# BMO2 2022

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.

Problem Four

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 $\angle APA'$. So I considering intersecting L with the lines $\ell_B,\ell_C$ given in the question, since this would create two cyclic quadrilaterals, denoted $BDPP_B, DCP_CP$ in the figure.

Let’s just focus on $BDPP_B$. 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 $\angle C$. In particular, by extending $P_BD$ to E on AC, we get a third cyclic quadrilateral $BECP_B$, and hence find out that $P_BD,P_BE\perp AC$.

So four lines go through D: AP, BC, and the perpendiculars from $P_B,P_C$ 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 $\ell_B,\ell_C$ which are given, then one involving $P_BD,P_BE$ being perpendicular to AC, and the same for $P_C$ 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 $P_B$ as we did above, and assert that the corresponding result also held for $P_C$.

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 $P_BE,P_CF,AT$ 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 $BECP_B, ACT'B$ cyclic gives $BQT'P_B$ cyclic (in an exact reversal of the argument in the prelude earlier). What we don’t know is that $P_B,T',P_C$ are collinear, but this follows since now we derive $\angle QT'P_B=\angle P_CT'Q=90^\circ$ 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.

Problem One

This question reduces to establishing whether

$m(m+k)+k=n^2$

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:

and here:

This is a good opportunity to reflect in more detail. Let $P(x)$ be a polynomial with integer coefficients, and we consider whether the image $P(\mathbb{Z}):= \{ P(n)\,:\, n\in \mathbb{Z}\}$ contains infinitely many squares.

Clearly if $P(x)=Q(x)^2$ 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 $x^2$ 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 $P(x)=2x^3$, we have P(n) a square whenever $n=2k^2$.
• Indeed, note that $P(x)=x$ has this property too!
• Amongst non-monic quadratics, note that $P(x)=nx^2+1$ works when n is not a square because of the theory of Pell’s equation.

We could have the alternative question: if $P(n)$ is always a square, does this imply $P(x)=Q(x)^2$. Indeed, one could generalise this and ask: if $P(\mathbb{Z})\subset R(\mathbb{Z})$ does this imply that $P(x)=R(Q(x))$ for some choice Q?

Our original setting with $R(x)=x^2$ is classical, due to Polya, and is a good problem for any students reading this. The case $R(x)=x^k$ 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).

Problem Two

Rearranging to get

$\frac{f(2xy)}{2y} = \frac{f(f(x^2)+x)}{f(x+1)}$

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 $f(2y)=Cy$, 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) $f(f(x^2)+x)$ 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.

Problem Three

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 $(x_1,x_2,\ldots,x_{50})$ with the 49-dimensional vector $(x_2-x_1,x_3-x_1,\ldots,x_{50}-x_1)$. 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 $\mathcal V$ denote the collection of vectors, and $A,B\subset \mathcal V$ some subsets, inducing sums $\Sigma_A,\Sigma_B$ respectively. Clearly $\Sigma_A=-\Sigma_B$ if $B=A^c$, and we might expect to be able to find $A\ne B$ such that $\Sigma_A=\Sigma_B$. I was hoping one could then move some vectors around (eg from $B\mapsto A^c$) to produce disjoint sets A,B for which $\Sigma_A=-\Sigma_B$, 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.

# BMO1 2021

The first round of the British Mathematical Olympiad was sat on Thursday by roughly 2000 pupils in the UK, and a significant number overseas on Friday.

For obvious reasons, much of the past 18 months has been dominated by logistical rather than academic problems, so it’s refreshing to be returning to the traditional format of this exam, even if the marking has been stayed online.

I wasn’t involved in setting in this paper, so here are some thoughts on the problems. These aren’t supposed to be official solutions, and it’s entirely possible that I’m missing the most natural approach or framing. Students reading this in future years are advised, as always, that such commentaries are normally more valuable after attempting and digesting the problem yourself first.

Note that 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.

Problem One

If one exhibits eighteen (= 6 x 3) explicit examples of relevant sums of consecutive positive odd numbers, one has finished the problem. An exhaustive search is impractical and uninteresting. However, beyond that, we would also like to find integers N which we are convinced have this property without needing to explicitly decompose them into the desired sum format.

Let’s study such a sum. We need an even number of odd numbers to end up with an even sum. Probably the algebraically neatest way is to set up such a sum symmetrically as:

$2n-k, 2n-(k-2),\ldots,2n-1,2n+1,2n+3,\ldots, 2n+k,$

for k an odd number. The sum of this sequence of consecutive odd numbers is 2n(k+1). So, we seek integers N which can be written as 2nm, where m is an even number, in as many ways as possible. Note that if we have an integer N which can be factorised as a product of two distinct even numbers we are forced to take the larger as 2n, and the smaller as m, since we require 2n-k to be strictly positive.

It’s equivalent to finding integers M which are less than 100 and have lots of factors (then multiplying by 4 to recover an N), and there are a few ways to do this, including now potentially by exhaustive search. It is useful to remember that is M can be factorised into primes as $M=p_1^{a_1}\ldots p_k^{a_k}$ then its factors can be described as the set:

$\{1,p_1,p_1^2,\ldots,p_1^{a_1}\}\times \{1,p_2,p_2^2,\ldots,p_2^{a_2}\}\times\cdots \times \{1,p_k,p_k^2,\ldots,p_k^{a_k}\},$

In particular, the number of such factors is $(a_1+1)(a_2+1)\ldots(a_k+1)$, independently of the choice of the primes. This gives a recipe for constructing all integers M<100 with at least twelve factors. We could take $(a_i)$ to be some permutation of (11), or (5,1), or (3,2), or (2,1,1), which give the following valid M:

$M=2^5\times 3 = 96, \; M=2^3\times 3^2 = 72, \; M=2^2\times 3\times 5 = 60,$ $M=2^2\times 3 \times 7 = 84,\; M=3^2\times 2 \times 5 = 90,$

and the corresponding valid N are 240, 288, 336, 360, and 384.

For what it’s worth, this problem took me a while. I spent a long time saying things like “we need numbers that can be written as 4n and as 8n and as…” and aiming to conclude “all multiples of 24 greater than 24 x ?? will work”, but this didn’t ever generate enough examples of one class.

Problem Two

In order to have won exactly 30% of the games so far, the number of games so far must be a multiple of 10, and the same is true for 70% success rate. The other success rates have similar constraints, but the multiples are smaller. After ten games, of course Arun cannot have simultaneously won both 30% and 70% of the total! So we need at least twenty games.

It feels as if we should be able to include the other success percentages so long as we can make 30% and 70% work. But can we do this in twenty games? This would require Arun to win 3 out of the first 10, and 14 out of the first 20 (or the exact opposite), and after some head-scratching, this is impossible, since it would require winning eleven out of the ten games between these observations. This is a nice observation, and we now know the answer must be at least thirty. Given that we have made some nifty, rigorous deductions already, one might speculate that this is the true answer. To verify this, we just have to give a construction, and there are a number of ways to achieve this, for example such that Arun wins

• One of the first two
• Two of the first five
• Three of the first ten
• Twelve of the first twenty
• Twenty-one of the first thirty

To get full credit, almost any justification that a construction exists for 30 games would be fine – but one must be given. Otherwise how do you know that the true bound isn’t 40 or 50 because there’s some other reason why 30 fails?

Problem Three

Facing this problem for the first time, there’s a temptation to try values smaller than 2021, and this may well be useful. If nothing else, such experiments may well lead to a conjecture of the general form of the answer. Personally, I find it relatively easy to make mistakes when dealing with small numbers, and would turn to this as a second strategy after trying more theoretical routes first.

First note that the condition $0\le n\le 11$ could be replaced by $n\ge 0$, since we are forced to include zero pieces of weight $2^{11}=2048$, and would similarly be forced to include zero pieces of weight $2^n$ for any $n\ge 11$. (%)

When counting a potentially complex set, we should start by seeing whether there is anything which we can control effectively. For example, I don’t think it is clear how to control the number of pieces of weight 64 in a configuration. However, powers of two lend themselves well to inductive arguments (if you take the powers of two and multiply them all by two, you still have powers of two!) and to studying parity (that is, whether a number is odd or even).

In this case, note that you must have either one or three of the pieces with weight one to ensure the total weight is odd. Similarly, if the goal had been 2022, we would need either zero or two of these pieces to ensure the total weight is even.

So that settles the number of pieces of weight one, and it remains to solve the problem for total weight 2018 and 2020 with the extra condition that we are no longer allowed to use any pieces of weight one. A way to capture this is to divide everything in the problem by two. So we are now trying to capture all the ways to generate weights 1009 and 1010.

We can reformulate this more generally, and assert that the number of configurations $f(n)$ to make total weight n satisfies the relations:

$f(2n)=f(2n+1)=f(n)+f(n-1).$ (*)

It would be quite tiresome to solve backwards in a binary search fashion, but perhaps this is where trying some small examples turns useful. Either by observation on the original problem, or by thinking about (*) with some initial condition like f(0)=f(1)=1, we find that $f(n)=\lfloor n/2\rfloor +1$. (That is, $f(n)=\frac{n+2}{2}$ when n is even, and $f(n)=\frac{n+1}{2}$ when n is odd.)

As a final aside, this question is an excellent exercise in using generating functions in an olympiad context. A couple of points are worth making:

1. It is important to decide whether you’re including f(0) as an example, much more so than in the original argument. As we’ll see, it works better if you do include it, and will avoid an awkward sequence of algebra steps.
2. Figuring out the answer, for example from small-ish cases, will make a generating function argument easier unless you are very fluent with manipulating rational functions. In this case, we would be studying $\sum_{n=0}^\infty f(n)x^n$, and trying to show that it is equal to

$\sum (\lfloor n/2\rfloor +1)x^n = (1+x)(1+2x^2+3x^4+\ldots) = \frac{(1+x)}{(1-x^2)^2}.$ (*)

Many readers who’ve played around a bit with generating functions may find it easier to go from the left to the right than vice versa.

Writing down a direct generating function for f(n) and showing it matches up with (*) remains a good exercise. It is helpful to bear in mind the discussion at (%).

Problem Four

Since I last blogged regularly, I have given a series of masterclass on Euclidean geometry three times, including twice in an online format.

Part of the point of this masterclass is to get younger students to think in more general terms about what they are doing when they are solving geometry problems, even when they are straightforward. In a competition, time is tight, and anything that gets the job done is good, but there’s still value in reflecting on these ideas here. Two ideas in the first session are:

• Attacking the conclusion coherently
• Clean angle chasing

and this problem offers a good example of both of these in practice.

