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Given a triangle ABC and MN a line parallel to BC with M on AB and N on AC, let BN and CM meet at a point P. Then the circumcircles of BMP and CNP intersect at two points P and Q. Show that $\angle BAQ = \angle PAC$.

I drew a little diagram of this on a napkin to stare at before going to sleep last night, and a solution beautifully unfolded over about 10-15 minutes of just thinking. Of course, I still don’t know if there is a solution involving an easy angle chase, but I thought I’d post my approach just to illustrate why thinking about geometry problems (and knowing a few facts) can lead to proofs far more easily than immediately starting to chase angles, or any other badly planned strategy.

1) MN is parallel to BC (so their point of intersection is at infinity). This is significant because P and A are also significant, so we have a big complete quadrilateral. Hence the line AP has to be a median of the triangle (by the harmonic property of the quadrilateral).

2) The condition we wish to prove is therefore equivalent to showing Q lies on the symmedian of the triangle through A.

3) Finally, a configuration involving circles intersecting as they do here indicates that Q is the centre of a spiral similarity taking BM to NC. Hence in particular the perpendicular distances of Q to AB,AC are in the ratio BM:NC = AB:AC which is the ratio described by a symmedian.

So we’re done.

Remark: Step (1) can be easily proved by doing a similar triangle argument (or doing two homotheties, if you like). However, the knowledge of the harmonic quadrilateral is important not because it makes the proof easier but because it makes finding it easier. In the IMO if I had time I would probably write out the Euclidean proof, but the idea came instantly because I saw a complete quadrilateral in the diagram. Also, step (2) is more a motivating step than necessary: you can just do step (3) and then prove that the distances of P to the sides are in the ratio AC:AB, so a similar triangles argument takes you home and you don’t need to use the words ‘isogonal conjugacy’ or ‘symmedian’. Some might consider this proof ‘using a sledgehammer to crack a nut’, but I think it tells me why it is true in relation to what I already know (in a much better way than if I’d slogged a trigbash).

Question: For $A \subset \mathbb{N}$ we define the density $\delta(A,N) = \frac{|A \cap \{1,2,...,N\}|}{N}$. How large can $\Delta(N) = \delta(A,N)$ be if $A$ contains no nontrivial arithmetic progressions?
Finding upper bounds on $\Delta(N)$ is probably very difficult. Szemeredi’s theorem, which is highly nontrivial, gives the bound $\Delta(N) = o(1)$. If there is a better bound, it will probably be harder to prove.
However, finding lower bounds is actually very possible and perhaps the ideal distraction for exam term. My first attempts in the CMS gave a bound of $\Delta(N) \geq N^{-c}$ for $c = \frac{1}{3}\log \frac{3}{2}$. I’ll try to optimise this tomorrow and over the course of the next month, but will not yet post any of my constructions to allow the reader the opportunity to try this problem.