Lin McMullin’s Theorem: an affine geometry assisted proof

I learned about this from Paysages Mathématiques, which in turn cites as source Theorem of the Day (March 15, 2025) by Robin Whitty.

One can read online Lin McMullin’s description of how he found the theorem, a lovely mix of serendipity and computer aided calculation. 

The statement if as follows: consider y=f(x) in the plane, the graph of a quartic (degree 4) polynomial with two flexes p and q. Let L be the line through p and q, and let u and v be the other two intersections of L with the graph.

Then the coordinates of u and v can be obtained from those of p and q via the formulas 

u= φp+(1-φ)q  and   v=(1-φ)p+ φq.

McMullin found this by having a computer algebra system do the calculations for him. Of course the most amazing thing here is that he thought about this problem and imagined the result could be interesting! And it’s great he didn’t have to compute it all by hand (he may have been able to do so without error, I certainly wouldn’t).

In a bluesky thread, I showed how to obtain the result with easy calculations, by taking advantage of affine geometry to reduce the problem to one specific degree 4 polynomial. I will reproduce it here.

Bonus: this has nothing to do with the real numbers, which can be replaced by any field containing 1/2 and √5, as long as we have two distinct flexes.

Notice that φ and -(1/φ) sum to 1, i.e. describe an affine combination. They are thus preserved by any affine change of coordinates:

X=ax+by+c

Y=dx+ey+f

Let f be the degree 4 polynomial we are interested in.

By Y=y+ax+b and X=x, we can assume f is a multiple of x²: write f=gx^2. By Y=ay, we can assume g is monic (coefficient if x² is 1). By X=ax+b, we can choose any monic g with distinct roots.

We now choose g so the flexes are 1 and -1; by symmetry switching x to -x, we must have g=x^2-u.

We have f"(x)= 12x²-2u, thus u=6 does the trick.The flexes are (1,-5) and (-1,-5) so the line thru them is y=-5. We look for other two solutions of x⁴-6x²=-5.

We get x²=5, hence x=±√5.

We now must find affine combinations of 1 and -1 yielding ±√5, and it's easy to check we get the claimed result.The key point of using affine geometry is to realize this problem has NO free parameters, i.e. up to coordinate change there is only one quartic polynomial with distinct flexes.

It's very cute! I totally enjoyed doing the calculations on the kitchen whiteboard while cooking lunch.

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