2023 AMC 10A Problems/Problem 19

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

Video Solution 1 by OmegaLearn

https://youtu.be/88F18qth0xI

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2023 AMC 10A Problems/Problem 19

Solution 1

Video Solution 1 by OmegaLearn