The proof of Radon's theorem is little more than an imaginative exercise in linear algebra, but (like the theorem) it always seems to me to reveal some of the structure of Rd. In any case, make sure you read what we're proving before reading the proof...

So suppose we're given d+2 points x0,...,xd+1Rd (the theorem is true for any n>=d+2 points, but obviously adding more points just makes it easier to prove). Then these points have some affine dependence, and WLOG (maybe after reordering the points) we may write

xd+1 = ∑i=0d cixi,
i=0d ci = 1.
Since their sum is positive, some of the ci's must be positive, and WLOG (again, after reordering the points) we may say c0,...,ck > 0 >= ck+1,...,cd.

Rewrite the affine dependence all coefficients separated by their signs:

xd+1 + ∑i=k+1d (-ci)xi = ∑i=0k cixi.
The point here is that all coefficients appearing are positive!

Now let S = ∑i=0k ci = 1 - ∑i=k+1d ci > 0. Write bi=ci/S for 0<=i<=k, bi=-ci/S for k<i<=d, and bd+1=1/S. Then all the bi's are positive, ∑i=0kbi = ∑i=k+1d+1bi = 1, and we have the equal convex combinations

i=0k bixi = ∑i=k+1d+1 bixi
Q.E.D.!

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