Descartes' rules of signs

Hard to believe that I don't remember seeing Descartes' rules of signs before -- you'd think this is impossible given the time I spent on those kinds of things in middle school. Anyways, it says:

If the terms of a single-variable polynomial with real coefficients are ordered by descending variable exponent, then the number of positive roots of the polynomial is either equal to the number of sign differences between consecutive nonzero coefficients, or is less than it by an even number. Multiple roots of the same value are counted separately.

Proof:

By induction on the number of positive roots. All zero coefficients are ignored here. The base case is a polynomial \(f(x)\) without any positive root. Clearly the first and last cofficients must have the same sign, as otherwise \(f(x)\) has a positive root between \(0\) and \(\infty\). So the base case is resolved.

Now, let \(g(x)=f(x)(x-a)\) with \(a>0\). It is not hard to observe that the number of sign changes increases at least \(1\), with parity altered.
Q.E.D.

Helly's theorem

This is the last of a small series of similar and basic results in convex geometry.

Helly's theorem: Let \(X_1,\ldots,X_n\) be a finite collection of convex subsets of \(R^d\). If the intersection of every \(d+1\) of these sets is nonempty, then the whole collection has a nonempty intersection.

Proof:
Below we show that if every \(k\geq d+1\) sets have nonempty intersection, then so do every \(k+1\) sets.

It suffices to consider only \(k+1\geq d+2\) convex sets \(X_1,\ldots,X_{k+1}\).

Let \(C_i=\bigcap_{v\neq i}X_v\)

Clearly if any two distinct \(C_i\) and \(C_j\) overlap then we are done, so assume they don't and pick a representative point \(p_i\) for \(C_i\) for each \(i\). Now apply Radon's theorem to \(p_1,\ldots,p_{k+1}\), i.e. without loss of generality there is a point \(x\) that is convex combination of \(p_1,\ldots,p_m\), and also convex combination of \(p_{m+1},\ldots,p_{k+1}\). For any \(i\in[1,m]\), the latter ensures \(x\in X_i\), and for any \(i\in[m+1,k+1]\), the former implies the same thing.
Q.E.D.