AMC10 2023 A

AMC10 2023 A · Q19

AMC10 2023 A · Q19. It mainly tests Linear equations, Transformations.

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 $|r-s|$?

由 $A(1, 2)$ 和 $B(3, 3)$ 形成的线段绕点 $P(r, s)$ 旋转成由 $A'(3, 1)$ 和 $B'(4, 3)$ 形成的线段。求 $|r-s|$？

(A) \frac{1}{4} \frac{1}{4}

(B) \frac{1}{2} \frac{1}{2}

(D) \frac{2}{3} \frac{2}{3}

(E) 1 1

Answer

Correct choice: (E)

正确答案：（E）

Solution

Due to rotations preserving an equal 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 $(3-r)^2+(3-s)^2=(4-r)^2+(3-s)^2$. Canceling $(3-s)^2$ from the second equation makes it clear that $r$ equals $3.5$. Substituting will yield \begin{align*}(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 \\ \end{align*}. Now $|r-s| = |3.5-4.5| = \boxed{\textbf{(E) } 1}$.

由于旋转保持等距，我们可以用距离公式暴力求解。$D(A, P) = D(A', P)$，且 $D(B, P) = D(B',P)$。于是我们平方方程得到： $(1-r)^2+(2-s)^2=(3-r)^2+(1-s)^2$，且 $(3-r)^2+(3-s)^2=(4-r)^2+(3-s)^2$。第二个方程消去 $(3-s)^2$ 后明显 $r = 3.5$。代入得到 \begin{align*}(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 \\ \end{align*}。现在 $|r-s| = |3.5-4.5| = \boxed{\textbf{(E)} 1}$。

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