Questionable Statistics

My four years in the Statistics Department have been characterised by a continual feeling that I don’t know enough about statistics. That said, I can’t help but notice how statistics are used in non-specialist contexts, and of course there are times when eyebrows might reasonably be raised. Andrew Gelman (who gave a very enjoyable departmental seminar on this topic in May) has a notable blog critiquing poor statistical methodology, especially in social science; and the blogosphere is packed with compilations of egregious statistical oxymorons (“All schools *must* be above average” etc).

I’m aiming for more of a middle ground here. I’ve picked three examples of articles I found interesting at some level over the past few months. In each of them, some kind of data presentation or experiment design arose, and in all cases, I think it resulted in more questions rather than more answers, not really in a good way, since in all cases I did actually want to know more about what was going on.

Extreme rounding errors

This Guardian article about inequality in career trajectories for teachers in schools serving more advantaged and more deprived areas raises a lot of interesting questions, and actually proposes answers to several of them. The problem is the double bar chart halfway down.

First a superficial moan. Rounding to the nearest integer is fine, indeed sensible in most contexts, but not when you are treating small numbers. Comparing the extreme bars, we could have 11.5 v 12.5 or we could have 10.5 v 13.5. The headlines would probably read “N% more teachers leave disadvantaged schools”. In the first case N is 9, in the second it’s 29. So it’s not really a superficial moan. Data is supposed to offer a measure of the effect under consideration, and a discussion of how treat this effect requires a veneer of reliability about the size of the effect. We don’t just need to be asking “how can we solve this”; we also need to be asking “is this worth solving?” I’m not saying the answer is no, but based on this graph alone the answer is not ‘definitely yes’. [1]

A question to ask in the context of this report is “why do more teachers leave the profession from schools in deprived areas?” But I think most people could speculate a broadly sensible answer to this. The data raises the more subtle question “why are the rates of leaving the profession so similar across all types of school?” Essentially I think this survey is a negative result – the effect just isn’t as strong as I’d have suspected, nor I imagine the authors nor the Guardian education editorship.

By contrast, even with the amateurish presentation, the rates of moving school clearly are significant and worth talking about based on this survey. Significant data demands further information though. Once an effect is significant, we need to know more details, like to what extent the teachers are moving to schools in the same band or to mostly towards the left end of the graph. Again though, there’s no comment on the magnitude of this effect. Assuming the rate is given per year [2], then among teachers who do not leave the profession, the average tenure of a given job is more than ten years, even in the most deprived category. Maybe this is all very heavy-tailed, and so the average isn’t a good thing to be using. But this doesn’t seem like an unreasonable number to me? If the claim is that this is a major problem for the teaching profession, then we should be told what the numbers are in vaguely comparable industries.

The final thing is pretty bad. Three paragraphs under the graph, it is claimed that the rate of leaving is 70% higher in most- than least-deprived categories. We decided that N was somewhere between 9 and 29. But not 70. I can only assume that this came from the ‘moving’ category instead…

[1] I’m assuming the sample pool was large enough that the effect is statistically significant, because there isn’t a direct link to the source data or even the source report unfortunately.

[2] and is it plausible that this is rate per year? Does the profession really experience ten percent annual turnover? Is it even plausible that the scale is the same? Maybe rate of leaving really is twice rate of moving school, and I have no idea about anything, but this seems questionable.

Many, many categories

The bank Halifax commissions an annual survey about pocket money, and this year’s received widespread coverage. My attention was caught by this article on BBC news. The short summary is that they surveyed about 1,200 children from about 600 families (so in many cases a direct comparison between siblings could have been possible), and it sounds like they restricted to the age range 8-15, and asked how much pocket money they received, and whether they wanted more.

The clickbait summary was that boys receive about 13% more than girls. While it doesn’t give the exact numbers, it also mentions that more boys than girls thought they deserved more money. A psychologist conjectures that this latter effect might explain the former effect, and indeed this seems plausible. The problem is that elsewhere in the article, it says that in the corresponding survey in 2015, boys receive on average only 2% more than girls.

We therefore have to consider a couple of things. 1) Is it plausible that the actual average pocket money rates (among the whole population) have fluctuated in such a gender-dependent way in 12 months? 2) If not, then can we still say something useful, even if our confidence in the methodology of the survey is reduced?

