EGMO 2016 Paper II

Posted on April 14, 2016 by dominicyeo

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

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

Question 4

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

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

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

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

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

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

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

Question 5

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

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

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

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

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

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

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

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

Question 6

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

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

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

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

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

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

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

EGMO 2016 Paper I

Posted on April 13, 2016 by dominicyeo

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

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

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

Question 1

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

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

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

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

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

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

Idea 1: We can write

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

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

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

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

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

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

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

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

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

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

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

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

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

Question 3

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

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

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

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

Claim A: No point having three in a row.

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

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

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

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

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

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

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

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

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

EGMO 2015

Posted on April 21, 2015 by dominicyeo

It’s been a while since I last wrote anything substantial. There have been some posts in the pipeline, but mainly I’ve been working hard on technical things that don’t translate very well into blog posts, and when I haven’t been working on those, have felt like doing non-maths.

Anyway, among other things which happened recently were the UK’s IMO selection camp in Cambridge during the last week of March, and the fourth European Girls’ Mathematical Olympiad in Belarus this past weekend. At the former, I was quite busy organising various things, and so gave a session based on some of my favourite problems about points and lines that I’ve used in the past. My hope was that with each problem in turn the students would actively invest time in thinking about whether the ideas each other were having seemed likely to be profitable. The aim was that being critical about your approach is a necessary skill once you start having lots of ideas for approaches.

This is hard to practise at school just by reading around, whether regarding competition material or generally. In the competition world, official solutions often contain unmotivated magic. This is reasonable, since they are supposed to be a minimal elementary demonstration of the problem’s validity. Motivation is a luxury which space and time sometimes doesn’t permit. And the solutions you find on, for example, Mathlinks often give the impression that the solvers know how to do 25,000 specific types of problem, and the sole task is to identify which type the latest one belongs to. Relating problems to ones you’ve seen before is important, but can hide, or at least dilute the actual ideas in some cases. Knowing that a specific inequality is a special case of a big machine allows you to claim a ‘solution’ but doesn’t help you identify the relevant ideas.

Later in the camp, a conversation arose concerning to what extent the younger staff found these elementary methods and problems easier now that they had experienced various levels of higher education in mathematics than when they were at school. It’s a complicated question, and I don’t think there’s a simple answer. I think the students might suspect that doing a university degree teaches you ‘advanced techniques’ which immediately simplify some of these problems. In rare examples this can be the case, but the majority of the time, I personally feel the only advantage I have is perhaps better instincts for whether something is or isn’t going to work.

Probably a good test would be Euclidean geometry. Adult olympiad staff typically come in two flavours: those who used to be good at geometry and feel out of practice; and those who weren’t good at geometry and certainly had no inclination to remedy this after they left school. I’m probably closer to the first category and I definitely feel out of practice, but also have minimal inclination to remedy this. Nonetheless, on the rare occasions I try these questions (and it’s not going to be for 4.5 hours at this stage…) my progress rate is probably comparable to when I was in sixth form. I’ve no idea how to turn this into a more concrete assessment, but there must be something about doing abstract maths all the time that prevents you forgetting how to do this, so presumably it should be at least as helpful in the types of problem with non-zero overlap with things I actually do. I won’t discuss geometry in the rest of this post, but I did also enjoy the geometry questions – it’s simply that I feel anything I have to say about them would be less useful than saying nothing.

In any case, I enjoyed trying the problems from Belarus in between bouts of typing, and thought I would write some brief comments about how I decided whether my ideas were good or not. To achieve this, I’ve looked at my rough, and will tell you the ideas which didn’t work, as well as some of the ones which did. I’ve paraphrased the statements slightly to avoid having too many LaTeX tags.

WARNING: what follows will spoil questions {2,4,5} if you haven’t already looked at them, but would like to.

