AMC10 2016 B

AMC10 2016 B · Q20

AMC10 2016 B · Q20. It mainly tests Coordinate geometry, Transformations.

A dilation of the plane—that is, a size transformation with a positive scale factor—sends the circle of radius $2$ centered at $A(2,2)$ to the circle of radius $3$ centered at $A'(5,6)$. What distance does the origin $O(0,0)$ move under this transformation?

平面上的一次伸缩变换（即缩放因子为正的比例变换）把以 $A(2,2)$ 为圆心、半径为 $2$ 的圆变换为以 $A'(5,6)$ 为圆心、半径为 $3$ 的圆。在该变换下，原点 $O(0,0)$ 移动了多远距离？

(A) 0 0

(B) 3 3

(D) 4 4

(E) 5 5

Answer

Correct choice: (C)

正确答案：（C）

Solution

Answer (C): The scale factor for this transformation is $\frac{3}{2}$. The center of the dilation, $D$, must lie along ray $A'A$ (with $A$ between $A'$ and $D$), and its distance from $A$ must be $\frac{2}{3}$ of its distance from $A'$. Because $A$ is 3 units to the left of and 4 units below $A'$, the center of the dilation must be 6 units to the left of and 8 units below $A$, placing it at $D(-4,-6)$. The origin is $\sqrt{(-4)^2+(-6)^2}=2\sqrt{13}$ units from $D$, so the dilation must move it half that far, or $\sqrt{13}$ units. Alternatively, note that the origin is 4 units to the right of and 6 units above $D$, so its image must be 6 units to the right of and 9 units above $D$; therefore it is located at $(2,3)$, a distance $\sqrt{2^2+3^2}=\sqrt{13}$ from the origin.

答案（C）：该变换的伸缩因子为$\frac{3}{2}$。位似中心$D$必须在射线$A'A$上（$A$位于$A'$与$D$之间），且$D$到$A$的距离必须是其到$A'$距离的$\frac{2}{3}$。因为$A$在$A'$的左侧3个单位、下方4个单位，所以位似中心必须在$A$的左侧6个单位、下方8个单位，因此$D(-4,-6)$。原点到$D$的距离为$\sqrt{(-4)^2+(-6)^2}=2\sqrt{13}$，因此位似会将其移动一半的距离，即$\sqrt{13}$。或者注意到原点在$D$的右侧4个单位、上方6个单位，因此其像必须在$D$的右侧6个单位、上方9个单位；所以它位于$(2,3)$，到原点的距离为$\sqrt{2^2+3^2}=\sqrt{13}$。

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