I think the answer to the first question is definitely ‘no’. So we come to the second question. Firstly, how could this have arisen? Well, there’s no indication how the children were chosen, but equally it’s hard to think of a way of choosing them that would increase artificially the boys’ pocket money. In this last sentence, I’m deliberately suggesting the null hypothesis that boys and girls are the same. This is misleading. For now, our null hypothesis is that 2015 should be the same as 2016. Indeed, it’s equally plausible that this year’s summary data was ‘more correct’, and so we are asking why the difference was excessively compressed in 2015.

This is entirely a matter of effect size, and it’s to the article’s credit that so much information is given further down the page that we can actually make the following remarks quantitatively. At the bottom, they show the average pay (across genders) in different regions. There is a lot of variation here. The rate of pay in East Anglia is only 60% what it is in London. At some level, I’m surprised it isn’t even higher. The article tells us that children between the ages of eight and fifteen were surveyed. By what rate did your pocket money change in that interval? Surely it doubled at the very least? Weekly pocket money at age eight is for sweets and an occasional packet of Pokemon cards (substitute 21st century equivalent freely…). Teenagers need to pay for outings and clothes etc, so there’s no reason at all why this should be comparable. For the purposes of definiteness in the next calculation, let’s say that fifteen-year-olds get paid four times as much as eight-year-olds on average, which I think is rather conservative.

So here’s a plausible reason for the fluctuations. Suppose between 2015 and 2016, the survey substituted 10 eight-year-old boys from the Midlands for 10 fifteen-year-old boys from London. The expectation change in the average amongst the 600 boys surveyed is

\frac{10}{600} \times (\sim 6.5) \times (4 \times (0.6)^{-1}) \approx 0.7,

in pounds, which is on the order of the actual change from 2015 to 2016. Choosing the participants in a non-uniform way seems like too basic a mistake to make, but mild fluctuations in the distribution of ages and locations, as well as the occasional outlier (they want to use the mean, so one oligarch’s daughter could make a noticeable difference in a sample size of 600) seems more plausible as an explanation to me than a general change in the population rates. Choosing participants in a uniform way is just hard when there are loads and loads of categories.

I’m not really saying that this survey is bad – they have to report what they found, and without a lot more information, I have no idea whether this year’s is more or less plausible than last year’s. But if you add the phrase “especially in 2016” to everything the psychologist says, it suddenly seems a bit ridiculous. So it would be worth making the remark that even if this effect is statistically significant, that doesn’t mean the effect size is large relative to lots of other less controversial effect sizes visible in the data.

Comparing precious metals

I’ve recently returning from this year’s International Mathematical Olympiad, and now we are well into the swing of its sweatier namesake in Rio. At both events, there have been many opportunities to observe how different people with different ambitions and levels of extroversion display different levels of pleasure at the same medal outcomes. In the light of this, I was drawn to this article, which has been getting quite a lot of traction online.

The basic premise is simple. Silver medallists are less happy than bronze medallists (when – unlike at the IMO – there is only one of each medal), and it’s not hard to come up with a cheap narrative: silver-winners are disappointed to have missed out on gold; bronze-winners are glad to have got a medal at all. This is all pretty convincing, so maybe there’s no need actually to collect any data, especially since asking a crowd of undergraduates to rate faces on the 1-10 agony-ecstacy scale sounds like a bit of an effort. Let’s read a couple of sentences, describing a follow-up study on judo medallists at the Athens Olympics:

Altogether, they found that thirteen of the fourteen gold medal winners smiled immediately after they completed their winning match, while eighteen of the twenty-six bronze medalists smiled. However, none of the silver medalists smiled immediately after their match ended.

You might wonder how come there are almost twice as many bronzes as golds. Well, very full details about the judo repechage structure can be found here, but the key information is that all bronze-medallists won their final match, as, naturally, did the gold-medallists. The silver-medallists lost their final match, ie the gold-medal final. So this study is literally reporting that highly-driven competitive sportsmen are happier straight after they win, than straight after they lose. This doesn’t have much to do with medals, and isn’t very exciting, in my opinion. I would, however, be interested to know what was eating the one gold medallist who didn’t smile immediately.

This isn’t really a gripe about the statistics, more about the writing. They are testing for an effect which seems highly plausible, but is branded as surprising. The study for which they give more methodological details in fact seems to be testing for a related effect, which is not just plausible, but essentially self-evident. So naturally I want to know more about the original study, which is the only place left for there to be any interesting effects, for example if they looked at athletics events which don’t have a binary elimination structure. But we aren’t told enough to find these interesting effects, if they exist. How annoying. The only thing left is to think about is my bronze medal at IMO 2008. They said eight years would be enough for the wounds to heal but I’m still not smiling.

