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

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-Antifreeze5420
 
-Antifreeze5420
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==Solution 2==
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Due to rotations preserving distance, we have that <math>BP = B^\primeP</math>, as well as <math>AP = A^\primeP</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.
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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>.
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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>.
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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>.
  
 
== Video Solution 1 by OmegaLearn ==
 
== Video Solution 1 by OmegaLearn ==
 
https://youtu.be/88F18qth0xI
 
https://youtu.be/88F18qth0xI

Revision as of 19:31, 9 November 2023

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 $|r-s|$?

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

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 $|r-s| = |3.5-4.5| = 1$.

-Antifreeze5420

Solution 2

Due to rotations preserving distance, we have that $BP = B^\primeP$ (Error compiling LaTeX. Unknown error_msg), as well as $AP = A^\primeP$ (Error compiling LaTeX. Unknown error_msg). 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 $|r - s| = |3.5 - 4.5| = \boxed{1}$.

Video Solution 1 by OmegaLearn

https://youtu.be/88F18qth0xI

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