其中任意三个点可以组成一个三角形,且面积均不超过
1。证明:存在一个面积不超过
4的三角形,把这N点完全盖住。分别过A, B, C作BC, AC 和AB的平行线。三条线围成的三角形即为一解。 |
如果有无穷多个点结论如何? |
当然要加上一个条件,这些点在一个有限区域内。 |
当然在有限区域内,因为每个三角形面积小于 1。可用有限复盖定里,推出其中有限个点联成的三角形能复盖这无限个三角形。 |
请看题目:其中任意三个点可以组成一个三角形,如果这个条件没有,有限点也不成呀! |
不过对有限点没关系。 |
The area of the constructed triangle is four times that of ABC. |
You get a triangle whose area is larger than ABC. |
I know that. The triangle is 4 times as much as ABC. But why it can cover all the points? |
must be able to explain this. |
设大三角形为 A'B'C',A'B'//AB,如果P点在A'B'的外侧,那么三角形PAB 的面积>ABC,因为P到AB的垂直距离比C到AB的垂直距离远(由于A'B'//AB). |
You can take a sequence of the triangles and find a limit with its area equal to the least upper bound. But I could not see how to use 有限复盖定里. Also if the points are not on the same line, it must be in a bounded region. (Bounded by a parallelogram.) |
没有漏洞能逃得过你的眼睛。 把每个极限点都计入内如何? |
That should work. |
Do you mean this? It does not look right. |
欢迎光临 珍珠湾ART (http://zzwav.com/) | Powered by Discuz! X3 |