RMM 2017 – UK Team Blog

This is the customary and slightly frivolous account of a trip to Bucharest for the ninth edition of the Romanian Master of Mathematics, an annual competition for school students, widely recognised as the hardest of its kind.

I discuss the problems in two previous posts (here and here), and there is also a pdf with fewer pictures, which includes both the discussion and this diary, as well as some more formal comments about the competition itself, the results, and thanks.

Wednesday 22 February

Did you know that trains in Moldova use different width tracks to trains in Romania? Well, I didn’t know either, but I found out at 1am today, as my wagon lit from Chisinau was painstakingly jacked up to allow the transfer from ex-Soviet gauge to Western gauge. Outside, a man in a smart uniform and epaulettes shouted loudly and continuously at a group of men in smart uniforms without epaulattes. When their task was done, four sets of border and custom checks remained before the opportunity for another visit to the samovar, and finally a chance to sleep.

All of which is to say that I have arrived at maths competitions in better mental shape than 6am today at Gara de Nord. The UK students have a more conventional itinerary, but their flight from Luton doesn’t arrive until mid-afternoon. After my first Haifa ‘winter’, I’m craving pork and snow, and find both in the mountain town of Sinaia, an hour away by train in Transylvania. I also find a bear. The bear seems very scared.

I return in time to meet the UK students as well as James and MT. Some of our contestants are now into their fourth year of attending international competitions, and the labour of finding them fresh material resembles Hercules against the hydra, but some problems on combinatorial geometry with convexity seem to have kept everyone entertained on the flight. Dinner is at the Moxa campus of the University of Economics, and features chicken with one of two possible carbohydrates, as in fact do the next six meals. However, today is Thomas’s 18th birthday, and so his parents have arranged a delicious cake, which elicits considerably more enthusiasm. On the short walk back to our meeting, we notice it is possible both to buy fireworks and get a tattoo among other options, so Thomas is spoiled for choice about how to take advantage of his majority.

The team’s activities remain a mystery to James and me though, as we have to join the other leaders for the first meeting, to receive the proposed problems. We spend some time thinking about them separately then together, and our initial impression is that it’s a very suitable paper, that hopefully our team will enjoy.

Thursday 23 February

The leaders meet to finalise the choice and statement of the problems. With a bit more time this morning, I’ve solved Q1, Q2, Q5, and proved Q3 once I’d looked up the correct bound. James eats conics for breakfast and shows me a glorious range of interpretations of Q4. We feel happy that our students will have a chance at all of these, while Q6 may prove more restricting. Either way, it’s clearly an appropriate set for this competition, and is approved quickly. So it’s time to finalise the English version of the paper, or finalize the American version. Many alternatives to the word sieve are proposed. Andrea from Italy is clearly already craving home comforts, but his suggestion of cheese grater is not taken up. This time I’m sorting the LaTeX, so get to settle the commas, but also take the blame for inconsistently spacing the rubric between the two papers. I’m sure everyone noticed.

While all this has been happening, the students have been at a lecture by Sergiu Moroianu at the Institute of Mathematics. Joe Benton gives an account of what they learned in the longer pdf version of this report.

For all the charms of Chipping Norton, I sense MT is enjoying the grittier nature of Bucharest Sector 1, and has been shepherding the students round various sites in between attempts at practice problems. I join them for a brief visit to a geology museum. I am very cynical, but it slightly exceeds my expectations, and is infinitely better than the nearby Museum of the Romanian Peasant, which currently ties with the Hanoi Ethnology Museum as my least favourite olympiad excursion of all time.

The opening ceremony is held in the grand hall of the university, and includes several welcoming and thoughtful speeches from the Mayor of Bucharest and the headteacher of Tudor Vianu, the school which hosts this competition every year. Each team briefly presents themselves on stage. Joe and Neel have accumulated a large collection of UK flags from previous competitions, and should hereby consider themselves publicly shamed for forgetting their promise to bring them. It is over soon, and while the students enjoy a quiet evening and an early night, the leaders have to finalise markschemes for all the problems. The walk back takes us through Victory Square, and past the protesters whose fires and slogans have been on front pages around the world in the past months. It’s an interesting time, and the atmosphere of this city feels very different from my first visit, for the inaugural edition of this competition in 2008.

Friday 24 February

The first day of the contest starts at 9am. The British students seem fairly relaxed, and hopefully are aiming high. Contestants may ask questions of clarification during the first 30 minutes. Rosie does this, and I send my reply to her two queries back via the courier. Five minutes later it is returned to me with the explanation that the student does not understand the answer. Even under competition pressure this seems unlikely, given that my answers are, respectively ‘yes’, and putting a ring around one of three options she has listed. It turns out that actually the student courier did not understand what to do with the answer, and the situation is quickly corrected.

We approve more markschemes. The US deputy leader Po-Shen and I share our views on the challenge of correctly finding the bound in Q3, and our suggestion that this instead be worth 2 points is upheld. Various further discussions fill the morning, and we return just in time to meet the students at the end of the exam. Harvey claims all three problems with a relaxed grin, while Joe claims all three problems with the haunted look of a man whose twelfth espresso of the day has just worn off. Alexander and Thomas say that they spent most of the time making sure their solutions to Q1 were totally watertight, which, given the intricacy of the arguments, was clearly a very sensible strategy.

To provide a distraction, if not actually a break from time-pressured problem-solving, I’ve booked a pair of escape rooms for the UK students later in the afternoon. Bucharest is the home of these games, where the aim is to solve themed puzzles as part of a story in time to escape a locked room. I join one of the rooms, where there are some theatrical reveals involving wrenches, and clues hidden in combination-locked cabinets, where ability to add three-digit numbers proves useful. Someone’s carrying voice means we get to enjoy some of the drama and twists of the other room too. Anyway, this proved an ideal way to avoid useless post-mortems, and I highly recommend Vlad and his pair of rooms.

Later, James and I get to look at the students’ work from this morning. Their assessments are pretty accurate. Harvey’s solutions to everything are beautiful, while Neel’s bounding argument in Q2 is certainly the most vulgar (and, in fact, unnecessary) calculation of the year so far. Joe’s solution to Q3 bears such obvious resemblence to an official solution that his uncharacteristic abundance of small errors probably won’t matter, including the memorable set A_i\backslash\{i\}, where the two is mean different things. Some of the team might reflect that a moment of casualness in checking the n=2 case on Q2 is a frustrating way to lose a potential mark, but when I compare notes with James, it sounds like the slow and steady approach to Q1 has indeed paid off for everyone, so hopefully it will not be too painful to agree the scores tomorrow.

Saturday 25 February

It’s the second day of the competition, and the UK team look bright-eyed and positive at breakfast. They aren’t the only ones under pressure this morning, as James and I must settle the scores from yesterday’s questions with local markers, known as coordinators. It’s hard to guess in how much detail one will have to explain your contestants’ scripts, so it is safer to prepare almost line-by-line. On this occasion though, perhaps we have over-prepared, as every meeting ends quickly with offers of 7/7 exactly where we were hoping, and indeed in a couple of places where we were not hoping. The markschemes are very clear about certain omissions which carry a point deduction, so to ensure fairness and consistency, we insist that two scores are moved down. I’m confident that any British student would prefer an honourable 41/42 than an accidental 42/42.

No-one’s going to be scoring 41 nor 42 unless they solve the extremely challenging geometry Q6, and as we meet our students afterwards, it turns out they have not managed any progress there. However, they claim an almost full set of solutions to Questions 4 and 5, which, if accurate, is a very good return. Everyone is in a good mood, and after I explain a couple of approaches to Q6, no-one seems too disappointed that they didn’t spot these.

There are various schedules floating around, listing multiple locations and times for lunch, but our space-time trajectory intersects none of them, so we follow the Chinese team to a recommended cheap Szechuan restaurant round the corner. Various circle theorems are explored via the Lazy Susan, and there is a grand reveal of the marks we’ve recently confirmed. There’s time for another pair of escape rooms while the second day scripts arrive. As Rosie remarks, two in two days can lead to excessive outside-the-box thinking. Sometimes a radiator really isn’t a sinister prop, a device for encoding five-digit numbers, or a clue to a Templar tunnel; it’s just a radiator. Otherwise we’d be cold.

When the scripts arrive, as expected the cupboard is pretty bare on Q6. If there were marks for quantity, Neel might get some, and if there were marks for most uses of esoteric theory in a single page, Alexander might get one. No set of scripts for an international-level medium combinatorics problem will ever be perfect, but our Q5s come close. It’s therefore not a long evening, and we can join the students for dinner with the American team. For most of them it’s their first visit to Europe, and there’s much comparing of culture and maths training programmes. There’s also a long discussion of whether it’s sensible to teach maths in primary school. Those present who have small children or younger siblings weigh in on the mysteries of the ‘grid method’, and whether toddlers implicitly understand commutativity, even if they can’t spell it.

Sunday 26 February

The UK leaders gather early in the ‘philosophical anti-cafe’ opposite Vianu school, to ponder the final scripts with a coffee and a view of an artfully-arranged folio of Spinoza. James has a loyalty card here. Unfortunately two of our students have clear algebraic errors in Q4, but apart from that everything is very straightforward. Though following last night’s conversation, we note that maybe a revision clinic on mathematical spelling might prove useful. Anonymous student X thinks there’s one L in ‘ellipse’, counterbalanced by anonymous student Y who thinks there are two in ‘column’. The word ‘parallel’ comes in many disguises.

Of course, the coordinators couldn’t care less about that, and they don’t even mind Neel’s two-cases-at-once inductive step, so again we get what we ask for on Q5 immediately, and on Q4 in the time it takes James to draw a lozenge tiling representing Thomas’s shearing argument. For Q6, it turns out there clearly is a mark for most uses of esoteric theory in a single page, so Alexander gets it. They show us a diagram with over a hundred lines which suggests that the exotic equivalence he claims is actually true. There we go. Overall, the quality of our written solutions has been extremely high. It feels like I say this every time now, but it isn’t idle propaganda. We remember the horrors that used to emerge occasionally, and the effort to make this improvement permanent feels well worth it.