We’ll start with attacking the conclusion coherently. We’ve got to prove that a triangle is equilateral. How can we achieve this? There are a number of ways:

• Prove all three lengths are equal
• Prove that all three angles are equal
• Prove that two angles are equal to 60 degrees.
• Prove that one angle is equal to 60, and (any) two lengths are equal
• Prove that one angle is equal to 60, and the other two angles are equal

It’s important to keep all of these in mind as one explores the configuration. It’s genuinely unclear at the very start which of these characterisations of equilateral triangles will be most appropriate. Note though, that the configuration is symmetric in the roles of D and E, so that, for example, proving that the angle at D is 60 would be sufficient, since by symmetry the angle at E would also be 60, and we would conclude.

As regards clean angle chasing, the main moral here is to avoid angle arithmetic unless you are convinced it’s necessary. Ideally one keeps track of what could be determined in terms of other quantities, and only attempts the algebra/arithmetic once you are sure that it will work. It certainly isn’t helpful to have diagrams littered with angle measures like $240^\circ - \theta - \frac{\alpha}{2}$. Equality is the best thing to aim for. Try and demonstrate and notate equal angles wherever possible.

For example, here, even before introducing C,D,E, there is plenty of structure which might useful. Essentially every length in the configuration involving $(O_1,O_2,A,B)$ is equal, and all the angles are 60 or 120.

So we bear this in mind. I would consider proving that $\angle DO_1E=60^\circ$ with the following figure.

There are number of ways to complete the argument. Here are a couple of more obscure ones that illustrate my approach.

Firstly, let’s ignore the result that $\angle DO_1E=60^\circ$ and proceed directly to $\angle O_2DO_1=60^\circ$. It suffices to show that the blue angles are equal, which is equivalent to demonstrating that $ADO_2O_1$ is cyclic.

To achieve this, we would like to show that the angle $\angle AO_2D$ is equal to the red angle we’ve already discussed, and this follows by chasing round, using that $O_1$ is the centre of triangle ABC, and then the fact that $AO_2D$ is the external angle opposite $\angle ABC$ in cyclic quadrilateral $ABCO_2$. Indeed, note that this use of an equal external angle rather than ‘opposite angle is 180 – […]’ is the textbook example of clear angle chasing, where focusing on equality rather than arithmetic massively cleans up many diagrams.

As a second, rather overblown way to complete the argument, let’s show that the angles at D and E are equal, without calculating them. Since $O_1D,O_1E$ are angle bisectors, we have two pairs of equal angles as shown.

It would suffice to show that these measures are themselves equal. This is saying that line $ECO_2D$ is the external angle bisector of $\angle ACB$. And this is true, since $O_2$ is the arc-midpoint of AB on $\Gamma_1$!

I hope these observations are interesting. But, to reiterate, this level of complexity is unnecessary, and arguments by congruent triangles or direct angle chasing are perfectly satisfactory too!

Problem Five

The punchline of this problem is that m(N) is an integer precisely if N is a triangle number, from which it is quick to count the number of such N.

But how to arrive that this punchline? It is reasonably clear that the N-set with minimal mean must have the form $(1,2,\ldots,k, N)$. One could proceed by setting up inequalities involving the means of $(1,2,\ldots,k-1,N)$ and $(1,2,\ldots,k,k+1,N)$ to establish a relationship between k and N.

But there is a more direct route. A useful, and very intuitive result about means is the following. If you have a set of numbers with mean x, and you add into the set a new number y, then the mean of the new larger set is between x and y. In particular, if y>x, then the new mean is larger than x.

In particular, consider the mean of $(1,2,\ldots,k,N)$. If it is less than k, then removing k gives a smaller mean. If it is greater than k+1, then adding k+1 gives a smaller mean. So since the mean is an integer, it must be equal to k or k+1. That is, we have

$\frac{1+2+\ldots+k+N}{k+1}=k\text{ or }k+1.$

These lead to $N=\frac{k(k+1)}{2}\text{ or }\frac{(k+1)(k+2)}{2}$.

Problem Six

There must be some significance to the choice of 71 terms, and the 999,999 but it’s hopeless to try and see this immediately. In this scenario, it really is helpful to play around with semi-small examples.

Note that the second term is 1 regardless of whether we view this as P for PREVIOUS or S for SUM. However, if we then have an S, our sequence starts 1,1,2,2… or 1,1,2,4,… and these are bad, as once we have two even consecutive even numbers, we are doomed to see even numbers forever.

So once we have a common factor, that common factor is retained. How would we introduce a factor of, say, 11? Well, if you take 1,1,1,1,1,1,1,1,1,1,1 then take an S, you get 11, and afterwards you get either 11 or 22.

This gives a clue of the right way to set up the problem. For definiteness, we consider the second term to be an S. If you then consider a sequence of Ss and Ps, and write down the number of Ps between successive Ss (which may be zero) as a list $p_1,p_2,p_3,\ldots$, it holds that $\sum p_i=70$, and that each string of $p_i$ Ps introduces a factor of $p_i+1$, leading to

$\prod (p_i+1)=999,999=3^3\times 7\times 11\times 13\times 37.$

Note that $2\times 3 + 6+10+12+36=70$, so there exists a choice of the $p_i$s that satisfies these relations simultaneously. (And if we had replaced 71 by a smaller choice, this would have been impossible.)

The actual answer to the problem is obtained by studying how to rearrange the blocks of SPPP…s, but the main challenge lies in reducing to this form.

I thought this was a hard problem for BMO, and was impressed to see some solutions under exam conditions.

# BMO1 2017 – Questions 5 and 6

The first round of the British Mathematical Olympiad was sat yesterday. The questions can be found here and video solutions here. My comments on the first four questions are in the previous post.

Overall, I didn’t think any of the questions on this paper were unusually difficult by the standard of BMO1, but I found everything slightly more time-consuming than typical. I thought Question 5 was a great problem, and I tried lots of things unsuccessfully, first, and so wanted to discuss it in slightly more technical language. For Question 6 I made a decisive mistake, which I’ll explain, and which cost a lot of time. But in general, my point is that the back end of the paper was a little fiddlier than normal, and required longer written solutions, and perhaps many students might have had less time than expected to attack them anyway after details earlier in the paper.

Question Five

As I said before, I thought this question was quite challenging. Not because the solution is particularly exotic or complicated, but because there were so many possible things that might have worked. In my opinion it would not have been out of place at the start of an IMO paper, because it’s perfectly possible to have enough good ideas that eliminating the ones that don’t work takes an hour, or hours. Even though it slightly spoils the flow of the solution, I’m particularly trying to emphasise the tangents that didn’t work, mostly for reassurance to anyone who spent a long time struggling.

I was thinking about this question in terms of a 2Nx2N board, where N is even, and for the given question equal to 100. I spent a while thinking that the bound was 8N-4, corresponding to taking the middle two rows and the middle two columns, but not the 2×2 square which is their intersection. If you think of a comb as a ‘handle’ of 1xN cells, with an extra N/2 alternating cells (say, ‘teeth’) bolted on, then it’s clear this construction works because there’s never space to fit in a handle, let alone the teeth.

I couldn’t prove that this was optimal though. A standard way to prove a given bound K was optimal would be to produce a tiling on the board with K combs, where every cell is included in exactly one comb. But this is clearly not possible in this situation, since the number of cells in a comb (which is 150) does not divide the total number of cells on the board.

Indeed, the general observation that if you take a comb and a copy of the comb rotated by 180, the teeth of the second comb can mesh perfectly with the teeth of the first comb to generate a 3xN unit. I wasted a moderate amount of time pursuing this route.

[Note, it will be obvious in a minute why Iâ€™m writing â€˜shadedâ€™ instead of â€˜colouredâ€™.]

But in motivating the construction, I was merely trying to shade cells so that they intersected every possible 1xN handle, and maybe I could prove that it was optimal for this. In fact, I canâ€™t prove itâ€™s optimal because it isnâ€™t optimal â€“ indeed itâ€™s clear that a handle through one of the middle rows intersects plenty of shaded cells, not just one. However, with this smaller problem in mind, it didnâ€™t take long to come up with an alternative proposal, namely splitting the board into equal quarters, and shading the diagonals of each quarter, as shown.

It seems clear that you canâ€™t fit in a 1xN handle, and any sensible tiling with 1xN handles contains exactly one shaded cell, so this shading (with 4N shaded cells) is optimal. But is it optimal for a comb itself?

Consider a shading which works, so that all combs include a shaded cell. Itâ€™s clear that a comb is contained within a 2xN block, and in such a 2xN block, there are four possible combs, as shown.

You need to cover all these combs with some shading somewhere. But if you put the shaded cell on a tooth of comb A, then you havenâ€™t covered comb B. And if you put the shaded cell on the handle of comb A, then you havenâ€™t covered one of comb C and comb D. You can phrase this via a colouring argument too. If you use four colours with period 2×2, as shown

then any comb involves exactly three colours, and so one of them misses out the colour of the shaded cell. (I hope itâ€™s clear what I mean, even with the confusing distinction between â€˜shadedâ€™ and â€˜colouredâ€™ cells.)

Certainly we have shown that any 2xN block must include at least two shaded cells. And thatâ€™s pretty much it. We have a tiling with 2N copies of a 2xN block, and with at least two shaded cells in each, that adds to at least 4N shaded cells overall.

Looking back on the method, we can identify another way to waste time. Tiling a board, eg a chessboard with dominos is a classic motif, which often relies on clever colouring. So itâ€™s perhaps lucky that I didn’t spot this colouring observation earlier. Because the argument described really does use the local properties of how the combs denoted A-D overlap. An attempt at a global argument might start as follows: we can identify 2N combs which don’t use colour 1, and tile this subset of the grid with them, so we need to shade at least 2N cells from colours {2,3,4}. Similarly for sets of colours {1,3,4}, {1,2,4}, and {1,2,3}. But if we reduce the problem to this, then using roughly 2N/3 of each colour fits this global requirement, leading to a bound of 8N/3, which isn’t strong enough. [1]

Question Six

A word of warning. Sometimes it’s useful to generalise in problems. In Q5, I was thinking in terms of N, and the only property of N I used was that it’s even. In Q4, we ignored 2017 and came back to it at the end, using only the fact that it’s odd. By contrast, in Q2, the values did turn out to be important for matching the proof bounds with a construction.

You have to guess whether 300 is important or not here. Let’s see.

I have a natural first question to ask myself about the setup, but some notation is useful. Let $a_1,a_2,\ldots,a_{300}$ be the ordering of the cards. We require that $\frac{a_1+\ldots+a_n}{n}$ is an integer for every $1\le n\le 300$. Maybe the values of these integers will be important, so hold that thought, but for now, replace with the divisibility statement that $n | a_1+\ldots+a_n$.

