AMC10 2014 A

AMC10 2014 A · Q18

AMC10 2014 A · Q18. It mainly tests Coordinate geometry, Transformations.

A square in the coordinate plane has vertices whose $y$-coordinates are 0, 1, 4, and 5. What is the area of the square?

坐标平面上的一个正方形，其顶点的 $y$ 坐标为 0、1、4 和 5。该正方形的面积是多少？

(A) 16 16

(B) 17 17

(D) 26 26

(E) 27 27

Answer

Correct choice: (B)

正确答案：（B）

Solution

Let the square have vertices $A, B, C, D$ in counterclockwise order. Without loss of generality assume that $A = (0, 0)$ and $B = (x, 1)$ for some $x > 0$. Because $D$ is the image of $B$ under a 90° counterclockwise rotation about $A$, the coordinates of $D$ are $(-1, x)$, so $x = 4$. Therefore the area of the square is $(AB)^2 = 4^2 + 1^2 = 17$. Note that $C = (3, 5)$, and $ABCD$ is indeed a square.

设正方形顶点为 $A, B, C, D$，按逆时针顺序。不妨设 $A = (0, 0)$，$B = (x, 1)$，$x > 0$。因为 $D$ 是 $B$ 绕 $A$ 逆时针旋转 90° 后的像，所以 $D$ 的坐标为 $(-1, x)$，故 $x = 4$。因此正方形面积为 $(AB)^2 = 4^2 + 1^2 = 17$。注意 $C = (3, 5)$，$ABCD$ 确实是正方形。

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