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

Revision as of 17:48, 10 November 2023

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 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 $(3-r)^2+(3-s)^2=(4-r)^2+(3-s)^2$ . Cancelling $(3-s)^2$ from the second equation makes it clear that r equals 3.5. Now substituting will yield $(2.5)^2+(2-s)^2=(-0.5)^2+(1-s)^2$ . $6.25+4-4s+s^2=0.25+1-2s+s^2$ $2s = 9$ , $s = 4.5$ . 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 $2x-\frac{5}{2} = 3.5$ . Solving for x, we get that $x = \frac{9}{2}$ . 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

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 1 by OmegaLearn

https://youtu.be/88F18qth0xI

@@ Line 57: / Line 57: @@
 ~aydenlee & wuwang2002
+==Solution==
+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 1 by OmegaLearn ==

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

Page

Toolbox

Search