I don’t think it’s worth playing with small examples until I have a better idea whether the answer is 5 or 295. So the natural first question is: “what does it mean to have $(a_1,\ldots,a_{n-1})$ such that you can’t pick a suitable $a_n$?”

It means that there is no integer k in $\{1,\ldots,300\}\backslash\{a_1,\ldots,a_{n-1}\}$ such that $n\,\big|\,(a_1+\ldots+a_{n-1})+k$, which for now we write as

$k\equiv -(a_1+\ldots+a_{n-1})\,\mod n.$

Consider the congruence class of $-(a_1+\ldots+a_{n-1})$ modulo n. There are either $\lfloor \frac{300}{n}\rfloor$ or $\lceil \frac{300}{n}\rceil$ integers under consideration in this congruence class. If no such k exists, then all of the relevant integers in this congruence class must appear amongst $\{a_1,\ldots,a_{n-1}\}$. At this stage, we’re trying to get a feel for when this could happen, so lower bounds on n are most relevant. Therefore, if we get stuck when trying to find $a_n$, we have

$\lfloor \frac{300}{n} \rfloor\text{ or }\lceil \frac{300}{n}\rceil \le n-1,$ (*)

which is summarised more succinctly as

$\lfloor \frac{300}{n} \rfloor \le n-1.$ (**)

[Note, with this sort of bounding argument, I find it helpful to add intermediate steps like (*) in rough. The chance of getting the wrong direction, or the wrong choice of $\pm 1$ is quite high here. Of course, you don’t need to include the middle step in a final write-up.]

We can check that (**) is false when $n\le 17$ and true when $n\ge 18$. Indeed, both versions of (*) are true when $n\ge 18$.

So we know the minimum failure length is at least 17. But is there a failing sequence of length 17? At a meta-level, it feels like there should be. That was a very natural bounding argument for 17 (which recall corresponds to $n=18$), and it’s easy to believe that might be part of an official solution. If we achieve equality throughout the argument, that’s most of the way to a construction as well. It won’t be so easy to turn this argument into a construction for $n\ge 19$ because there won’t be equality anywhere.

We have to hope there is a construction for $n=18$. What follows is a description of a process to derive (or fail to derive) such a construction. In a solution, one would not need to give this backstory.

Anyway, in such a construction, let $\alpha\in\{1,2,\ldots,18\}$ describe the congruence class modulo 18 which is exhausted by $\{a_1,\ldots,a_{17}\}$. I’m going to hope that $\alpha=18$ because then the calculations will be easier since everything’s a multiple of 18. We haven’t yet used the fact that for a problem, we need $\alpha\equiv-(a_1+\ldots+a_{n-1})$. We definitely have to use that. There are 16 multiples of 18 (ie relevant integers in the congruence class), so exactly one of the terms so far, say $a_j$, is not a multiple of 18. But then

$0 \equiv 0+\ldots+0+a_j+0+\ldots+0,$

which can’t happen. With a bit of experimentation, we find a similar problem making a construction using the other congruence classes with 16 elements, namely $\alpha\in \{13,14,\ldots,18\}$.

So we have to tackle a different class. If $\alpha\le 12$ then our sequence must be

$\alpha,18+\alpha,2\times 18 +\alpha, \ldots, 16\times 18 + \alpha,$

in some order. In fact, let’s add extra notation, so our sequence is

$(a_1,\ldots,a_{17}) = (18\lambda_1+ \alpha,\ldots,18\lambda_{17}+\alpha),$

where $(\lambda_1,\ldots,\lambda_{17})$ is a permutation of {0,…,16}. And so we require

$k \,\big|\, 18(\lambda_1+\ldots+\lambda_k) + k\alpha,$ (%)

for $1\le k\le 17$. But clearly we can lop off that $k\alpha$, and could ignore the 18. Can we find a permutation $\lambda$ such that

$k \,\big|\, \lambda_1+\ldots+\lambda_k.$

This was where I wasted a long time. I played around with lots of examples and kept getting stuck. Building it up one term at a time, I would typically get stuck around k=9,10. And I had some observations that in all the attempted constructions, the values of $\frac{\lambda_1+\ldots+\lambda_k}{k}$ were around 8 and 9 too when I got stuck.

I became convinced this subproblem wasn’t possible, and decided that would be enough to show that n=18 wasn’t a possible failure length. I was trying to show the subproblem via a parity argument (how must the $a_i$s alternate odd/even to ensure all the even partial sums are even) but this wasn’t a problem. Then I came up with a valid argument. We must have

$\lambda_1+\ldots+\lambda_{17}=136= 16\times 8 + 8\quad\text{and}\quad 16\,\big|\,\lambda_1+\ldots+\lambda_{16},$

which means $\lambda_1+\ldots+\lambda_{16}$ must be 128 = 15×8 + 8, ie $\lambda_{17}=8$. But then we also have $15\,\big|\, \lambda_1+\ldots+\lambda_{15}$, which forces $latex\lambda_{16}=8$ also. Which isn’t possible.

If this then hadn’t wasted enough time, I then tried to come up with a construction for n=19, for which there are lots more variables, and took a lot more time, and seemed to be suffering from similar problems, just in a more complicated way. So I became convinced I must have made a mistake, because I was forced down routes that were way too complicated for a 3.5 hour exam. Then I found it…

What did I do wrong? I’ll just say directly. I threw away the 18 after (%). This made the statement stronger. (And in fact false.) Suppose instead I’d thrown away a factor of 9 (or no factors at all, but it’s the residual 2 that’s important). Then I would be trying to solve

$k\,\big|\,2(\lambda_1+\ldots+\lambda_k).$

AndÂ now if you experiment, you will notice that taking $\lambda_1=0,\lambda_2=1,\lambda_3=2,\ldots$ seems to work fine. And of course, we can confirm this, using the triangle number formula for the second time in the paper!

This had wasted a lot of time, but once that thought is present, we’re done, because we can go straight back and exhibit the sequence

$(a_1,\ldots,a_{17}) = (1, 18+1,2\times 18 +1,\ldots, 16\times 18 +1).$

Then the sum so far is congruent to -1 modulo 18, but we have exhausted all the available integers which would allow the sum of the first 18 terms to be a multiple of 18. This confirms that the answer to the question as stated is 17.

At the start, I said that we should be cautious about generalising. In the end, this was wise advice. We definitely used the fact that 18 was even in the stage I over-reduced the first time. We also used the fact that there was at least one value of $\alpha$ with an ‘extra’ member of the congruence class. So I’m pretty sure this proof wouldn’t have worked with 288 = 16×18 cards.

Footnotes

[1] â€“ If shading were a weighted (or continuous or whatever you prefer) property, ie that each cell has a quantity of shading given by a non-negative real number, and we merely demand that the total shading per comb is at least one, then the bound 8N/3 is in fact correct for the total shading. We could look at a 2xN block, and give 1/3 shading to one cell of each colour in the block. Alternatively, we could be very straightforward and apply 2/3N shading to every cell in the grid. The fact that shading has to be (in this language) zero or one, imposes meaningful extra constraints which involve the shape of the comb.

# BMO1 2017 – Questions 1-4

The first round of the British Mathematical Olympiad was sat yesterday. The questions can be found here. I recorded some thoughts on the questions while I was in Cyprus, hence the nice Mediterranean sunset above. I hope this might be useful to current or future contestants, as a supplement to the concise official solutions available. It goes without saying that while these commentaries may be interesting at a general level, they will be much more educational to students who have at least digested and played around with the questions, so consider trying the paper first. Video solutions are available here. These have more in common with this blog post than the official solutions, though inevitably some of the methods are slightly different, and the written word has some merits and demerits over the spoken word for clarity and brevity.

The copyright for these questions lies with BMOS, and are reproduced here with permission. Any errors or omissions are obviously my own.

I found the paper overall quite a bit harder than in recent years, or at least harder to finish quickly. I’ve therefore postponed discussion of the final two problems to a second post, to follow shortly.

Question One

A recurring theme of Q1 from BMO1 in recent years has been: “itâ€™s possible to do this problem by a long, and extremely careful direct calculation, but additional insight into the setup makes life substantially easier.”

This is the best example yet. It really is possible to evaluate Helen’s sum and Phil’s sum, and compare them directly. But itâ€™s easy to make a mistake in recording all the remainders when the divisor is small, and itâ€™s easy to make a mistake in summation when the divisor is large, and so it really is better to have a think for alternative approaches. Making a mistake in a very calculation-heavy approach is generally penalised heavily. And this makes sense intellectually, since the only way for someone to fix an erroneous calculation is to repeat it themselves, whereas small conceptual or calculation errors in a less onerous solution are more easily isolated and fixed by a reader. Of course, it also makes sense to discourage such attempts, which arenâ€™t really related to enriching mathematics, which is the whole point of the exercise!

Considering small divisors (or even smaller versions of 365 and 366) is sometimes helpful, but here I think a ‘typical’ divisor is more useful. But first, some notation will make any informal observation much easier to turn into a formal statement. Corresponding to Helen and Phil, let h(n) be the remainder when n is divided by 365, and p(n) the remainder when n is divided by 366. I would urge students to avoid the use of ‘mod’ in this question, partly because working modulo many different bases is annoying notationally, partly because the sum is not taken modulo anything, and partly because the temptation to use mod incorrectly as an operator is huge here [1].

Anyway, a typical value might be n=68, and we observe that 68 x 5 + 25 = 365, and so h(68)=25 and p(68)=26. Indeed, for most values of n, we will have p(n)=h(n)+1. This is useful because

$p(1)+p(2)+\ldots+p(366) - \left(h(1)+h(2)+\ldots+h(365)\right)$

$= \left(p(1)-h(1)\right) + \ldots+\left(p(365)-h(365)\right) + p(366),$

and now we know that most of the bracketed terms are equal to one. We just need to handle the rest. The only time it doesn’t hold that p(n)=h(n)+1 is when 366 is actually a multiple of n. In this case, p(n)=0 and h(n)=n-1. We know that 366 = 2 x 3 x 61, and so its divisors are 1, 2, 3, 6, 61, 122, 183.