Meanwhile, to fill the day, the students have gone to Sinaia. Two of their guides went with them to help with tickets at the station, apparently under the impression that never having taken a train before wouldn’t be an obstacle to this role. Either way, they made it, and following my request for material for this report, I receive a trickle of presentable photos, though there is talk afterwards of some rather more informal versions which are apparently not suitable. The Transylvanian winter is thawing, but slowly and messily, and Harvey reports that several of the group spent more time horizontal than vertical. Irrespective of their preferred axis, there’s no comment on whether they saw my bear, or any other bear. But since my bear was scared of me, one wonders what it would make of MT’s telling-off face? (Last seen by me during the notorious ‘balcony incident’ at a summer school in 2005, but hardly forgotten.)

The students return in time for confirmation of the results and their medals. As so often, there is pleasure that we have done so well collectively, mixed with mild disappointment for those who ended up just short of a boundary, and that the UK was so close to placing first. Because of the strength of the invited countries, earning a medal of any colour is a very worthwhile achievement, and so Rosie is impressively sanguine about missing out so narrowly in such an unfortunate manner. Alexander was closer than it appears, and could have two more opportunities to take part.

The closing ceremony at Vianu school proceeds rapidly. There is the usual challenge of photographing the students receiving their prizes, but this time is easy. Thomas is about a foot taller than everyone else on the stage, while Neel is flanked by almost the entire Russian team, but his chutzpah trumps their numerical advantage, with laughter all round. Joe claims this year’s gold medal is substantially weightier. He hasn’t brought his previous pair, so the chance to verify this and recreate a Mark Spitz moment goes begging.

It’s 7pm, and UK student enthusiasm for the closing disco (not my words) is about as high as MT’s enthusiasm to chaperone the closing disco. Instead we find a Middle Eastern restaurant, and it’s refreshing to eat hummus in a place which doesn’t claim to be the ‘best in Israel’ though I don’t think Abu Said in Akko will be rushing to steal the recipe. Po-Shen outlines his vision of a year-long maths camp. I think present company are tired enough after five days here. Some are interested to view, if not actually participate in, the protests in Victory Square, but it seems tonight is a quiet one and nothing is being burned, so late-night cards and a perusal of each others’ scripts will have to do.

Monday 27th February

The rest of the group have a flight back to London later today which apparently cost 99p per person before tax. I don’t know how much less the 5am option was, but I think it’s probably worth it. My own flight is truly at 5am tomorrow and I plan to stay up all night. The students return to school tomorrow, doubtless to receive a glorious mix of adulation and apathy. Harvey requests whether next year this trip can be timed differently so that he can miss the whole of his local Eisteddfod, rather than just one day. I promise to ask the organisers, say goodbye, then head for the hills on a train journey long enough to write the entirety of this report.

3am at Bucharest airport, and thoughts can now turn to the future. Many of us will meet in five weeks’ for another round of mathematics in the more tranquil setting of Cambridge. Meanwhile, I certainly enjoyed, admittedly through red eyes, the entertainment of a flight to Israel where baggage size regulations are actually enforced at the boarding gate, and apparently everyone else made it back safely too.

RMM 2017 – Problems 2, 3 and 6

In the previous post, I discussed Problems 1, 4 and 5 from this year’s Romanian Master of Mathematics competition. In this post, I discuss the harder problems (modulo my subjective appreciation of difficulty).

Problem 2

Determine all positive integers n satisfying the following condition: for every monic polynomial P of degree at most n with integer coefficients, there exists a positive integer k \leq n, and (k+1) distinct integers x_1,\ldots,x_{k+1} such that

P(x_1) + P(x_2) + \cdots + P(x_k) = P(x_{k+1}).

Parsing this question deserve at least a moment. Straight after a first reading, I find it worth writing down any key quantifiers which I might forget later. Here, it’s the words at most. If you want to show the statement holds for n=2, you need to investigate monic polynomials with degree zero, one and two. You should also make sure that any instances of x_i really are always distinct.

This matters in competitions! Two of our contestants failed to get the mark for showing n=2 works, precisely because of not checking the linear case, and a third could have lost it for using examples which are sometimes not distinct. On hard papers, one mark actually is the difference between triumph and frustration. And of course it matters outside competitions too, since small cases are exactly what your reader might examine first, to check they understand the problem posed, so it’s not a good place for awkward errors.

I started by trying to show that it couldn’t possibly happen that every polynomial with degree at most n had this property, for some combinatorial reason. For example, that if every set of distinct integers could only be a solution set for a small number of polynomials, then we would end up with not enough polynomials. But I couldn’t make this work at all; every bound ended up heavily in the wrong direction.

The next natural question is, does a typical polynomial of degree at most n have this property? But choosing a typical polynomial is hard, so in fact I asked, do the simplest polynomials of degree at most n have this property? I think the simplest polynomials of degree at most n are \{1,x,x^2,\ldots,x^n\}. Under what circumstances does

x_1^m + \ldots x_k^m = x_{k+1}^m, (1)

have solutions in distinct integers? Famously, when k=2 and m\ge 3 this is a very very hard problem indeed. So the first point is that it though it might be useful to use Fermat’s Last Theorem, it would be foolish to pursue a strategy which, if successful, would have a proof of FLT as a sub-problem. At least, it would be foolish if the aim was to finish this strategy within a few hours.

So my main comment on this question is meta-mathematical. If lots of attempts at general arguments don’t work, there must be some special example that does it. And what properties do I want this special example to have? Maybe one might have thought of this from scratch, but my motivation came from (1) in the case m=p-1. Then, by Fermat’s Little Theorem, all the summands are equal to 1 or 0 modulo p. If k>p, then after discounting any uniform factors of p, we obtain a congruence equation which is, in informal terms,

\left(0\text{ or }1\right)+\ldots+\left(0\text{ or }1\right) \equiv \left(0\text{ or }1\right).

This looks really promising because it’s quite restrictive, but it’s still just a bit annoying: there are quite a few solutions. But it does give us the right idea, which is to find a polynomial P for which P(x)\equiv 1 modulo n. The equation 1+\ldots+1\equiv 1 modulo n has solutions only if the number of summands on the LHS is 1 modulo n. So in this context, this reduces to showing that P is, additionally, injective on the integers, ie that P(x)=P(y) only when x=y.

It’s a nice exercise to show the existence of polynomials which are constant modulo n, and a good problem to work out how to force injectivity. If a polynomial is increasing everywhere, then it is certainly injective, and so the problem ends up being slightly easier in the case where the degree is odd than when the degree is even, but this is a nice conclusion to a nice problem, so I’ll save it for any interested readers to finish themselves.

Problem 3

Let n be an integer greater than 1 and let X be an n-element set. A non-empty collection of subsets A_1,\ldots, A_k of X is tight if the union A_1 \cup \dots \cup A_k is a proper subset of X and no element of X lies in exactly one of the A_is. Find the largest cardinality of a collection of proper non-empty subsets of X, no non-empty subcollection of which is tight.

Note. A subset A of X is proper if A\neq X. The sets in a collection are assumed to be distinct. The whole collection is assumed to be a subcollection.

By Neel Nanda:

If |X|=n, there are 2^n possible subsets, so at first glance the answer could be a variety of things, from a linear to an exponential function of n, each of which would suggest a different approach. So the first step is to conjecture an answer, and by examining small cases it seems impossible to do better than 2n-2. There are several natural constructions for this bound, such as n subsets of size (n-1) and (n-2) subsets of size 1, so we guess this to be our answer (which later turn out to be right!).

From here, a solution is deceptively simple, though empirically the five full solutions in the contest show that it was by no means easy to find. We proceed by induction on the size of X, and want to show that any collection of subsets S has size at least (2n-2). By assumption all subcollections are not tight, so if the union of a subcollection is not the whole set X, then there is an element which appears in exactly one subset. This is a useful result, so we’d like to force a subcollection whose union is not the whole set X.

One way to guarantee that the union of a subcollection is not X is by taking the subcollection of all subsets not containing some element b. So there is some element c which appears in only one subset not containing b. If we choose b so that it’s the element contained in the fewest subsets of S, c is in at least as many subsets of S, but in only one subset not containing b. This means that at most one subset containing b doesn’t contain c. This is useful, because after removing at most 2 subsets (the coefficient of n in 2n-2, importantly!), we now have that every subset in S either contains both b and c or neither. This means that we can replace the pair (b,c) with a new element d, to get a new collection of subsets S’ of a set X’, of size n-1, so by induction |S| \le |S'|+2\le 2n-2.

There is also the case where all subsets contain b, but we can create an equivalent collection of subsets of X \ {b} by removing b from all subsets. So again by induction we are done.

Problem 6

Let ABCD be any convex quadrilateral and let P, Q, R, S be points on the segments AB, BC, CD, and DA, respectively. It is given that the segments PR and QS dissect ABCD into four quadrilaterals, each of which has perpendicular diagonals. Show that the points P, Q, R, S are concyclic.

I thought this problem was extremely hard. The official solution starts with a ‘magic lemma’, that isn’t quite so magic if you then read how it’s used. The overall claim is that PQ, RS and AC are concurrent (or parallel), and this is proved using the fact that the radical axis of the two circles with diameters PQ and RS also passes through this point of concurrency. Hunting for key properties of subsets of points in the diagram is an important skill in hard olympiad geometry, since it exactly reflects how problem-setters produce the problems. All the more so when there is lots of symmetry in the construction. But this is a hard example – there are a lot of potentially relevant subsets of the configuration.

When you’re really stuck with how to get involved in a synthetic configuration, you might consider using coordinates. Some of the UK students have been reading some chapters of a book (Euclidean Geometry in Mathematical Olympiads by Evan Chen. I’ve only had my own copy for a couple of days, but my initial impression is very positive – it fills a gap in the literature in a style that’s both comprehensive and readable.) focusing on various analytic approaches, so James and I felt it was safer to make sure we knew what the best settings were, and how far we could take them.

