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

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<math>6.25+4-4s+s^2=0.25+1-2s+s^2</math>
 
<math>6.25+4-4s+s^2=0.25+1-2s+s^2</math>
 
<math>2s = 9</math>, <math>s = 4.5</math>.
 
<math>2s = 9</math>, <math>s = 4.5</math>.
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Now <math>|r-s| = |3.5-4.5| = 1</math>.
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Now <math>|r-s| = |3.5-4.5| = \boxed{\textbf{(E) } 1}</math>.
  
 
-Antifreeze5420
 
-Antifreeze5420
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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>.
 
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{1}</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
 
-DEVSAXENA

Revision as of 21:23, 9 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

Video Solution 1 by OmegaLearn

https://youtu.be/88F18qth0xI

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

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