EGMO 2016 Paper II

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

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

Question 4

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

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

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

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

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

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

20160414_093104

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

Question 5

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

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

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

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

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

20160414_100244

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

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

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

Question 6

EGMO2016 Q6

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

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

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

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

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

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

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

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Rhombus Tilings and a Nice Bijection

I want to write a short post giving an example of what seems to me to be a rather nice proof without words. Like all the best proofs without words, they require some words to set everything up, and then even the proof itself is enhanced with a few words.

The goal is a bijection between two combinatorial objects. The first is the family of rhombus tilings. Perhaps the easiest way to define these is to give an example.

DSC_2215 - Copy

As you can see, we have tiled a hexagon with rhombi. The tiles are allowed to be in any of the three possible orientations. It matters that the angles of the hexagon are 120, as we want it to be possible to squeeze rhombi into a corner in two different ways (ie either a single tile or two tiles together), and thus the rhombi should also have angles 120 and 60. The hexagon does not have to be equilateral, as in this example, but obviously all the side lengths should be an integer multiple of the side length of the rhombus, which without loss of generality we may take to be 1.

The other combinatorial object is the class of plane partitions. We again give an example:

4 3 3 2 1
4 2 2 2
3 2 2 1
2 1 1

Notice that all the rows and columns are weakly decreasing. One observation worth making is that the diagonals gives a family of so-called interlaced partitions. In any case, we want to establish this bijection. First I show the idea, that is the proof without words bit. Then I’ll clarify exactly how to make the bijection work.

The first step is to colour a rhombus tiling with a different colour for each orientation, as shown.

DSC_2216 - Copy

The next step is the proof without words bit. We now look at the diagram as if we were looking into a stack of cubes arranged in the positive orthant of R^3. The colouring makes this much more visually arresting. Black rhombi correspond to the (visible) top sides of cubes, while blue and red faces point out in the x and y directions respectively. The key observation is that a rhombus tiling means we can see at least one face of every cube. Otherwise we would need some smaller rhombi to account for the way that some cubes will be partially hidden between taller but closer piles. So if make a note of the heights of all the piles, we should get a plane partition.

After reordering our definition of plane partition, so it is weakly increasing left-right and down-up, corresponding to the x and y axes drawn on the above diagram, the given rhombus tiling should give the following plane partition:

1 2 3
0 1 3
0 0 2

The only thing we need to sort out is precisely how the dimensions of the hexagon restrict the choice of plane partition. Note that we could keep the heights exactly the same but get a different tiling by adding an extra row of red oriented rhombi above the top-left part, and an extra row of blue oriented rhombi above the top-right part. The point is that this would give us a bigger hexagon.

The first observation is that the dimensions of the plane partition correspond to two of the side lengths of the hexagon, indeed the bottom two sides. The third length of the hexagon corresponds to the maximum possible height (ie z component) of the region we are looking at. This is therefore an upper bound on the heights of the stacks.

So we can conclude our bijective argument. There is a bijection between rhombus tilings of the hexagon with side lengths X, Y, Z and plane partitions with dimension X x Y, (where entries are allowed to be zero) where the largest element (which is by definition also the top-left element, or top-right in our re-definition) is at most Z.

It seems there are plenty of interesting questions to be asked about both deterministic and random tilings and plane partitions, based on talks in Marseille. For now though, I feel ill-qualified even to read about such things, so will leave it at that for today.