EGMO 2019

Posted on May 15, 2019 by dominicyeo

Last week, we held our annual IMO training and selection camp in the lovely surroundings of Trinity College, Cambridge. Four of our students have subsequently spent this week in Kiev, for the ninth edition of the prestigious European Girls’ Mathematical Olympiad.

The UK team, none of whom had attended the competition before, and all of whom remain eligible to return at least once, did extremely well, placing fourth out of the official European countries, and earning three silver medals, and one gold medal, with complete or almost-complete solutions to all six problems between them. More details are available on the competition website.

In this post, I’m going to discuss some of the non-geometry problems. As a special treat, the discussion of Question 6 is led by Joe Benton, which is only fitting, since he wrote it. You can find the first day’s three problems here, and the second day’s problems here. The geometry problems are treated in a separate post.

Problem One

Triple equalities are hard to handle directly, so generally one starts with a single equality, for example the left hand one $a^2b+c = b^2c+a$ , after noting that the setup is cyclic (but not symmetric) in the variables, under cycling $a\to b \to c\to a$ .

Some quick notes:

The given equations are inhomogeneous, meaning that not every term has the same degree in terms of {a,b,c}. In complicated equations, normally we want to cancel terms, and we certainly can’t cancel terms with different degrees! So a good first idea is to find a way to make the equations homogeneous, for example using the condition, which is itself inhomogeneous. For example, one could replace $c$ with $c(ab+bc+ca)$ , and it’s clear this might lead to some cancellation. In this case, the algebra is more straightforward if one rearranges and factorises first to obtain $a(1-ab)=c(1-bc)$ , from which the condition gives $a(bc+ca)=c(ab+ac)$ .
Shortly after this, we find $a^2c=ac^2$ , which means $a=c$ unless at least one of these terms is equal to zero.
This case distinction should not be brushed over lightly as a tiny detail. This is the sort of thing that can lose you significant marks, especially if the omitted case turns out to be more involved than the standard case.
This is a good example of planning your final write-up. The case distinction $\{a,c\ne 0\}, \{a\text{ or }c=0\}$ is really clumsy. What about b? We have already said that the setup is cyclic in (a,b,c), and it’s annoying to throw this aspect away. It really would be much much to have the overall case distinction at the start of your argument, for example into $\{a,b,c \ne 0\}$ and $\{a\text{ or }b\text{ or }c=0\}$ , because then in the latter case you could assume that $a=0$ without loss of generality, then see what that implied for the other variables.
In this setting, one really should check that the claimed solutions satisfy both the condition and the triple-equality. This is a complicated family of simultaneous equations, and even if you’re lucky enough to have settled on an argument that is reversible, it’s much easier to demonstrate that everything is satisfied as desired than actually to argue that the argument is reversible.

Problem Two

Maybe I speak for a number of people when I say I’m a little tired of domino tiling problems, so will leave this one out.

Problem Five

I liked this problem. A lot. It did feel like there were potential time-wasting bear traps one might slip into, and perhaps these comments are relevant:

This feels like quite a natural problem to ask. In particular, the task specified by (A)+(B) is much more elegant than the RHS of (C). So unless you have immediate insight for the RHS of (C), it makes sense to start with the task, and aim for any bound similar to (C).
There are a lot of transformations one could make, for example, repeatedly removing n from one of the $a_n$ s, or re-ordering them conveniently somehow, which shouldn’t affect any solution. You can also attack the RHS of (C) in a number of ways (perhaps with a case split for n odd or even?). As with involved but easy angle chases discussed in the companion post, it’s probably better to hold this in mind than instantly to do all of these, since it will just obscure the argument if you end up not using the result of the simplification!
One should always try small examples. Here, it’s worth trying examples large enough to get a sense of what’s happening, because I think the crucial observation is that (unless you were very very unlucky) there are lots of suitable sequences $B=(b_i)$ . (*)
Certainly the case where the $(a_i)$ s themselves form a complete n-residue system is ‘easiest’ and certainly straightforward. One might say that the case where the $(a_i)$ are equal (or at least congruent) is ‘hardest’ (in that $(b_i-a_i)$ might have to take the largest values in some sense), but is also straightforward. There’s certainly the option of trying to rigorise this, but I don’t know whether this can be made to work.

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BMO1 2018

Posted on December 1, 2018 by dominicyeo

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

I hope any students who sat the paper enjoyed at least some of the questions, and found it challenging! The following commentaries on the problems are not official solutions, and are not always full solutions at all, but contain significant steps of solutions, so would be best saved until after you have attempted the problems, if you are planning to do so. I’ve written quite a lot about Q5 because I found it hard (or at least time-consuming) and somewhat atypical, and I’ve written a lot about Q6 because there was a lot to say. I hope at least some of this is interesting to some readers of all levels of olympiad experience.

Question 1

A list of five two-digit positive integers is written in increasing order on a blackboard. Each of the five integers is a multiple of 3, and each digit {0,1,…,9} appears exactly once on the blackboard. In how many ways can this be done? (Note that a two-digit number cannot begin with zero.)

It’s a trope of BMO1 that the first question must be doable by some sort of exhaustive calculation or listing exercise. Of course, that is rarely the most efficient solution.

However, there is normally a trade-off between eliminating all listing, and reducing to a manageable task.

The key observation here is that writing the integers in increasing order is really just a way to indicate that order of the choices doesn’t matter. Even if that seems counter-intuitive. The question wants to know how many ways to choose these five numbers. The order of choice doesn’t matter since we’re going to put them in ascending order on the blackboard anyway.

You want to make your choices with as much independence as possible. So it would, for example, be a bad idea to choose the smallest number first. How many possibilities are there where the smallest number is 24? What about 42? What about 69? These are all different, and some are zero, so will make the computation very taxing.

However, you might notice that the digits {0,3,6,9} have to go together to form two numbers, and the rest have to pair up with one digit from {1,4,7} and one from {2,5,8}. You might know that an integer is divisible by 3 precisely if its digit sum is divisible by 3, but in this context you wouldn’t lose too much time by simply listing everything! These tasks are now completely separate, so you can take the number of ways to pair up {0,3,6,9} and multiply by the number of ways to pair up {1,4,7} and {2,5,8}. You need to take care over the ordering. It does (obviously) matter which is the first digit and which is the second digit in a number!

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MOG 2018

Posted on October 18, 2018 by dominicyeo

I’m teaching a lecture course at Technion this coming semester, which is going to limit opportunities for travel somewhat for the next three months or so. It was therefore nice to take advantage of the new ultra-cheap Luton-Tel Aviv connection on Wizzair to make a whistlestop weekend trip to Cambridge. It was lovely to catch up with friends, colleagues and former students, including some people who fit into all those categories.

Among my various mathematical diversions over the weekend, about thirty of us spent most of Sunday marking the 2018 edition of the UK’s Mathematical Olympiad for Girls (MOG). You can find the paper without solutions here, and the paper with official solutions and commentary here.

I have had several blog posts on probabilistic, competitive and educational topics queued up over this summer which has alternated between extreme busyness and lethargy, but this felt like a good opportunity to break the drought and both start and finish a post on the same day!

I’m going to write some thoughts about Question 4, mostly about part a), which I hope might be useful to future students who perhaps find this post while preparing for next year’s edition of the competition. Here’s the statement:

Each of 100 houses in a row are to be painted white or yellow. The residents are quite particular and request that no three neighbouring houses are all the same colour.

a) Explain why no more than 67 houses can be painted yellow.

b) In how many ways may the houses be painted if exactly 67 are painted yellow?

Of all the questions on the paper, based on the fraction of the scripts which I saw, I suspect Q4a) is the one for which the most candidates will have scored significantly fewer points than they’d been hoping for.

I think this illustrates some useful guidance for maths competitions, and also illuminates some differences between maths and other subjects which candidates might not have thought about.

Many students wrote that because three houses in a row cannot be yellow, the way to get as many yellow houses as possible is to use a YYWYYW… pattern. Since 100 is not a multiple of three, you have to work out what to do with the final house, after you’ve coloured 99 using the pattern, and sometimes you can colour it yellow, and sometimes you can’t.

Finding and proving results about maxima

Before focusing on the qualities of this argument, it’s worth returning to the question statement. It’s in two parts, and (assuming the answer to the second part isn’t “none“) then the overall goal is to find the maximum number of yellow houses, and to count the number of ways to achieve this maximum number, presumably by characterising them somehow. If the question had just asked us to find the maximum, and had not told us what it is, we would have to:

Decide on a value M, and show that there is a configuration with exactly M yellows.
Show that no value >M of yellows is possible.

