2023 AMC 10A Problems/Problem 19

1 Problem
2 Solution 1
3 Solution 2
4 Solution 3 (Coordinate Geometry)
5 Solution 4
6 Solution 5 (Overkill Trigonometry)
7 Video Solution by Little Fermat
8 Video Solution by Math-X (First fully understand the problem!!!)
9 Video Solution
10 Video Solution by
11 Video Solution 1 by OmegaLearn
12 Video Solution by SpreadTheMathLove
13 Video Solution
14 Video Solution by TheBeautyofMath
15 See Also

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)

Solution 5 (Overkill Trigonometry)

First, we need to find the coordinates of point $ P(r, s) $ such that rotating the line segment from $ A(1,2) $ to $ B(3,3) $ about $ P $ maps it to the line segment from $ A'(3,1) $ to $ B'(4,3) $. After finding $ r $ and $ s $, we will compute $ |r - s| $.

When a point $ (x, y) $ is rotated about another point $ (r, s) $ by an angle $ \theta $, the new coordinates $ (x', y') $ are given by:

\begin{cases} x' = r + (x - r) \cos \theta - (y - s) \sin \theta \\ y' = s + (x - r) \sin \theta + (y - s) \cos \theta \end{cases}

We apply the rotation to points $ A $ and $ B $:

For $ A(1,2) $ mapping to $ A'(3,1) $:

\begin{cases} 3 = r + (1 - r) \cos \theta - (2 - s) \sin \theta \quad \text{(1)} \\ 1 = s + (1 - r) \sin \theta + (2 - s) \cos \theta \quad \text{(2)} \end{cases}

For $ B(3,3) $ mapping to $ B'(4,3) $:

\begin{cases} 4 = r + (3 - r) \cos \theta - (3 - s) \sin \theta \quad \text{(3)} \\ 3 = s + (3 - r) \sin \theta + (3 - s) \cos \theta \quad \text{(4)} \end{cases}

Now, we subtract equations to eliminate variables:

Subtract equation (1) from equation (3):

$(4 - 3) = [r + (3 - r) \cos \theta - (3 - s) \sin \theta] - [r + (1 - r) \cos \theta - (2 - s) \sin \theta]$

Simplifying yields:

$1 = 2 \cos \theta - \sin \theta \quad \text{(A)}$

Similarly, subtract equation (2) from equation (4):

$(3 - 1) = [s + (3 - r) \sin \theta + (3 - s) \cos \theta] - [s + (1 - r) \sin \theta + (2 - s) \cos \theta]$

Simplifying further:

$2 = 2 \sin \theta + \cos \theta \quad \text{(B)}$

Now, using equations (A) and (B):

\begin{cases} 2 \cos \theta - \sin \theta = 1 \\ \cos \theta + 2 \sin \theta = 2 \end{cases}

We now express $ \cos \theta $ from the second equation:

$\cos \theta = 2 - 2 \sin \theta$

Substituting this into the first equation, we get:

$2(2 - 2 \sin \theta) - \sin \theta = 1 \\ 4 - 4 \sin \theta - \sin \theta = 1 \\ -5 \sin \theta = -3 \\ \sin \theta = \frac{3}{5}$

Now, we find $\cos \theta$ as follows:

$\cos \theta = 2 - 2 \left( \frac{3}{5} \right) = \frac{4}{5}$

Substituting $\cos \theta$ and $\sin \theta$ back into equations (1) and (2):

Equation (1):

$3 - r = (1 - r) \left( \frac{4}{5} \right) - (2 - s) \left( \frac{3}{5} \right)$

Multiply both sides by 5:

$5(3 - r) = 4(1 - r) - 3(2 - s)$

Simplify:

$15 - 5r = 4 - 4r - 6 + 3s \\ 15 - 5r = -2 - 4r + 3s \\ -5r + 4r - 3s = -2 - 15 \\ -r - 3s = -17 \quad \text{(C)}$

Equation (2):

$1 - s = (1 - r) \left( \frac{3}{5} \right) + (2 - s) \left( \frac{4}{5} \right)$

Multiply both sides by 5:

$5(1 - s) = 3(1 - r) + 4(2 - s)$

Simplifying more:

$5 - 5s = 3 - 3r + 8 - 4s \\ 5 - 5s = 11 - 3r - 4s \\ -5s + 4s + 3r = 11 - 5 \\ 3r - s = 6 \quad \text{(D)}$

From equations (C) and (D):

$\begin{cases} -r - 3s = -17 \\ 3r - s = 6 \end{cases}$

We then multiply the first equation by 3:

$-3r - 9s = -51$

Add this to the second equation multiplied by 9:

$27r - 9s = 54$

Subtract:

$(27r - 9s) - (-3r - 9s) = 54 - (-51) \\ 30r = 105 \\ r = \frac{105}{30} = \frac{7}{2} = 3.5$

Now find $ s $:

$3(3.5) - s = 6 \\ 10.5 - s = 6 \\ s = 4.5$

Therefore, we have the following:

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

~dbnl

Video Solution by Little Fermat

https://youtu.be/h2Pf2hvF1wE?si=8YX-h2OqRiF8j2y4&t=4238 ~little-fermat

Video Solution by Math-X (First fully understand the problem!!!)

https://youtu.be/GP-DYudh5qU?si=UCyIECgDXCoSakc0&t=5831

~Math-X

Video Solution

https://youtu.be/aYHwBpdWkAk

Video Solution by

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

2023 AMC 10A (Problems • Answer Key • Resources)
Preceded by Problem 18	Followed by Problem 20
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2023 AMC 10A Problems/Problem 19

Contents

Problem

Solution 1

Solution 2

Solution 3 (Coordinate Geometry)

Solution 4

Solution 5 (Overkill Trigonometry)

Video Solution by Little Fermat

Video Solution by Math-X (First fully understand the problem!!!)

Video Solution

Video Solution by

Video Solution 1 by OmegaLearn

Video Solution by SpreadTheMathLove

Video Solution

Video Solution by TheBeautyofMath

See Also