Then, in the big expression above, seven of the 365 bracketed terms are not equal to 1. So 358 of them are equal to one. The remaining ones are equal to 0, -1, -2, -5, -60, -121, -182 respectively. There are shortcuts to calculate the sum of these, but itâ€™s probably safer to do it by hand, obtaining -371. Overall, since p(366)=0, we have

$p(1)+p(2)+\ldots+p(366) - \left(h(1)+h(2)+\ldots+h(365)\right)$

$= -371 + 358 + 0 = -13.$

So, possibly counter-intuitively, Helen has the larger sum, with difference 13, and we didnâ€™t have to do a giant calculation…

Question Two

Suppose each person chooses which days to go swimming ‘at random’, without worrying about how to define this. Is this likely to generate a maximum or minimum value of n? I hope itâ€™s intuitively clear that this probably wonâ€™t generate an extreme value. By picking at random we are throwing away lots of opportunity to force valuable overlaps or non-overlaps. In other words, we should start thinking about ways to set up the swimming itinerary with lots of symmetry and structure, and probably weâ€™ll eventually get a maximum or a minimum. At a more general level, with a problem like this, one can start playing around with proof methods immediately, or one can start by constructing lots of symmetric and extreme-looking examples, and see what happens. I favour the latter approach, at least initially. You have to trust that at least one of the extreme examples will be guess-able.

The most obvious extreme example is that everyone swims on the first 75 days, and no-one swims on the final 25 days. This leads to n=75. But we’re clearly ‘wasting’ opportunities in both directions, because there are never exactly five people swimming. I tried a few more things, and found myself simultaneously attacking maximum and minimum, which is clearly bad, so focused on minimum. Just as a starting point, let’s aim for something small, say n=4. The obstacle is that if you demand at most four swimmers on 96 days, then even with six swimmers on the remaining four days, you donâ€™t end up with enough swimming having taken place!

Maybe you move straight from this observation to a proof, or maybe you move straight to a construction. Either way, I think itâ€™s worth saying that the proof and the construction come together. My construction is that everyone swims on the first 25 days, then on days 26-50 everyone except A and B swim, on days 51-75 everyone except C and D swim, and on days 76-100 everyone except E and F swim. This exactly adds up. And if you went for the proof first, you might have argued that the total number of swim days is 6×75 = 450, but is at most 4n + 6(100-n). This leads immediately to $n\ge 25$, and I just gave the construction. Note that if you came from this proof first, you can find the construction because your proof shows that to be exact you need 25 days with six swimmers, and 75 days with four swimmers, and itâ€™s natural to try to make this split evenly. Anyway, this clears up the minimum.

[Less experienced contestants might wonder why I was worried about generating a construction despite having a proof. Remember we are trying to find the minimum. I could equally have a proof for $n\ge 10$ which would be totally totally valid. But this wouldnâ€™t show that the minimum was n=10, because that isnâ€™t in fact possible (as weâ€™ve seen), hence itâ€™s the construction that confirms that n=25 is the true minimum.]

Itâ€™s tempting to go back to the drawing board for the maximum, but itâ€™s always worth checking whether you can directly adjust the proof youâ€™ve already given. And here you can! We argued that

$450\le 4n + 6(100-n)$

to prove the minimum. But equally, we know that on the n days we have at least five swimmers, and on the remaining days, we have between zero and four swimmers, so

$450 \ge 5n + 0\times (100-n),$ (*)

which gives $n\le 90$. If we have a construction that attains this bound then we are done. Why have I phrased (*) with the slightly childish multiple of zero? Because it’s a reminder that for a construction to attain this bound, we really do need the 90 days to have exactly five swimmers, and the remaining ten days to have no swimmers. So it’s clear what to do. Split the first 90 days into five groups of 15 days. One swimmer skips each group. No-one swims in the final ten days, perhaps because of a jellyfish infestation. So we’re done, and $25\le n\le 90$.

At a general level, it’s worth noting that in the story presented, we found an example for the minimum which we turned into a proof, and then a proof for the maximum, which we then analysed to produce a construction.

Note that similar bounding argumentsÂ would apply if we fiddled with the numbers 5, 75 and 100. But constructions matching the bounds might not then be possible because the splits wouldn’t work so nicely. This would make everything more complicated, but probably not more interesting.

Question Three

It’s understandable that lots of students attempting this paper might feel ill-at-ease with conventional Euclidean geometry problems. A good first rule of thumb here, as in many settings, is “donâ€™t panic!”, and a more specific second rule of thumb is “even if you think you can calculate, try to find geometric insight first.”

Here, it really does look like you can calculate. A configuration based on a given isosceles triangle and a length condition and a perpendicular line is open to several coordinate approaches, and certainly some sensible trigonometry. Itâ€™s also very open to organised labelling of the diagram. You have three equal lengths, and a right-angle, as shown.

The key step is this. Drop the perpendicular from A to BC, and call its foot D. That alone really is the key step, as it reduces both parts of the question to an easy comparison. It’s clear that the line AD splits the triangle into two congruent parts, and thus equal areas and perimeters. So it is enough to show that triangle BMN has the same area as triangle ABD, and that their outer-perimeters (ie the part of its perimeter which is also the perimeter of ABC) are the same.

But theyâ€™re congruent, so both of these statements are true, and the problem is solved.

My solution could be as short as two or three lines, so for the purposes of this post all that remains is to justify why you might think of the key step. Here are a few possible entry routes:

• You might notice that line AD induces the required property for triangle ABD.
• You might try to find a triangle congruent to AMN, and come up with D that way.
• Thereâ€™s already a perpendicular in the question so experimenting with another one is natural, especially since the perpendicular from A has straightforward properties.
• AMN is a right angle, and so constructing D gives a cyclic quadrilateral. We didnâ€™t use that directly in the proof above, but constructing cyclic quadrilaterals is usually a good idea.
• If you were trying a calculation approach, you probably introduced the length AD, or at least the midpoint D as an intermediate step.

On the video, Mary Teresa proposes a number of elegant synthetic solutions with a few more steps. You might find it a useful exercise to try to come up with some motivating reasons like the bullet points above to justify her suggestion to reflect A in M as a first step.

Question Four

I wasn’t paying enough attention initially, and I calculated $a_2=0\text{ or }2$. This made life much much more complicated. As with IMO 2017 Q1, if trying to deduce general behaviour from small examples, itâ€™s essential to calculate the small examples correctly!

Once you engage your brain properly, you find that $a_2=0 \text{ or }3$, and of course $a_2=0$ is not allowed, since it must be positive. So $a_2=3$, and a similar calculation suggests $a_3=1\text{ or }6$. Itâ€™s clear that the set of values for $a_{k+1}$ depends only on $a_k$, so if you take $a_3=1$, then youâ€™re back to the situation you started with at the beginning. If you choose to continue the exploration with $a_3=6$, you will find $a_4=2\text{ or }10$, at which point you must be triggered by the possibility that triangle numbers play a role here.

As so often with a play-around with small values, you need to turn a useful observation into a concrete statement, which could then be applied to the problem statement. It looks like in any legal sequence, every term will be a triangle number, so we only need to clarify which triangle number. An example of a suitable statement might be:

Claim: If $a_n=T_k=\frac{k(k+1)}{2}$, the k-th triangle number, then $a_{n+1}=T_{k-1}\text{ or }T_{k+1}$.

There are three stages. 1) Checking the claim is true; 2) checking the claim is maximally relevant; 3) proving it. In this case, proving it is the easiest bit. Itâ€™s a quick exercise, and Iâ€™m omitting it. Of course, we canâ€™t prove any statement which isnâ€™t true, and here we need to make some quick adjustment to account for the case k=1, for which we are forced to take $a_{n+1}=T_{k+1}$.

The second stage really concerns the question “but what if $a_n\ne T_k$?” While there are deductions one could make, the key is that if $a_1$ is a triangle number, the claim we’ve just made shows that $a_n$ is always a triangle number, so this question is irrelevant. Indeed the claim further shows that $a_{2017}\le T_{2017}$, and also that $a_{2017}=T_k$ for some odd value of k. To be fully rigorous you should probably describe a sequence which attains each odd value of k, but this is really an exercise in notation [2], and it’s very obvious they are all attainable.

In any case, the set of possible values is $\{T_1,T_3,\ldots,T_{2017}\}$, which has size 1009.

Final two questions

These are discussed in a subsequent post.

Footnotes

[1] â€“ mod n is not an operator, meaning you shouldnâ€™t think of it as ‘sending integers to other integers’, or ‘taking any integer, to an integer in {0,1,…,n-1}’. Statements like 19 mod 5 = 4 are useful at the very start of an introduction to modular arithmetic, but why choose 4? Sometimes itâ€™s more useful to consider -1 instead, and we want statements like $a^p\equiv a$ modulo p to make sense even when $a\ge p$. 19 = 4 modulo 5 doesnâ€™t place any greater emphasis on the 4 than the 19. This makes it more like a conventional equals sign, which is of course appropriate.

[2] â€“ Taking $a_n=T_n$ for $1\le n\le k$, and thereafter $a_n=T_k$ if k is odd, and $a_n=T_{k+1}$ if k is even will certainly work, as will many other examples, some perhaps easier to describe than this one, though make sure you don’t accidentally try to use $T_0$!

# BMO2 2017

The second round of the British Mathematical Olympiad was taken yesterday by about 100 invited participants, and about the same number of open entries. To qualify at all for this stage is worth celebrating. For the majority of the contestants, this might be the hardest exam they have ever sat, indeed relative to current age and experience it might well be the hardest exam they ever sit. And so I thought it was particularly worth writing about this year’s set of questions. Because at least in my opinion, the gap between finding every question very intimidating, and solving two or three is smaller, and more down to mindset, than one might suspect.

A key over-arching point at this kind of competition is the following: the questions have been carefully chosen, and carefully checked, to make sure they can be solved, checked and written up by school students in an hour. That’s not to say that many, or indeed any, will take that little time, but in principle it’s possible. That’s also not to say that there aren’t valid but more complicated routes to solutions, but in general people often spend a lot more time writing than they should, and a bit less time thinking. Small insights along the lines of “what’s really going on here?” often get you a lot further into the problem than complicated substitutions or lengthy calculations at this level.

So if some of the arguments below feel slick, then I guess that’s intentional. When I received the paper and had a glance in my office, I was only looking for slick observations, partly because I didn’t have time for detailed analysis, but also because I was confident that there were slick observations to be made, and I felt it was just my task to find them.

Anyway, these are the questions: (note that the copyright to these is held by BMOS – reproduced here with permission.)

Question One

I immediately tried the example where the perpendicular sides are parallel to the coordinate axes, and found that I could generate all multiples of 3 in this way. This seemed a plausible candidate for an answer, so I started trying to find a proof. I observed that if you have lots of integer points on one of the equal sides, you have lots of integer points on the corresponding side, and these exactly match up, and then you also have lots of integer points on the hypotenuse too. In my first example, these exactly matched up too, so I became confident I was right.