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.

P1000371_compressed

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

Tuesday 10th May

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

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

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

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

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

Balkan MO 2016 – UK Team Blog Part One

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

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

Wednesday 4th May

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

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

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

Thursday 5th May

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

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

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

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

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

Friday 6th May

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

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

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

P1000166_compressed

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

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

Saturday 7th May

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

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

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

Turan’s Theorem

Turan’s theorem gives bounds on the number of edges required in a graph on a fixed number of vertices n to guarantee it contains a complete graph of size r+1. Equivalently, an upper bound on the number of edges in a K_{r+1}-free graph. For some of the applications and proofs, it may be more natural to look instead at the complement graph, for which the theorem becomes a statement about the existence or otherwise of an independent set of size r+1.

Rather than give an expression for the bound immediately, it is more natural to consider the Turan graph T(n,r), the maximal graph on n vertices without a copy of K_{r+1}. This is constructed by dividing the vertices into r classes with ‘as equal size as possible’. That is, some classes have size \lfloor \frac{n}{r}\rfloor and others have size \lfloor \frac{n}{r}\rfloor +1. Then connect any pair of vertices which are not in the same class by an edge. This gives a complete r-partite graph on these classes. Since any collection of r+1 vertices contains at least two in the same class, it can’t contain a K_{r+1}. Note that the complement of the complete r-partite graph is the union of r disjoint complete graphs on the classes.

There are a number of ways to enumerate the edges in T(n,r), and some can get quite complicated quite quickly. After a moderate amount of thought, this is my favourite. Let n=\ell r+k, so T(n,r) has k classes of size (l+1) and (r-k) classes of size l. Pick an ordered pair of vertices uniformly at random. (So picking the same vertices is indeed an option, and is counted twice.) Then the probability they are the same class is

\frac{k}{r}\cdot\frac{\ell+1}{n}+\frac{r-k}{r}\cdot \frac{\ell}{n} = \frac{1}{r}.

So the probability they are in different classes is \frac{r-1}{r}, and we can treat all of the 2n^2 ordered pairs in this way, noting a) that we count everything twice; and b) we know a priori that we don’t have loops, so the fact that we’ve included these in the count doesn’t matter. We end up with the enumeration (1-\frac{1}{r})\frac{n^2}{2} for the edges in T(n,r).

A standard proof

For both proofs, I find it slightly easier to work in the complement graph, where we are aiming for the largest number of edges with an independent set of size (r+1). Suppose we have a graph with the minimal number of vertices such that there’s no independent set of given size. Suppose also that there is an edge joining vertices v and w, such that d(v)> d(w). Then if we change v’s neighbourhood \Gamma(v) so that it becomes the same as \Gamma(w), (that is, we replace v with a copy of w, and maintain the original edge vw), then it is easily checked that we still do not have an independent set of that size, but fewer edges.

Note that by attempting to make the neighbourhoods of connected vertices equal, we are making the graph look more like a union of complete components. We can do a similar trick if we have three vertices u,v,w such that there are edges between u and v and v and w, but not u and w. Then we know the degrees of u,v,w are the same by the previous argument, and so it can again be checked that making \Gamma(u),\Gamma(w) the same as \Gamma(v), and adding the edge uw reduces the number of edges, and maintains the non-existence of the independent set.

The consequence of this is that we’ve shown that the minimum can only be attained when presence of edges is an equivalence relation (ignoring reflexivity). Thus the minimum is only attained for a union of at most r complete graphs. Jensen (or any root-mean-square type inequality) will then confirm that the true minimum is attained when the sizes of the r components are as equal as possible.

A probabilistic proof

The following probabilistic proof is courtesy of Alon and Spencer. The motivation is that in the (equality) case of a union of complete graphs, however we try to build up a maximal independent set, we always succeed. That is, it doesn’t matter how we choose which vertex (unconnected to those we already have) to add next – we will always get a set of size r. This motivates a probabilistic proof, as an argument in expectation will have equality in the equality case, which is always good.

Anyway, we build up an independent set in a graph by choosing uniformly at random a vertex which is not connected to any we have so far, until this set of vertices is empty. It makes sense to settle the randomness at the start, so give the vertices a uniform random labelling on [n], and at each stage, choose the independent vertex with minimal label.