Note that, apart from deciding on the value of M, which really has to come at the start of the argument, you can do these steps in either order. And if you want to find the number of configurations achieving the maximum, you have to:

Decide on a value M, and count the number of configurations with exactly M yellows, and check that this is positive!
Show that no value >M of yellows is possible.

Again, you can perform these steps in either order. If the question is posed like this (with M unknown), you have no idea as you are attempting the paper whether finding the value of M is supposed to be easy or supposed to be a really significant part of the attack on the problem, so starting your solution with a sentence like “I’m going to show that M is the maximum” and continuing by giving your proof an exoskeleton like “First, I’ll show that M+1 (or >M) is impossible. [Do this. Then leave a gap.] Now, I’ll count the number of configurations with M yellows.” This leaves no ambiguity that you are doing the problem in the right way, and for someone reading it, it might be the case that the validity of your argument is almost instantly clear once the rough outline is established.

In any case, on MOG Q4, the question has been split into two parts deliberately, and you have been told what the value of the maximum is. In fact, the order is the opposite to what I’ve proposed in the bullet points above. But this just serves to emphasise that, even when or especially when the maximum is already given, the order of attack is not important. The two bullet points might be completely unrelated mathematical arguments, but they are certainly structurally unrelated.

Structuring your approach to MOG Q4

But, to return to MOG Q4, the first part is asking you to show that 68 or more yellow houses is not possible under the rules. The second part is where you are supposed to establish what all the 67 yellow configurations are, and on this occasion it may turn out that you re-use ideas established in part a) during the enumeration of part b).

Anyway, I’m going to repeat an earlier paragraph. Many students wrote that because three houses in a row cannot be yellow, the way to get as many yellow houses as possible is to use a YYWYYW… pattern. Since 100 is not a multiple of three, you have to work out what to do with the final house, after you’ve coloured 99 using the pattern, and sometimes you can colour it yellow, and sometimes you can’t.

Now that we’ve established some general principles, we can see why some students may have got themselves into knots that weren’t directly relevant to the question.

The blue comment “sometimes you can colour it yellow, and sometimes you can’t” is just not relevant because at this stage we are trying to show that 68 (or more) is not possible. If at any point it becomes a discussion of whether one class of configurations leads to 66 or 67, this can’t be relevant to showing that 68 is not possible.
The green sentence “because three houses in a row cannot be yellow, the way to get as many yellow houses as possible is to use a YYWYYW… pattern” is, as written, true, but dangerous. Because it’s not precise. As becomes clear in part b), there are actually lots of ways to get as many yellow houses as possible, and while all of them have this pattern most of the time, it is certainly not the case that you use this pattern for the first 99, and then make a decision about the final house. (Since this only gives one configuration.)
The fact that 100 is not a multiple of 3 is crucial to the problem. If it had asked about 99 houses, you would have had more room just to work with a breakdown into 33 groups of three houses, and some of the argument would have been simpler. So an argument saying that 100/3 = 66.66…, which we can round up to 67 is not really explaining what’s going on. Of course we have to have a whole number of houses, but why you round up rather than down is all about the rules of colouring.
Finally, the notion of an ‘iterated maximum’ is a generally erroneous argument. In such an argument, you argue that the maximum way to attack a subproblem is X, and once you have X, clearly the maximal way to finish is Y, so the maximum total is X+Y. This is wrong, because of course it’s possible there might be a non-maximum way X*>X to attack the subproblem, from which you can finish in Y*<<Y, so the total ends up being lower overall.
In this example, the problem is not that the green statement is an erroneous iterated maximum. It’s just sounds like one. It can easily be made into a non-erroneous argument. But as soon as you start with a pattern, but then consider how to add something onto the end of it, you have essentially fallen into this trap unless you very carefully explain why taking the maximum for the first 99 houses plus the maximum for the final house must give a maximum overall.

Because the condition is a local one (meaning it tells you about rules governing two or three adjacent houses, not a global one affecting all houses in the street), it’s good to use a local argument. In the example:

(YYW)(YYW)…(YY*W)(YW*Y)…(YWY)(Y),

notice that it is not true that in every group of three consecutive houses, two are Yellow. (Look between the *asterisks*.) However, it is true that in every group of three consecutive houses, at most two are Yellow, and so in particular if we split as shown above, there are at most two yellows in all the three-brackets, and at most one at the end, giving at most 67 in total. The fact that the pattern changes from YYW to YWY does not affect this argument.

If you set it up this way, then this gives you a good headstart on part b). You might notice that the pattern can change from YYW to YWY (as in the previous example), but not the other way round. And what about WYY? Can this ever come up? You need to justify both aspects of this, and then clarify how to count the number of places where the pattern could change, having now established that all configurations have this form.

Other comments

The question setters produced a beautiful paper, and it was clear that one of their goals was to use multiple parts sensibly so as to suggest entrance routes into the problem. Many many students successfully used the factorisations discussed in the first parts of Q1, and even those who didn’t succeed seemed mostly to be trying to find differences of squares, rather than prime factors out of nowhere.

This principle applied elsewhere though, and in particular on Q3. I think some students thought Q3a) was supposed to involve three calculations, but in fact it required just giving a combinatorial justification for why the set of objects counted by b can be split into two, where one subset is counted by c and one by d. This argument, which might have been familiar to some through knowledge of Pascal’s triangle, applies equally well at other squares in the box, and also in three dimensions. If you don’t have the gap in the middle of the house, you can immediately apply this argument to Ghastly in part b). But in fact, even with the gap, you can make this work, because you are only really counting across *places you could have been just before in your path*, and the gap means that sometimes this is smaller than it would be without the gap.

There were other ways to do the question, such as counting all paths, and all paths through the gap, but all attempts using a b=c+d+e argument seemed quick and successful.

A couple of other tips to bear in mind on these sorts of questions.

When you’re deep in a calculation, it’s easy to forget that you started with a problem of some kind, and you should always go back to check that your answer makes sense in the context of the problem! Even if fits the algebra, you should not have any of the following (among many other examples that can emerge after a small conceptual or calculation error):
- A negative radius for a circle
- A non-integer number of configurations
- A probability >1
You should also check for potential symmetries. While it’s dangerous to use symmetry excessively until you’re sure the situation genuinely is symmetric, it can be your friend in reducing errors. For example, maybe you found the final count in the far corner for Ghastly the ghost to be eg 18+16+12? But the three adjacent-to-far-corner cells are clearly equivalent from the starting point. Any difference between them can only be a feature of the argument you are trying. So they should have the same path count, and probably you made a small error in one of them.
It’s also worth checking your answers feel roughly right for magnitude. In particular, again on Q3 , if you get a very large answer for part b), you should try to check against other options. How many total paths are there? (Ie, ignoring the gap.) If your answer is much bigger than this, you must have made a small mistake, probably multiplying where you should have been adding. It might well, for example, not feel right if you end up showing that the number of paths is greater than the number of atoms in the universe. This applies equally well for Q4. Note that 67! is an extremely large number, and is larger than 2^100, which is the total number of colourings without observing any of the consecutive rules, let alone the maximum condition.
As a final point, it was good to see that many candidates were making a bona fide effort to get as many marks as possible. I would say, however, that just writing down a speculative answer is often less helpful than a single sentence explaining what you are trying to do. Pretty much any comment on patterns and a division into cases for Q4 would be useful, even if it wasn’t completely correct, whereas a guess at eg $\binom{67}{33}$ or similar for the final answer is probably less valuable, and draws attention to what you haven’t done, rather than any progress that you really have made.

Anyway, I hope that was interesting at some level to some girls potentially interested in taking MOG or any other challenging mathematics in the future.

Balkan MO 2018 – UK Team Blog

Posted on May 15, 2018 by dominicyeo

The Balkan Mathematical Olympiad is a competition for secondary school students organised annually by eleven countries in Eastern Europe on a rotating basis. The 2018 edition was held near Belgrade, Serbia over the period 7-12 May 2018. The UK was grateful to be invited as a guest nation.

Our participation is arranged by the UK Maths Trust, as part of a broader programme to introduce the country’s most enthusiastic young mathematicians to regular problem-solving, challenging mathematics, and several annual opportunities to participate in competitions. For the Balkan MO, we have a self-imposed rule that students may attend at most once, so that as many as possible might enjoy the experience of an international competition.

