.5| = \boxed{\textbf{(E) } 1}</math>.
 +
 
 +
<asy>
 +
pair A=(1,2);
 +
pair B=(3,3);
 +
pair A1=(3,1);
 +
pair B1=(4,3);
 +
dot("A",A,NW);
 +
dot("B",B,S);
 +
dot("A'",A1,S);
 +
dot("B'",B1,E);
 +
draw(A--A1);
 +
draw(B--B1);
 +
draw((3.5,0)--(3.5,6),BeginArrow(5),EndArrow(5));
 +
draw((1,-0.5)--(4,5.5),BeginArrow(5),EndArrow(5));
 +
pair P=(3.5,4.5);
 +
dot("P",P,NW);
 +
</asy>
 +
 
 +
~aydenlee & wuwang2002
 +
 
 +
==Solution 4==
 +
 
 +
We use the complex numbers approach to solve this problem.
 +
Denote by <math>\theta</math> the angle that <math>AB</math> rotates about <math>P</math> in the counterclockwise direction.
 +
 
 +
Thus, <math>A' - P = e^{i \theta} \left( A - P \right)</math> and <math>B' - P = e^{i \theta} \left( B - P \right)</math>.
 +
 
 +
Taking ratio of these two equations, we get
 +
<cmath>
 +
\[
 +
\frac{A' - P}{A - P} = \frac{B' - P}{B - P} .
 +
\]
 +
</cmath>
 +
 
 +
By solving this equation, we get <math>P = \frac{7}{2} + i \frac{9}{2}</math>.
 +
Therefore, <math>|s-t| = \left| \frac{7}{2} - \frac{9}{2} \right| = \boxed{\textbf{(E) 1}}</math>.
 +
 
 +
~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)
 +
 
 +
== Video Solution by Power Solve (easy to understand!) ==
 +
https://youtu.be/aYHwBpdWkAk
 +
 
 +
== Video Solution by CosineMethod [🔥Fast and Easy🔥]==
 +
 
 +
https://www.youtube.com/watch?v=fIzCR4x4x-M
 +
 
 +
== Video Solution 1 by OmegaLearn ==
 +
https://youtu.be/88F18qth0xI
 +
 
 +
==Video Solution by SpreadTheMathLove==
 +
https://www.youtube.com/watch?v=oHMJwiEwOS0
 +
 
 +
==Video Solution==
 +
 
 +
https://youtu.be/va3ZCFeKfzg
 +
 
 +
~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)
 +
== Video Solution by TheBeautyofMath==
 +
https://youtu.be/Q_4uMxMbQlI
 +
 
 +
~IceMatrix
 +
==See Also==
 +
{{AMC10 box|year=2023|ab=A|num-b=18|num-a=20}}
 +
{{MAA Notice}}

Revision as of 17:40, 28 July 2024

Problem

The line segment formed by $A(1, 2)$ and $B(3, 3)$ is rotated to the line segment formed by $A'(3, 1)$ and $B'(4, 3)$ about the point $P(r, s)$. What is $|r-s|$?

$\textbf{(A) } \frac{1}{4} \qquad \textbf{(B) } \frac{1}{2} \qquad \textbf{(C) } \frac{3}{4} \qquad \textbf{(D) } \frac{2}{3} \qquad \textbf{(E) } 1$

Solution 1

Due to rotations preserving an equal distance, we can bash the answer with the distance formula. $D(A, P) = D(A', P)$, and $D(B, P) = D(B',P)$. Thus we will square our equations to yield: $(1-r)^2+(2-s)^2=(3-r)^2+(1-s)^2$, and $(3-r)^2+(3-s)^2=(4-r)^2+(3-s)^2$. Canceling $(3-s)^2$ from the second equation makes it clear that $r$ equals $3.5$.

Substituting will yield

\begin{align*}(2.5)^2+(2-s)^2 &= (-0.5)^2+(1-s)^2 \\ 6.25+4-4s+s^2 &= 0.25+1-2s+s^2 \\ 2s &= 9 \\ s &=4.5 \\ \end{align*}.

Now $|r-s| = |3.5-4.5| = \boxed{\textbf{(E) } 1}$.