Question 2 – A domino is a 2 × 1 or 1 × 2 tile. Determine in how many ways exactly n^2 dominoes can be placed without overlapping on a 2n × 2n chessboard so that every 2 × 2 square contains at least two uncovered unit squares which lie in the same row or column.

The wording of the question prompted me to consider splitting the board naturally into n^2 2 x 2 squares. I then thought about this ‘at least’ in the question, and concluded that for these 2 x 2 squares, it should be ‘exactly’.

I tried doing an unusual colouring, when I coloured half the black squares green, and half blue and tried to show that either only greens or only blues were covered, but this was clearly not true, or fixable. I then tried to do something similar for the other set of 2 x 2 squares (those whose vertices have (odd, odd) coordinates). Roughly speaking, if too few of the outer cells on the original board are covered, you can’t achieve the bounds on these odd inner squares. But this didn’t really give any useful information.

However, it did get me thinking about how what happens in the far top-left affects the rest of the board, and from there most of the ideas came quickly. I described a 2 x 2 square as N, E, S, W depending on which ‘half’ of the square was covered. Then if a square is N, all the squares directly above it must be also be N (*).

I then made two mistakes, and if we’re looking for reasons where experience is helpful, it was probably here, as I spotted them fairly quickly, rather than wasting ages and ages.

First, I decided that either all squares were {N,E} or all were {S,W} which seemed plausible when I had been focusing on the top-left. This gave an answer of $2 \binom{2n}{n}$ , but after a bit more thought is clearly not symmetric enough.

Second, I thought condition (*) might be the only constraint, along with similar ones for E, S, W naturally too. I tried to count these using a similar method of enumerating lines between these regions, and I realised I didn’t know how to do these sensibly, for example if it looked like this:

This led to another bit of meta-maths that I probably wouldn’t have considered if I was 16. Namely, that the idea of counting these monotone lines across the 2n x 2n grid was too nice not to be useful. Also, if I couldn’t see a way to adapt it for this more complicated setup, my setup was probably wrong. This thought was useful, and then by thinking about the interface between {N,E} and {S,W}, then the other way round, it made sense that the configuration could be parameterised by two monotone lines between different pairs of corners, giving an answer of $\binom{2n}{n}^2$ .

So, if it’s possible to give a reason why I could do this, it’s probably because I had an arsenal of ways to check an answer for plausibility, which saved wasting time on something wrong, and also because I trusted that the answer would be nice, which saved wasting time on something which was also wrong, and would probably have been very complicated to resolve.

Question 4 – Determine whether there exists an infinite sequence $a_1,a_2,\ldots$ of positive integers satisfying $a_{n+2}=a_{n+1}+\sqrt{a_{n+1}+a_n}$ .

So, my first reaction was ‘no way’. Why? Because everything’s determined by the first two terms, and I couldn’t think of any reason why a cool choice of the first two terms would force all of the sums $a_{n+1}+a_n$ to be perfect squares. It seemed just about possible that we could arbitrarily large finite sequences with this property. (Though this also turns out to be impossible.)

I imagine many contestants might have felt similarly and spent a while playing round with quadratic residues to get a sense for exactly how hard it is to make this work for the initial terms. But I was suspicious of the form of the recurrence. We know that if it had been defined linearly, then the terms would grow exponentially, but it’s natural to ask roughly how fast they grow in this example, even relaxing the restriction that the terms be integers.

The first observation was that they are indeed (strictly) growing! But how fast? Are there actually enough squares that every $a_{n+1}+a_n$ can be a different one? Note that the squares themselves satisfy a similar recurrence relation $(N+1)^2 = N^2 + 2\sqrt{N^2} + 1 > N^2 + \sqrt{ N^2 + N^2}$ . So this seemed like a very good idea, and my instinct was that this should work, and I felt glad that I hadn’t pursued the quadratic residues approach.