The non-geometry problems of the contest are discussed at length in this blog post, and the geometry problem which appeared as Q1 is discussed at considerable length, along with some background on harmonic ranges, in this blog post. A full report encapsulating all these aspects, is available here.

This post covers the non-mathematical aspects of the contest, which was enjoyed by all the UK students.

Problem selection

The programme of this competition is a scaled down version of the IMO. The leaders gather in suburban Belgrade on Monday night to select four problems from a shortlist compiled by the organisers. To recreate the students’ experience, it makes sense to start by trying these without reference to solutions. Some of the questions are UK submissions, so I can briefly astonish my colleague Vesna with almost instant fluency, before admitting that I wrote or edited the corresponding solutions.

Making the choice occupies Tuesday morning. As always, it feels slightly like a shot in the dark, as one night is not really sufficient to get a feeling for twenty problems, especially the hardest ones. In the end, there was clearly a unique good hard problem, but unfortunately it had to be rejected because it was too similar to a recent problem from a well-known source. Some of us have been investing considerable energy in finding natural Euclidean arguments to the geometry problem chosen as Q3, but once Greek leader Silouanos outlines the role of harmonic ranges, it is hurriedly moved to Q1. I think the resulting set of four questions are attractive, but with a rather compressed difficulty range, and certainly not in the right order for the UK students, whose geometric toolkits probably don’t yet include the ideas needed to access the `easy’ solutions.

In any case, it’s interesting to discuss with the leaders from some of the eleven Balkan full member countries. Our opinions differ concerning which styles of problem give an advantage to extensively-trained problems. I personally feel that Q2 and Q3 are accessible even to students (or adults!) without much mathematical background, whereas here is a prevailing view that no problem with combinatorial flavour is ever ‘easy’. By contrast, many of the ideas required for a short solution to either Q1 or Q4 might be considered obscure even by serious olympiad enthusiasts, though feature on the school curriculum, at least for the most able children, in many of these countries.

We have to finalise the wording of the problems, and there are many many proposed improvements to Q2 and Q3. The final problem, unsurprisingly, requires considerably less attention. That’s our job done for the British delegation, while the other leaders get to work producing versions in their own languages, including Bosnian and Serbian, the (non-)differences between which can happily fill one dinner’s worth of interesting conversation.

The contest

On Wednesday morning, we are transferred to the contestant site, in the rolling hills just outside the south-east city limits of Belgrade. An extremely brief opening ceremony takes place in a room slightly smaller than the number of people attending the competition. The UK team look happy enough perched on a table. Two local violinists play Mozart with a gypsy flourish, before Teodor von Burg, a former Serbian olympiad star and graduate of Exeter College, Oxford, speaks briefly about the usual cliches of such speeches, and the additive paradox of wishing everyone good luck before a competition, then ends rapidly to avoid indulging such cliches himself.

After the contestants fan out to various exam rooms spread through the hotel, the contest begins and they are allowed to ask queries about the problems for 30 minutes. Many many students ask ‘what does exactly the same route mean?’ and ‘what if Alice and Bob play forever?’, but some variety is provided when Aron shares his detailed dilemma about the exact usage of carbon paper. (FAO future UK students: this is not to become a habit, please…)

After Monday’s 2am start, I am overdue a nap. There has been some room-swapping, and mine is reserved for ‘Professor Mr Jill Parker’. Whomever the bed truly belongs to, I leave it in time to meet the team outside the exam with Jill and Vesna. As we’d predicted, many are enthusiastic about Q2 and Q3, but have been frustrated by the geometry. Tom crowd-sources an investigation to recover a result about the incentre claimed by Alex, who perhaps now regrets, in his rush to move to other questions, not offering more of such details himself. No-one claims anything beyond observations in the number theory, so we suggest they keep thinking about it through the afternoon.

A brief excursion

Agnijo and Nathan had done their research on Belgrade, and had asked about the possibility of visiting the Nikola Tesla museum. The team have a guide, Sandra, a maths undergraduate, and I’m extremely impressed that she and some of her colleagues are able to organise a visit downtown and guided tour of this museum at essentially no notice for them, along with Italy, Bosnia and Azerbaijan. Vesna and I diverge to make a start on marking in a cafe, rejoining in time for the museum, where Giles apparently learns what ‘Azerbaijan’ is, and we all learn about Tesla’s extraordinary life story, and get to see the original Tesla coil (briefly) in action. Agnijo and Tom have been primed with fluorescent tubes, which do indeed glow as lightning surges between the century-old coil and its crowning sphere. Other exhibits, including highlights from Tesla’s wardrobe (pre-dating \emph{geek chic}, it would seem), and an imitation ticket from Belgrade to New York, are perhaps less fascinating.

But the roar of $10^6$ Volts is still in our ears as we stroll across the city centre, where Alex confidently identifies several churches as the orthodox cathedral they’d visited earlier, and eyes are drawn to the faded but strident protest banners outside the parliament. We choose a restaurant in bohemian Skadarska street, where prices are low, and availability of protein and itinerant accordion players is high. The team are trying to be polite about their hotel’s food, but I sense this variation is welcome. Giles pokes gingerly at a deep-fried pork slab, which erupts with multiple cheeses. The ‘Serbian sword‘ could be retitled ‘as many meat items on a stick as possible (plus 1/8 of a pepper)’.

We return to Avala feeling sleepily satisfied. Tom and Agnijo discuss the GCSE question ‘prove using algebra that the product of two odd numbers is odd’, and whether you can or should prove it without algebra. The taxis clearly sense our post-prandial vulnerability, and operate a creative attitude to receipts, and to powers of ten. But this round of ambiguous paperwork and mathematical corrections is just the prelude for Vesna and myself, who have a cosy night in with the scripts.

Coordination

At a competition, the leaders of each team study their own students’ work, and agree an appropriate mark with a team of local coordinators. The UK has an easier workload: we do not have to provide translations, since our students write in English, though some of them might like to note that in a question about parity, mixing up the words `odd’ and `even’ as if flipping a coin does make it harder to convince the reader you know what you’re talking about.

We start with 9am geometry, where the coordinators are proposing giving Aron 8 or 9 out of 10 as part of a crusade against citing configurational properties as ‘well-known’. Aron has, in fact, outlined a proof of his (fairly) well-known fact, and if the proposal is to award 6 or 7 without this, then the marking team’s entire day is guaranteed to be a continuous series of wars. I think the penny drops shortly after our meeting, and Aron gets upgraded to 10/10 at 9.30. Unfortunately, what remains of the crusade will deny Alex any credit at all for his unjustified claim about the incentre, despite its role in an appealing synthetic solution.

The middle two questions have a wide range of arguments. The British work on Q3 is actually pretty good, and even in the two scripts with small corners missing have organised their cases very clearly, and the coordinators (who initially want to give all full marks) can see that the students already had the ingredients to fix their minor errors. Q2 is more challenging. Once we have worked out where the good bit begins, Nathan’s solution is clearly superb, and once we’ve worked out which of his mysterious side-comments to ignore, Giles has all but the final details of a really imaginative solution, and Agnijo is flawless. Aron seems keen to make an even number of really confusing mistakes on this paper, so on this question has mixed up ‘horizontal’ and ‘vertical’ as if flipping a coin, though the coordinators are more sympathetic than I would have been. Tom claims that his solution is ‘very poorly written’, which is very far from the case, but after rolling back and forth through his logic a few times, we agree that a couple of cases of q are inadvertently missing.

The students return from their short excursion in time to hear their scores before dinner, and though Alex is a bit disappointed about the non-acceptance of his ‘lemma’, everyone is broadly pleased with themselves, as they should be. I get my first experience of the infamous hotel salad, which the students had previously described as ‘vinegar topped with lettuce’, which is roughly accurate, though the rest is nice enough. Agnijo is worried the main course includes beef, but is satisfied with the supposedly vegan alternative, namely a grilled fish.

The Balkan countries take the table of scores a bit more seriously than we do, and so this year’s celebratory table is sipping Bulgarian cognac washed down with Romanian tears, though this wholesome rivalry shouldn’t distract from the hugely impressive seven perfect scores from those countries’ contestants (plus four from the others). The competition at the adjacent table seems to be the relative merits of Serbian, Macedonian and Montenegrin wine and rakija. Meanwhile, the UK students have made plenty of new friends to induct into their favourite card games, and some Albanians, Bosnians and Turks seem a) very keen to practise their excellent English, and b) appropriately baffled by the rules, and lack of rules.