Thus, a vertex v will be chosen for the independent set if, and only if, it has a smaller label than all of its neighbours, that is, with probability \frac{1}{1+d(v)}. So the expected size of the independent set constructed in this fashion is

\sum_{v\in V(G)} \frac{1}{1+d(v)}\ge \frac{V}{1+\bar d} = \frac{V}{1+\frac{2E}{V}}.

One can chase through the expressions to get the bound we want back.

Olympiad example

The reason I was thinking about Turan’s theorem was a problem which the UK IMO squad was discussing. It comes from an American selection test (slightly rephrased): given 100 points in the plane, what is the largest number of pairs of points with \ell_1 distance in (1,2]?

The key step is to think about how large a collection of points can have this property pairwise. It is easy to come up with an example of four points which work, and seemingly impossible to come up with an example with five points. To prove this, I found it easiest to place a point at the origin, then explicitly work with coordinates relative the basis (1,1),(1,-1) for fairly obvious reasons in this metric.

Anyway, once you are convinced that you can’t have five points with this property pairwise, you are ready to convert into a graph-theoretic statement. Vertices correspond to points, and edges link pairs of points whose distance is in (1,2] as required. We know from the previous paragraph that there is no copy of K_5 here, so Turan’s theorem bounds the number of edges, ie the number of suitable pairs.

It also tells us under what sort of circumstances the bound is attained, and from this, it’s natural to split the 100 points into four groups of 25, for example by taking four points which satisfy the condition pairwise (eg a diamond around the origin), and placing each group very near one of the points.

Extensions and other directions

The existence of a complete subgraph is reminiscent of Ramsey theory, which in one case is a symmetric version of Turan’s theorem. In Turan, we are adding enough edges to force a complete subgraph, while in the original form of Ramsey theory, we are asking how large the graph needs to be to ensure that for any edge configuration, either the original graph or the complement graph includes a complete subgraph. It makes a lot more sense to phrase this in terms of colours for the purpose of generalisation.

A natural extension is to ask about finding copies of fixed graphs H other than the complete graph. This is the content of the Erdos-Stone theorem. I’d prefer to say almost nothing rather than be vague, but the key difference is that the bound is asymptotic in the number of vertices rather than exact. Furthermore, the asymptotic proportion of vertices depends on the chromatic number of H, which tells you how many classes r are required to embed H in a (large if necessary) r-partite graph. So it is perhaps unsurprising that the limiting proportions end up matching the proportion of edges in the Turan graphs, namely r-1/r as r varies, which leaves the exact scaling open to further investigation in the case where H is bipartite (hence has chromatic number 2).

EGMO 2016 Paper II

Continuing from yesterday’s account of Paper I, this is a discussion of my thoughts about Paper II of EGMO 2016, happening at the moment in Busteni, Romania. This is not an attempt to describe official solutions, but rather to describe the thought process (well, a thought process) of someone tackling each question. I hope it might be interesting or useful, but for students, it will probably be more useful after at least some engagement with the problems. These are excellent problems, and reading any summary of solutions means you miss the chance to hunt for them yourself.

In actual news, you can follow the scoreboard as it is updated from Romania here. Well done to the UK team on an excellent performance, and hope everyone has enjoyed all aspects of the competition!

Question 4

Circles \omega_1,\omega_2 with the same radius meet at two points X_1,X_2. Circle \omega is externally tangent to \omega_1 at T_1, and internally tangent to \omega_2 at T_2. Prove that lines X_1T_1,X_2T_2 meet on \omega.

Thought 1: I’m not the biggest fan of geometry ever, but I thought this looked like a nice problem, because it’s only really about circles, so I figured it probably wouldn’t require anything too exotic.

Thought 2: I bet lots of people try inversion. But the equal radius condition means I’m probably happy with the circles as they are. I hope lots of people don’t try to place the diagram in some co-ordinate system, even if it possible to do it sensibly (eg by making \omega the reference circle).

Thought 3: The labelling of X_1,X_2 is unrelated to the rest of the indexing. So the intersection of X_1T_2,X_2T_1 should also lie on \omega, and possibly has some relationship (antipodal?) to the point I actually need to find out. But I can’t think of any reason why it’s easier to prove two points lie on a circle than just one, so let’s leave this as a thought rather than an idea.