You almost certainly want the intersection of PR and QS to be your origin. I wanted to set up the configuration using the language of vectors, referenced by (P,Q,R,S). This was because PQ\perp BO and so on, hence \mathbf{b}\cdot (\mathbf{q}-\mathbf{p})=0 and so on. An alternative is to use complex numbers, which makes this condition a bit more awkward, but is more promising for the conclusion. Concyclity is not a natural property in vectors unless you can characterise the centre of the circle, but can be treated via cross-ratios in \mathbb{C}. You also have to decide whether to describe the collinearity of A, B and P by expressing \mathbf{p}=\lambda_{\mathbf{p}} \mathbf{a}+(1-\lambda_{\mathbf{p}})\mathbf{b}, or via something more implicit. There definitely are not four degrees of freedom here, since specifying A certainly defines at most one valid set of (B,C,D), so one is mindful we’ll have to eliminate many variables later. We also have to account for fact that \mathbf{r} is a negative scalar multiple of \mathbf{p}, and it’s not clear whether it’s better to break symmetry immediately, or use this towards the end of a calculation.

The point of writing this was that if your initial thought was ‘this looks promising via coordinate methods’, then I guess I agree. But there’s a difference between looking promising, and actually working, and there are lots of parameterisation options. It’s certainly worth thinking very carefully about which to choose, and in this case, challenging though they were, the synthetic or synthetic-trigonometric methods probably were better.

RMM 2017 – Problems 1, 4 and 5

I’ve recently taken a UK team to the 2017 edition of the Romanian Master of Mathematics competition in Bucharest. The British students did extremely well and we all enjoyed ourselves mathematically and generally. The customary diary may appear shortly, but this time I want to focus mainly on the questions, since that is after all the main point of these competitions! I hope that what follows is interesting, and at slightly education to potential future students.

I’ve split this into two posts based on my opinion on difficulty, which is subjective but probably correlates fairly positively with most people’s. The account of Q1 is guest-written by two British students, based on their solutions during the competition.

Problem 1

a) Prove that every positive integer n can be written uniquely in the form

n = \sum_{j=1}^{2k+1} (-1)^{j-1} 2^{m_j},

where k\geq 0 and 0 \leq m_1 < m_2 < \cdots < m_{2k+1} are integers. This number k is called the weight of n.

b) Find (in closed form) the difference between the number of positive integers at most 2^{2017} with even weight and the number of positive integers at most 2^{2017} with odd weight.

Rosie Cates and Neel Nanda:

a) We are trying to express n in terms of powers of 2, so it seems sensible to write in binary. As 2^{m_1} is the smallest power of 2, this term is responsible for the last 1 in the binary representation of n. Let $letx x = n – 2^{m_1}$ (ie n with the last 1 removed from its binary expansion). Now if we pair up terms in the sum to get

x = (2^{m_{2k}+1} - 2^{m_{2k}}) + \ldots + (2^{m_3} - 2^{m_2}),

we can see that each bracket looks like 11…100…0 when written in binary. Also, the condition that m_i < m_{i+1} is equivalent to ensuring that we do not break any strings of consecutive 1s that were in the binary expansion of x (so for example 111110 = 110000 +1110 is not allowed). So writing x in the desired form is the same as writing it as the sum of numbers of the form 11…100\ldots 0 without breaking any strings of 1s. For example

1110100110 = 1110000000 + 100000 + 110.

Clearly there is exactly one way of doing this for every x, so (as each n has exactly one x) there is exactly one way to do it for each n as well.

This approach allows k to be understood differently. Write n in binary and remove the last 1; now count the number of groups of consecutive 1s. This is equal to k.

b) The second half of the problem becomes a lot simpler with the observation that n\leq 2^{m_{2k+1}}, as

n=2^{m_{2k+1}}-(2^{m_{2k}}-2^{m_{2k-1}})-\ldots-(2^{m_2}-2^{m_1}),

and the sequence m_n is increasing, so each bracket is positive. As each sequence of (m_n)s corresponds uniquely to an integer, this means we just want to count sequences of (m_n)s with greatest term at most 2017. The sequence is increasing, so each sequence corresponds to a subset of {0, 1, …, 2017} of size (2k+1). There are \binom{2018}{2k+1} subsets of size (2k+1), so the question reduces to finding a closed form for \sum_{k=0}^{1008} (-1)^k {{2018}\choose{2k+1}}.

This is reminiscent of a classic problem in combinatorics: using the binomial theorem to evaluate sums of binomial coefficients weighted by powers. The best example is

\sum_{k=0}^n (-1)^k \binom{n}{k} =(1-1)^n=0,

but here rather than (-1) we want something whose square is $(-1)$, so we consider the complex number i. Using the same ideas, we get that

\sum_{r=0}^{2018} i^r \binom{2018}{r}=(1+i)^{2018},

which contains what we want, but also binomial coefficients with even r. But if r is even, i^r is real, and if r is odd, i^r is imaginary. So the sum we want appears as the imaginary part, that is

\mathrm{Im}\left((1+i)^{2018}\right)=\mathrm{Im}\left((\sqrt{2} \cdot e^{\frac{i\pi}{4}})^{2018}\right)=2^{1009}.

Dominic: note that in both parts, the respective authors find slightly more than what they were required to. That is, respectively, the interpretation of k, and a bound on m_{2k+1}. The latter is an excellent example of the general notion that sometimes it is better to use a stronger statement than what you actually require in an induction argument (here for existence). The stronger statement (which you guess from playing with examples) makes the inductive step easier, as it’s then clear that the new term you get is distinct from the terms you already have.

Problem 4

In the Cartesian plane, let \mathcal G_1, \mathcal G_2 be the graphs of the quadratic functions f_1(x) = p_1x^2 + q_1x + r_1, f_2(x) = p_2x^2 + q_2x + r_2, where p_1 > 0 > p_2. The graphs \mathcal G_1, \mathcal G_2 cross at distinct points A and B. The four tangents to \mathcal G_1, \mathcal G_2 at~A and B form a convex quadrilateral which has an inscribed circle. Prove that the graphs \mathcal{G}_1 and \mathcal{G}_2 have the same axis of symmetry.

This question is quite unusual for an olympiad of this kind, and I was initially skeptical, but then it grew on me. Ultimately, I was unsurprised that many contestants attacked entirely with coordinate calculations. If you use this strategy, you will definitely get there in the end, but you have to accept that you aren’t allowed to make any mistakes. And because of the amount of symmetry in the configuration, even if you make a mistake, you might still get the required answer, and so not notice that you’ve made a mistake. But I decided I liked it because various levels of geometric insight either reduced or removed the nastier calculations.

Typically, one could gain geometric insight by carefully observing an accurate diagram, but an accurate parabola is hard to draw. However, even from a vague diagram, we might guess the key intermediate property of the configuration, which is that the line joining the other two points in the quadrilateral is parallel to the y-axis. This means that they have the same x-coordinate, and indeed this x-coordinate must in fact be the same for any parabola through A and B, so it is reasonable to guess that it is \frac{x_A+x_B}{2}, the mean of the x-coordinates of A and B.

Since you know this is the goal, it’s not too bad to calculate the equations of the tangent lines directly, and demonstrate this algebraically. But I was determined to use the focus-directrix definition of a parabola. Either recall, or digest the interesting new fact that a parabola may be defined as the locus of points which are the same distance from a fixed point P (the focus), and a fixed line \ell (the directrix). Naturally, the distance to the line is perpendicular distance.

To ensure the form given in the statement where y is a quadratic function of x, in this setting the directrix should be parallel to the x-axis. To define the tangent to the parabola at A, let A’ be the foot of the perpendicular from A onto \ell, so AA’=PA. I claim that the tangent at A is given by the perpendicular bisector of A’P. Certainly this passes through A, and it is easy to convince yourself that it can’t pass through any other point B on the parabola, since BA’> PB, as A’ is on \ell but is not the foot of the perpendicular form B to \ell. This final observation is truly a lot more obvious if you’re looking at a diagram.

We now want to finish geometrically too. In our quadrilateral, one diagonal is parallel to the y-axis, and it will suffice to show that the existence of an incircle implies that A and B must have the same y-coordinate. We have just shown A and B are the same (horizontal) distance from the other diagonal. So certainly if they have the same y-coordinate, then the quadrilateral is a kite, and the sums of opposite sides are equal, which is equivalent to the existence of an incircle. One could then finish by arguing that this ceases to be true if you move one of A and B in either direction, or by some short explicit calculation if such a perturbation argument leaves you ill at ease.

Question 5

Fix an integer n \geq 2. An n x n  sieve is an n x n array with n cells removed so that exactly one cell is removed from every row and every column. A stick is a 1 x k or k x 1 array for any positive integer k. For any sieve A, let m(A) be the minimal number of sticks required to partition A. Find all possible values of m(A), as A varies over all possible n x n sieves.

This is a fairly classic competition problem, and while in my opinion the statement isn’t particularly fascinating, it’s interesting that it admits such a wide range of approaches.

As ever, you need to start by playing around with the setup, and guessing that the answer is 2n-2, and not thinking `it can’t possibly be the same answer as Q3??’ Then think about reasons why you couldn’t do better than 2n-2. My very vague reason was that if you only use horizontal sticks, the answer is clearly 2n-2, and the same if you only use vertical sticks. But it feels like you can only make life harder for yourself if you try to use both directions of sticks in lots of places. Note that some sort of argument involving averaging over stick lengths is definitely doomed to fail unless it takes into account the Latin square nature of the location of holes! For example, if you were allowed to put all the holes in the first row, m(A) would be n-1.

Induction is tempting. That is, you remove some number of sticks, probably those corresponding to a given hole, to reduce the board to an (n-1)x(n-1) configuration. If you do this, you need to be clear about why you can remove what you want to remove (in particular, the number of sticks you want to remove), and whether it’s qualitatively different if the hole in question lies on the border of the board. In all of these settings, you want to be careful about 1×1 sticks, which it’s easy inadvertently to count as both horizontal and vertical. This is unlikely to affect the validity of any argument (just picking either an arbitrary or a canonical direction if it’s 1×1 should be fine) but does make it much harder to check the validity.