Round and about

The bulk of Friday is set aside for an excursion. Our destination is Valjevo, a town two hours’ drive west of Belgrade, which represents some sort of historical home for the Serbian maths enrichment community. We gather in their gloriously rococo hall to listen to an in-depth presentation concerning many aspects of daily life at Valjevo Grammar School. The nearby research institute in leafy Petnica offers a more science-focused perspective. The students get to tour some labs, though they don’t get to practise for their upcoming A-level or Highers physics by trying any experiments. Nathan, however, finalises his solution to Q4 from the contest, which seems a good use of time, and which you can read earlier in this report. Aron asks me to solve what seems a challenging geometry question in my head. I cannot. A stamp-sized freehand diagram on a napkin doesn’t help either.

Vesna was a regular visitor to Petnica as a teenage olympiad contestant, and she has briefed me on the charms of a nearby cave, apparently a regular choice for planned and unplanned excursions during her selection camps in the 90s. The UK group plans to sneak away from the third phase of the tour to find this cave, but we are foiled because the third phase of the tour is indeed a visit to the cave. This involves a short walk, during which Agnijo is harassed by the world’s least threatening dog. The temperature is pushing 30C, but Aron is worried about sunburn, so is reluctant to remove his polar fleece. He gently roasts, while Alex tells us some horror stories from his experience as a Wimbledon ballboy during the 2016 heatwave. The cave provides cool relief, and is indeed giant, with plenty of sub-caves underneath the looming stalactites.

It turns out we are in the less impressive half of the cave. The students want to climb to the more impressive upper cave. It may be more impressive, but it is also considerably darker, and I admire Giles’ and Nathan’s tenacity to find out exactly how far a distant rocky staircase extends into the gloom temporarily illuminated by a phone torch. That concludes the adventure, and we return to Belgrade coated in varying quantities of cave detritus. The return journey affords great views of the distant mountains towards the Bosnian and Montenegrin borders, though Tom is keen to use the time to make a start on coordinating the multi-author student report. Unable to avoid eavesdropping on the discussion, sounds like it will be a substantial document when completed…

Finishing up

Back in Avala, the closing ceremony takes place during dinner, and is informal. Jury chair Zoran Kadelburg awards the certificates; chief organiser Miljan presents the medals; and Miljan’s wife notices and steps into the essential role of helping the medallists flip their newly-acquired prizes in front of any flags they might be carrying for the waiting photographers. This one-at-a-time low-key arrangement was actually very nice for everyone, and our four medallists enjoyed their moments.

It is a balmy evening, so we drift outside again. Aron is random-walking, hunting for the WiFi sweetspot so he can download the punchline to our colleague Sam’s claimed complex solution to Q1 before Nathan finishes rounding up new players for the next round of card games; while Giles and Alex disappear off towards the most distant unlit car park with a troupe of guides and Bosnians and a volleyball. At the leaders’ table, Vesna and the other Balkan residents give a collective hollow laugh on hearing that I have elected to travel to the Montenegrin Alps by bus. But that ten hour experience starts tomorrow, outside the remit of this report, which will end here, with some pictures of mountains.

Balkan MO 2018 – Questions 2,3,4

Posted on May 15, 2018 by dominicyeo

The Balkan Mathematical Olympiad is a competition for secondary school students organised annually by eleven countries in Eastern Europe on a rotating basis. The 2018 edition was held near Belgrade, Serbia over 7-12 May 2018. The UK was grateful to be invited as a guest nation.

A full account of the competition can be found in this report and this blog report. This blog post concerns the three non-geometry questions on the paper. For typical able British schoolchildren, the order of the questions might not be ideal, and so this post talks about the problems in a non-standard order. A post about the geometry Q1, and a discussion of some of the background, including an introduction to harmonic ranges may be found here.

Problem Three

Alice and Bob play the following game. They start with two non-empty piles of coins. Taking turns, with Alice playing first, each player chooses a pile with an even number of coins and moves half of the coins of this pile to the other pile. The game ends if a player cannot move, in which case the other player wins.

Determine all pairs (a,b) of positive integers such that if initially the two piles have a and b coins, respectively, then Bob has a winning strategy.

Clearly, the game ends when both piles are odd. If one pile a is odd, and the other b is even, then only one move is possible, namely ending up a+b/2 and b/2. It’s not possible that both of these are odd, so further analysis would be required. However, we might notice from this that if a is even, and b is 2 modulo 4, then there are two possible moves, but at least one of them ends up with both piles now being odd.

So when the official solution starts with the sentence ‘let $v_2(a)$ be the exponent of the largest power of two dividing (this is often called a valuation, and is a useful property to consider in many contexts.) a‘, this is not magic, but a natural response to a preliminary investigation along the lines of the previous paragraph.

One should then consider some cases. It is clear that Bob wins if (a,b) are both odd, that is $v_2(a)=v_2(b)=0$ , and in our preliminary exploration we established that Alice wins if $a\equiv b\equiv 2$ modulo 4, that is $v_2(a)=v_2(b)=1$ . It’s not too hard to establish from here that if $v_2(a)=v_2(b)$ , then Bob wins iff this common valuation is even, and Alice wins when it’s odd. It’s also worth noting that this holds irrespective of the players’ choices of moves.

To finish the problem, we now have to classify the remaining cases, and prove what happens in these cases. From the final preliminary observation, we know that Alice wins if $v_2(a)=1$ and $v_2(b)\ge 1$ , but it seems like the game might go on for ever if both players aim to avoid losing when starting from $v_2(a)=0$ and $v_2(b)\ge 1$ . One can try some more small examples, or move straight to a conjecture, but the parity (That is, whether a number is odd or even) of $\min(v_2(a),v_2(b))$ determines the outcome. In neither case does Bob win, but Alice wins when this minimal valuation is odd, and the game continues forever if it’s even, and if you haven’t already, you should try proving this by considering how the valuations could change on any move.

As a slight alternative, especially once you know the answer and have observed that the outcome is invariant under multiplying both a,b by four (and so $v_2(a)\mapsto v_2(a)+2$ ), one could attempt the following argument. Introduce the notation $(a_t,b_t)$ for the pile sizes at time $t\ge 0$ , so $(a,b)=(a_0,b_0)$ . We know the outcomes in all cases where $\min(v_2(a),v_2(b))\le 2$ . So if we start the original game G from a pair (a,b) satisfying $\min(v_2(a),v_2(b))\ge 2$ , we could consider an alternative game G’ whose rule for winning instead says that we wait for the first time t such that Alice is to move and $\min(v_2(a_t),v_2(b_t))\le 2$ . Then we declare the winner (or non-winner) to be the outcome of the original game G started from $(a_t,b_t)$ . While the outcome profile is obviously the same as the original game G, we can claim that playing G from (a,b) is the same as playing G’ from (4a,4b), and thus derive the entire outcome profile by induction.

The details required to establish this claim are easy but numerous, and certainly need to be present in a full solution, which explains Alex’s unfortunate mark for this problem despite having this sophisticated and workable idea. Finishing the details would be an excellent exercise for anyone aiming to tighten up their combinatorial clarity at this level.

Problem Two

Let q be a positive rational number. Two ants are initially at the same point X in the plane. In the n-th minute (n=1,2, … ) each of them chooses whether to walk due north, east, south, or west, and then walks $q^n$ metres in this direction. After a whole number of minutes, they are at the same point in the plane (not necessarily X), but have not taken exactly the same route within that time. Determine all possible values of q.

The answer is that only q=1 is possible, and the majority of approaches will eliminate all but a finite number of potential values first, then study the cases q=2 and q=1/2 separately. Even though it might seem obvious, remember that you have to provide an example for q=1 too!

This is really a question about polynomials, where the variable is q. So for example, if ant A follows the path NNESWN, then its coordinates after the sixth minute are

$(x^6_A,y^6_A)=\left(q^3-q^5,\, q+q^2-q^4+q^6\right).$

So if we want to prove it’s impossible for $(x^n_A,y^n_A)=(x^n_B,y^n_B)$ for some different length-n paths, we could first focus on just one coordinate, say the x-coordinate. But note that $x_A^n-x_B^n$ is a polynomial in q with degree at most n, where all the coefficients are {-2,-1,0,1,2}. So if the ants are in the same place at time n, then q is a root of this polynomial.

