1998 AMC 12 第 19 题

Question

有多少个三角形的面积为

10

，且其三个顶点分别为

(-5,0)

、

(5,0)

和

(5\cos\theta,5\sin\theta)

，其中

\theta

是某个角？

How many triangles have area

10

and vertices at

(-5,0),

(5,0),

and

(5\cos\theta,5\sin\theta)

for some angle

\theta?

0

2

4

6

8

固定底边长为

10

，第三个顶点的高为

|5\sin\theta|

The fixed base has length

10

, and the third vertex has height

|5\sin\theta|

数出圆上满足所得正弦方程的不同点

Count the distinct points on the circle satisfying the resulting sine equation

面积为

\frac12(10)|5\sin\theta|=25|\sin\theta|

。所以

|\sin\theta|=\frac{2}{5}

。圆上有四个相应的点，每个点都给出一个不同的三角形。因此答案为

4

，正确答案是 C。

The area is

\frac12(10)|5\sin\theta|=25|\sin\theta|.

Thus

|\sin\theta|=\frac{2}{5}.

There are four corresponding points on the circle, and each gives a distinct triangle. Hence the answer is

4,

making C correct.

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