AMC8 1996

AMC8 1996 · Q22

AMC8 1996 · Q22. It mainly tests Area & perimeter, Counting in geometry (lattice points).

The horizontal and vertical distances between adjacent points equal 1 unit. The area of triangle $ABC$ is

相邻点之间的水平和垂直距离均为1单位。三角形 $ABC$ 的面积是

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

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

(D) 1 1

(E) $\frac{5}{4}$ $\frac{5}{4}$

Answer

Correct choice: (B)

正确答案：（B）

Solution

Answer (B): From the total area of 12, subtract the area of the four surrounding polygons whose areas are indicated in the diagram. Thus the area of the remaining triangle $ABC$ is $$12-6-3-0.5-2=0.5=\frac{1}{2}$$ Note. Points whose coordinates are integers are called lattice points. According to Pick’s Theorem, if there are $I$ lattice points in the interior of a triangle and $B$ lattice points on the boundary, then the area of the triangle is $I+\frac{B}{2}-1$. In this problem, $I=0$ and $B=3$. Therefore the area of the triangle is $0+\frac{3}{2}-1=\frac{1}{2}$.

答案（B）：从总面积 12 中减去图中标明的四个周围多边形的面积。因此剩余三角形 $ABC$ 的面积为 $$12-6-3-0.5-2=0.5=\frac{1}{2}$$ 注：坐标为整数的点称为格点。根据皮克定理，若三角形内部有 $I$ 个格点、边界上有 $B$ 个格点，则三角形面积为 $I+\frac{B}{2}-1$。在本题中，$I=0$ 且 $B=3$。因此三角形面积为 $0+\frac{3}{2}-1=\frac{1}{2}$。

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