The second and final round of this year’s British Mathematical Olympiad took place on Thursday.
Here are some thoughts on the problems. I wasn’t involved in choosing the problems, although I did write Q4. I’ll say a bit more about this geometry problem than about the other three, so I’ve put this first.
The copyright to the problems is retained by BMO, and I’m reproducing here with permission. The paper can be found in its original format here.
Let me explain how I devised this problem. Let A’ be the point diametrically opposite A on the circumcircle of triangle ABC. (Sometimes called the antipode of A.) Now consider any line L through A’, and let P be the point where it meets the circumcircle again. I was particularly thinking about lines L where P lay on the same side of BC as A’, as depicted.
Because AA’ is a diameter, there is a right angle . So I considering intersecting L with the lines given in the question, since this would create two cyclic quadrilaterals, denoted in the figure.
Let’s just focus on . This cyclic quadrilateral gives us two equal angles in many different ways. But a particularly nice pair is the one shown, because ACPB is also cyclic, which confirms that the angle measure is . In particular, by extending to E on AC, we get a third cyclic quadrilateral , and hence find out that .
So four lines go through D: AP, BC, and the perpendiculars from to the opposite sides of the triangle.
How to turn this into a question? The steps outlined above are relatively clear because
- the right angle at P is essentially given;
- there are four other right angles, two involving which are given, then one involving being perpendicular to AC, and the same for and AB, which are to be deduced,
so the cyclic quadrilaterals just pop out directly. This would then be too straightforward for BMO2, and would also waste the fact that four lines meet at D. It would suffice to work with as we did above, and assert that the corresponding result also held for .
To move towards the form of the BMO2 question, we consider taking P=A’, so that line L is actually the tangent to the circumcircle. We still have the quadruple concurrency at D, but now D lies on AA’. It seems almost certain that we wouldn’t have this concurrency if we built the figure using the tangent from any point other than A’. So this will be the configuration for the problem.
As the writer, I know that the four lines meet when T=A’, so I know I can force T=A’ by insisting that three lines meet, and then set the problem asking to prove that the fourth goes through the same point. One could have stated that meet, and ask to prove that the meeting point lies on BC. I preferred the one I chose. Reflecting on the difficulty of the problem:
- It is hard to draw a diagram which looks right! Essentially you have to draw a few, and if T ends up close to A’ on one, the diagram will look right.
- Crucially, we now only have four right angles in the figure, so only one pair of (symmetric) cyclic quads can be read off immediately
I think the cleanest way to attack the problem is to extend AQ to meet the circumcircle at T’, with the goal of showing T’=T. Then cyclic gives cyclic (in an exact reversal of the argument in the prelude earlier). What we don’t know is that are collinear, but this follows since now we derive from this new cyclic quad. Consequently T’=T.
There are many routes for coming up with olympiad problems in geometry and other topics, and the scenario I’ve described isn’t particularly typical for myself and certainly isn’t necessarily standard for other problem setters. Nonetheless, I hope that’s an interesting insight into the origin story of a problem like this.
This question reduces to establishing whether
has infinitely many solutions in (m,n). The technique for proving via ‘square sandwiching’ that this is normally not possible has been discussed previously on this blog here:
This is a good opportunity to reflect in more detail. Let be a polynomial with integer coefficients, and we consider whether the image contains infinitely many squares.
Clearly if for Q another polynomial with integer coefficients, then this holds. Square sandwiching establishes that the answer is no when P(x) is a monic quadratic (meaning that the coefficient of is one) unless P(x) is the square of a linear polynomial. The same holds if the leading coefficient is a square. Note that the general case where P(x) is negative except on a finite range (eg if P has even degree and negative leading coefficient) is also immediate.
But there are several cases where the answer is yes:
- For example, when , we have P(n) a square whenever .
- Indeed, note that has this property too!
- Amongst non-monic quadratics, note that works when n is not a square because of the theory of Pell’s equation.
We could have the alternative question: if is always a square, does this imply . Indeed, one could generalise this and ask: if does this imply that for some choice Q?
Our original setting with is classical, due to Polya, and is a good problem for any students reading this. The case is handled by Polya and Szego. The case of general R is more challenging, and though it can be treated without heavier concepts, is probably best left until one knows some undergraduate algebra. Readers looking for a paper may consult Davenport, Lewis and Schinzel Polynomials of certain special types (Acta Arith. 1964).
Rearranging to get
is a good first step, as y appears only on the LHS. As y varies, the LHS is therefore constant, and so we deduce that , ie f is linear on the even integers.
There were then a number of ways to proceed, bearing in mind one had to study the odd integers and solve for C. Taking x even in the given equation is a good next step, because (after some careful argument about C and divisibility by powers of 2) is already known, leading to an expression for f(x+1), ie some control over the odd integers.
There were a few possibilities for how to piece the odds and evens together, and extract the constant C. It’s worth mentioning that if the problem had studied functions from the non-negative integers to the non-negative integers, then C=0 would have been a possibility. You might like to consider whether this could lead to extra functions satisfying the given equation.
I personally found this quite difficult. We have a large number of 50-dimensional integer vectors with entries bounded by [0,2022] such that their sum is (n,n,…,n) and we are tasked to show that we can split the sum to obtain (m,m,…,m) and (n-m,n-m,…,n-m).
It seemed helpful to replace each vector with the 49-dimensional vector . We now have a large number of 49-dimensional integer vectors with entries bounded by [-2022,2022] with sum equal to the zero vector. Our task is to show that we can partition into two smaller sums with sum zero. (*)
It seems likely that the role of 2022 is not important, and will just determine how large n must be. I was hoping there might be a quick proof along the following lines: Let denote the collection of vectors, and some subsets, inducing sums respectively. Clearly if , and we might expect to be able to find such that . I was hoping one could then move some vectors around (eg from ) to produce disjoint sets A,B for which , but I couldn’t see any way to achieve this.
This doesn’t mean it’s impossible though of course! And perhaps a contestant will find a successful argument along these lines during the olympiad.
In fact, it seems that a careful induction (on the number of cards, ie the dimension of the vectors) is the best way to proceed, and I will leave that there.