Idea 1: I drew a terrible diagram on the back of a draft of my abstract, and for once, this was actually kind of helpful. Forget about radii being equal, one of them wasn’t even a circle. Anyway, while drawing in the later points, I was struggling to make it look convincingly like all the lengths which were supposed to be equal were in fact equal. So the idea was: almost all the segments in the diagram (once I’ve defined the circle centres O_{\omega_1} etc) have one of two lengths (the radii of \omega_1,\omega – red and green-ish in the diagram below), and with this in mind I can forget about the circles. We’ve got a rhombus, which is even better than a parallelogram, which is itself a really useful thing to have in a configuration. Another consequence of the proliferation of equal lengths is that almost all triangles are isosceles, and we know that similarity of isosceles triangles is particularly easy, because you only have to match up one angle.

Idea 2: How to prove it? We have to prove that two lines and a circle concur. This is where I actually need to stop and think for a moment: I could define the point where the lines meet and try to show it’s on the circle, or intersect one line with the circle, and show it’s on the other line. Idea 1 basically says I’m doing the problem using lengths, so I should choose the way that fits best with lengths.

20160414_093104

If I define the point P where X_2T_2 meets the circle (this was easier to draw in my diagram), then I know PO_\omega=T_2 O_\omega and so on. Then there were loads of isosceles triangles, and some of them were similar, which led to more parallel lines, and from this you could reverse the construction in the other direction to show that P also lay on the other line.

Question 5

Let k, n be integers such that k\ge 2,\, k\le n\le 2k-1. Place rectangular k x 1 or 1 x k tiles on an n x n chessboard in the natural way with no overlap until no further tile can be placed. Determine the minimum number of tiles that such an arrangement may contain.

Idea 1: It took me a while to parse the question. Minimum over what? I rephrased it in my head as: “to show the answer is eg n+5, I need to show that whenever you place n+4 tiles legally, you can’t add another. I also need to show that you can place n+5 such that you can’t add another.” This made life a lot easier.

Thought 1: What goes wrong if you take n=2k and beyond? Well, you can have two horizontal tiles on a given row. I’m not really sure how this affects the answer, but the fact that there is still space constraint for n<2k is something I should definitely use.

Diversion: I spent a while thinking that the answer was 4 and it was a very easy question. I spent a bit more time thinking that the answer was n, and it was a quite easy question, then realised that neither my construction nor my argument worked.

Thought 2: can I do the cases n=k,or 2k-1 or k+1? The answers were yes, unsure, and yes. The answer to k+1, which I now felt confident was actually four, was helpful, as it gave me a construction for k+2, …, 2k-1 that seemed good, even though it was clearly not optimal for 2k-1. Therefore, currently my potential answer has three regimes, which seemed unlikely, but this seemed a good moment to start trying to prove it was optimal. From now on, I’m assuming I have a configuration from which you can’t add another block.

20160414_100244

Idea 2: About this diagram, note that once I’ve filled out the top-left (k+1)x(k+1) sub-board in this way, there are still lots of ways to complete it, but I do have to have (n-k-1) horizontal and (n-k-1) vertical tiles roughly where I’ve put them. Why? Because I can’t ‘squeeze in’ a vertical tile underneath the blue bit, and I can’t squeeze in a horizontal tile to the right of the blue bit. Indeed, whenever I have a vertical block, there must be vertical blocks either to the left or to the right (*) (or possibly both if we’re near the middle). We need to make this precise, but before doing that, I looked back at where the vertical blocks were in the proposed optimum, and it turns out that all but (k-1) columns include a vertical block, and these (k-1) columns are next to each other.

This feels like a great idea for two reasons: 1) we’ve used the fact that n<2k at (*). 2) this feels very pigeonhole principle-ish. If we had fewer tiles, we’d probably have either at least k columns or least k rows without a vertical (or, respectively, horizontal) tile. Say k columns don’t include a vertical tile – so long as they are next to each other (which I think I know) we can probably include a horizontal tile somewhere in there.

So what’s left to do? Check the previous sentence actually works (maybe it’s full of horizontal tiles already?), and check the numerics of the pigeonhole bound. Also work out how the case n=2k-1 fits, but it seems like I’ve had some (/most) of the good ideas, so I stopped here.

