I came up with a geometry problem that appears to be really surprising.
1) Consider a unit circle (a circle with radius 1 unit). Place $n$ randomly placed points on this circle. Then there exists a point on this circle such that the sum of the distances from this point to all the other points is exactly $n$.
This generalizes in an obvious way.
2) Consider a unit sphere (a sphere with radius 1 unit). Place $n$ randomly placed points on this sphere. Then there exists a point on this sphere such that the sum of the distances from this point to all the other points is exactly $n$.
So in particular, if you have 2010 points on a unit circle or sphere then there is this special point so that the distance sum is precisely 2010.
Note, this problem can be generalized to higher dimensional spheres also. But let us not go there because it is impossible to visualize.
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proof?
ReplyDeleteUse sub-harmonic functions. If $$x_1,...,x_n$$ are the n-points on the unit circle define $$f:D\to D$$ (where D is the closed unit disk) by $$ f(x) = ||x-x_1|| + ... + ||x-x_n||$$. Notice that $$f(0) = ||x_1|| + ... + ||x_n|| = n$$, therefore $$f(0)=n$$. Now since f is a sub-harmonic function on in interior of D and continous to its boundary it follows by the maximum-modulos principle that there is a point p on the boundary of D such that $$f(p)>n$$. Now use a minimum argument to also argue why there has to be a point q on the boundary of D such that $$f(q)<n$$. Since f when restricted to the boundary of D is a continous function is follows by the intermediate value theorem there got to be a point w such that $$f(w)=n$$ exactly.
ReplyDeleteOK. That was Chinese to me. I bow before your superior knowledge.
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