AMC8 2026

AMC8 2026 · Q13

AMC8 2026 · Q13. It mainly tests Coordinate geometry, Geometry misc.

The figure below shows a tiling of $1 \times 1$ unit squares. Each row of unit squares is shifted horizontally by half a unit relative to the row above it. A shaded square is drawn on top of the tiling. Each vertex of the shaded square is a vertex of one of the unit squares. In square units, what is the area of the shaded square?

下图显示了由 $1 \times 1$ 单位正方形组成的铺砌。每一行单位正方形相对于上一行水平移动半个单位。在铺砌上画出了一个阴影正方形。阴影正方形的每个顶点都是某个单位正方形的顶点。该阴影正方形的面积（单位为平方单位）是多少？

(A) 10 10

(B) 21/2 21/2

(D) 11 11

(E) 34/3 34/3

Answer

Correct choice: (A)

正确答案：（A）

Solution

Label the vertices $A$, $B$, $C$, $D$, in clockwise order. Then, we use coordinate geometry to easily solve this. Let point $A$ be on $(0, 0)$. Then, point $B$ is 1 unit up and 3 right of $(0, 0)$, or $(1, 3)$. The figure is said to be a square. and so we just calculate the distance from $(0, 0)$ to $(1, 3)$. Using the distance formula, we have $\sqrt{(1-0)^2 + (3-0)^2} = \sqrt{(1+9)} = \sqrt{10}$. The area is just this value squared, or $\boxed{10}$.

将顶点依顺时针方向标记为 $A$, $B$, $C$, $D$。接着使用坐标几何法进行简单求解。设点 $A$ 位于 $(0, 0)$。则点 $B$ 由于向上 1 个单位且向右 3 个单位，坐标为 $(1, 3)$。题中已知这是一个正方形，因此只需计算点 $(0, 0)$ 到 $(1, 3)$ 的距离。使用距离公式，得到 $\sqrt{(1-0)^2 + (3-0)^2} = \sqrt{1 + 9} = \sqrt{10}$。面积即为边长的平方，结果为 $\boxed{10}$。

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