Elementary Techniques used in the IMO (International Mathematical Olympiad) - Invariants

I know this is much for a day, but I will be going overseas tomorrow for a long time, so I thought that It be best for me to post all the instalments and that my reader can pace himself/herself when reading through these posts.

This is the 55 th instalment, following the 4th.

At the request of Priyansh Sangule, we will do some colouring proofs today.

Example 4: (IMO 2004) Define a hook to be a figure made up of 6 unit squares as shown in the diagram

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or any of the tiles by applying rotations and reflections to the diagram. Determine all m×n m\times n rectangles that can covered be hooks such that:

  • the rectangle is covered with hooks and no overlaps

  • no part of the hook covers an area outside the rectangle.

Solution: Yet again, do not fret that we are about to attempt a considerably recent IMO problem.

We hate hooks because they are in a way disjoint. We like tiles that are nice 2×3 2 \times 3 rectangles or things of the sort. So it is pretty motivated to consider pairs of hooks due to the hole, as marked out (with a black dot) in the following diagram:

tagtext tagtext

So necessarily if tile X X cover tile Ys Y's "hole", then necessarily tile Y Y also cover the "hole" of tile X X . We have the following 2 2 pairs:

tagtext tagtext

We have our first breakthrough for this problem: 12mn 12 \mid mn . Consider a 2×6 2 \times 6 board. Clearly we cannot tile it with the pairs of tiles. However, we can obviously tile a 4×6 4 \times 6 board. We might now conjecture that we must have (WLOG) that 4m 4 \mid m .

Let us consider the case where they are both even but 4m,n 4 \nmid m,n . So we want to find a colouring that makes use of the divisibility by 4 4 condition. Well, the most direct way of incorporating it this is to add a 1 1 to each cell of the board that belongs to a column or row divisible by 4 4 . Basically what we want to do next can then be simplified to proving that each of the pairs of hook only cover an odd number of 1 1 . (Do you see why this is sufficient?). Anyways, here's the board with the labelling:

tagtext tagtext

So each pair of hook clearly can only cover an odd number of 1 1 by doing a simple case-bashing (I leave the reader to convince himself). So there must be an even number of tile \rightarrow 4m 4 \mid m (WLOG).

Now, let us summarise what we have:

  • 12mn 12 \mid mn

  • 4m 4 \mid m

  • m,n1,2,5 m,n \neq 1,2,5 (do you see why?)

(i) So if 3n 3 \mid n then we can clearly tile the whole board using the 3×4 3 \times 4 rectangles formed from the pair of hooks.

(ii) If 12m 12 \mid m and by point 3 3 , m6 m \geq 6 then by a simple induction we can show that m m can be represented in the form 3a+4b 3a + 4b . The motivation behind this step is that we want to be able to split up the board into smaller sections of case (i) above, because intuitively the 3×4 3 \times 4 rectangles is more predictable than the second kind of pairing. Then all we do is that we split up the board into a a strips of 3×m 3 \times m and b b strips of 4×m 4 \times m which both can be tiled. □

Look forward to the next post, in roughly 1 1 and a half weeks time, for more on colouring!

Note by Anqi Li
7 years, 4 months ago

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