Then I tried another example ( (0,0), (1,1), (-1,1) ) which has four integer points, and could easily be generalised to give any multiple of four as the number of integer points. But I was convinced that this matching up approach had to be the right thing, and so I continued, trusting that I’d see where this alternate option came in during the proof.

Good setup makes life easy. The apex of the isosceles triangle might as well be at the origin, and then your other vertices can be $(m,n), (n,-m)$ or similar. Since integral points are preserved under the rotation which takes equal side to the other, the example I had does generalise, but we really need to start enumerating. The number of integer points on the side from (0,0) to (m,n) is G+1, where G is the greatest common divisor of m and n. But thinking about the hypotenuse as a vector (if you prefer, translate it so one vertex is at the origin), the number of integral points on this line segment must be $\mathrm{gcd}(m+n,m-n) +1$.

To me, this felt highly promising, because this is a classic trope in olympiad problem-setting. Even without this experience, we know that this gcd is equal to G if m and n have different parities (ie one odd, one even) and equal to 2G if m and n have the same parity.

So we’re done. Being careful not to double-count the vertices, we have 3G integral points if m and n have opposite parities, and 4G integral points if m and n have the same parity, which exactly fits the pair of examples I had. But remember that we already had a pair of constructions, so (after adjusting the hypothesis to allow the second example!) all we had to prove was that the number of integral points is divisible by at least one of 3 and 4. And we’ve just done that. Counting how many integers less than 2017 have this property can be done easily, checking that we don’t double-count multiples of 12, and that we don’t accidentally include or fail to include zero as appropriate, which would be an annoying way to perhaps lose a mark after totally finishing the real content of the problem.

Question Two

(Keen observers will note that this problem first appeared on the shortlist for IMO 2006 in Slovenia.)

As n increases, obviously $\frac{1}{n}$ decreases, but the bracketed expression increases. Which of these effects is more substantial? Well $\lfloor \frac{n}{k}\rfloor$ is the number of multiples of k which are at most n, and so as a function of n, this increases precisely when n is a multiple of k. So, we expect the bracketed expression to increase substantially when n has lots of factors, and to increase less substantially when n has few factors. An extreme case of the former might be when n is a large factorial, and certainly the extreme case of the latter is n a prime.

It felt easier to test a calculation on the prime case first, even though this was more likely to lead to an answer for b). When n moves from p-1 to p, the bracketed expression goes up by exactly two, as the first floor increases, and there is a new final term. So, we start with a fraction, and then increase the numerator by two and the denominator by one. Provided the fraction was initially greater than two, it stays greater than two, but decreases. This is the case here (for reasons we’ll come back to shortly), and so we’ve done part b). The answer is yes.

Then I tried to do the calculation when n was a large factorial, and I found I really needed to know the approximate value of the bracketed expression, at least for this value of n. And I do know that when n is large, the bracketed expression should be approximately $n\log n$, with a further correction of size at most n to account for the floor functions, but I wasn’t sure whether I was allowed to know that.

But surely you don’t need to engage with exactly how large the correction due to the floors is in various cases? This seemed potentially interesting (we are after all just counting factors), but also way too complicated. An even softer version of what I’ve just said is that the harmonic function (the sum of the first n reciprocals) diverges faster than n. So in fact we have all the ingredients we need. The bracketed expression grows faster than n, (you might want to formalise this by dividing by n before analysing the floors) and so the $a_n$s get arbitrarily large. Therefore, there must certainly be an infinite number of points of increase.

Remark: a few people have commented to me that part a) can be done easily by treating the case $n=2^k-1$, possibly after some combinatorial rewriting of the bracketed expression. I agree that this works fine. Possibly this is one of the best examples of the difference between doing a problem leisurely as a postgraduate, and actually under exam pressure as a teenager. Thinking about the softest possible properties of a sequence (roughly how quickly does it grow, in this case) is a natural first thing to do in all circumstances, especially if you are both lazy and used to talking about asymptotics, and certainly if you don’t have paper.

Question 3

I only drew a very rough diagram for this question, and it caused no problems whatsoever, because there aren’t really that many points, and it’s fairly easy to remember what their properties are. Even in the most crude diagram, we see R and S lie on AC and AD respectively, and so the conclusion about parallel lines is really about similarity of triangles ARS and ACD. This will follow either from some equal angles, or by comparing ratios of lengths.

Since angle bisectors by definition involve equal angles, the first attack point seems promising. But actually the ratios of lengths is better, provided we know the angle bisector theorem, which is literally about ratios of lengths in the angle bisector diagram. Indeed

$\frac{AR}{RC}=\frac{AQ}{CQ},\quad \frac{AS}{SD}=\frac{AP}{PD},$Â Â Â Â  (1)

and so it only remains to show that these quantities are in fact all equal. Note that there’s some anti-symmetry here – none of these expressions use B at all! We could for example note that AP/PD = BP/PC, from which

$\left(\frac{AS}{SD}\right)^2 = \frac{AP.BP}{PC.PD},$Â Â Â Â  (2)

and correspondingly for R and Q, and work with symmetric expressions. I was pretty sure that there was a fairly well-known result that in a cyclic quadrilateral, where P is the intersection of the diagonals

$\frac{AP}{PC} = \frac{AD.AB}{DC.BC},$Â Â Â Â  (3)

(I was initially wondering whether there was a square on the LHS, but an example diagram makes the given expression look correct.)

There will be a corresponding result for Q, and then we would be almost done by decomposing (2) slightly differently, and once we’d proved (3) of course. But doing this will turn out to be much longer than necessary. The important message from (3) is that in a very simple diagram (only five points), we have a result which is true, but which is not just similar triangles. There are two pairs of similar triangles in the diagram, but they aren’t in the right places to get this result. What you do have is some pairs of triangles with one pair of equal angles, and one pair of complementary angles (that is, $\theta$ in one, and $180-\theta$ in the other). This is a glaring invitation to use the sine rule, since the sines of complementary angles are equal.

But, this is also the easiest way to prove the angle bisector theorem. So maybe we should just try this approach directly on the original ratio-of-lengths statement that we decided at (1) was enough, namely $\frac{AQ}{CQ}=\frac{AP}{PD}$. And actually it drops out rapidly. Using natural but informal language referencing my diagram

$\frac{AP}{PD} = \frac{\sin(\mathrm{Green})}{\sin(\mathrm{Pink})},\quad\text{and}\quad \frac{AQ}{CQ}= \frac{\sin(\mathrm{Green})}{\sin(180-\mathrm{Pink})}$

and we are done. But whatever your motivation for moving to the sine rule, this is crucial. Unless you construct quite a few extra cyclic quadrilaterals, doing this with similar triangles and circle theorems alone is going to be challenging.

Remark: If you haven’t seen the angle bisector theorem before, that’s fine. Both equalities in (1) are a direct statement of the theorem. It’s not an intimidating statement, and it would be a good exercise to prove either of these statements in (1). Some of the methods just described will be useful here too!

Question 4

You might as well start by playing around with methodical strategies. My first try involved testing 000, 111, … , 999. After this, you know which integers appear as digits. Note that at this stage, it’s not the same as the original game with only three digits, because we can test using digits which we know are wrong, so that answers are less ambiguous. If the three digits are different, we can identify the first digit in two tests, and then the second in a further test, and so identify the third by elimination. If only two integers appear as digits, we identify each digit separately, again in three tests overall. If only one integer appears, then we are already done. So this is thirteen tests, and I was fairly convinced that this wasn’t optimal, partly because it felt like testing 999 was a waste. But even with lots of case tries I couldn’t do better. So I figured I’d try to prove some bound, and see where I got.

A crucial observation is the following: when you run a test, the outcome eliminates some possibilities. One of the outcomes eliminates at least half the codes, and the other outcome eliminates at most half the codes. So, imagining you get unlucky every time, after k tests, you might have at least $1000\times 2^{-k}$ possible codes remaining. From this, we know that we need at least 9 tests.

For this bound to be tight, each test really does need to split the options roughly in two. But this certainly isn’t the case for the first test, which splits the options into 729 (no digit agreements) and 271 (at least one agreement). Suppose the first test reduces it to 729 options, then by the same argument as above, we still need 9 tests. We now know we need at least 10 tests, and so the original guess of 13 is starting to come back into play.

We now have to make a meta-mathematical decision about what to do next. We could look at how many options might be left after the second test, which has quite a large number of cases (depending on how much overlap there is between the first test number and the second test number). It’s probably going to be less than 512 in at least one of the cases, so this won’t get us to a bound of 11 unless we then consider the third test too. This feels like a poor route to take for now, as the tree of options has branching at rate 3 (or 4 if you count obviously silly things) per turn, so gets unwieldy quickly. Another thought is that this power of two argument is strong when the set of remaining options is small, so it’s easier for a test to split the field roughly in two.

Now go back to our proposed original strategy. When does the strategy work faster than planned? It works faster than planned if we find all the digits early (eg if they are all 6 or less). So the worst case scenario is if we find the correct set of digits fairly late. But the fact that we were choosing numbers of the form aaa is irrelevant, as the digits are independent (consider adding 3 to the middle digit modulo 10 at all times in any strategy – it still works!).

This is key. For $k\le 9$, after k tests, it is possible that we fail every test, which means that at least $(10-k)$ options remain for each digit, and so at least $(10-k)^3$ options in total. [(*) Note that it might actually be even worse if eg we get a ‘close’ on exactly one test, but we are aiming for a lower bound, so at this stage considering an outcome sequence which is tractable is more important than getting the absolute worst case outcome sequence if it’s more complicated.] Bearing in mind that I’d already tried finishing from the case of reduction to three possibilities, and I’d tried hard to sneak through in one fewer test, and failed, it seemed sensible to try k=7.

After 7 tests, we have at least 27 options remaining, which by the powers-of-two argument requires at least 5 further tests to separate. So 12 in total, which is annoying, because now I need to decide whether this is really the answer and come up a better construction, or enhance the proof.

Clearly though, before aiming for either of these things, I should actually try some other values of k, since this takes basically no time at all. And k=6 leaves 64 options, from which the power of two argument is tight; and k=5 leaves 125, which is less tight. So attacking k=6 is clearly best. We just need to check that the 7th move can’t split the options exactly into 32 + 32. Note that in the example, where we try previously unseen digits in every position, we split into 27 + 37 [think about (*) again now!]. Obviously, if we have more than four options left for any digit, we are done as then we have strictly more than 4x4x4=64 options. So it remains to check the counts if we try previously unseen digits in zero, one or two positions. Zero is silly (gives no information), and one and two can be calculated, and don’t give 32 + 32.

