1985 AMC 12 第 14 题

Question

一个凸多边形恰有三个内角为钝角。这样的多边形最多有多少条边？

Exactly three of the interior angles of a convex polygon are obtuse. What is the maximum number of sides of such a polygon?

4

5

6

7

8

用

180^\circ

限定三个钝角，并用

90^\circ

限定其余每个内角

Bound the three obtuse angles by

180^\circ

and every other angle by

90^\circ

将该上界与内角和

(n-2)180^\circ

比较

Compare that upper bound with the interior-angle sum

(n-2)180^\circ

对于一个

n

边形，三个钝角之和小于

3\cdot180^\circ

，其余

n-3

个角之和至多为

(n-3)90^\circ

。因此

\begin{aligned} (n-2)180^\circ &\lt3(180^\circ)\\ &\quad{}+(n-3)90^\circ\text{，} \end{aligned}

化简得

n\lt7

。六边形可以达到，例如其内角为

150^\circ

、

150^\circ

、

150^\circ

、

90^\circ

、

90^\circ

、

90^\circ

。所以最大边数为

6

。

因此正确答案是 C。

For an

n

-gon, the three obtuse angles have total less than

3\cdot180^\circ,

while the other

n-3

angles have total at most

(n-3)90^\circ.

Hence

\begin{aligned} (n-2)180^\circ &\lt3(180^\circ)\\ &\quad{}+(n-3)90^\circ, \end{aligned}

which simplifies to

n\lt7.

Six sides are attainable, for example with interior angles

150^\circ,

150^\circ,

150^\circ,

90^\circ,

90^\circ,

and

90^\circ.

Thus the maximum is

6.

Therefore the correct answer is C.

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