-Antifreeze5420

Solution 2

Due to rotations preserving distance, we have that $BP = B^\prime P$, as well as $AP = A^\prime P$. From here, we can see that P must be on the perpendicular bisector of $\overline{BB^\prime}$ due to the property of perpendicular bisectors keeping the distance to two points constant.

From here, we proceed to find the perpendicular bisector of $\overline{BB^\prime}$. We can see that this is just a horizontal line segment with midpoint at $(3.5, 3)$. This means that the equation of the perpendicular bisector is $x = 3.5$.

Similarly, we find the perpendicular bisector of $\overline{AA^\prime}$. We find the slope to be $\frac{1-2}{3-1} = -\frac12$, so our new slope will be $2$. The midpoint of $A$ and $A^\prime$ is $(2, \frac32)$, which we can use with our slope to get another equation of $y = 2x - \frac52$.

Now, point P has to lie on both of these perpendicular bisectors, meaning that it has to satisfy both equations. Plugging in the value of $x$ we found earlier, we find that $y=4.5$. This means that $|r - s| = |3.5 - 4.5| = \boxed{\textbf{(E) } 1}$.

-DEVSAXENA

Solution 3 (Coordinate Geometry)

To find the center of rotation, we find the intersection point of the perpendicular bisectors of $\overline{AA^\prime}$ and $\overline{BB^\prime}$.

We can find that the equation of the line $\overline{AA^\prime}$ is $y = -\frac{1}{2}x + \frac{5}{2}$, and that the equation of the line $\overline{BB^\prime}$ is $y = 3$.

When we solve for the perpendicular bisector of $y = -\frac{1}{2}x + \frac{5}{2}$, we determine that it has a slope of 2, and it runs through $(2, 1.5)$. Plugging in $1.5 = 2(2)-n$, we get than $n = \frac{5}{2}$. Therefore our perpendicular bisector is $y=2x-\frac{5}{2}$. Next, we solve for the perpendicular of $y = 3$. We know that it has an undefined slope, and it runs through $(3.5, 3)$. We can determine that our second perpendicular bisector is $x = 3.5$.

Setting the equations equal to each other, we get $y = \frac{9}{2}$. Therefore, $|r - s| = |3.5 - 4.5| = \boxed{\textbf{(E) } 1}$.

[asy] pair A=(1,2); pair B=(3,3); pair A1=(3,1); pair B1=(4,3); dot("A",A,NW); dot("B",B,S); dot("A'",A1,S); dot("B'",B1,E); draw(A--A1); draw(B--B1); draw((3.5,0)--(3.5,6),BeginArrow(5),EndArrow(5)); draw((1,-0.5)--(4,5.5),BeginArrow(5),EndArrow(5)); pair P=(3.5,4.5); dot("P",P,NW); [/asy]

~aydenlee & wuwang2002

Solution 4

We use the complex numbers approach to solve this problem. Denote by $\theta$ the angle that $AB$ rotates about $P$ in the counterclockwise direction.

Thus, $A' - P = e^{i \theta} \left( A - P \right)$ and $B' - P = e^{i \theta} \left( B - P \right)$.

Taking ratio of these two equations, we get \[ \frac{A' - P}{A - P} = \frac{B' - P}{B - P} . \]

By solving this equation, we get $P = \frac{7}{2} + i \frac{9}{2}$. Therefore, $|s-t| = \left| \frac{7}{2} - \frac{9}{2} \right| = \boxed{\textbf{(E) 1}}$.

~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)

Video Solution by Power Solve (easy to understand!)

https://youtu.be/aYHwBpdWkAk

Video Solution by CosineMethod [🔥Fast and Easy🔥]

https://www.youtube.com/watch?v=fIzCR4x4x-M

Video Solution 1 by OmegaLearn

https://youtu.be/88F18qth0xI

Video Solution by SpreadTheMathLove

https://www.youtube.com/watch?v=oHMJwiEwOS0

Video Solution

https://youtu.be/va3ZCFeKfzg

~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)

Video Solution by TheBeautyofMath

https://youtu.be/Q_4uMxMbQlI

~IceMatrix

See Also

2023 AMC 10A (Problems • Answer Key • Resources)
Preceded by
Problem 18 		Followed by
Problem 20
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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

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