Joe exhibited directly a construction of 2n-2 cells which must be covered by different sticks. This approach lives or dies by the quality of the written argument. It must look general, even though any diagram you draw must, almost by definition, correspond to some particular case. Alternatively, since the problem is set on a grid, the cells correspond naturally to edges of a bipartite graph, where classes correspond to rows and columns. The holes form a perfect matching on this bipartite graph. But, as Harvey observed, if you split the rows and columns in two, on either side of the relevant hole (or not in the 2+2 cases where the hole is at the border), you have a (2n-2)+(2n-2) bipartite graph, and a perfect matching here corresponds to a set of cells which must be covered by different sticks. This is an ingenious idea, and if you’ve recently met Hall’s Marriage Theorem, which gives a verifiable criterion for the existence of such a perfect matching, there are few better uses of your next ten minutes than to check whether Hall’s condition a) should hold; b) can be proven to hold in this setting.

Antichains in the grid

In the previous post on this topic, we discussed Dilworth’s theorem on chains and antichains in a general partially ordered set. In particular, whatever the size of the largest antichain in a poset, it is possible to partition the poset into exactly that many chains. So for various specific posets, or the directed acyclic graphs associated to them, we are interested in the size of this largest antichain.

The following example turned out to be more interesting than I’d expected. At a conventional modern maths olympiad, there are typically three questions on each paper, and for reasons lost in the mists of time, each student receives an integer score between 0 and 7 per question. A natural question to ask is “how many students need to sit a paper before it’s guaranteed that one will scores at least as highly as another on every question?” (I’m posing this as a straight combinatorial problem – the correlation between scores on different questions will be non-zero and presumably positive, but that is not relevant here.)

The set of outcomes is clearly \{0,1,\ldots,7\}^3, with the usual weak domination partial order inherited from \mathbb{R}^3. Then an antichain corresponds to a set of triples of scores such that no triple dominates another triple. So the answer to the question posed is: “the size of the largest antichain in this poset, plus one.”

In general, we might ask about \{1,2,\ldots,n\}^d, again with the weak domination ordering. This directed graph, which generalises the hypercube as well as our example, is called the grid.

Heuristics for the largest antichain

Retaining the language of test scores on multiple questions is helpful. In the previous post, we constructed a partition of the poset into antichains, indexed by the elements of some maximal chain, by starting with the sources, then looking at everything descended only from sources, and so on. (Recall that the statement that this is possible was referred to as the dual of Dilworth’s theorem.) In the grid, there’s a lot of symmetry (in particular under the mapping x\mapsto n+1-x in every coordinate), and so you end up with the same family of antichains whether you work upwards from the sources or downwards from the sinks. (Or vice versa depending on how you’ve oriented your diagram…) The layers of antichains also have a natural interpretation – each layer corresponds to a given total score. It’s clear a priori why each of these is an antichain. If A scores the same as B overall, but strictly more on the first question, this must be counterbalanced by a strictly lower score on another question.

So a natural guess for the largest antichain is the largest antichain corresponding to some fixed total score. Which total score should this be? It ought to be the middle layer, that is total score \frac{(n+1)d}{2}, or the two values directly on either side if this isn’t an integer. My intuition was probabilistic. The uniform distribution on the grid is achieved by IID uniform distributions in each coordinate, which you can think of as a random walk, especially if you subtract off the mean first. It feels that any symmetric random walk should have mode zero or next-to-zero. Certainly this works asymptotically in a rescaled sense by CLT, and in a slightly stronger sense by local CLT, but we don’t really want asymptotics here.

When I started writing the previous paragraph, I assumed there would be a simple justification for the claim that the middle layer(s) was largest, whether by straight enumeration, or some combinatorial argument, or even generating functions. Perhaps there is, and I didn’t spot it. Induction on d definitely works though, with a slightly stronger hypothesis that the layer sizes are symmetric around the median, and monotone on either side of the median. The details are simple and not especially interesting, so I won’t go into them.

From now on, the hypothesis is that this middle layer of the grid is the largest antichain. Why shouldn’t it, for example, be some mixture of middle-ish layers? (*) Well, heuristically, any score sequence in one layer removes several possibilities from a directly adjacent layer, and it seems unlikely that this effect is going to cancel out if you take some intermediate number of score sequences in the first layer. Also, the layers get smaller as you go away from the middle, so because of the large amount of symmetry (coordinates are exchangeable etc), it feels reasonable that there should be surjections between layers in the outward direction from the middle. The union of all these surjections gives a decomposition into chains.

This result is in fact true, and its proof by Bollobas and Leader, using shadows and compression can be found in the very readable Sections 0 and 1 of [1].

Most of the key ideas to a compression argument are present in the case n=2, for which some notes by Leader can be found here, starting with Proof 1 of Theorem 3, the approach of which is developed over subsequent sections. We treat the case n=2, but focusing on a particularly slick approach that does not generalise as successfully. We also return to the original case d=3 without using anything especially exotic.

Largest antichain in the hypercube – Sperner’s Theorem

The hypercube \{0,1\}^d is the classical example. There is a natural correspondence between the vertices of the hypercube, and subsets of [d]. The ordering on the hypercube corresponds to the ordering given by containment on \mathcal{P}([d]). Almost by definition, the k-th layer corresponds to subsets of size k, and thus includes \binom{d}{k} subsets. The claim is that the size of the largest antichain is \binom{d}{\lfloor d/2 \rfloor}, corresponding to the middle layer if d is even, and one of the two middle layers if d is odd. This result is true, and is called Sperner’s theorem.

I know a few proofs of this from the Combinatorics course I attended in my final year at Cambridge. As explained, I’m mostly going to ignore the arguments using compression and shadows, even though these generalise better.

As in the previous post, one approach is to exhibit a covering family of exactly this number of disjoint chains. Indeed, this can be done layer by layer, working outwards from the middle layer(s). The tool here is Hall’s Marriage Theorem, and we verify the relevant condition by double-counting. Probably the hardest case is demonstrating the existence of a matching between the middle pair of layers when d is odd.

Take d odd, and let d':= \lfloor d/2\rfloor. Now consider any subset S of the d’-th layer \binom{[d]}{d'}. We now let the upper shadow of S be

\partial^+(S):= \{A\in \binom{[d]}{d'+1}\,:\, \exists B\in S, B\subset A\},

the sets in the (d’+1)-th layer which lie above some set in S. To apply Hall’s Marriage theorem, we have to show that |\partial^+(S)|\ge |S| for all choice of S.

We double-count the number of edges in the hypercube from S to \partial^+(S). Firstly, for every element B\in S, there are exactly d’ relevant edges. Secondly, for every element A\in\partial^+(S), there are exactly d’ edges to some element of \binom{[d]}{d'}, and so in particular there are at most d’ edges to elements of S. Thus

d' |S|=|\text{edges }S\leftrightarrow\partial^+(S)| \le d' |\partial^+(S)|,

which is exactly what we require for Hall’s MT. The argument for the matching between other layers is the same, with a bit more notation, but also more flexibility, since it isn’t a perfect matching.

The second proof looks at maximal chains. Recall, in this context, a maximal chain is a sequence \mathcal{C}=B_0\subset B_1\subset\ldots\subset B_d where each B_k:= \binom{[d]}{k}. We now consider some largest-possible antichain \mathcal{A}, and count how many maximal chains include an element A\in\mathcal{A}. If |A|=k, it’s easy to convince yourself that there are \binom{d}{r} such maximal chains. However, given A\ne A'\in\mathcal{A}, the set of maximal chains containing A and the set of maximal chains containing A’ are disjoint, since \mathcal{A} is an antichain. From this, we obtain

\sum_{A\in\mathcal{A}} \binom{d}{|A|} \le d!. (**)

Normally after a change of notation, so that we are counting the size of the intersection of the antichain with each layer, this is called the LYM inequality after Lubell, Yamamoto and Meshalkin. The heuristic is that the sum of the proportions of layers taken up by the antichain is at most one. This is essentially the same as earlier at (*). This argument can also be phrased probabilistically, by choosing a *random* maximal chain, and considering the probability that it intersects the proposed largest antichain, which is, naturally, at most one. Of course, the content is the same as this deterministic combinatorial argument.

Either way, from (**), the statement of Sperner’s theorem follows rapidly, since we know that \binom{d}{|A|}\le \binom{d}{\lfloor d/2\rfloor} for all A.

Largest antichain in the general grid

Instead of attempting a proof or even a digest of the argument in the general case, I’ll give a brief outline of why the previous arguments don’t transfer immediately. It’s pretty much the same reason for both approaches. In the hypercube, there is a lot of symmetry within each layer. Indeed, almost by definition, any vertex in the k-th layer can be obtained from any other vertex in the k-th layer just by permuting the labels (or permuting the coordinates if thinking as a vector).

The hypercube ‘looks the same’ from every vertex, but that is not true of the grid. Consider for clarity the n=8, d=3 case we discussed right at the beginning, and compare the scores (7,0,0) and (2,2,3). The number of maximal chains through (7,0,0) is \binom{14}{7}, while the number of maximal chains through (2,2,3) is \binom{7}{2, 2,3}\binom{14}{4,5,5}, and the latter is a lot larger, which means any attempt to use the second argument is going to be tricky, or at least require an extra layer of detail. Indeed, exactly the same problem arises when we try and use Hall’s condition to construct the optimal chain covering directly. In the double-counting section, it’s a lot more complicated than just multiplying by d’, as was the case in the middle of the hypercube.

Largest antichain in the d=3 grid

We can, however, do the d=3 case. As we will see, the main reason we can do the d=3 case is that the d=2 case is very tractable, and we have lots of choices for the chain coverings, and can choose one which is well-suited to the move to d=3. Indeed, when I set this problem to some students, an explicit listing of a maximal chain covering was the approach some of them went for, and the construction wasn’t too horrible to state.

[Another factor is that it computationally feasible to calculate the size of the middle layer, which is much more annoying in d>3.]

[I’m redefining the grid here as \{0,1,\ldots,n-1\}^d rather than \{1,2,\ldots,n\}^d.]

The case distinction between n even and n odd is going to make both the calculation and the argument annoying, so I’m only going to treat the even case, since n=8 was the original problem posed. I should be honest and confess that I haven’t checked the n odd case, but I assume it’s similar.

