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标题: Prove or disprove [打印本页]

作者: yma16 时间: 2005-9-13 17:26
标题: Prove or disprove

that you can find some points on the plane as vertices of concave polygons such that

all the polygons have exactly two inner angles with degree > 180.

Note: This question is from an older question.

It is easy to do for exactly one inner angles with degree > 180. When the number is bigger than one, it may not have solution.

BTW, you can alway draw a concave polygon of n vertices (n>3) with n-3 inner angles of degree > 180.

作者: 乱弹 时间: 2005-9-13 19:24
标题: disprove

I can disprove it but the current version is not simple: I need three cases.

作者: fzy 时间: 2005-9-13 19:47
标题: 回复：disprove

I thoutght it was not hard to disprove. Of course I have not thought about all the details:

First we need to have at least three interior points. (Your problem below.) And we know we always have at least three exterior points (in the convex hull). Then we can "concave in" from three different gaps of the exterior. (Probably need more argument about how to do this.) Each gap would produce at least one concave angle.

作者: 乱弹 时间: 2005-9-13 20:04
标题: Yes. Not very hard. My three cases

(1) The convex hull of the inner points is colinear. (Disprove by my problem below)

(2) The convex hull of the inner points is a triangle.

(Disprove by fzy's idea: concave in. )

(3) The convex hull of the inner points is a polygon of more than three vertices.

(The inner points form a polygon P with at least four angles<180. Break one (some) edge of P, connect it with outer points smartly, got a new polygon Q. At least two inner angles of <180 of P become outer angles of Q; besides, the connection of P and outer points produce another outer angle <180. )

(2) and (3) can be combined and proved using fzy's idea. It is tedious to write down all details.

作者: 乱弹 时间: 2005-9-13 20:50
标题: For cases (2) and (3)（图）

Basically, if we do not allow three points colinear, we can find the above picture, such that all other points are in the blue region. All points except B form a polygon P with EM and MD being two edges. Replace curve AMC with ABC, we get a polygon we need.

作者: 乱弹 时间: 2005-9-14 00:52
标题: When allowing three points colinear

It is a bit complicated (need to explain more carefully how to divide points). I incorrectly guessed the answer to the original problem is yes.

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