Question 6

EGMO2016 Q6

I don’t actually want to say very much about this, because I didn’t finish off all the details. I want to talk briefly in quite vague terms about what to do if you think this problem looks scary. I thought it looked a bit scary because it looked similar to two number theoretic things I remember: 1) primes in arithmetic progressions. This is very technical in general, but I can remember how to do 3 mod 4 fairly easily, and 1 mod 4 with one extra idea; 2) if a square-free number can be written as a sum of two squares, this controls its factors modulo 4.

Vague Ideas: It seemed unlikely that this would involve copying a technical argument, so I thought about why I shouldn’t be scared. I think I shouldn’t be scared of the non-existence part. Often when I want to show there are no integer solutions to an equation, I consider showing there are no solutions modulo some base, and maybe this will be exactly what I do here. I’ll need to convert this statement about divisibility into an equation (hopefully) and check that n\equiv 3,4 modulo 7 doesn’t work.

For the existence of infinitely many solutions, maybe I’d use Chinese Remainder Theorem [1], or I’ll reduce it to something that I know has lots of solutions (eg Pythagoras), or maybe I can describe some explicit solutions?

Actual Idea 1: We’ve got n^2+m | n^4, but this is a very inefficient statement, since the RHS is a lot larger than the LHS, so to be useful I should subtract off a large multiple of the LHS. Difference of two squares is a good thing to try always, or I could do it manually. Either way, I get n^2+m | m^2 which is genuinely useful, given I know m=1,2, …, 2n, because the RHS is now comparable in size to the LHS, so I’ve narrowed it down from roughly n^2 possibilities to just three:

n^2+m=m^2,\quad 2(n^2+m)=m^2,\quad 3(n^2+m)=m^2. (*)

I’m going to stop now, because we’ve turned it into a completely different problem, which might be hard, but at least in principle this is solvable. I hope we aren’t actually scared of (*), since it looks like some problems we have solved before. I could handle one of these in a couple of lines, then struggled a bit more with the other pair. I dealt with one by recourse to some theory, and the final one by recourse to some theory after a lot of rearranging which I almost certainly got wrong, but I think I made an even number of mistakes rather than an odd number because I got the correct solution set modulo 7. Anyway, getting to (*) felt like the majority of the ideas, and certainly removed the fear factor of the Q6 label, so to fit the purpose of this discussion I’ll stop here.

[1] During one lunch in Lancaster, we were discussing why Chinese Remainder Theorem is called this. The claim was that an ancient Chinese general wanted to know the size of their army but it was too big to count, so had them arrange themselves into columns of various sizes, and counted the remainders. The general’s views on the efficiency of this algorithm remain lost in the mists of time.

EGMO 2016 Paper I

We’ve just our annual selection and training camp for the UK IMO team in Cambridge, and I hope it was enjoyed by all. I allotted myself the ‘graveyard slot’ at 5pm on the final afternoon (incidentally, right in the middle of this, but what England fan could have seen that coming in advance?) and talked about random walks on graphs and the (discrete) heat equation. More on that soon perhaps.

The UK has a team competing in the 5th European Girls Mathematical Olympiad (hereafter EGMO 2016) right now in Busteni, Romania. The first paper was sat yesterday, and the second paper is being sat as I write this. Although we’ve already sent a team to the Romania this year (where they did rather well indeed! I blame the fact that I wasn’t there.), this feels like the start of the olympiad ‘season’. It also coincides well with Oxford holidays, when, though thesis deadlines loom, I have a bit more free time for thinking about these problems. Anyway, last year I wrote a summary of my thoughts and motivations when trying the EGMO problems, and this seemed to go down well, so I’m doing the same this year. My aim is not to offer official solutions, or even outlines of good solutions, but rather to talk about ideas, and how and why I decided whether they did or didn’t work. I hope some of it is interesting.

You can find the paper in many languages on the EGMO 2016 website. I have several things to say about the geometry Q2, but I have neither enough time nor geometric diagram software this morning, so will only talk about questions 1 and 3. If you are reading this with the intention of trying the problems yourself at some point, you probably shouldn’t keep reading, in the nicest possible way.

Question 1

[Slightly paraphrased] Let n be an odd positive integer and x_1,\ldots,x_n\ge 0. Show that

\min_{i\in[n]} \left( x_i^2+x_{i+1}^2\right) \le \max_{j\in[n]} 2x_jx_{j+1},

where we define x_{n+1}=x_1 cyclically in the natural way.