Insisting on converting q into $\frac{a}{b}$ at an early stage is a sort of intellectual comfort blanket that’s probably going to distract from the main insight. But at this stage, we do need to introduce this, and argue that if $q=\frac{a}{b}$ in lowest terms, then q cannot be a root of such a polynomial if either a or b is at least 3. Proving this yourself is definitely a worthwhile exercise. Remember to use that a and b are coprime! (With an additional idea, you can reduce instead to a polynomial with coefficients in {-1,0,+1}, from which you can finish even faster.)

To reduce the number of cases left, we can show that there are examples for q iff there are examples for 1/q, arguing either via the polynomial description (much easier with q rather than $\frac{a}{b}$ again here), or more combinatorially in terms of reversed ant paths.

To finish the problem we have to eliminate one of the possibilities q=1/2 and q=2 (as one follows from the other by the previous paragraph). For q=2, we should study the first time at which the ants diverge, but life will be easier if we argue that we may assume that this happens on the first step. Now, we study the first couple of moves.

If one ant moves horizontally and the other moves vertically on the first move, then what can you say about the parity of each ant’s coordinates after the first step, and indeed after all future steps? This will show that they cannot ever meet.
Otherwise, assume that both ants move horizontally, one East, one West. Since we can’t use parity, but powers of two are deeply involved, it makes sense to consider using congruence modulo 4. Indeed, after this first step, the ants’ x-coordinates are not congruent modulo 4 (since one is 1 and the other is -1).
- If they both move vertically on the second step, or both move horizontally on the second step, this remains the case. (One should check both options for the horizontal case.) Thereafter, all moves have length divisible by 4, and so this property holds forever, and so the ants do not meet.
- If one moves horizontally, and one moves vertically on the second step, what can you say about the ants’ y-coordinates modulo something relevant?

If you want to study q=1/2 instead, you might observe by trying some examples that if the ants head off in different directions, there is a real sense that they become too far apart to get back together using the future allowed moves. This motivates considering some sort of distance argument. The interplay between the coordinates is not really suited to standard Euclidean distance, since the ants can’t walk in a diagonal direction (which is what will mostly determine the Euclidean distance). Instead, it’s worth studying $d_n(A,B):=|x^n_A - x^n_B| + |y^n_A-y^n_B|$ (which is sometimes called $\ell_1$ -distance, or Manhattan distance or taxicab distance.) What is $d_1(A,B)$ , and can you control $d_n(A,B)-d_{n-1}(A,B)$ strongly enough to show that $d_n(A,B)$ is always strictly positive? If you can, perhaps you can draw an analogy with the argument for q=2 as a final insight into the workings of this interesting question?

Problem Four

Find all primes p and q such that $3p^{q-1}+1$ divides $11^p+17^p$ .

None of the UK students solved this problem during the competition, but several managed it during some free time the following morning. Nathan’s solution, lightly paraphrased, will follow shortly.

In a question like this, you don’t know how many of the details will be crucial. Is the choice of {3,11,17} going to be important? How will we use the fact that q is prime? You probably can’t answer these meta-questions immediately. It also looks like standard motifs of subtracting multiples of $3p^{q-1}+1$ from $11^p+17^p$ is not going to make life easier. Nathan’s approach is to study the possible factors of $11^p+17^p$ , focusing on prime power factors. Once he has a rich enough understanding of potential such factors, he can then study whether they combine to form $3p^{q-1}+1$ , which turns out to be very restrictive, leaving only a handful of cases to eliminate by hand.

Nathan writes: We can quickly eliminate the possibility that p=2, and so now assume we have a solution where p is odd.

Claim I: None of 8, 49 or 11 divide $3p^{q-1}+1$ .

Proof: It’s enough to show that they do not divide $11^p+7^p$ . The non-divisibility of 11 is clear. For 8, note that $11^p\equiv 1,3$ and $17^p\equiv 1$ modulo 8, and so $11^p+17^p\equiv 2,4 \not\equiv 0$ .

To handle 49, we rewrite $11^p+17^p$ as $11^p-(-17)^p$ and we have the valuation $v_7(11 - (-17))=v_7(28)=1$ . So when we lift the exponent (see later), we find

$v_7\left(11^p-(-17)^p\right) = 1+v_7(p).$

So if $49\mid 11^p+17^p$ , then the LHS is at least two, and so $v_7(p)\ge 1$ . But then p=7 is the only option, for which certainly $49\nmid 3p^{q-1}+1$ . The claim is now proved.

So we may now write

$3p^{q-1}+1 = 2^a7^b \prod r_i^{e_i},$ (1)

where $r_i$ are primes not equal to {2,7,11}, and $a\in\{1,2\},\, b\in\{0,1\}$ .

Claim II: each $r_i\equiv 1$ modulo p.

Proof: As before $r_i\mid 11^p-(-17)^p$ , and since $r_i\ne 11$ , 11 has a multiplicative inverse modulo $r_i$ , and so indeed there exists t such that $11t \equiv -17$ modulo $r_i$ . Using this in the divisibility relation:

$r_i\mid 11^p-(-17)^p \equiv 11^p - (11t)t^p \equiv 11^p(1-t^p) \quad\iff\quad r_i\mid 1-t^p.$

The order of t modulo $r_i$ then divides p, so is either 1 or p. If this order is 1, then $t\equiv 1$ , but then, modulo $r_i$ , $11\equiv -17$ , so $r_i\mid 28$ , which we have excluded already. So the order is p, and thus $p\mid r_i-1$ , as we claimed.

Going back to (1), we have

$1\equiv 3p^{q-1}+1 = 2^a7^b\prod r_i^{e_i} \equiv 2^a 7^b\quad \mod p,$

and so $p\mid 2^a7^b-1$ . But remember that $a\in\{1,2\}$ and $b\in\{0,1\}$ , so there are only a handful of cases to check. Each of the other cases requires a line or two to eliminate, so do try this yourself! In the end, though, we see that (a,b)=(2,0) or (2,1), both leading to p=3 are the only possibilities. Returning to the original question, we just have to check possible solutions to $3^q+1 \mid 11^3+17^3=2^2\cdot 7\cdot 223$ , which we can do manually (for example by checking all prime $q\le 7$ ), to find that the only solution is (p,q)=(3,3).

Dominic: As part of this solution, Nathan uses the lifting the exponent lemma to control $v_7(11^p-(-17)^p)$ . This example is simple enough that it’s probably easiest to go directly. Since p is odd, we can factorise

$11^p+17^p= 28\cdot \left( 11^{p-1}- 11^{p-2}\cdot 17 + 11^{p-3}\cdot 17^2-\ldots + 17^{p-1}\right).$

Can you come up with an argument for why 7 cannot divide the second factor? Some of the notation Nathan used elsewhere in his solution may be useful! If you can, then you’ve shown that $v_7(11^p+17^p)=1$ .

The general statement of the lemma relates $v_p(x^n-y^n)$ to $v_p(n)$ and $v_p(x-y)$ , which explains why Nathan converts $+17^p$ to $-(-17)^p$ , though it makes little difference to the proof. You can find statements of this lemma, which has become relatively well-known recently in this community (and which is sometimes attributed to Mihai Manea), in many places on the internet and in modern books. The proof is very similar in the general case to the special case discussed previously. It’s worth remembering that the case p=2 always requires extra care (and indeed a different statement). This distinction comes from the fact that the simultaneous congruence equations $x+y\equiv 0$ and $x-y\equiv 0$ modulo $n$ have two pairs of solutions when $2\mid n$ .

It’s worth noting also that in a solution like Nathan’s where different ranges of options are excluded one after the other, this clear organisation into claims is of huge benefit to the reader, irrespective of how much text is or isn’t included as a prelude.

EGMO 2018

Posted on April 18, 2018 by dominicyeo

Last week the UK held its annual training and selection camp in Cambridge. This week, four of the students have been attending the European Girls’ Mathematical Olympiad. 2018 is the seventh edition of this prestigious competition, and is being held in Florence.

You can find very many details about the competition, and observe the UK’s excellent performance (with particular congratulations to Emily, who obtained a perfect score) at the competition website. A short article about the team in the TES can be found here.

In this post, I’m going to muse on some of the problems. You can find the two papers here and here.

