1996 AMC 8 第 22 题

Question

相邻点之间的水平距离和竖直距离都等于

1

个单位。三角形

ABC

的面积是

The horizontal and vertical distances between adjacent points equal

1

unit. The area of triangle

ABC

is

\dfrac{1}{4}

\dfrac{1}{2}

\dfrac{3}{4}

1

\dfrac{5}{4}

给

A

、

B

、

C

赋坐标，再用行列式面积公式。

Assign coordinates to

A,

B,

and

C

and use the determinant formula for area

也可以用 Pick 定理，面积

= I + \tfrac{B}{2} - 1

，其中内部点

I = 0

，边界点

B = 3

。

Alternatively use Pick’s theorem, area

= I + \tfrac{B}{2} - 1,

with

I = 0

interior points and

B = 3

boundary points

取

A = (0, 0)

、

B = (3, 2)

、

C = (4, 3)

，由行列式面积公式可得

\frac12\left\lvert 3\cdot3-2\cdot4\right\rvert=\frac12\text{。}

等价地，用 Pick 定理，内部格点数为零，边界格点数为

3

，面积为

0 + \tfrac32 - 1 = \tfrac12

。

所以正确答案是 B。

Taking

A = (0, 0),

B = (3, 2),

and

C = (4, 3),

the determinant formula gives

\frac12\left\lvert 3\cdot3-2\cdot4\right\rvert=\frac12.

Equivalently, by Pick’s theorem with no interior lattice points and

3

boundary points, the area is

0 + \tfrac32 - 1 = \tfrac12.

Thus, the correct answer is B .

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