Proof without words part two

[jay_shark]

Here is a nice little problem which some may find difficult unless you have experience with these types of problems . It has a very nice visual solution and you don't need any words to justify this . See if you can figure this one out .

There are n points in the plane so that no two pairs are the same distance apart. Each point is connected to the nearest point by a line. Show that no point is connected to more than 5 points.

[RocketManJames]

Here's my attempt. No words, but I colored a few things. And, I am aware that the image doesn't satisfy the constraint of no same distance pairs, but I think that's the point.

http://i63.photobucket.com/albums/h158/RocketManJames/MyAttempt.jpg

If it's not right, am I on the right track?

-RMJ

[jay_shark]

You are definitely on the right track . You may need to explain a few things in words but it shouldn't be more than a few sentences .

[Duke]

http://i29.photobucket.com/albums/c299/krizazy/triangle.png

[holmansf]

I don't understand the problem. If no two pairs are the same distance apart then every point has exactly one closest point, and so is connected to only one point. Or do you consider points connected transitively (point A connected to point B connected to point C implies A connected to C)? In this case it seems easy to find an example of arbitrarily many points connected- just a bunch of points all on the same line, getting closer together.

[RocketManJames]

[ QUOTE ]
You are definitely on the right track . You may need to explain a few things in words but it shouldn't be more than a few sentences .

[/ QUOTE ]

Here was the idea behind my non-words. Image 1 shows a perfect fit of a point and its 6 connected neighbors. Image 2 tries to show that perturbation of any of the neighbor points would result in the breakage of some connection to the center point (either the connection of the perturbed point is broken, or the connection made by one of its neighbors would be, depending on the nature of the disturbance).

-RMJ

[holmansf]

Nevermind the last post, I get it now.

[holmansf]

I don't have the necessary technology to get a picture in here, but I think a proof would essentially consist of these two things (maybe you can imagine them drawn as a picture /images/graemlins/smile.gif

1) The largest angle of an non-equalateral triangle must be larger than 60 degrees.

2) 360/6 = 60 .

[gumpzilla]

[ QUOTE ]
I don't have the necessary technology to get a picture in here, but I think a proof would essentially consist of these two things (maybe you can imagine them drawn as a picture /images/graemlins/smile.gif

1) The largest angle of an non-equalateral triangle must be larger than 60 degrees.

2) 360/6 = 60 .

[/ QUOTE ]

Oooh. Nice.

[jay_shark]

Right Holmansf .

This is pretty much how I did it . If you prove it's not possible for 6 , then it follows that it's not possible for n>6 .

It's much easier to draw it out but I can explain it in words . Suppose it's possible for a single point to have 6 lines from it . Of the 6 lines , one of them is the largest . Draw a circle with radius equal to this distance which means that all of the other 5 points must be contained within the circle . Now divide the circle into 6 equal sectors of 60 degrees each . From the pigeonhole principle , 5 of the points must be contained in one of the 4 sectors .But this would be a contradiction because then we can find two points that are closer to each other .