So this is a slightly fiddly end to the solution, and relies upon having good control over what you’re trying to do, and what tools you currently have. The trick to solving this is resisting calculations and case divisions that are very complicated. In the argument I’ve proposed, the only real case division is right at the end, by which point we are just doing an enumeration in a handful of cases, which is not really that bad.

# BMO1 2016 – the non-geometry

Here’s a link to yesterday’s BMO1 paper, and the video solutions for all the problems. I gave the video solution to the geometric Q5, and discuss aspects of this at some length in the previous post.

In these videos, for obvious educational reasons, there’s a requirement to avoid referencing theory and ideas that aren’t standard on the school curriculum or relatively obvious directly from first principles. Here, I’ve written down some of my own thoughts on the other problems in a way that might add further value for those students who are already have some experience at olympiads and these types of problems. In particular, on problems you can do, it’s worth asking what you can learn from how you did them that might be applicable generally, and obviously for some of the harder problems, it’s worth knowing about solutions that do use a little bit of theory. Anyway, I hope it’s of interest to someone.

Obviously we aren’t going to write out the whole list, but there’s a trade-off in time between coming up with neat ideas involving symmetry, and just listing and counting things. Any idea is going to formalise somehow the intuitive statement ‘roughly half the digits are odd’. The neat ideas involve formalising the statement ‘if we add leading zeros, then roughly half the digits are odd’. The level of roughness required is less in the first statement than the second statement.

Then there’s the trade-off. Trying to come up with the perfect general statement that is useful and true might lead to something like the following:

‘If we write the numbers from 0000 to N, with leading zeros, and all digits of N+1 are even, then half the total digits, ie 2N of them, are odd.’

This is false, and maybe the first three such things you try along these lines are also false. What you really want to do is control the numbers from 0000 to 1999, for which an argument by matching is clear, and gives you 2000 x 4 / 2 = 4000 odd digits. You can exploit the symmetry by matching k with 1999-k, or do it directly first with the units, then with the tens and so on.

The rest (that is, 2000 to 2016) can be treated by listing and counting. Of course, the question wants an actual answer, so we should be wary of getting it wrong by plus or minus one in some step. A classic error of this kind is that the number of integers between 2000 and 2016 inclusive is 17, not 16. I don’t know why the memory is so vivid, but I recall being upset in Year 2 about erring on a problem of this kind involving fences and fenceposts.

As with so many new types of equation, the recipe is to reduce to a type of equation you already know how to solve. Here, because {x} has a different form on different ranges, it makes sense to consider the three ranges

$x\in[0,1/25],\, x\in[1/25,1/8],\, x\in [1/8,\infty),$

as for each of these ranges, we can rewrite $5y\{8y\}\{25y\}$ in terms of standard functions without this bracket notation. On each range we can solve the corresponding equation. We then have to check that each solution does actually lie in the appropriate range, and in two cases it does, and in one case it doesn’t.

Adding an appropriately-chosen value to each side allows you to factorise the quadratics. This might be very useful. But is it an invitation to do number theory and look at coprime factors and so on, or is a softer approach more helpful?

The general idea is that the set of values taken by any quadratic sequence with integer coefficients and leading coefficient one looks from a distance like the set of squares, or the set $\{m(m+1), \,m\in\mathbb{N}\}$, which you might think of as ‘half-squares’ or ‘double triangle numbers’ as you wish. And by, ‘from a distance’ I mean ‘up to an additive constant’. If you care about limiting behaviour, then of course this additive constant might well not matter, but if you care about all solutions, you probably do care. To see why this holds, note that

$n^2+2n = (n+1)^2 - 1,$

so indeed up to an additive constant, the quadratic on the LHS gives the squares, and similarly

$n^2 - 7n = (n-4)(n-3)-12,$

and so on. To solve the equation $n^2=m^2+6$, over the integers, one can factorise, but another approach is to argue that the distance between adjacent squares is much more than 6 in the majority of cases, which leaves only a handful of candidates for n and m to check.

The same applies at this question. Adding on 9 gives

$n^2-6n+9 = m^2 + m -1,$

which is of course the same as

$(n-3)^2 = m(m+1)-1.$

Now, since we now that adjacent squares and ‘half-squares’ are more than one apart in all but a couple of cases, we know why there should only be a small number of solutions. I would call a method of this kind square-sandwiching, but I don’t see much evidence from Google that this term is generally used, except on this blog.

Of course, we have to be formal in an actual solution, and the easiest way to achieve this is to sandwich $m(m+1)-1$ between adjacent squares $m^2$ and $(m+1)^2$, since it is very much clear-cut that the only squares which differ by one are zero and one itself.

I really don’t have much to say about this. It’s not on the school curriculum so the official solutions are not allowed to say this, but you have to use that all integers except those which are 2 modulo 4 can be written as a difference of two squares. The easiest way to show this is by explicitly writing down the appropriate squares, treating the cases of odds and multiples of four separately.

So you lose if after your turn the running total is 2 modulo 4. At this point, the combinatorics isn’t too hard, though as in Q1 one has to be mindful that making an odd number of small mistakes will lead to the wrong answer! As in all such problems, it’s best to try and give a concrete strategy for Naomi. And it’s best if there’s something inherent in the strategy which makes it clear that it’s actually possible to implement. (Eg, if you claim she should choose a particular number, ideally it’s obvious that number is available to choose.)

One strategy might be: Naomi starts by choosing a multiple of four. Then there are an even number of multiples of four, so Naomi’s strategy is:

• whenever Tom chooses a multiple of four, Naomi may choose another multiple of four;
• whenever Tom chooses a number which is one (respectively three) modulo 4, Naomi may choose another which is three (respectively one) modulo 4.

Note that Naomi may always choose another multiple of four precisely because we’ve also specified the second condition. If sometimes Tom chooses an odd number and Naomi responds with a multiple of four out an idle and illogical sense of caprice, then the first bullet point would not be true. One can avoid this problem by being more specific about exactly what the algorithm is, though there’s a danger that statements like ‘whenever Tom chooses k, Naomi should choose 100-k’ can introduce problems about avoiding the case k=50.

I started this at the train station in Balatonfured with no paper and so I decided to focus on the case of just m, m+1 and n, n+2. This wasn’t a good idea in my opinion because it was awkward but guessable, and so didn’t give too much insight into actual methods. Also, it didn’t feel like inducting on the size of the sequences in question was likely to be successful.

If we know about the Chinese Remainder Theorem, we should know that we definitely want to use it here in some form. Here are some clearly-written notes about CRT with exercises and hard problems which a) I think are good; b) cite this blog in the abstract. (I make no comment on correlation or causality between a) and b)…)

CRT is about solutions to sets of congruence equations modulo various bases. There are two aspects to this , and it feels to me like a theorem where students often remember one aspect, and forget the other one, in some order. Firstly, the theorem says that subject to conditions on the values modulo any non-coprime bases, there exist solutions. In many constructive problems, especially when the congruences are not explicit, this is useful enough by itself.

But secondly, the theorem tells us what all the solutions are. There are two stages to this: finding the smallest solution, then finding all the solutions. Three comments: 1) the second of these is easy â€“ we just add on all multiples of the LCM of the bases; 2) we don’t need to find the smallest solution â€“ any solution will do; 3) if you understand CRT, you might well comment that the previous two comments are essentially the same. Anyway, finding the smallest solution, or any solution is often hard. When you give students an exercise sheet on CRT, finding an integer which is 3 mod 5, 1 mod 7 and 12 mod 13 is the hard part. Even if you’re given the recipe for the algorithm, it’s the kind of computation that’s more appealing if you are an actual computer.

Ok, so returning to this problem, the key step is to phrase everything in a way which makes the application of CRT easy. We observe that taking n=2m satisfies the statement â€“ the only problem of course is that 2m is not odd. But CRT then tells us what all solutions for n are, and it’s clear that 2m is the smallest, so we only need to add on the LCM (which is odd) to obtain the smallest odd solution.

# BMO1 2016 Q5 – from areas to angles

For the second year in a row Question 5 has been a geometry problem; and for the second year in a row I presented the video solution; and the for the second year in a row I received the question(s) while I was abroad. You can see the video solutions for all the questions here (for now). I had a think about Q5 and Q6 on the train back from a day out at Lake Balaton in Western Hungary, so in keeping with last year’s corresponding post, here are some photos from those sunnier days.

I didn’t enjoy this year’s geometry quite as much as last year’s, but I still want to say some things about it. At the time of writing, I don’t know who proposed Q5, but in contrast to most geometry problems, where you can see how the question might have emerged by tweaking a standard configuration, I don’t have a good intuition for what’s really going on here. I can, however, at least offer some insight into why the ‘official’ solution I give on the video has the form that it does.

The configuration given is very classical, with only five points, and lots of equal angles. The target statement is also about angles, indeed we have to show that a particular angle is a right-angle. So we might suspect that the model approach might well involve showing some other tangency relation, where one of the lines AC and BC is a radius and the other a tangent to a relevant circle. I think it’s worth emphasising that throughout mathematics, the method of solving a problem is likely to involve similar objects to the statement of the problem itself. And especially so in competition problems â€“ it seemed entirely reasonable that the setter might have found a configuration with two corresponding tangency relations and constructed a problem by essentially only telling us the details of one of the relations.

There’s the temptation to draw lots of extra points or lots of extra lines to try and fit the given configuration into a larger configuration with more symmetry, or more suggestive similarity [1]. But, at least for my taste, you can often make a lot of progress just by thinking about what properties you want the extra lines and points to have, rather than actually drawing them. Be that as it may, for this question, I couldn’t initially find anything suitable along these lines [2]. So we have to think about the condition.

But then the condition we’ve been given involves areas, which feels at least two steps away from giving us lots of information about angles. It doesn’t feel likely that we are going to be able to read off some tangency conditions immediately from the area equality we’ve been given. So before thinking about the condition too carefully, it makes sense to return to the configuration and think in very loose terms about how we might prove the result.

How do we actually prove that an angle is a right-angle? (*) I was trying to find some tangency condition, but it’s also obviously the angle subtending by the diameter of a circle. You could aim for the Pythagoras relation on a triangle which includes the proposed right-angle, or possibly it might be easier to know one angle and two side-lengths in such a triangle, and conclude with some light trigonometry? We’ve been given a condition in terms of areas, so perhaps we can use the fact that the area of a right-angled triangle is half the product of the shorter side-lengths? Getting more exotic, if the configuration is suited to description via vectors, then a dot product might be useful, but probably this configuration isn’t.