Thought 1: this is a very nice statement. Obviously when i and j are equal, the inequality holds the other way round, and so it’s interesting and surprising that constructing a set of pairs of inequalities in the way suggested gives a situation where the ‘maximum minimum’ is at least the ‘minimum maximum’.

Thought 2: what happens if n is actually even? Well, you can kill the right-hand-side by taking at least every other term to be zero. And if n is even, you really can take every other term to be even, while leaving the remaining terms positive. So then the RHS is zero and the LHS is positive.

The extension to this thought is that the statement is in danger of not holding if there’s a lot of alternating behaviour. Maybe we’ll use that later.

Idea 1: We can write

2(x_i^2+x_{i+1}^2)=(x_i+x_{i+1})^2 + |x_i-x_{i+1}|^2, \quad 4x_ix_{i+1}=(x_i+x_{i+1})^2 - |x_i-x_{i+1}|^2,

which gives insight into ‘the problem multiplied by 2’. This was an ‘olympiad experience’ idea. These transformations between various expressions involving sums of squares turn out to be useful all the time. Cf BMO2 2016 question 4, and probably about a million other examples. As soon as you see these expressions, your antennae start twitching. Like when you notice a non-trivial parallelogram in a geometry problem, but I digress. I’m not sure why I stuck in the absolute value signs.

This was definitely a good idea, but I couldn’t find a way to make useful deductions from it especially easily. I tried converting the RHS expression for i (where LHS attains minimum) into the RHS expression for any j by adding on terms, but I couldn’t think of a good way to get any control over these terms, so I moved on.

Idea 2: An equality case is when they are all equal. I didn’t investigate very carefully at this point whether this might be the only equality case. I started thinking about what happens if you start with an ‘equal-ish’ sequence where the inequality holds, then fiddle with one of the values. If you adjust exactly one value, then both sides might stay constant. It seemed quite unlikely that both would vary, but I didn’t really follow this up. In any case, I didn’t feel like I had very good control over the behaviour of the two sides if I started from equality and built up to the general case by adjusting individual values. Or at least, I didn’t have a good idea for a natural ordering to do this adjustment so that I would have good control.

Idea 3: Now I thought about focusing on where the LHS attains this minimum. Somewhere, there are values (x,y) next to each other such that x^2+y^2 is minimal. Let’s say x\le y. Therefore we know that the element before x is at least y, and vice versa, ie we have

\ldots, \ge y, x, y, \ge x,\ldots.

and this wasn’t helpful, because I couldn’t take this deduction one step further on the right. However, once you have declared the minimum of the LHS, you are free to make all the other values of x_i smaller, so long as they don’t break this minimum. Why? Because the LHS stays the same, and the RHS gets smaller. So if you can prove the statement after doing this, then the statement was also true before doing this. So after thinking briefly, this means that you can say that for every i, either x_{i-1}^2+x_i^3 or x_i^2+x_{i+1}^2 attains this minimum.

Suddenly this feels great, because once we know at least one of the pairs corresponding to i attains the minimum, this is related to parity of n, which is in the statement. At this point, I was pretty confident I was done. Because you can’t partition odd [n] into pairs, there must be some i which achieves a minimum on both sides. So focus on that.

Let’s say the values are (x,y,x) with x\le y. Now when we try to extend in both directions, we actually can do this, because the values alternate with bounds in the right way. This key is to use the fact that the minimum x^2+y^2 must be attained at least every other pair. (*) So we get

\ldots, \le x,\ge y,x,y,x,\ge y,\le x,\ldots.

But it’s cyclic, so the ‘ends’ of this sequence join up. If n\equiv 1 modulo 4, we get \ge y,\ge y next to each other, which means the RHS of the statement is indeed at least the LHS. If n\equiv 3 modulo 4, then we get \le x,\le x next to each other, which contradicts minimality of x^2+y^2 unless x=y. Then we chase equality cases through the argument (*) and find that they must all be equal. So (after checking that the case x\ge y really is the same), we are done.

Thought 3: This really is the alternating thought 2 in action. I should have probably stayed with the idea a bit longer, but this plan of reducing values so that equality was achieved often came naturally out of the other ideas.

Thought 4: If I had to do this as an official solution, I imagine one can convert this into a proof by contradiction and it might be slightly easier, or at least easier to follow. If you go for contradiction, you are forcing local alternating behaviour, and should be able to derive a contradiction when your terms match up without having to start by adjusting them to achieve equality everywhere.

