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

Revision as of 20:47, 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 $|r-s|$ ?

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

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 $BP = B^\prime P$ , 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 $|r - s| = |3.5 - 4.5| = \boxed{1}$ .

-DEVSAXENA

Video Solution 1 by OmegaLearn

https://youtu.be/88F18qth0xI

@@ Line 1: / Line 1: @@
+==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>?
-<math>\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</math>
+<math>\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</math>
 ==Solution 1==

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