BMO1 2019

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The first round of the British Mathematical Olympiad was sat yesterday. The paper can be found here, and video solutions here. Copyright for the questions is held by BMOS. They are reproduced here with permission.

I hope any students who sat the paper enjoyed at least some of the questions, and found it challenging! The following brief commentaries on the first three problems are not official solutions, and are not always full solutions at all, but contain significant steps of solutions, so would be best saved until after you have attempted the problems, if you are planning to do so.

Question 1

Show that there are at least three prime numbers p less than 200 for which p+2, p+6, p+8, p+12 are all prime. Show also that there is only one prime number q for which q+2, q+6, q+8, q+12, and q+14 are all prime.

It is a major open problem of considerable interest in mathematics to study primes p for which p+2 is also a prime. You are warmly encouraged to have a look at the video of this talk by oucolleague Vicky Neale which discusses this topic in an accessible fashion.

Fortunately, the statement of the second half of this BMO1 problem is not open, but the wider context is a sign that we should be looking for a simple reason why the numbers {q, q+2, q+6, q+8, q+12, q+14} cannot all be prime very often. A candidate for a simple reason is that one of them must be divisible by eg 6 (since there are six numbers), but of course this is not true, eg when q=3, or any other odd value. However, it is the case that one of these numbers must be divisible by 5. You could show this in a few ways:

  • studying what the last digit could be (a number is divisible by 5 precisely when it’s final digit is 0 or 5);
  • studying the remainder of each of the numbers under division by 5;
  • these are in fact pretty much the same thing, only the first option cycles with period ten, whereas the second cycles with period five.

In particular, the only chance that this collection consists of six prime numbers is when one of them is 5, since this is the only multiple of 5 which is a prime! So we need to check q=3, and q=5, and the latter leads to a solution, while q=3 does not, since q+6=9 is not prime.

In fact, the question is true even if we remove one of the terms {q+2, q+6, q+8, q+12, q+14} from consideration. To check that you understand the argument, see if you can work out which term(s) could be removed without breaking the solution?

For the first part, you will probably need to write down some integers less than 200 and check which are and aren’t prime, and maybe you have some shortcuts to narrow down the pool of candidates for p, but this doesn’t have to appear on your final solution. It really is enough to write down your three proposals for p, perhaps indicating what {p, p+2, p+6, p+8, p+12} are in these cases as a sanity check!

Question 2

A sequence of integers a_1,a_2,a_3,\ldots satisfies the relation:

4a_{n+1}^2 - 4a_na_{n+1}-a_n^2 - 1 = 0 (*)

for all positive integers n. What are the possible values of a_1?

We can ‘solve’ the equation (*) by viewing it as a quadratic in the variable a_{n+1}, with coefficients involving a_n. We end up showing that

a_{n+1}=\frac{a_n+1}{2}\text{ or }\frac{a_n-1}{2}. (**)

Note that this is a problem if a_n is even, since neither of these possibilities for the value of a_{n+1} is actually an integer. So certainly a_1 cannot be even, since then we can’t even construct a sequence of length 2, let alone an infinite sequence!

Crucially, one of the options in (**) is odd, and one of the options is even. It’s helpful to think of trying to construct the sequence one term at a time starting from a_1, where at each step we must ‘pick the odd option next’ as otherwise we’d be stuck at an even number, and couldn’t extend the sequence.

The exact wording for how you write this argument isn’t essential. What’s important is that you are clear about what you are trying to assert. Hopefully you are trying to assert something like:

For every odd choice of a_1, there does exist an (infinite) sequence satisfying (*).

Why is this true? Because you can construct the sequence one term at a time. It’s not really necessary to set this up as a formal induction. Rather, you just need to tell the reader that you know how to ensure you never get stuck, namely by always choosing the odd option for a_{n+1}, so that you do indeed end up with an infinite sequence.

Question 3

Two circles S_1,S_2 are tangent at P. A common tangent, not through P, touches S_1 at A, and S_2 at B. Points C and D, on S_1,S_2 respectively, are outside triangle APB, such that P lies on line CD.

Prove that AC is perpendicular to BD.

Here’s a diagram for this problem, lifted from the video. Cropped Dominic is pointing out where AC and BD meet, which I’ve chosen to call X.

As so often in olympiad problems, especially in geometry, the key to success is choosing the right characterisation of the property you are required to prove. In this case, proving that the angle at X is a right angle is awkward via a direct calculation since we don’t really know anything about X except its literal definition. However, an alternative is to show that the angles at C and at D sum up to 90. This is a much better contender for a useful characterisation since we know a lot more about C and D.

For example, C and D both lie on a circle, so are eligible for using circle theorems. Since we have several tangents, the alternate segment theorem might well be useful here. Indeed, if can apply it twice for each of C and D, and in doing so, characterise all the angles of triangle APB in terms of the angles at C and D:

There’s no harm in being informal at this stage. Within triangle APB, we know that the sum of two copies of the red angle and two copies of the black angle is 180 degrees, and so we can read off that the sum of the angles at C and at D is 90 degrees.

For the video presentation, I added the common tangent at P, which did turn out to be useful, and the centres and radii of the circles, which didn’t turn out to be useful. It’s generally a good idea to be thoughtful and flexible about what to include on a diagram. You can probably imagine why it might have been a distraction if we’d had lots of extra lines like BC or PX added on the figure, without any evidence they would be useful to the solution.

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