So when n is even, there are two middle layers namely \frac{3n}{2}-2, \frac{3n}{2}-1 (corresponding to total score 10 and total score eleven in the original problem). I calculated the number of element in the \frac{3n}{2}-1 layer by splitting based on the value of the first coordinate. I found it helpful to decompose the resulting sum as

\sum_{k=0}^{n-1} = \sum_{k=0}^{\frac{n}{2}-1} + \sum_{k=\frac{n}{2}}^{n-1},

based on whether there is an upper bound, or a lower bound on the value taken by the second coordinate. This is not very interesting, and I obtained the answer \frac{3n^2}{4}, and of course this is an integer, since n is even.

Now to show that any antichain has size at most \frac{3n^2}{4}. Here we use our good control on the chain coverings in the case d=2. We note that there is a chain covering of the (n,d=2) grid where the chains have 2n-1, 2n-3,…, 3, 1 elements (%). We get this by starting with a maximal chain, then taking a maximal chain on what remains etc. It’s pretty much the first thing you’re likely to try.

Consider an antichain with size A in the (n,d=3) grid, and project into the second and third coordinates. The image sets are distinct, because otherwise a non-trivial pre-image would be a chain. So we have A sets in the (n,d=2) grid. How many can be in each chain in the decomposition (%). Well, if there are more than n in any chain in (%), then two must have been mapped from elements of the (n,d=3) grid with the same first coordinate, and so satisfy a containment relation. So in fact there are at most n image points in any of the chains of (%). So we now have a bound of n^2. But of course, some of the chains in (%) have length less than n, so we are throwing away information. Indeed, the number of images points in a given chain is at most

\max(n,\text{length of chain}),

and so the number of image points in total is bounded by

n+\ldots+n+ (n-1)+(n-3)+\ldots+1,

where there are n/2 copies of n in the first half of the sum. Evaluating this sum gives \frac{3n^2}{4}, exactly as we wanted.

References

[1] – Bollobas, Leader (1991) – Compressions and Isoperimetric Inequalities. Available open-access here.

The Envelope ‘Paradox’

At the recent IMO in Hong Kong, there were several moments where the deputy leaders had to hang around, and I spent some of these moments discussing the following problem with Stephen Mackereth, my counterpart from New Zealand. He’s a mathematically-trained philosopher, so has a similar level of skepticism to me, but for different reasons, regarding supposed paradoxes in probability. Because, as we will see shortly, I don’t think this is a paradox in even the slightest fashion, I think there’s probably too much written about this on the internet already. So I’m aware that contributing further to this oeuvre is hypocritical, but we did the thinking in HKUST’s apparently famous Einstein Cafe, so it makes sense to write down the thoughts.

[And then forget about it for eight weeks. Oops.]

The ‘Paradox’

Here’s the situation. A cryptic friend gives you an envelope containing some sum of money, and shows you a second envelope. They then inform you that one of the envelopes contains twice as much money as the other. It’s implicit in this that the choice of which is which is uniform. You have the option to switch envelopes. Should you?

The supposed paradox arises by considering the amount in your envelope, say X. In the absence of further information, it is equally likely that the other envelope contains X/2 as 2X. Therefore, the average value of the other envelope is

\frac12 \left(\frac{X}{2}+2X \right)= \frac54 X > X.

So you should switch, since on average you gain money. But this is paradoxical, since the assignment of larger and smaller sums was uniform, so switching envelope should make no difference.

Probabilistic setup

This is not supposed to be a problem on a first-year probability exercise sheet. It’s supposed to be conducive to light discussion. So saying “I won’t engage with this problem until you tell me what the probability space is” doesn’t go down terribly well. But it is important to work out what is random, and what isn’t.

There are two sources of randomness, or at least ignorance. Firstly, there is the pair of values contained in the envelopes. Secondly, there is the assignment of this pair of values to the two envelopes. The second is a source of randomness, and this problem is founded on the premise that this second stage is ‘symmetric enough’ to smooth over any complications in the first stage. If we think that probability isn’t broken (and that’s what I think), then the answer is probably that the second stage isn’t symmetric enough.

Or, that the first stage isn’t very well-defined. In what follows, I’m going to make the second stage very symmetric, at the expense of setting up the first stage in what seems to me a reasonable way using the conventional language of probability theory to record our ignorance about the values in play.

So what’s the first stage? We must have a set of possible pairs of values taken by the envelopes. Let’s call this A, so

A\subset \mathbb{A}:=\{(x,2x)\,:\, x\in (0,\infty)\}.

Maybe we know what A is, but maybe we don’t, in which we should take A=\mathbb{A}, on the grounds that any pair is possible. Suppose that your friend has chosen the pair of values according to some distribution on \mathbb{A}, which we’ll assume has a density f, which is known by you. Maybe this isn’t the actual density, but it serves perfectly well if you treat it as *your* opinion on the likelihood. Then this actually does reduce to a problem along the lines of first-year probability, whether or not you get to see the amount in your envelope.

Suppose first that you do get to see the amount, and that it is x. Then the conditional probabilities that the pair is (x/2,x) or (x,2x) are, respectively

\frac{f(x/2,x)}{f(x/2,x)+f(x,2x)},\quad \frac{f(x,2x)}{f(x/2,x)+f(x,2x)}.

So you can work out your expected gain by switching, and decide accordingly. If you don’t know the value in your envelope, you can still work out the probability that it is better (in expectation) to switch, but this isn’t really a hugely meaningful measure, unless it is zero or one.

It’s worth noting that if you can view inside your envelope, and you know A has a particular form, then the game becomes determined. For example, if

A\subset \{(n,2n), n\text{ an odd integer}\},

then life is very easy. If you open your envelope and see an odd integer, you should switch, and if you see an even integer you shouldn’t.

We’ll return at the end to discuss a case where it is always better to switch, and why this isn’t actually a paradox.

Improper prior and paradox of resampling when \mathbb{E}=\infty

For now though, let’s assume that we don’t know anything about the amounts of money in the envelopes. Earlier, we said that “in the absence of further information, it is equally likely that the other envelope contains X/2 as 2X”. In the language of a distribution on \mathbb{A}, we are taking the uniform measure. Of course this not a distribution, in the same way that there isn’t a uniform distribution on the positive reals.

However, if this is your belief about the values in the pair of envelopes, what do you think is the mean value of the content of your envelope? Well, you think all values are equally likely. So, even though this isn’t a distribution, you pretty much think the value of your envelope has infinite expectation.

[This is where the philosophy comes in I guess. Is (expressing uniform ignorance about the content of the other envelope given knowledge of your own) the same as (expressing uniform ignorance of both envelopes at the beginning)? I think it is, even though it has a different consequence here, since the former can be turned into a proper distribution, whereas the latter cannot.]

Let’s briefly consider an alternative example. It’s fairly easy to conjure up distributions which are almost surely finite but which have infinite expectation. For example \mathbb{P}(X=2^k)=2^{-k} for k=1,2,…, which is the content of the *St. Petersburg paradox*, another supposed paradox in probability, but one whose resolution is a bit more clear.

Anyway, let X and Y be independent copies of such a distribution. Now suppose your friend offers you an envelope containing amount X. You look at the value, and then you are offered the opportunity to switch to an envelope containing amount Y. Should you?

Well, if expectation is what you care about, then you definitely should. Because with probability one, you are staring at a finite value in your envelope, whereas the other unknown envelope promises infinite expectation, which is certainly larger than the value that you’re looking at.

Is this also a paradox? I definitely don’t think it is. The expectation of the content of your envelope is infinite, the expected gain is infinite with probability one, which is consistent with the expected content of the other envelope being infinite. [Note that you don’t want to be claiming that the expectation of X-Y is zero.]

An example density function

As an exercise that isn’t necessarily hugely interesting, let’s assume that f, the distribution of the smaller of the pair, is \mathrm{Exp}(\lambda). So the mean of this smaller number is 1/\lambda. Then, conditional on seeing x in my envelope, the expected value of the number in the other envelope is

\frac{\frac{x}{2} e^{-\lambda x/2} + 2x e^{-\lambda x}}{e^{-\lambda x/2}+ e^{-\lambda x}}. (*)

Some straightforward manipulation shows that this quantity is at least x (implying it’s advantageous to switch) precisely when

e^{-\lambda x/2}\ge \frac12.

That is, when x\le \frac{2\log 2}{\lambda}. The shape of this interval should fit our intuition, namely that the optimal strategy should be to switch if the value in your envelope is small enough.

The point of doing this calculation is to emphasise that it ceases to be an interesting problem, and certainly ceases to be a paradox of any kind, once we specify f concretely. It doesn’t matter whether this is some true distribution (ie the friend is genuinely sampling the values somehow at random), or rather a perceived likelihood (that happens to be normalisable).

What if you should always switch?

The statement of the paradox only really has any bite if the suggestion is that we should always switch. Earlier, we discussed potential objections to considering the uniform prior in this setting, but what about other possible distributions f which might lead to this conclusion?

As at (*), we can conclude that when f(x)+f(x/2)>0, we should switch on seeing x precisely if

f(x)\ge 2f\left(\frac{x}{2}\right).

Therefore, partitioning the support of f into a collection of geometric sequences with exponent 2, it is clear that the mean of f is infinite if everything is integer-valued. If f is real-valued, there are some complications, but so long as everything is measurable, the same conclusion will hold.

So the you-should-switch-given-x strategy can only hold for all values of x if f has infinite mean. This pretty much wraps up my feelings. If the mean isn’t infinite, the statement of the paradox no longer holds, and if it is infinite, then the paradox dissolves into a statement about trying to order various expectations, all of which are infinite.

Conclusions

Mathematical summary: it’s Bayes. Things may be exchangeable initially, but not once you condition on the value of one of them! Well, not unless you have a very specific prior.

Philosophical summary: everything in my argument depends on the premise that one can always describe the protagonist’s prior opinion on the contents of the pair of envelopes with a (possibly degenerate) distribution. I feel this is reasonable. As soon as you write down \frac12 \cdot\frac{x}{2} + \frac12 \cdot2x, you are doing a conditional expectation, and it’s got to be conditional with respect to something. Here it’s the uniform prior, or at least the uniform prior restricted to the set of values that are now possible given the revelation of your number.

Second mathematical summary: once you are working with the uniform prior, or any measure with infinite mean, there’s no reason why

\mathbb{E}\left[X|Y\right]>Y,

with probability one (in terms of Y) should be surprising, since the LHS is (almost-surely) infinite while the RHS is almost surely finite, despite having infinite mean itself.