Problem Two

Part a) would have been more immediate if the set A had been defined as

$A:= \left\{\frac{k+1}{k} \,:\, k=1,2,3,\ldots\right\},$

as this is instantly suggestive of a telescoping product such as

$7 = \frac{7}{6}\cdot \frac{6}{5}\cdot\ldots \cdot \frac{2}{1}.$

I found part b) to be one of the most difficult sections of the paper. It builds on the idea that given an expression for x as a product of elements of A, and an expression for y as a product of elements of A, we can combine these (ie take the product of the products!) to get an expression for xy as a product of elements of A. This yields $f(xy)\le f(x)+f(y)$ , and so the task is to show that sometimes this isn’t the most efficient way to proceed.

I tried a couple of things, like trying to bound f(p) below when p is a prime. This wasn’t ludicrous, as one would certainly need to use a term $\frac{kp}{kp-1}$ somewhere in the product so that it is divisible by p. However, this didn’t go anywhere, and nor did investigating f(n!). Perhaps others had more success down these avenues.

But as a general rule, if an abstractly-defined function is typically hard to calculate, then classes where you can calculate it are likely to be extra valuable. Here, powers of two make life particularly easy. We have $2\in A$ , and so $2^n=2\times 2\times\ldots\times 2$ is a valid product. And we can’t possibly achieve $2^n$ as a product of fewer terms than this, because 2 is the largest element of A. So $f(2^n)=n$ . Note that this is already way better than the bound we would have achieved from our argument in part a), which yields $f(m)\le m-1$ .

My next observation was that a similar argument and a natural construction gives $f(2^n+1)=n+1$ . But this can be extended so that when $2^n+1\le m\le 2^{n+1}$ , we have $f(m)\ge n+1$ and in fact there is equality exactly for

$m= 2^n+1, 2^n+2, 2^n+4,\ldots, 2^n+2^{n-1},2^{n+1}.$ (*)

In particular, note that all of these are even except $2^n+1$ . It may well be the case that we don’t need this extension, but before you’ve solved the problem you don’t know how much you’ll have to extend each idea!

I had a feeling that this meant $f(2^n+1)$ was a strong candidate to satisfy the requirements, and that almost any factorisation would work. I suspect many students at this point did some refinement of choice of n, but I want to stay abstract, and use the extended observation (*). Since $2^n+1$ is certainly not always prime, let’s focus on the infinitely many values n where it has a factorisation as $2^n+1 = ab$ , and consider whether a or b can achieve equality at (*). We’d better introduce the notation

$2^\alpha<a<2^{\alpha+1},\quad 2^\beta<b<2^{\beta+1}.$

So $ab> 2^{ab}+2^a+2^b+1$ , and so $\alpha+\beta>n$ . But similarly, $ab< 2^{\alpha+1}2^{\beta+1}$ , so $\alpha+\beta<n+2$ . We obtain

$\alpha+\beta+1=n,$

which is kind of what I’d been hoping for when I started this analysis. Now, we have

$f(a)\ge \alpha+1,\quad f(b)\ge \beta+1,$

$\Rightarrow\quad f(a)+f(b)\ge \alpha+\beta+2 = n+1,$ (**)

with equality precisely if $a,b$ both satisfy the equality conditions at (*). But $a,b$ are odd, and so we have equality at (**) precisely if $a=2^\alpha+1,b=2^\beta+1$ . So we have a resolution to the problem whenever $2^n+1$ can be non-trivially factorised in any way other than $2^n+1=(2^\alpha+1)(2^\beta+1)$ , so we have a rich (and certainly infinite) class of suitable (x,y).

Problem Three

An obvious remark. The jury will never choose contestant i if she has fewer than i contestants in front of her unless they are forced to. They are only forced to if everyone has this property. So we ignore the second dashed bullet point, as it just tells us when the process ends. And with a little further thought, the process ends exactly when the contestants are in correct order.

I suspect part a) of this may end up featuring on future examples of our interactive write-up clinic, where students are challenged to produce technically-correct arguments for obvious but awkward mini-problems. The location of contestant $C_n$ is probably a good starting point.

For part b), you have to find an optimal construction, and prove that it’s optimal. At national and junior olympiad level, students often forget that they have to supply both of these components. At international level, the challenge is to work out which one gives the entry into the problem. I would say that about two-thirds of the time, either the optimal construction is very obvious, or is best attacked after you’ve had some insight into the bound. For this question (and again it’s just my opinion), I felt it was all about the construction. I made absolutely no progress by aiming for bounds. Whereas the construction offers plenty of insight into how to prove the bounds, and so once I had it, I found it quick.

The usual rules apply. An optimal construction is going to have to be neat to describe. It’s very unlikely to have millions of cases. Intuitively, it seems reasonable that starting the contestants in reverse order gives the jury the greatest possible ‘elbow room’ to squeeze moves into the procedure. Perhaps you tried to prove this directly, by coupling a procedure starting from arbitrary order with a corresponding procedure starting from reverse order? Well, I found that very hard, and perhaps you did too.

However, that doesn’t mean it’s the wrong construction! The key question is, what to do about contestant $C_n$ ? Well, essentially nothing. This contestant can never be moved. So when do we start allowing other contestants to pass her? It seemed natural to let the other contestants $C_1,\ldots,C_{n-1}$ do as much as possible among themselves first. That is

$\mathbf{C_n},C_{n-1},\ldots,C_2,C_1 \quad \Rightarrow\Rightarrow\Rightarrow \quad \mathbf{C_n}, C_1,C_2,\ldots,C_{n-1},$

where $\Rightarrow\Rightarrow\Rightarrow$ denotes lots of moves. At this point, what to do next stood out for me, namely that one could use $\mathbf{C_n}$ at the start to put all the others back into reverse order, while moving $\mathbf{C_n}$ to the back. That is

$\mathbf{C_n},C_1,C_2,\ldots,C_{n-1}\quad\Rightarrow \quad C_1,\mathbf{C_n},C_2,\ldots,C_{n-1} \quad\Rightarrow\quad C_2,C_1,\mathbf{C_n},C_3,\ldots,C_{n-1}$

$\Rightarrow\Rightarrow \quad C_{n-1},C_{n-2},\ldots,C_2,C_1,\mathbf{C_n}.$

You might have tried other things first, but once you notice this, you just know it has to be right. It’s just too elegant a construction, and it looks like the sort of thing one prove will be optimal, because the overall process

$\mathbf{C_n},C_{n-1},\ldots,C_n\quad \Rightarrow\Rightarrow\Rightarrow \quad \mathbf{C_n},C_1,C_2,\ldots,C_{n-1}$

$\Rightarrow\Rightarrow\quad C_{n-1},\ldots,C_2,C_1,\mathbf{C_n}\quad\Rightarrow\Rightarrow\Rightarrow\quad C_1,C_2,\ldots,C_{n-1},\mathbf{C_n},$

incorporates the corresponding process for n-1 (twice, in fact) and so induction is very accessible both for calculating the total number of moves. We conjecture that this is indeed the optimal process, and under this assumption, with $f(n)$ the number of moves, we would have $f(n) = f(n-1) + (n-1) + f(n-1)$ , from the three stages of the process, from which, after looking at small values,

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

I started by saying that the construction was the hard part of this problem. Well, that doesn’t mean the bound is easy. But at least with a construction in hand, you can make observations that might inform a bounding argument:

observation 1: contestant $C_n$ never jumps;
observation 2: in the optimal construction, by induction $C_{n-1}$ doesn’t jump in the outer phases, so in fact jumps only once, ie during the middle phase;
observation 3: contestant $C_{n-2}$ doesn’t jump very often, and in fact can only jump after at least one of $C_{n-1}$ and $C_n$ have ended up in front of her. Since we’ve established that $C_{n-1},C_n$ don’t jump very often themselves, this gives a bound on the number of times $C_{n-2}$ .

There is still work to do here, and errors with $\pm 1$ could easily creep in. But I still hold fast to my original claim that the construction was the hard part here. Or at least, the rough form of the construction. I guess it’s possible that one would have started making observations like the three above without a construction in mind, but I think it’s unlikely. Anyway, I leave for you the final details of the bounding argument, which involves formally transcribing observation 3, proving it, then generalising it to jumps of $C_{n-3}$ and so on.

Problem Four

One of the exercises I have been setting to UK students recently is to produce short solution digests, which manifest any key ideas of the solution abstractly and briefly enough to resonate in the future. I’m a little tired of domino tiling problems, so I’ll do one of these here. This will be slightly longer than if I were not writing for a (small) audience.