From here we were basically home. I asked whether the sequence grows faster than the sequence $(\frac{n^2}{2})$ , and the answer was no as

$a_{n+2}\le a_{n+1}+ \sqrt{2 a_{n+1}} \le (\sqrt{a_{n=1}} + \frac{1}{\sqrt 2})^2,$

so if (after translating indices) $a_n = \frac{x^2}{2}$ , we have $a_{n+1}\le \frac{(x+1)^2}{2}$ . This is clearly a problem or at best a very tight constraint if all the $\{a_n+a_{n+1}\}$ have to be perfect squares, so even though we aren’t completely finished, I am confident I can be in one or two lines, with a bit of care. Fiddling with small values of n looked like it would work, or showing that looking at a large enough initial subsequence that the effect of the first two terms dissipated, we must by the pigeonhole principle have $a_{n+2}+a_{n+1}=(k+1)^2,\, a_{n+1}+a_n=k^2$ , which is enough by a parity argument, using the original statement.

This final bit looks and feels a bit messy, but by this stage it really is just finding any argument to justify why a sequence which grows at most as fast as $n^2$ can’t actually be $n^2$ eventually.

Probably the reason I did this quickly was because I trusted my instinct for ‘no’, and also because there seemed a quick way to qualify *roughly* how this sequence grew. Sensibly approximating things, and deciding whether it’s appropriate to approximate them is definitely something I feel I’ve got better at during my PhD, so I guess this was helpful, then just try and throw back the important condition that the elements were integers at the very end.

Question 5 – Anastasia partitions the integers [1,2m] into pairs. Boris tries to choose one integer from each pair so that the sum is n. Given n and m, prove Anastasia can select pairs so that Boris can’t succeed.

So I wasted some thought time by imagining that n was fixed and trying to come up with ideas for the choice of pairs which might avoid n. It struck me that there was no reason why a a typical (uniformly chosen) pairing should avoid any particular n unless this value was particularly big or small.

How big or how small? Well Boris can always choose the bigger element of a pair, so the way to make the minimum maximum is to pair as (1,2), (3,4), … , (2m-1,2m). Conveniently, this also achieves the maximum minimum. These can be calculated as $m^2,m(m+1)$ respectively. Suddenly this seems great, because we’ve actually solved the problem for a huge range of n, ie everything not within between these extrema.

The key step here was to start considering special pairings, chosen to give a reduced set of possible sums. Once we’ve had this idea, it makes sense to consider other different special pairings. The reason we got a small set of possible sums is that there’s lots of overlap. We can achieve lots of overlap by looking at the difference between elements in a pair, and making as many of these the same as possible. For, consider pairs (a,a+d), (b,b+d). Then it’s the same to take a and b+d as to take a+d and b, so we get the same sums in lots of different ways.

The other way to have as many differences the same as possible is to go for (1,m+1), (2,m+2), … , (m,2m). Conveniently, we can parameterise the sums now because at each choice, we decide whether to add an extra m or not, so the sum must be 1+2+…+m, plus some multiple of m. So we can defeat Boris, except when n is $\binom{m}{2}+km$ .

This is a good point to stop because what followed was basically book-keeping. We only have to consider a couple of specific cases when m is odd, and one when m is even, that happen not to fall into either of the categories we can now deal with. It wasn’t too hard to corrupt the examples we already have to deal with these.

The key here was responding to initial failure by trying to find any control at all over n. Perhaps given enough time I would have started trying special pairings? Equally, I might have tried small examples, which would not really have been hugely helpful for this question. In any case, trying to eliminate very small or very large n luckily worked well, as a) I didn’t need to use the word ‘very’ twice in the end; and b) the idea of looking at choices of pairings to minimise the number of sums quickly gave other useful deductions.

I also really enjoyed Question 3, though was suspicious that I’d used a bound a great deal weaker than one in the question. Along the way, I was trying something confusing and not especially useful that led me into some interesting theory about Latin squares, which I might write something about later in the week.

Eventually Almost Everywhere

A blog about probability and olympiads by Dominic Yeo

Tag Archives: chessboard problem

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