Choose some point A inside a regular polygon. The sum of length of line segments connecting A to the vertices of the polygon achieves minimum when A is the center of the polygon.

Can somebody give a proof of this? Thanks!!

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- Jun 11th 2011, 08:05 AMgodelproofregular polygon
Choose some point A inside a regular polygon. The sum of length of line segments connecting A to the vertices of the polygon achieves minimum when A is the center of the polygon.

Can somebody give a proof of this? Thanks!! - Jun 11th 2011, 09:38 AMTKHunny
Is it valid? Your personal exploration would be of most benefit to you.

Prove it for a triangle. Does it work?

Pove it for a square. A Rhumbus. A convex quadrilateral. A concave quadrilateral. (That last one presents some LOS problems.)

What say you? Do you believe it should be provable? - Jun 11th 2011, 07:18 PMgodelproof
Why, it should! Let's consider convex ones, say B1B2...Bn. A friend of mine told me his idea: suppose A's not the center of the polygon. Rotate the polygon by an angle of 2pi/n around it's center, then A becomes A'. So if A minimizes the sum of lines connecting to vertices, so does A'. Let this sum be L. Now choose the midpoint O of AA'. We have OBi < (ABi+A'Bi)/2. Sum both sides over i we find the sum of length connecting O to vertices is less than L. So A can't be minimizing L if it's not the center, where A=A'=O.

- Jun 11th 2011, 07:51 PMTKHunny
Yeah, forget I said anything about concave polygons. I missed "regular".

I like it. It not only completes the proof, but also create an algorithm (albeit infinite) for finding the center. - Jun 11th 2011, 08:25 PMgodelproof
Indeed. And here's one without algorithm. When he first told me about the problem, my maths teacher gave a entertaining "proof": Find n long thin strings. Tie one end of them together (call the knot A). Tie the other ends each to a plumb weighing 1N. For a horizontally placed regular polygon board, let each string go through the a small hole at each vertex, with its plumb hanging below the board. Now if A's at the center O, sum of forces on it is zero so it stays there. Otherwise, sum of forces in the $\displaystyle \vec{AO}$ direction is positive (using vectors it's easy to prove this!). So by moving A to O plumbs will have done work and their potential energy decreases! That means the sum of length of strings below the board will increase and the sum of length inside the board decreases! and this completes the proof.

- Jun 11th 2011, 08:27 PMAlso sprach Zarathustra
EDIT: late in 2 minutes.

A question...

Is it enough to prove the following for making the conclusion in OP's question?

Say we have regular polygon A1A2...An with center O. Let denote $\displaystyle \vec{OA_i}=\vec{a_i}$ for $\displaystyle 1\leq i\leq n$. Then $\displaystyle \sum_{i=1}^{n}\vec{a_i}=0$ - Jun 11th 2011, 08:34 PMgodelproof
- Jun 11th 2011, 08:39 PMAlso sprach Zarathustra
- Jun 11th 2011, 08:57 PMgodelproof
N

- Jun 11th 2011, 09:02 PMgodelproof
- Jun 11th 2011, 09:12 PMgodelproof
- Jun 12th 2011, 01:10 AMgodelproof
and here's the proof using vectors for Zarathustra's question, in the attachment. Interestingly, I suddenly realize that this together with the the argument in #5 also allows us to compare the sums of connections for any two points in the polygon: the one farther away from the center O has a larger sum. And the same comparison can be made even when one or both points are outside the polygon! (since there's no requirement in #5 and #12 for A to be inside the polygon)

The problem has a nice and simple structure.

EDIT: the contours are not circles... so we cannot compare arbitary points. The best #5 and #12 allow us is to compare 2 points if they and the center O lie on a straight line.