IMO 2016 Diary – Part Four

A pdf of this report is also available here.

Thursday 14th July

I have now spent a while thinking about square-free n in Q3 after rescaling, and I still don’t know what the markscheme should award it. I therefore request that Joe and Warren receive the same score as each other, and any other contestant who has treated this case. In my opinion this score should be at most one, mainly as a consolation, but potentially zero. However, we are offered two, and after they assure me this is consistent, I accept.

There is brief but high drama (by the standards of maths competitions) when we meet Angelo the Australian leader, who confirms that he has just accepted one mark for almost the same thing by his student Johnny. A Polish contestant in a similar situation remains pending, so we all return for a further meeting. I’m unconvinced that many of the coordinators have read all the scripts in question, but they settle on two for everyone, which is consistent if generous. The only drama on Q5 is the ferocious storm that sets in while I’m making final notes in the plaza. Again though, coordinator Gabriele has exactly the same opinion on our work as Geoff and I, apart from offering an additional mark for Lawrence’s now slightly damp partial solution.

And so we are finished well before lunch, with a total UK score of 165 looking very promising indeed. I’m particularly pleased with the attention to detail – Jacob’s 6 on Q4 is the only mark ‘dropped’, which is brilliant, especially since it hasn’t come at the expense of the students’ usual styles. We’ll have to wait until later to see just how well we have done.

It would be nice to meet the students to congratulate them in person, but they are with Jill on the somewhat inaccessible Victoria Peak, so instead I take a brief hike along the trail down the centre of HK Island, ending up at the zoo. This turned out to be free and excellent, though I couldn’t find the promised jaguar. There was, however, a fantastic aviary, especially the striking flock of scarlet ibis. A noisy group of schoolchildren are surrounding the primates, and one lemur with an evil glint in his eye swings over and languidly starts an activity which elicits a yelp from the rather harried teacher, who now has some considerable explaining to do.

With 1000 people all returning to UST at roughly 6.30, dinner is not dissimilar to feeding time at the zoo, and afterwards various leaders lock horns during the final jury meeting. Two countries have brought an unresolved coordination dispute to the final meeting, and for the first time since I became deputy leader, one of them is successful. Congratulations to the Koreans, who now have a third student with a highly impressive perfect score. Andy Loo and Geoff chair the meeting stylishly and tightly, and although there are many technical things to discuss, it doesn’t drag for too long. Eventually it’s time to decide the medal boundaries, and the snazzy electronic voting system makes this work very smoothly. I feel the gold and bronze cutoffs at 29 and 16 are objectively correct, and the 50-50 flexibility at silver swings towards generosity at 22. We can now confirm the UK scores as:

UKscoresThis is pretty much the best UK result in the modern era, placing 7th and with a medal tally tying with the famous food-poisoning-and-impossible-geometry IMO 1996 in India. But obviously this is a human story rather than just a 6×6 matrix with some summary statistics, and Harvey in particular is probably looking at the world and thinking it isn’t fair, while Warren’s gold is the ideal end to his four years at the IMO, two of which have ended one mark short. The American team are pretty keen to let everyone know that they’ve placed first for the second year in succession, and their remarkable six golds will hopefully allow scope for some good headlines. There is much to talk about, celebrate and commiserate, and this continues late into the night.

Friday 15th July

Our morning copy of the IMO Newsletter includes an interview with Joe, with the headline ‘Meh’. Frank Morgan has rather more to say, which is good news, since he’s delivering the IMO lecture on Pentagonal Tilings. He discusses the motivation of regular tilings where the ratio Perimeter/Area is minimised, starting from questions about honeycombs raised by the Roman author Varro! We move onto more mathematical avenues, including the interesting result of L’Huilier that given a valid set of angles, the associated polygon with minimal Perimeter/Area has an incircle, and the corresponding result for in-n-spheres in higher dimension. A brief diversion to the beach on the way home is punctuated with attempts to project the hyperbolic plane onto the sand.

The day’s main event is the closing ceremony, held at the striking Hong Kong Convention Centre. As usual, the adults and our students have been vigorously separated for the journey. As I arrive, it seems the UK boys have been directing a massed gathering behind the EU flag on stage, while the non-European teams are divided into two sides in a giant paper aeroplane dogfight. All attempts by the organisers to quash this jocularity are being ignored, and after bringing everyone here two hours early, I have minimal sympathy. Geoff sits on a secluded bench, and agrees to the many selfie requests from various teams with regal if resigned tolerance.

The ceremony is started by a fantastically charismatic school brass band, and proceeds with some brief speeches, and more astonishing drumming. Then it’s time to award the medals. Lawrence and Jacob get to go up together among the clump of 24-scorers, while Kevin from Australia does an excellent job of untangling his flag and medal while keeping hold of the ubiquitous cuddly koala. Neel has been threatened with death if he appears on stage again with an untucked shirt, but no direction is required for his and Warren’s smiles as they receive the gold medallists’ applause.

P1010513 (3)Afterwards, there is a closing banquet. We get to join British coordinators James and Joseph for a climate-defying carrot soup, followed by a rare diversion onto Western carbohydrates accompanying what is, for many of us, a first taste of caviar. Both Geoff and the American team are forced to make speeches at no notice. It is all generally rather formal, and fewer photographs are taken than usual. An attempt to capture Joe and Harvey looking miserable results in one the biggest grins of the evening. The UK and Australian teams have a thousand stickers and micro-koalas to give out as gifts, and some of the attempts at this descend into silliness. All clothing and body parts are fair game, and Jacob makes sure that Geoff is fully included. The UK and Australian leaders, variously coated, retreat from the carnage to the relative safety of our top-floor balcony as the IMO drifts to an end, until midnight, when it seems sensible to find out what the students are up to.

Saturday 16th July

This is what the students are up to. When we arrived at UST last week, everyone was given food vouchers to redeem at the campus’s various restaurants. Very very many of these are left over, and, despite the haute cuisine on offer earlier, people are hungry. They have therefore bought McDonalds. And I mean this literally. Animated by Jacob and American Michael, they have bought the entire stock of the nearest branch. If you want to know what 240 chicken nuggets looks like, come to common room IX.1, because now is your chance. Fortunately our team have made many friends and so after the Herculean task (I make no comment on which Herculean labour I feel this most resembles) of getting it to their common room, pretty much the entire IMO descends to help. Someone sets up a stopmotion of the slow erosion of the mountain of fries, while the usual card games start, and a group around a whiteboard tries to come up with the least natural valid construction for n=9 on Q2. Around 3.30am everything is gone, even the 30 Hello Kitties that came with the Happy Meals, and we’re pre-emptively well on the way to beating jetlag.

I wake up in time to wave Geoff off, but he’s been bumped to an earlier bus, so the only thing I see is Lawrence and colleagues returning from a suicidal 1500m round the seaside athletics track. Our own departure is mid-morning, and on the coach the contestants are discussing some problems they’ve composed during the trip. They’ll soon be able to submit these, and by the sounds of it, anyone taking BMO and beyond in 2018 has plenty to look forward to. Jacob has already mislaid his room key and phone, and at the airport he’s completed the hat-trick by losing one of the two essential passport insert pages. Fortunately, it turns out that he’s lost the less essential one, so we can clear security and turn thoughts towards lunch.

Jill has given me free licence to choose our dim sum, so the trip ends with pork knuckle and chicken feet. Our aim is to stay awake for the whole flight, and Neel helps by offering round copies of a Romanian contest from 2010, while I start proof-reading. By the time they finish their paper, many rogue commas have been mercilessly expunged. It should be daylight outside, but the windows are all shut, and by the ninth hour time starts to hang drowsily in a way that combinatorial geometry cannot fix, and so the mutual-waking-up pact kicks in, aided by Cathay Pacific’s unlimited Toblerone. Winding through Heathrow immigration, Joe unveils his latest airport trick of sleeping against vertical surfaces. We diverge into the non-humid night.

Reflection

IMG_0468 original (2)There’s a great deal more to life and mathematics than problem-solving competitions, but our contestants and many other people have worked hard to prepare for IMO 2016 over the past months (and years). So I hope I’m allowed to say that I’m really pleased for and proud of our UK team for doing so well! The last three days of an IMO are very busy and I haven’t had as much time as I’d have liked to talk in detail about the problems. But I personally really liked them, and thought the team showed great taste in choosing this as the British annus mirabilis in which to produce lots of beautiful solutions.

But overall, this is really just the icing on the cake of a training progamme that’s introduced lots of smart young people to each other, and to the pleasures of problem-solving, as well as plenty of interesting general mathematics. I have my own questions to address, and (unless I’m dramatically missing something) these can’t be completed in 4.5 hours, but as ever I’ve found the atmosphere of problem discussion totally infectious, so I hope we are doing something right.

Lawrence and Warren are now off to university. I’m sure they’ll thrive in every way at this next stage, and hopefully might enjoy the chance to contribute their energy and expertise to future generations of olympiad students. The other four remain eligible for IMO 2017 in Brazil, and while they will doubtless have high personal ambitions, I’m sure they’ll also relish the position as ideal role models for their younger colleagues over the year ahead. My own life will be rather different for the next two years, but our camp for new students is held in my no-longer-home-town Oxford in a few weeks’ time, and I’m certainly feeling excited about finding some new problems and doing as much as possible of the cycle all over again!

IMO 2016 Diary – Part Three

Sunday 10th July

I’m awake at 6am and there’s nothing to do, so take a short run along the edge of the bay. I meet an old lady singing along to a walkman (yes, really) while doing taichi. She encourages me to join and it seems rude to refuse. Suffice it to say I’m as grateful no video evidence exists as she should be that no audio recording was made. Six-hundred mathematicians queueing for powdered eggs seems like an unwelcome start to the day, so we are self-catering. The guides have been commanded to show every student how to find their place in the exam hall, and I approve of Allison’s contempt for the triviality of this task.