Double-counting the total value by rows/columns and by dominos shows there are $\frac{2kn}{3}$ dominos in a balanced configuration. When n=3, we can achieve k=1, and by tiling copies of this down the main diagonal, can extend to $3\,|\,n$ . For $3\not|\,n$ , we must have $3\,|\,k$ ie $k\ge 3$ , but in fact k=3 is achievable, by tiling the main diagonal with copies of small boards for which k=3 can be constructed with a bit of trial-and-error.

The double-counting idea at the start is the nice part of the problem. The construction is a bit annoying, but saving ourselves work by building up large examples from copies of small examples is a useful motif to have in mind.

Problem Six

This question has lots of clues in the statement. It would, for example, be pretty surprising if the answer were ‘no’ to part b) given the setup in part a).

My confession is that I wasted lots of time on part a) not using the option m=0, which was foolish given that it’s clued from part b) that one needs to use the option m=0. My thought had been to consider some integer y, and ask which integers x were banned (if we were aiming for contradiction in part a)). For part a), it gets harder if t is smaller, so I found it helpful to think of t as $\epsilon\ll 1$ . Anyway, if you struggled on part a), maybe worth reviewing whether you were definitely trying to solve part a), and not accidentally using the setup that really addressed part b)!

Some people have shown me solutions to part a) that carry an air of magic, by placing all the key steps (such as (*) to follow) in the language of the original setup. Let’s try to be cleaner. The key is to consider m=0. Since m=0 is included, we know that whenever x<y, we must have

$\epsilon y \le x \le (1-\epsilon)y.$ (*)

Maybe you just have a gut feeling that this can’t be possible if you have enough xs and ys? But either way, choosing to focus on (*) is the key step, because once you know you have to prove the result based on this, it’s not too hard. I prefer addition to multiplication, so one might as well take logs (since does it really look like we’re going to use heavily the integer property now?) to obtain

$\alpha\le |z_i - z_j|\le \beta,$

for all $z_i,z_j$ in some large finite collection, where $0<\alpha<\beta$ . You should now have a strong gut feeling that this is impossible. You have an arbitrarily large collection of real numbers which have to be close to each other pairwise, but never too close pairwise. How to finish the argument is a matter of taste.

For part b), assuming we’re aiming for the answer ‘yes’, we probably want to construct it one step at a time, and you want to think of $t\approx \frac12$ to make life as hard as possible.

Now, suppose you have $x_1,x_2,\ldots,x_n$ so far. What next? Well if we could have

$x_{n+1} \equiv \frac{x_i}{2}\,\mod x_i,$

for all $i=1,\ldots,n$ , that would be perfect. We can often solve sets of coupled residue equations like this using the Chinese Remainder Theorem. (Recall of course that the solutions themselves are rarely important – the fact that they exist is enough!) A couple of things go wrong with trying this crudely:

If $x_i$ is odd, then $\frac{x_i}{2}$ is not an integer…
If we correct this by asking for $x_{n+1}\equiv \lfloor\frac{x_i}{2}\rfloor\,\mod x_i$ , then there’s a chance we might still be within the t-window around a multiple of $x_i$ .
Unless we are going to make complicated demands on the residues, to use CRT it would be easier if all the $x_i$ s were coprime.

One option is to give up. But actually all these objections can be handled with fairly small alterations. Can you see how the second objection can be overcome by an appropriate choice of $x_1$ ? Remember that t is fixed right at the start, and cannot be equal to 1/2. Is the third objection actually an objection any more? If it is, can we fix it?

Anyway, I guess P6 was my favourite non-geometry question on the paper, though, that’s far from relevant. P5 was pretty neat too, but who knows whether a follow-up geometry post will materialise soon.

Balkan MO 2017 – Qs 1, 3 and 4

Posted on May 5, 2017 by dominicyeo

The UK is normally invited to participate as a guest team at the Balkan Mathematical Olympiad, an annual competition between eleven countries from South-Eastern Europe. I got to take part in Rhodes almost exactly ten years ago, and this year the competition was held in Ohrid, in Macedonia. There’s one paper, comprising four questions, normally one from each of the agreed olympiad topic areas, with 4.5 hours for students to address them. The contest was sat this morning, and I’m going to say quite a bit about the geometric Q2, and a little bit about Qs 1 and 3 also. In all cases, this discussion will include most of a solution, with some commentary, so don’t read these if you are planning to try the problems yourself.

I’m not saying anything about Q4, because I haven’t solved it. (Edit: I have solved it now, so will postpone Q2 until later today.)

Question One

Find all ordered pairs of positive integers (x,y) such that

$x^3+y^3=x^2+42xy+y^2.$

The first thought is that if either of x or y is ‘large’, then the LHS is bigger than the RHS, and so equality can’t hold. That is, there are only finitely many solutions. The smallest possible value of y is, naturally, 1, and substituting y=1 is convenient as then $y^2=y^3$ , and it’s straightforward to derive $x=7$ as a solution.

Regarding the non-existence of large solutions, you can make this precise by factorising the LHS as

$(x+y)(x^2-xy+y^2) = x^2+42xy+y^2.$

There are 44 terms of degree two on the RHS, and one term of degree in the second bracket on the LHS. With a bit of AM-GM, you can see then that if $x+y>44$ , you get a contradiction, as the LHS will be greater than the RHS. But that’s still a lot of possibilities to check.

It struck me that I could find ways to reduce the burden by reducing modulo various primes. 2, 3 and 7 all divide 42, and furthermore cubes are nice modulo 7 and squares are nice modulo 3, so maybe that would bring the number of possibilities down. But my instinct was that this wasn’t the right way to use the fact that we were solving over positive integers.

The second bracket in the factorisation looks enough like the RHS, that it’s worth exploring. If we move $x^2-xy+y^2$ from the right to the left, we get

$(x+y-1)(x^2-xy+y^2) = 43xy.$ (1.1)

Now it suddenly does look useful that we are solving over positive integers, because 43 is a prime, so has to appear as a factor somewhere on the LHS. But it’s generally quite restrictive that $x^2-xy+y^2 | 43xy$ . This definitely looks like something that won’t hold often. If x and y are coprime, then certainly $x^2-xy+y^2$ and $y$ are coprime also. But actually if x and y have a non-trivial common factor d, we can divide both sides by $d^2$ , and it still holds. Let’s write

$x=dm,\quad y=dn,\quad\text{where }d=\mathrm{gcd}(x,y).$

Then $m^2 -mn+n^2$ really does divide 43, since it is coprime to both m and n. This is now very restrictive indeed, since it requires that $m^2-mn+n^2$ be equal to 1 or 43. A square-sandwiching argument gives $m^2-mn+n^2=1$ iff $m=n=1$ . 43 requires a little bit more work, with (at least as I did it) a few cases to check by hand, but again only has one solution, namely $m=7, n=1$ and vice versa.

We now need to add the common divisor d back into the mix. In the first case, (1.1) reduces to $(2d-1)=43$ , which gives $(x,y)=(22,22)$ . In the second case, after cancelling a couple of factors, (1.1) reduces to $(8d-1)=7$ , from which $(x,y)=(7,1),(1,7)$ emerges, and these must be all the solutions.

The moral here seemed to be that divisibility was a stronger tool than case-reduction. But that was just this question. There are other examples where case-reduction is probably more useful than chasing divisibility.

Question Three

Find all functions $f:\mathbb{N}\rightarrow\mathbb{N}$ such that

$n+f(m) \,\big|\, f(n)+nf(m)$

for all $m,n\in\mathbb{N}$ .

What would be useful here? There are two variables, and a function. It would be useful if we could reduce the number of variables, or the number of occurences of f. We can reduce the number of variables by taking m=n, to get

$n+f(n) \,\big|\, f(n) [1+n].$ (3.1)

From this, we might observe that $f(n)\equiv 1$ is a solution. Of course we could analyse this much more, but this doesn’t look like a 10/10 insight, so I tried other things first.

In general, the statement that $a\,|\,b$ also tells us that $a\,|\, b-ka$ . That is, we can subtract arbitrary multiples of the divisor, and the result is still true. A recurring trope is that the original b is elegant, but an adjusted b-ka is useful. I don’t think we can do the latter, but by subtracting $n^2 +nf(m)$ from the problem statement, we get

$n+f(m) \,\big|\, n^2-f(n).$ (3.2)

There’s now no m on the RHS, but this relation has to hold for all m. One option is that $f(n)=n^2$ everywhere, then what we’ve deduced always holds since the RHS is zero. But if there’s a value of n for which $f(n)\ne n^2$ , then (3.2) is a very useful statement. From now on, we assume this. Because then as we fix n and vary m, we need $n+f(m)$ to remain a divisor of the RHS, which is fixed, and so has finitely many divisors. So $f(m)$ takes only finitely many values, and in particular is bounded.

