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I was playing around with GeoGebra tonight and stumbled across something really surprising.

You probably drew some parabolas when you were in school, using coordinate geometry, but a parabola can also be defined in purely geometrical terms, without using coordinates. It is the set of points whose distance for a fixed point, called its focus, is the same as its perpendicular distance from a fixed line called its directrix. A parabola is defined as soon as you specify its focus and its directrix.

It occurred to me to wonder what would happen if we took a point in the interior of a triangle and drew three parabolas, taking this point as the common focus, but using each of the three sides of the triangle in turn as directrix.

So I tried this and I got a very interesting figure.

P is the common focus AB, BC, and CA the directrices, and I have labelled the parabolas r, s and t. Each pair of parabolas defines two crossing points: I and J, E and F, and G and F in the diagram.

Now for the surprises. If we draw lines through each of these three pairs, each line passes through a vertex of the triangle, and all three meet at the point I have labelled O.

This is surprising enough, but more surprising still these lines bisect the angles of the triangle/ It is well known that the angle bisectors of a triangle meet at a point called the incentre, because it is the centre of the unique circle, the incircle, inscribed in the triangle and tangent to its sides.

The point O is a fixed point of the triangle. which means the position of P, the common focus, makes no difference to the three lines. Moving P will change the parbolas, and hence the points at which the parabolas intersect, but the intersection points will always lie on the same three lines.

Here are a couple of examples where P is in different places.

So that's what I discovere, I have no idea (yet) how to prove it, and it's about 99% likely thatit's a well-known result, but if not I shall claim it as Walker's Theorem.

An old story tells of a Maths professor, who at a certain point in his lecture told the class, "The proof of this is obvious", but then began to have doubts. After gazing at the blackboard for some time he said, "Hmmm. Perhaps it's not obvious. I'll have think about it and let you know in next week's lecture."

The following week the class assembled, and after a new minutes the professor arrived. Gazing round the lecture theatre, he said "Since our last lecture I have been thinking about that problem in every waking hour, but I simply could not see it. But you will be glad to know that on my way here today I finally saw the answer."

"It is obvious."

Well, I have been thinking for several days about how to prove the Conway Circle Theorem, a rather nice result usually credited to the brilliant and original mathematician John Horton Conway, who sadly died in the recent Covid Pandemic. I could fairly quickly see a proof but it had a rather messy feel to it, it wasn't anything I would want to explain on this blog. It wasn't what I would call elegant. I felt there must be a nicer proof, one that would make the proof of the theorem, well, obvious.

After much chewing the problem over I think I have a proof which does make the theorem fairly obvious. But I'd better tell you what the theorem is before we go any further. Consider the following diagram.



As shown in the diagram, the three sides of triangle ABC have each been extended at both ends, AB by distances a and b, BC by distances b and c, and CA by distance c and a. This gives the six points G, H, I, J, K, L.  Conway's theorem says that (rather surprisingly) these six points lie on a circle, see below:



How to prove this? I'm going to use some well-known properties of isosceles triangles, triangles with two sides equal in length.

In triangle PQR sides PQ and QR are equal. The line bisecting the base PR at right angles, the perpendicular bisector, must pass through the apex Q, and it bisects the angle PQR. Moreover any point, for example Y, that lies on the perpendicular bisector is equidistant from points P and Q, as shown by the dotted lines.

Going back to our original diagram, let's join up the six points to make a hexagon, and then draw in the perpendicular bisectors of its sides, shown dotted:


We see that each side of the hexagon is the base of a triangle, which by the method use to construct the six  points must be isosceles. Each perpendicular bisector passes through a vertex of triangle ABC and bisects the angle there, and is fact the shared perpendicular bisector of a pair of the hexagon's sides lying opposite one another.

It's well known that the three lines bisecting the angles of a triangle meet in a single point, called the incentre. Because the three dotted lines bisect the angles of ABC this mean they meet at its incentre.

Because the incentre is on the perpendicular bisector of HI it must be equidistant from H and I. Similarly because it is on the perpendicular bisector of IJ it must be equidistant from I and J. By the same argument, it must be equidistant from J and K; from K and L; from L and G; and from G and H. Thus it is equidistant from all six points and is the centre of a circle that passes through them all.