The main event of the day is the opening ceremony, held at the Queen Elizabeth stadium in the centre of Hong Kong Island. To no-one’s surprise, this involves a lot of time waiting around in the stifling UST plaza, which the students use to take a large number of photographs. The UK and Australian boys are smartly turned out as usual, but the polyester blazers are rather ill-suited to this tropical conditions, so we invoke Red Sea rig until air conditioning becomes available. The Iceland team are particularly keen to seek out the English members for reasons connected to a football match of which Neel proudly claims total ignorance. I picked up an EU flag for next-to-nothing last Friday, and now Jacob and Warren prove very popular as they circulate inviting our (for now) European colleagues to join us behind the stars.

The deputies are segregated in an upper tier and obliged to watch a rehearsal of the parade. Some of the organisers have a confused interpretation of the IMO roles. I still have some of the uniform with me, but an official says it is literally impossible for me to give it to the team. She is small and Joe Benton can catch flying ties as well as colds, so it turns out to be literally entirely possible, but for my trouble I get called ‘a very bad boy’.

Many hours after we left our rooms, the ceremony starts, and is actually very good, with a handful of well-chosen speeches, a mercifully quickfire parade of teams, and musical interludes from a full symphony orchestra, with various traditional and non-traditional percussion. The new IMO song Every day in love we are one involves a B section accompanied by a melange of watercooler bottles, but despite its catchy conclusion about maths, friendship and beyond, I suspect it may not trouble the top of the charts.

Monday 11th July

It’s the morning of the first IMO paper, and you can feel both the excitement and the humidity in the air. Some of our boys are looking a bit under the weather, but we know from past experience that the adrenaline from settling down in a room of 600 young contestants who’ve been preparing for exactly this can carry them through anything. I skip an excursion in order to receive a copy of the contest paper. Security is tight, and the deputies who have chosen this option are locked in a lecture theatre for two hours, and our bathroom visits monitored with commendable attention to detail. I guess that the combinatorial second problem is most likely to provoke immediate discussion, so I spend my time working through the details of the argument, just in time to meet our contestants when their 4.5 hours are up.

Q3 has been found hard by everyone, and Q2 has been found hard by other countries. Harvey’s kicking himself for drawing the wrong diagram for the geometry, an error that is unlikely to improve Geoff’s mood when he receives the scripts later today. Apart from that, we have a solid clutch of five solutions to each of the first two problems, and various nuggets of progress on the final problem, which is an excellent start. Several of the team are itching to keep trying to finish Q3, but the campus is likely to be annoying hotbed of spurious gossip all day, so Allison and I take them out. The very convenient MTR takes us under the harbour while the students and I debate the usefulness of the square-free case, and how well it is preserved under rescaling so that the circumcentre is a lattice point.

As we emerge above ground, Jacob is entranced by the live-action Finding Dory playground at Causeway Bay, and we toy with buying a pig’s trotter from a nearby market, but not even Lawrence is feeling adventurous enough with another exam tomorrow. We travel over to Kowloon via double-decker tram and ferry, and fortified by ice cream, take lots of photographs of the unique HK skyline, where even the giant waterfront office towers are dwarfed by Victoria Peak, which the contestants will visit while I’m marking. On our return journey, some of the team are impressed by the HK rush hour, indicating that they’ve clearly never tried to change line at Leicester Square around 6pm on a Friday…

Tuesday 12th July

Another morning, another trek uphill to a 4.5 hour exam. Time passes rapidly, especially now I’ve worked out how to order coffee without the ubiquitous condensed milk. The security arrangements concerning the deputies’ copies of the paper have been increased even further, but the IMO photographers have outdone themselves, and published on Instagram some pictures of the exam room with a level of crispness such that it’s clear the paper includes no geometry, and after finally getting hold of a proper hard copy, it looks like a paper which the UK team should really enjoy.

As so often after IMO papers, there is a range of reactions. Lawrence is unsure whether he presented his exemplar polynomial in a form that actually works. Joe knows and I know that he could easily have got at least 35 on these papers, but after over-meta-thinking himself on Q5, this isn’t his year. Like Aeneas gazing on the ruins of Troy, sunt lacrimae rerum, but also plans for new foundations. By contrast, Harvey has atoned for yesterday’s geometric lapse with what sounds like a perfect score today. Warren and Neel seem to be flying overall, and are doing a good job of keeping their excitement under control while the others muse. There’s plenty to think about, and Geoff has now arrived bearing yesterday’s scripts and several novels’ worth of anecdotes from the leaders’ site.

Before getting down to business, it feels sensible to walk off the Weltschmerz, and provide an outlet for joy in the nearby Clearwater Bay country park. There’s a long trail all over the New Territories, and we join it for a brief but purposeful stroll up through the light jungle and along the ridge. We’re confident we didn’t find the global maximum, but we find a couple of local maxima with great views out around the coastline, which seems to have Hausdorff dimension slightly greater than 1. We see some enormous spiders (though the Australians are substantially less impressed) before ending up an uncontroversial minimum where Jill has bedded in with merciful bottles of water on the beach. To say we are sticky doesn’t even begin to cover it but, crucially, we are no longer consumed by the morning’s events.

The UK boys are now masters of the complicated UST food court ordering process, and Warren endears himself to Geoff by producing a steaming bowl of spicy ramen as if by magic. The contestants have a ‘cultural night’, which apparently includes a greater number of hedge fund representatives than one might have expected. For me, it’s a night in with Geoff, green tea and the scripts for Q2. Joe and Neel have filled fourteen pages between them checking a construction in glorious detail, a step which Harvey has described in its entirety with the words ‘glue them together’. Overall, they are complicated but precise, and I have few concerns, so it’s only necessary to burn the candle at one end.

Wednesday 13th July

It’s time for coordination, where Geoff and I agree the UK marks with a team of local and international experts. The scheduling has assigned us the Q1 geometry early in the morning, which is a clear case of five perfect solutions, so we move to Q2. Coordinator Stephan seems very well-prepared for the UK scripts, so again we are finished in a matter of minutes. This allows us to bring forward our discussion of Q4. Jacob has made several small errors, all of which could be fixed by attacking his script with a pair of scissors and some glue. I believe the mark scheme should award this 4+2, and coordinator Juan thinks it should be 5+1. We are both open to each other’s interpretations, and have at least basic proficiency in addition, so again there is little need for debate.

The early evening brings the main challenge of the day, Q6, at which the UK has excelled. Our frogmaster Geoff has listed marks for five of our attempts, but the final script belonging to Joe has generated only the comment ‘magical mystery tour’. His solution to part a) diverges substantially from the most natural argument, and indeed involves wandering round the configuration, iteratively redirecting lines [1]. I am eventually convinced by the skeleton of the argument, though unconvinced I could complete the details in the finite time available.

We discuss the script with Lisa Sauermann, who explains some of the main challenges [2]. After a short pause for thought, we’re convinced by Lisa’s suggestion of equivalence with a point on the conventional markscheme. It would have been nice to have had more time to think about the subtleties myself, but this was some really interesting maths and we pack up for the day feeling very impressed with the quality of coordination here so far.

We and the coordinators are also very impressed with the quality of Harvey’s art. As a result, we now have an answer to the question ‘What should you do if you finish the IMO two hours early?’ Harvey’s answer at least is to draw a diagram of the Q6 configuration in the case n=3, where at each of the intersection points with the outer boundary stands a member of the current UK team. Precisely UNKs 1, 3 and 5 are wearing a frog. The real life sextet have been taken by Allison to Disneyland today, so some are potentially now wearing a princess. But while the contestants can let it go now, it’s off to work I go, as there’s still two sets of scripts left to ponder.

Harvey Q6[1] The mechanism for this redirection is neither canonical nor explained, and even in the best setup I can come up with in an hour or so of trying a huge class of diagrams, exactly half of the indices in the resulting calculation are off by \pm 1. The pressure of IMO Day Two can indeed derail even the most well-prepared contestants.

[2] There is a non-trivial difficulty when the area enclosed by our path is concave, as then some intersection points on the path arise from lines which are also part of the path. Handling the parity of such points looks easy once you’ve been shown it, but is definitely not obvious.

IMO 2016 Diary – Part Two

Wednesday 6th July

After starting the third exam, Mike, Jo and I go for a walk through some of the smaller villages on the other side of the ridge. Along the way, we pick up a bunch of local rascals who ask us, via their English-speaking henchman, many questions about basketball, and the colour of Mike’s shoes. Jo asks him why they aren’t in school, but this remains shrouded in mystery. Partly as a means of escape, we take a detour through a grove of the famous Tagaytay pineapples, which are indeed a striking crimson just before they ripen fully. I’m nervous about beard tanlines so am looking for a barber, but it seems I’m one of only two people in the Philippines with facial hair. (The other is Neel, who is adamant that his school approves of the ‘Wild man of Borneo’ look.)

We return to find that the students have been issued with cake. Its icing is impossible to manage without a fork. It is also entirely purple, and Lawrence describes it as ‘tasting of air’. None of this has distracted the UK students, who all solve the first two problems perfectly, which bodes well for the IMO itself, now less than a week away. To fill some time and provide a brief variation from the constant problem-solving, I give a talk about correlation and graphs, based on a subsubsubsection of my thesis and for now, fortunately no-one finds any logical holes.

Thursday 7th July

To add variety, today the two teams have set each other a paper, which they will spend the afternoon marking. It transpired late last night that the hotel has no means of printing or photocopying documents, and we haven’t brought copies of tomorrow’s final exam. So today’s paper has been painstakingly written on whiteboards, and some of the adults set off round Tagaytay in search of a working printer. The mode of transport is the ‘tricycle’, a small motorcycle with one place behind the rider and two in a bone-shaking pillion enclosed within a lace curtain. Availability of tricycles is infinite, availability of photocopiers is positive but small, and availability of printers is zero. We’ll be going for the handwritten, personal touch.

Both teams have chosen their papers so as to get some fiddly answers, and both teams have helped the exercise by writing some rubbish. Mostly it is all correct on close inspection, but much requires serious digestion, and gives the students at least a flavour of what Andrew and I have to endure on a daily basis. The Australians have rephrased a combinatorial problem in terms of Neel wandering through security checks at an airport, and the UK boys have proved that whatever happens here, a gold medal at the International Metaphor-Extending Olympiad seems inevitable.