The conclusion should be that it’s not obvious what sort of geometry we’re going to need to do to solve the problem. Maybe everything will come out from similar triangles with enough imagination, but maybe it won’t. So that’s why in the video, I split the analysis into an analysis of the configuration itself, and then an analysis of the area condition. What really happens is that we play with the area condition until we get literally anything that looks at all like one of the approaches discussed in paragraph (*). To increase our chances, we need to know as much about the configuration as possible, so any deductions from the areas are strong.

The configuration doesn’t have many points, so there’s not much ambiguity about what we could do. There are two tangents to the circle. We treat APC with equal tangents and the alternate segment theorem to show the triangle is isosceles and that the base angles are equal to the angle at B in ABC. Then point Q is ideally defined in terms of ABC to use power of a point, and add some further equal angles into the diagram. (Though it turns out we don’t need the extra equal angle except through power of a point.)

So we have some equal angles, and also some length relations. One of the length relations is straightforward (AP=CP) and the other less so (power of a point $CQ^2 = AQ\cdot BQ$). The really key observation is that the angle-chasing has identified

$\angle PAQ = 180 - \angle \hat C,$

which gives us an alternative goal: maybe it will be easier to show that PAQ is a right-angle.

Anyway, that pretty much drinks the configuration dry, and we have to use the area condition. I want to emphasise how crucial this phase in for this type of geometry problem. Thinking about how to prove the goal, and getting a flavour for the type of relation that comes out of the configuration is great, but now we need to watch like a hawk when we play with the area condition for relations which look similar to what we have, and where we might be going, as that’s very likely to be the key to the problem.

We remarked earlier that we’re aiming for angles, and are given areas. A natural middle ground is lengths. All the more so since the configuration doesn’t have many points, and so several of the triangles listed as having the same area also have the same or similar bases. You might have noticed that ABC and BCQ share height above line AQ, from which we deduce AB=BQ. It’s crucial then to identify that this is useful because it supports the power of a point result from the configuration itself. It’s also crucial to identify that we are doing a good job of relating lots of lengths in the diagram. We have two pairs of equal lengths, and (through Power of a Point) a third length which differs from one of them by a factor of $\sqrt{2}$.

If we make that meta-mathematical step, we are almost home. We have a relation between a triple of lengths, and between a pair of lengths. These segments make up the perimeter of triangle APQ. So if we can relate one set of lengths and the other set of lengths, then we’ll know the ratios of the side lengths of APQ. And this is excellent, since much earlier we proposed Pythagoras as a possible method for establish an angle is a right-angle, and this is exactly the information we’d need for that approach.

Can we relate the two sets of lengths? We might guess yes, that with a different comparison of triangles areas (since we haven’t yet used the area of APC) we can find a further relation. Indeed, comparing APC and APQ gives CQ = 2PC by an identical argument about heights above lines.

Now we know all the ratios, it really is just a quick calculation…

[1] â€“ I discussed the notion of adding extra points when the scripts for the recording were being shared around. It was mentioned that for some people, the requirement to add extra points (or whatever) marks a hard division between ‘problems they can do’ and ‘problem they can’t do’. While I didn’t necessarily follow this practice while I was a contestant myself, these days the first thing I do when I see any angles or an angle condition in a problem is to think about whether there’s a simple way to alter the configuration so the condition is more natural. Obviously this doesn’t always work (see [2]), but it’s on my list of ‘things to try during initial thinking’, and certainly comes a long way before approaches like ‘place in a Cartesian coordinate system’.

[2] â€“ Well, I could actually find something suitable, but I couldn’t initially turn it into a solution. The most natural thing is to reflect P in AC to get P’, and Q in BC to get Q’. The area conditions [AP’C]=[ABC]=[BCQ’] continue to hold, but now P’ and B are on the same side of AC, hence P’B || AC. Similarly AQ’ || BC. I see no reason not to carry across the equal length deductions from the original diagram, and we need to note that angles P’AC, ACP’, CBA are equal and angles Q’AB and BAC are equal. In the new diagram, there are many things it would suffice to prove, including that CP’Q’ are collinear. Note that unless you draw the diagram deliberately badly, it’s especially easy accidentally to assume that CP’Q’ are collinear while playing around, so I wasted quite a bit of time. Later, while writing up this post, I could finish it [3].

[3] â€“ In the double-reflected diagram, BCQ’ is similar to P’BA, and since Q’C=2P’C = P’A, and Q’B=AB, you can even deduce that the scale factor is $\sqrt{2}$. There now seemed two options:

• focus on AP’BC, where we now three of the lengths, and three of the angles are equal, so we can solve for the measure of this angle. I had to use a level of trigonometry rather more exotic than the Pythagoras of the original solution, so this doesn’t really serve purpose.
• Since BCQ’ is similar to P’BA and ABQ’ similar to CP’A, we actually have Q’BCA similar to AP’BC. In particular, $\angle CBP' = \angle ACB$, and thus both are 90. Note that for this, we only needed the angle deductions in the original configuration, and the pair of equal lengths.
• There are other ways to hack this final stage, including showing that BP’ meets AQ’ at the latter’s midpoint, to give CP’Q’ collinear.

# Balkan MO 2016 – UK Team Blog Part Two

This short blog recordsÂ the UK team at the Balkan Mathematical Olympiad 2016, held in Albania. The first part is here. A more mathematical version of this report, with commentaries on the problems, will appear at the weekend.

Sunday 8th May

Gerry and I are separated by 15km, so we can’t work together until this morning, when I also get a chance to see the UK team at their base in Vore, before they are whisked off to a beach. We now have the chance to work on the geometry together, which includes two sensible trigonometric arguments, and a nice synthetic proof only with reference to an inverted diagram. We quickly decide that this isn’t a major error, and aim to schedule our meetings as quickly as possible.

The coordinators for questions 3 and 4 seem very relaxed, and we quickly get exactly what we deserve, plus a spurious extra point for Michael because he used the phrase ‘taxicab metric’ in his rough. Thomas’s trigonometry, especially its bold claim that ‘by geometry, there are no other solutions’ when an expression becomes non-invertible, seems not to have been read entirely critically. Michael’s inverted diagram is briefly a point of controversy, but we are able to get 9 rather than the 7 which was proposed, absurdly for an argument that was elegant and entirely valid in the correct diagram up to directed angles. Question 1 is again rapid, as the coordinators say that the standard of writing is so clear that they are happy to ignore two small omissions. It transpires after discussion with, among others, the Italian leader, that such generosity may have been extended to some totally incorrect solutions, but in the final analysis, everything was fair.

So we are all sorted around 11.30am with a team score of 152, a new high for the UK at this competition. This is not necessarily a meaningful or consistent metric, but with scores of {20,21,22,29,30,30} everyone has solved at least two problems, and the three marks lost were more a matter of luck than sloppiness. Irrespective of the colours of medals this generates, Gerry and I are very pleased. We find a table in the sun, and I return to my introduction while we await progress from the other countries’ coordinations, and our students’ return.

This does not happen rapidly, so I climb the hill behind the hotel up a narrow track. A small boy is standing around selling various animals. Apparently one buys rabbits by the bucket and puppies by the barrel in Albania. Many chickens cross the road, but key questions remain unanswered. From the summit, there is a panorama across the whole Tirana area, and the ring of mountains encircling us. One can also see flocks of swifts, which are very similar to swallows, only about twice as large, and their presence in any volume makes no comment on the arrival of the British summer.

The students return mid-afternoon, and are pleased with their scores. Jamie explains their protracted misadventure with a camp bed in their ‘suite’ of rooms, and Jacob shows off his recent acquisitions: a felt hat, and a t-shirt outlining the border of a ‘greater Albania’. The fact that they didn’t have his size seems not to have been a deterrent, but hopefully the snug fit will discourage him from sporting it in Montenegro, which might lead to a political incident.

Hours pass and time starts to hang heavily as dinner approaches, with no sign of the concluding jury meeting. Finally, we convene at 10pm to decide the boundaries. The chair of the jury reads the regulations, and implements them literally. There’s a clump of contestants with three full solutions, so the boundaries are unusually compressed at 17, 30 and 32. A shame for Thomas on 29, but these things happen, and three full solutions minus a treatment of the constant case for a polynomial is still something to be happy about. Overall, 4 bronze and 2 silvers is a pleasing UK spread, and only the second time we have earned a full set of medals at this competition. The leaders are rushed back to Tirana, but hopefully the teams have enough energy left for celebration!

Monday 9th May

Today is the tutti excursion, but on the way the leaders stop at the city hall to meet the mayor of Tirana. He is new to the office, reminiscent of a young Marlon Brando, and has a bone-crushing handshake. He improvises an eloquent address, and negotiates with flair the awkward silence which follows when the floor is opened for speeches in response. In the end, the Saudi leader and I both say some words of thanks on behalf of the guest nations, and soon we are back on our way south towards Greece. The Albanian motorway is smooth and modern, but we find ourselves competing for space with communist-era windowless buses and the occasional pedestrian leading by hand a single cow.

Our destination is Berat, known as the city of a thousand windows, and home to a hilltop castle complex from which none of the thousand windows are visible. The old orthodox cathedral is now a museum of icons and other religious art, and we get a remarkably interesting tour from a local guide. The highlight is a mosaic representation of the Julian calendar, and we discuss whether the symmetries built into the construction would be more conducive to a geometry or a combinatorics question.

Back in Tirana, we reconverge at the closing ceremony, held in the theatre at a local university for the arts. While we wait, there is a photo montage, featuring every possible Powerpoint transition effect, in which Jacob and his non-standard hat usage makes a cameo appearance. We are then treated to a speech by Joszef Pelikan, who wows the crowd by switching effortlessly into Albanian, and some highly accomplished dancing, featuring both classical ballet and traditional local styles.

The ministry have taken over some aspects of the organisation here, and there is mild chaos when it’s medal time. The leaders are called upon to dispense the prizes, though the UK is snubbed for alphabetic reasons. The end result is that forty students are on stage with neither medals nor any instructions to leave. Eventually it vaguely resolves, though it is a shame there is no recognition for the two contestants (from Serbia and Romania) who solved the final problem and thus achieved a hugely impressive perfect score.

As you can see, the UK team look extremely pleased with themselves, and Michael’s strategy to get to know all the other teams through the medium of the selfie is a storming success. A very large number of photographs are taken, and Thomas is not hiding in at least one of them. The closing dinner is back in Vore, which is very convivial and involves many stuffed vine leaves. Rosie suggests we retire somewhere quieter, but by the time we establish how to leave, she has instead dragged the rest of the team onto the dancefloor, where near-universal ignorance of the step pattern is no obstacle to enjoying the folk music. The DJ slowly transitions towards the more typical Year 11 disco playlist, and Jill feels ‘Hips don’t lie’ is a cue for the adults to leave.