Question 3

Let m be a positive integer. Consider a 4m x 4m grid, where two cells are related to each other if they are different but share a row or a column. Some cells are coloured blue, such that every cell is related to at least two blue cells. Determine the minimum number of blue cells.

Thought 1: I spent the majority of my time on this problem working with the idea that the answer was 8m. Achieved by taking two in each row or column in pretty much any fashion, eg both diagonals. This made me uneasy because the construction didn’t take advantage of the fact that the grid size was divisible by 4. I also couldn’t prove it.

Thought 2: bipartite graphs are sometimes useful to describe grid problems. Edges correspond to cells and each vertex set to row labels or column labels.

Idea 1: As part of an attempt to find a proof, I was thinking about convexity, and why having exactly two in every row was best, so I wrote down the following:

Claim A: No point having three in a row.

Claim B: Suppose a row has only one in it + previous claim => contradiction.

In Cambridge, as usual I organised a fairly comprehensive discussion of how to write up solutions to olympiad problems. The leading-order piece of advice is to separate your argument into small pieces, which you might choose to describe as lemmas or claims, or just separate implicitly by spacing. This is useful if you have to do an uninteresting calculation in the middle of a proof and don’t want anyone to get distracted, but mostly it’s useful for the reader because it gives an outline of your argument.

My attempt at this problem illustrates an example of the benefit of doing this even in rough. If your claim is a precise statement, then that’s a prompt to go back and separately decide whether it is actually true or not. I couldn’t prove it, so started thinking about whether it was true.

Idea 2: Claim A is probably false. This was based on my previous intuition, and the fact that I couldn’t prove it or get any handle on why it might be true. I’d already tried the case m=1, but I decided I must have done it wrong so tried it again. I had got it wrong, because 6 is possible, and it wasn’t hard from here (now being quite familiar with the problem) to turn this into a construction for 6m in the general case.

Idea 3: This will be proved by some sort of double-counting argument. Sometimes these arguments turn on a convexity approach, but when the idea is that a few rows have three blue cells, and the rest have one, this now seemed unlikely.

Subthought: Does it make sense for a row to have more than three blue cells? No. Why not? Note that as soon as we have three in a row, all the cells in that row are fine, irrespective of the rest of the grid. If we do the problem the other way round, and have some blues, and want to fill out legally the largest possible board, why would we put six in one row, when we could add an extra row, have three in each (maintaining column structure) and be better off than we were before. A meta-subthought is that this will be impossible to turn into an argument, but we should try to use it to inform our setup.

Ages and ages ago, I’d noticed that you could permute the rows and columns without really affecting anything, so now seemed a good time to put all the rows with exactly one blue cell at the top (having previously established that rows with no blue cell were a disaster for achieving 6m), and all the columns with one blue cell at the left. I said there were r_1,c_1 such rows and columns. Then, I put all the columns which had a blue cell in common with the r_1 rows next to the c_1 columns I already had. Any such column has at least three blues in it, so I said there were c_3 of these, and similarly r_3 rows. The remaining columns and rows might as well be r_0,c_0 and hopefully won’t matter too much.

From here, I felt I had all the ingredients, and in fact I did, though some of the book-keeping got a bit fiddly. Knowing what you are aiming for and what you have means there’s only one way to proceed: first expressions in terms of these which are upper bounds for the number of columns (or twice the number of columns = rows if you want to keep symmetry), and lower bounds in terms of these for the number of blue cells. I found a noticeable case-distinction depending on whether r_1\le 3c_3 and c_1\le 3r_3. If both held or neither held, it was quite straightforward, and if exactly one held, it got messy, probably because I hadn’t set things up optimally. Overall, fiddling about with these expressions occupied slightly more time than actually working out the answer was 6m, so I don’t necessarily have a huge number of lessons to learn, except be more organised.

Afterthought 2: Thought 2 said to consider bipartite graphs. I thought about this later while cycling home, because one can’t (or at least, I can’t) manipulate linear inequalities in my head while negotiating Oxford traffic and potholes. I should have thought about it earlier. The equality case is key. If you add in the edges corresponding to blue cells, you get a series of copies of K_{1,3}, that is, one vertex with three neighbours. Thus you have three edges for every four vertices, and everything’s a tree. This is a massively useful observation for coming up with a very short proof. You just need to show that there can’t be components of size smaller than 4. Also, I bet this is how the problem-setter came up with it…