Courtesy of Australian student Wilson, a penchant for fedoras has swept through the camp. Joe and Harvey look like extras in an ultra-budget production of Bugsy Malone. Our mock coordination has taken most of the afternoon, so we have not been tracking the imminent supertyphoon Nepartak as carefully as we ought, but at least the new headwear fashion offers some protection from the elements.

Friday 8th July

This morning is the final training exam, and in keeping with tradition is designated the Mathematical Ashes. Whichever team wins gets to keep an impressive urn, filled with the charred remains of some old olympiad scripts. The urn is quite heavy, so for the Cathay Pacific weight restrictions, it would be convenient for me if Australia won this year. The lack of an actual Ashes this year renders the competition all the more important in some people’s eyes, though if there were a test match here today, the covers would be on all day as the typhoon squalls.

The Ashes paper is the original Day Two paper [1] from IMO 2015. The problems are supposed to be secret until after this year’s IMO (exactly because of events like the one we are running) but the entire shortlist has been released overnight on the internet. Fortunately none of the students have been checking the relevant forum over breakfast, but ideally people will curb their admirable enthusiasm and follow the actual rules in future years. I mark the second problem, a fiddly recursive inequality, which invites many approaches, including calculus of varying rigour. Whatever the outcome, both teams have done a good job here.

For dinner, we are hosted by Dr Simon Chua, and some of his colleagues involved in the Philippines maths enrichment community, who suggested this location, and helped us set up this camp. We are treated to various Philippine dishes, including suckling pig and squid in its own ink, with a view of sunset across the lake as the storm clears. We’re very grateful to Simon, Joseph and their colleagues for tonight and their help and advice in advance.

We finish the marking after dinner, and the UK has consolidated our position on the third question, including a superb 21/21 for Warren on a genuinely hard paper, and we have won 82-74. It is late, everyone is tired, and there is packing required, so the celebrations are slightly muted, though it gives Jacob an excellent opportunity to lose his room key again, an alternative competition in which he is certainly the unique gold medallist. We transfer to the main event in Hong Kong early tomorrow, so it’s an early night all round.

[1] – Potted summary: some copies of this paper were accidentally released before they should have been, and so the paper had to be re-set.

Saturday 9th July

After a disturbed night, it has been vociferously recommended that we leave at 6am to beat the Manila traffic, with the result that we have four hours at the airport. I try to remain stoic, with difficulty. Joe practises sleeping on every available surfaces while the rest of us have a sudden enthusiasm to solve N8 from last year’s shortlist. It turns out the UKMT travel agent has outdone themselves, and booked half the group in premium economy, and half in regular economy, though the only real difference seems to be armrest width.

We are met in HK by Allison, our local guide for the week, and escorted onto coaches across from the airport on Lantau island to the University of Science and Technology in the New Territories. The check-in process is comprehensive: I sign and initial to confirm that they have correctly provided us with seven laptop sleeves, and then repeat for an infinite supply of other branded goods. Finally, we are allowed out to explore the spectacular campus, which stretches steeply down to Clearwater Bay. It is a novelty to take elevators up a total of 37 floors, and arrive on something called ‘Ground Floor’.

After a confusingly-managed dinner at the student cafeteria, a few of us head out to look at the nearby neighbourhood of Hang Hau. We pass the olympic velodrome, which gives Lawrence a good opportunity to explain gearing to those among his colleagues who do not naturally seek out applied mathematics. We return to find that Harvey decided to go to sleep before working out how to turn on his air conditioning. In humid hindsight, this was a poor strategy, as this was one of HK’s hottest days since records began. We have arrived back at the perfect time to watch the awesome thunderstorm from dry safety, which hopefully isn’t an omen of terrible things to follow in the contest, which starts on Monday.

IMO 2016 Diary – Part One

Friday 1st July

It’s my last morning as an Oxford resident, and I have to finish the final chapter of my thesis, move out of my flat, print twenty-four boarding passes, and hurtle round town collecting all the college and department stamps on my pre-submission form 3.03 like a Pokemon enthusiast. Getting to Heathrow in time for an early evening flight seems very relaxed by comparison, even with the requirement to transport two boxes of IMO uniform. Because I wasn’t paying very much attention when I signed off the order, this year we will be wearing ‘gold’, but ‘lurid yellow’ might be a better description. Hopefully the contestants might have acquired some genuinely gold items by the time we return to this airport in two weeks.

Saturday 2nd July

Our flight passes rapidly. I proved an unusual function was locally Lipschitz, watched a film, and slept for a while. Others did not sleep at all, though I suspect they also did not prove any functions were locally Lipschitz. The airport in Hong Kong is truly enormous; for once the signs advertising the time to allow to get to each gate have a tinge of accuracy. We have plenty of time though, and there is substantial enthusiasm for coffee as we transfer. Cathay Pacific approach me with a feedback form, which turns out to include 130 detailed questions, including one concerning the ‘grooming’ of the check-in staff, while we all collectively tackle an inequality from the students’ final sheet of preparatory problems.

Before long though, we have arrived in Manila, where Jacob is uncontrollably excited to receive a second stamp in his passport, to complement his first from Albania at the Balkan Olympiad last month. As we bypass the city, we get a clear view of the skyscrapers shrouded in smog across the bay, though the notorious Manila traffic is not in evidence today. We pass through the hill country of Luzon Island, the largest of the Philippines and get caught in a ferocious but brief rainstorm, and finally a weekend jam on the lakeside approach to Tagaytay, but despite these delays, the fiendish inequality remains unsolved. I’m dangerously awake, but most of the students look ready to keel over, so we find our rooms, then the controls for the air conditioning, then let them do just that.

Sunday 3rd July

We have a day to recover our poise, so we take advantage of morning, before the daily rain sets in, to explore the area. We’ve come to Tagaytay because it’s high and cool by Philippine standards, so more conducive to long sessions of mathematics than sweltering Manila. We follow the winding road down the ridge to the shore of Taal Lake, where a strange flotilla of boats is docked, each resembling something between a gondola and a catamaran, waiting to ferry us to Taal Volcano, which lies in the centre of the lake. The principal mode of ascent from the beach is on horseback, but first one has to navigate the thronging hordes of vendors. Lawrence repeatedly and politely says no, but nonetheless ends up acquiring cowboy hats for all the students for about the price of a croissant in Oxford.

Many of us opt to make the final climb to the crater rim on foot, which means we can see the sulphurous volcanic steam rising through the ground beside the trail. From the lip we can see the bright green lake which lies in the middle of the volcano, which is itself in the middle of this lake in the middle of Luzon island. To the excitement of everyone who likes fractals, it turns out there is a further island within the crater lake, but we do not investigate whether this nesting property can be extended further. After returning across the outer lake, we enjoy the uphill journey back to Tagaytay as it includes a detour for a huge platter of squid, though the van’s clutch seems less thrilled. Either way, we end up with a dramatic view of an electrical storm, before our return to the hotel to await the arrival of the Australians.

Monday 4th July

Morning brings the opportunity to meet properly the Australian team and their leaders Andrew, Mike and Jo. We’ve gathered in the Philippines to talk about maths, and sit some practice exams recreating the style of the IMO. The first of these takes place this morning, in which the students have 4.5 hours to address three problems, drawn from those shortlisted but unused for last year’s competition.

After fielding a couple of queries, I go for a walk with Jo to the village halfway down the ridge. On the way down, the locals’ glances suggest they think we are eccentric, while on the ascent they think we are insane. About one in every three vehicles is a ‘jeepney’, which is constructed by taking a jeep, extending it horizontally to include a pair of benches in the back, covering with chrome cladding, and accessorising the entire surface in the style of an American diner. We return to find that the hotel thinks they are obliged to provide a mid-exam ‘snack’, and today’s instalment is pasta in a cream sauce with salad, served in individual portions under cloches. Andrew and I try to suggest some more appropriate options, but we’re unsure that the message has got across.

I spend the afternoon marking, and the UK have started well, with reliable geometry (it appears to be an extra axiom of Euclid that all geometry problems proposed in 2015/2016 must include a parallelogram…) and a couple of solutions to the challenging number theory problem N6, including another 21/21 for Joe. Part of the goal of this training camp is to learn or revise key strategies for writing up solutions in an intelligible fashion. At the IMO, the students’ work will be read by coordinators who have to study many scripts in many languages, and so clear logical structure and presentation is a massive advantage. The discussion of the relative merits of claims and lemmas continues over dinner, where Warren struggles to convince his teammates of the virtues of bone marrow, a by-product of the regional speciality, bulalo soup.

Tuesday 5th July

The second exam happens, and further odd food appears. Problem two encourages solutions through the medium of the essay, which can prove dangerous to those who prefer writing to thinking. In particular, the patented ‘Agatha Christie strategy’ of explaining everything only right at the end is less thrilling in the realm of mathematics. It’s a long afternoon.

We organise a brief trip to the People’s Park in the Sky, based around Imelda Marcos’s abandoned mansion which sits at the apex of the ridge. In the canon of questionable olympiad excursions, this was right up there. There was no sign of the famous shoe collection. Indeed the former ‘palace’ was open to the elements, so the style was rather more derelicte than chic, perfect for completing your I-Spy book of lichen, rust and broken spiral staircases. Furthermore after a brief storm, the clouds have descended, so the view is reminiscent of our first attempt at Table Mountain in 2014, namely about five metres visibility. A drugged parrot flaps miserably through the gloom. Even the UK team shirts are dimmed.

There is a shrine on the far side of the palace, housing a piece of rock which apparently refused to be dynamited during the construction process, and whose residual scorchmarks resemble the Blessed Virgin Mary. A suggested prayer is written in Tagalog (and indeed in Comic Sans) but there is a man sitting on the crucial rock, and it’s not clear whether one has to pay him to move to expose the vision. Eventually it clears enough to get a tolerable set of team photos. Joe tries to increase the compositional possibilities by standing on a boulder, thus becoming ten times taller than the volcano, so we keep things coplanar for now. Harvey finds a giant stone pineapple inside whose hollow interior a large number of amorous messages have been penned. He adds

\mathrm{Geoff }\heartsuit\,\triangle \mathrm{s}

in homage to our leader, who has just arrived in Hong Kong to begin the process of setting this year’s IMO papers.

Balkan MO 2016 – UK Team Blog Part Two

This short blog records 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.

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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!