Can we solve this problem with analytic geometry and transfinite induction?

Question

Can we solve this problem with analytic geometry and transfinite induction? The solution must be with analytic geometry, not with euclidean geometry. Problem 4. Triangle ABC has circumcircle Ω and circumcenter O. A circle Γ with centre A intersects the segment BC at points D and E, such that B , D , E and C are all different and lie on line BC in this order. Let F and G be the points of intersection of Γ and Ω , such that A, F , B , C and G lie on Ω in this order. Let K be the second point of intersection of the circumcircle of the triangle BDF and AB segment. Let L be the second point of intersection of the circumcircle of triangle CGE and the segment CA. Suppose that the lines FK and GL are diferent and intersect at the point X. Prove that X is on the line AO .

Answer 1

It's very difficult to prove that a technique cannot be useful for a given problem. That said, I'm going to go out on a limb here and say: "no, transfinite induction is not useful here."

Let me try to explain why. Transfinite induction is the following result:

Suppose $\alpha$ is some ordinal, and $P$ is a statement such that $(i)$ $P(0)$ holds and $(ii)$ for every ordinal $\gamma<\alpha$, if $P(\beta)$ holds for all $\theta<\gamma$ then $P(\gamma)$ holds. (Note that $(i)$ is in fact redundant here.) Then $P(\gamma)$ holds for every ordinal $\gamma<\alpha$.

In particular, there are two things you want to look for in a problem to see that transfinite induction might be helpful:

• $(a)$ There is some way to view the "instances" of the problem as ordinals, and

• $(b)$ Under this viewpoint, instances with "smaller" ordinals are "simpler".

Now the axiom of choice seems to make point $(a)$ trivial: every set can be "ordinalized." So, for example, in the problem you mention the "instances" are configurations of points and circles in the plane; so why don't we just well-order that set and go from there?

Well, the answer is that this well-ordering is really ad hoc, and point $(b)$ won't be satisfied. Suppose I've shown that the statement holds for one instance of your problem; how do I show that it holds for the "next" instance? There is no obvious structure to the problem which lets you view one triangle as a "simpler" instance than another, let alone which does so in a way compatible with some well-ordering. So there is no "fingerhold" for transfinite induction to come into play.

Transfinite induction often seems magical at first glance, but it is a precise proof technique that requires specific hypotheses (just like standard induction!). Unless you see a reason for transfinite induction to work for a certain problem, you're unlikely to find it useful for that problem.