This ties to the observation that $f\equiv 1$ is a solution, which we made around (3.1), so let’s revisit that: (Note, there might be more elegant ways to finish from here, but this is what I did. Also note, n is no longer fixed as in previous paragraph.)

$n+f(n) \,\big|\, f(n) [1+n].$ (3.1)

Just to avoid confusion between the function itself, and one of the finite collection of values it might take, let’s say b is a value taken by f. So there are values of n for which

$n+b \,\big|\, b(1+n).$

By thinking about linear equations, you might be able to convince yourself that there are only finitely many solutions (in n) to this relation. There are certainly only finitely many solutions where LHS=RHS (well, at most one solution), and only finitely many where 2xLHS=RHS etc etc. But why do something complicated, when we can actually repeat the trick from the beginning, and subtract $b(n+b)$ , to obtain

$n+b \,\big|\, b^2-b.$

For similar reasons to before, this is a great deduction, because it means if $b\ne 1$ , then the RHS is positive, which means only finitely many n can satisfy this relation. Remember we’re trying to show that no n can satisfy this relation if $b\ne 1$ , so this is definitely massive progress!

If any of what’s already happened looked like magic, I hope we can buy into the idea that subtracting multiples of the divisor from the RHS is the only tool we used, and that making the RHS fixed gives a lot of information about the LHS as the free variable varies. The final step is not magic either. We know that f is eventually 1. If you prefer “for large enough n, $f(n)=1$ ,” since all other values appear only finitely often. I could write this with quantifiers, but I don’t want to, because that makes it seem more complicated than it is. We genuinely don’t care when the last non-1 value appears.

Anyway, since we’ve deduced this, we absolutely have to substitute this into something we already have. Why not the original problem statement? Fix m, then for all large enough n

$n+f(m) \,\big|\, 1+nf(m).$ (3.3)

To emphasise, (3.3) has to hold for all large enough n. Is it possible that f(m)=2? Again, it’s easy to convince yourself not. But, yet again, why not use the approach we’ve used so profitably before to clear the RHS? In fact, we already did this, and called it (3.2), and we can make that work [3.4], but in this setting, because f(m) is fixed and we’re working with variable large n, it’s better to eliminate n, to get

$n+f(m)\,\big|\, f(m)^2-1,$

again for all large enough n. By the same size argument as before, this is totally impossible unless f(m)=1. Which means that in fact $f(m)=1$ for all m. Remember ages ago we assumed that f(n) was not $n^2$ everywhere, so this gives our two solutions: $f(n)=1,\, f(n)=n^2$ .

Moral: choosing carefully which expression to work with can make life much more interesting later. Eliminating as many variables or difficult things from one side is a good choice. Playing with small values can help you understand the problem, but here you need to think about soft properties of the expression, in particular what happens when you take one variable large while holding another fixed.

[3.4] – if you do use the original approach, you get $n^2-1$ on the RHS. There’s then the temptation to kill the divisibility by taking n to be the integer in the middle of a large twin prime pair. Unfortunately, the existence of such an n is still just a conjecture…

Question Four

(Statement copied from Art of Problem Solving. I’m unsure whether this is the exact wording given to the students in the contest.)

On a circular table sit n>2 students. First, each student has just one candy. At each step, each student chooses one of the following actions:

(A) Gives a candy to the student sitting on his left or to the student sitting on his right.

(B) Separates all its candies in two, possibly empty, sets and gives one set to the student sitting on his left and the other to the student sitting on his right.

At each step, students perform the actions they have chosen at the same time. A distribution of candy is called legitimate if it can occur after a finite number of steps.
Find the number of legitimate distributions.

My moral for this question is this: I’m glad I thought about this on the bus first. What I found hardest here was getting the right answer. My initial thoughts:

Do I know how to calculate the total number of possibilities, irrespective of the algorithm? Fortunately yes I do. Marbles-in-urns = barriers between marbles on a line (maybe add one extra marble per urn first). [4.1]
What happens if you just use technique a)? Well first you can get into trouble because what happens if you have zero sweets? But fine, let’s temporarily say you can have a negative number of sweets. If n is even, then there’s a clear parity situation developing, as if you colour the children red and blue alternately, at every stage you have n/2 sweets moving from red children to blue and vice versa, so actually the total number of sweets among the red children is constant through the process.
What happens if you just use technique b)? This felt much more promising.
Can you get all the sweets to one child? I considered looking at the child directly opposite (or almost-directly opposite) and ‘sweeping’ all the sweets away from them. It felt like this would work, except if for some parity reason we couldn’t prevent the final child having one (or more, but probably exactly one) sweets at the crucial moment when all the other sweets got passed to him.

Then I got home, and with some paper, I felt I could do all possibilities with n=5, and all but a few when n=6. My conjecture was that all are possible with n odd, and all are possible with n even, except those when none of the red kids or none of the kids get a sweet. I tried n=8, and there were a few more that I couldn’t construct, but this felt like my failure to be a computer rather than a big problem. Again there’s a trade-off between confirming your answer, and trying to prove it.

Claim: If n is even, you can’t achieve the configurations where either the red children or the blue children have no sweets.

Proof: Suppose you can. That means there’s a first time that all the sweets were on one colour. Call this time T. Without loss of generality, all the sweets are on red at T. Where could the sweets have been at time T-1? I claim they must all have been on blue, which contradicts minimality. Why? Because if at least one red child had at least one sweet, they must have passed at least one sweet to a blue neighbour.

Now it remains to give a construction for all other cases. In the end, my proof has two stages:

Step One: Given a configuration, in two steps, you can move a candy two places to the right, leaving everything else unchanged.

This is enough to settle the n odd case. For the even case, we need an extra step, which really corresponds to an initial phase of the construction.

Step Two: We can make some version of the ‘sweeping’ move precise, to end up in some configuration where the red number of children have any number of sweets except 0 or n.

Step one is not so hard. Realising that step one would be a useful tool to have was probably the one moment where I shifted from feeling like I hadn’t got into the problem to feeling that I’d mostly finished it. As ever in constructions, working out how to do a small local adjustment, which you plan to do lots of times to get a global effect, is great. (Think of how you solve a Rubik’s cube for example.)

Step two is notationally fiddly, and I would think very carefully before writing it up. In the end I didn’t use the sweeping move. Instead, with the observation that you can take an adjacent pair and continually swap their sweets it’s possible to set up an induction.

Actual morals: Observing the possibility to make a small change in a couple of moves (Step one above) was crucial. My original moral does still hold slightly. Writing lots of things down didn’t make life easier, and in the end the ideas on the bus were pretty much everything I needed.

[4.1] – one session to a group of 15 year olds is enough to teach you that the canon is always ‘marbles in urns’ never ‘balls’ nor ‘bags’, let alone both.

RMM 2017 – UK Team Blog

Posted on March 15, 2017 by dominicyeo

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

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

Wednesday 22 February

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

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

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

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

Thursday 23 February

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

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

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

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

Friday 24 February

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

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

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

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

Saturday 25 February

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

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

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

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

Sunday 26 February

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

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

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

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

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

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

Monday 27th February

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

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

RMM 2017 – Problems 2, 3 and 6

Posted on March 14, 2017 by dominicyeo

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

Problem 2

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

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

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

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

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

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

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

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

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

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

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

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

Problem 3

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

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

By Neel Nanda:

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

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

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

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

Problem 6

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

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

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

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

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

RMM 2017 – Problems 1, 4 and 5

Posted on March 14, 2017 by dominicyeo

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

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

Problem 1

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

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

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

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

Rosie Cates and Neel Nanda:

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

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

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

1110100110 = 1110000000 + 100000 + 110.

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

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

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

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

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

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

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

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

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

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

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

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

Problem 4

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

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

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

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

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

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

Question 5

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

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

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

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

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

Eventually Almost Everywhere

A blog about probability and olympiads by Dominic Yeo

Tag Archives: olympiad

EGMO 2019

BMO1 2018

MOG 2018

Balkan MO 2018 – UK Team Blog

Balkan MO 2018 – Questions 2,3,4

EGMO 2018

Balkan MO 2017 – Qs 1, 3 and 4

RMM 2017 – UK Team Blog

RMM 2017 – Problems 2, 3 and 6

RMM 2017 – Problems 1, 4 and 5