Prove or disprove

yma16 · 发表于 2005-9-13 17:26:30

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.

乱弹 · 发表于 2005-9-13 19:24:26

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

fzy · 发表于 2005-9-13 19:47:15

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.

乱弹 · 发表于 2005-9-13 20:04:03

(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.

乱弹 · 发表于 2005-9-13 20:50:31

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.

乱弹 · 发表于 2005-9-14 00:52:53

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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Prove or disprove

disprove

回复：disprove

Yes. Not very hard. My three cases

For cases (2) and (3)（图）

When allowing three points colinear