Tuesday 10th May

Our flight leaves at 9pm so we have many hours to fill. It turns out that we have one of the shortest journeys. The Serbians have caught a bus at 3am, while the Cypriots are facing stopovers in Vienna and Paris! It is another beautiful day, so we hire a small van to take us to Lezhe, the hometown of our guide Sebastian, and the nearby beach at Shengjin.

We walk to the tip of the breakwater, and watch some fishermen hard at work, though apparently today is a lean catch. The buildings along the beachfront are a sequence of pastel colours, backing onto another sheer mountain, and we could easily be in Liguria. Jamie is revising for his A2-level physics and chemistry exams, which start at 9am tomorrow morning, and the rest of the team are trying to complete the shortlist of problems from IMO 2007. They progress through the questions in the sand, with a brief diversion as Jacob catches a crab in the shallows with his bare hands for no apparent reason.

After a fish-heavy lunch, we return to Vore, and I’ve run out of subsubsections to amend, so propose another walk into the hills. The animals we meet this time appear not to be for sale. Some scrabbling in the undergrowth is sadly not the longed-for bear or wolf. Many of its colleagues are loitering on the local saddle point, and our Albanian companion Elvis describes them as ‘sons of sheep’, while Renzhi confidently identifies them as cows. They are goats. There is a small but vigorous goatdog, who reacts with extreme displeasure to our attempt to climb to one viewpoint, so Gerry leads us off in another direction up the local version of the north face of the Eiger. We do emerge on the other side, dustier but with plenty of heavily silhouetted photographs.

Then the hour of departure, and time to say goodbye to the organisers, especially Adrian, Matilda and Enkel who have made everything happen, and in a wonderful spirit; and our guide Sebastian, who has set an impossibly high bar for any others to aspire to. We wish him well in his own exams, which start on Thursday! Albania has left a strong positive impression, and it will sit high on my list of places to explore more in the future, hopefully before too many others discover it. The airport affords the chance to spend the final Leke on brandy and figurines of Mother Teresa, and the flight the chance to finish problem N5, and discuss our geometry training regime with Rosie and Jacob as they work through some areal exercises.

2am is not a thrilling time to be arriving in Oxford, and 2.30am is not a thrilling time to be picking up solutions to past papers (and an even less thrilling time to discover that no such solutions have been handed in). But this has been a really enjoyable competition, at which the UK team were delightful company, and performed both strongly and stylishly at the competition, so it is all more than justified. We meet again at half-term in three weeks’ time to select the UK team for the IMO in Hong Kong, and hopefully explore some more interesting mathematics!

# Balkan MO 2016 – UK Team Blog Part One

The Balkan Mathematical Olympiad is a competition for high school students from eleven countries in Eastern Europe, hosting on an annually rotating basis. For the 33rd edition it was Albania’s turn to host, and the UK was invited to participate as a guest nation.

A report with more mathematics, less frivolity and minimal chronological monotonicity can be found [SHORTLY].

Wednesday 4th May

I put the finishing touches to another draft of another chapter of my thesis, cajole the Statistics Department printer into issueing eighteen tickets, six consent forms and a terrifyingly comprehensive insurance policy, and head for Gatwick to meet the team. The UK imposes a policy that we will only take anyone to the Balkan MO once, so as to maximise the number of students who get to experience an international competition. The faces aren’t entirely new though – all six attended our winter programme in Hungary over New Year and the recent selection camp in Cambridge. They are showing the right level of excitement: the level that suggests they will enjoy the competition but won’t lose their passports in the next thirty minutes. As a point of trivia, this UK team are all sixth-formers, which, after checking not very carefully, doesn’t seem to have happened for any UK team for a long time, possibly not since 2008 when I was a contestant.

As of February this year, it’s now possible to fly direct to Albania on British Airways, which is a major improvement on the alternatives featuring either a seven-hour layover in Rome, or a nailbiting twenty-five minute interchange in Vienna. A drawback of the diary format is the challenging requirement to say interesting things about flights. In this instance, my principal challenge is to find some leftover room in my seat, as my neighbour’s physique has the same level of respect for the constraining power of armrests as the sea for the battlements of a child’s sandcastle. Across the aisle, Renzhi and Thomas face the twin challenges of a sheet of functional equations I’ve collated, and the well-meaning attempts of cabin attendants and their own neighbours to discuss said functional equations.

Later, over dinner next to Mother Teresa Airport in Tirana, we discuss the role of mathematics in recent films. Based on a sample size of at most two, we decide that The Man who Knew Infinity’ is slightly better than The Imitation Game’, partly because the former had fewer mathematical errors, or at least mispronunciations, about which Gerry feels strongly.

Thursday 5th May

The drawback of the new BA route is that it doesn’t run on Thursdays, so we are actually almost a full day early. Morning brings a cloudless summer’s day, and views of the imposing mountains that encircle Tirana. The students have assembled a healthy collection of past problems that they are keen to attempt as practice, and it seems natural to attempt this in a slightly more interesting place than the hotel lobby for at least some of the day.

Our guide Sebastian waves his Blackberry and rapidly conjures up an excursion to Mount Dajti, a small resort two-thirds of the way up a small mountain accessed from suburban Tirana via cable car. We follow a sign that seems to point to the summit, but the trail has distinctly horizontal ambitions. We are rewarded nonetheless with some pleasant views over the mountain range down past enclosed cerulean lakes down to the Adriatic, and even beyond to Italy.

Gerry is concerned about whether our return route is actually taking us where we want to go. He is right to be concerned, but not for that reason. It is the correct direction, but through a military base. Despite this, we make it back to the top of the cable car in the correct number of pieces. There’s the chance to alter this with some diverting activities, namely horse-riding and target-shooting. The targets are balloons, mounted on a clothes line at roughly horse-head-height. We move along.

Several years of attending maths competitions has increased both my ability to solve problems in Euclidean geometry, and also my suspicion of anything with a title like ‘Museum of National History’. I’m going to have to adjust the latter, because the recently-opened Albanian version, called BunkART, was actually excellent. It was housed in the five-level 108-room bunker built into the mountain to protect Enver Hoxha from nuclear attack. The rooms detailed the recent, fragmented history of the country, and were interspersed with aggressively modern art installations. In one basement which used to house the isotope filters, we were treated to a video loop of blood dripping onto barbed wire set to Mahler’s 5th Symphony.

While some regional competitions have adopted the ‘benign dictatorship’ approach to choosing the problems, the Balkan MO still has a problem selection phase. So I separate from the students and spend a pleasant few hours playing around with some of the proposals in the rooftop lounge of the leaders’ hotel on a balmy night in central Tirana.

Friday 6th May

The task for today is to construct a paper. A committee has selected a shortlist of problems, and we have to narrow this down to four, with one from each topic area, with an appropriate range of difficulty. The shortlist definitely contains some gems and some anti-gems, and more thoughts about these can be found in the official report.

The only dramatic moment comes when the Greek leader flourishes a webpage and an old IMO shortlist problem, which does indeed contain a proposed question as a lemma, and so it is rejected. Partly as a result of this, a medium geometry problem is chosen quickly; and the hard combinatorics shortly after lunch, since everyone likes it, and no-one can propose a better alternative. Selecting the final two problems, from number theory and algebra produces several combinatorial challenges in its own right. A rather complicated, multi-round election takes place (in which the UK, as a guest nation, does not get a say), and the final two problems are chosen, and the paper is complete.

Interestingly, this matches exactly the ideal paper I’d been hoping for last night, but with the middle questions the other way round. I think the UK students will enjoy it, and I’ll be very pleased for anyone from any country who solves the final problem. It’s fascinating to talk to the leaders of Bosnia and Montenegro, who discuss in detail why their respective education systems mean they are confident their students will struggle much more with Q3.

In the middle of the selection process, there was a rapid transfer to the students’ site in Vore, 15km away, to attend a brief opening ceremony. There is a warm speech from the deputy minister for education, some brief dancing, and the parade of teams. The wholesaler had a bargain on quartered polo shirts, so, unlike the UK flag they are carrying, our team are invariant under both reflection and rotation.

I am summoned to be an expert on the usage of English to prepare the final version of the paper. I feel that the problem authors have done an excellent job, and there is little work to do except suggest some extra sentence breaks and delete some appearances of the word ‘the’. Pity then the other leaders who return to the Harry Fultz school to translate and approve all the versions in their respective languages. It’s midnight as a I write this, and no sign of their return…

Saturday 7th May

This is what we’ve all come for, as the contestants are transported into Tirana for the 4.5 hours of the competition paper. They are allowed to ask questions of clarification during the first half hour. Twenty-five minutes pass, and we are untroubled, so we smugly conclude we must have achieved a wording with total clarity. In fact, the exam is starting slightly late, and a mild deluge begins, mostly concerning the definition of ‘injective’. Both the era of UK students asking joke questions and UK students asking genuine questions have passed, so I am left in peace.

Somehow, Enkel Hysnelaj has single-handedly produced LaTeX markschemes for all four problems overnight, and these are discussed at some length, though it’s to his credit that they didn’t require even longer. The leaders and deputies are then wheeled off on an excursion. Our destination is Kruje, famous as the hometown of Skenderbeg, the Albanian national hero, and just before that is Fushe-Kruje, famous as the place where George W. Bush’s watch was stolen during an official visit. On the way up to the castle and museum we pass through a bazaar where there is the opportunity to buy a carpet, a felt hat, or a mug decorated with a picture of Enver Hoxha. I will be sure to drop some hints to the UK students about ideal choices of gift for Gerry.

The scripts will be arriving a bit later, so there’s the chance for a wander around Tirana in the early evening sun. My planned trip to the Museum of Secret Surveillance is sadly foiled since it hasn’t yet been opened, but there are several more statues of Skenderbeg to enjoy. The question of why he wears a goat head on his helmet remains open. Since dinner is a mere two hours after another meat-centric five course lunch, I turn my attention to the UK scripts which have just arrived. I glance at questions 1 and 4 and the latter is mostly bare while the former is pointedly well-written. The same applies to question 3. All of our nagging about clear written work has very much been rewarded here. As a personal bonus, I can therefore spare time for a late dinner. My attempt at ordering a quick snack results in about a kilo of ribs with the ubiquitous lemons, but will hopefully deflate slightly during coordination in the morning.