# 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 $$5$$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 $$m\times n$$ 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 $$2 \times 3$$ 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:

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

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

Let us consider the case where they are both even but $$4 \nmid m,n$$. So we want to find a colouring that makes use of the divisibility by $$4$$ condition. Well, the most direct way of incorporating it this is to add a $$1$$ to each cell of the board that belongs to a column or row divisible by $$4$$. 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 $$1$$. (Do you see why this is sufficient?). Anyways, here's the board with the labelling:

tagtext

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

Now, let us summarise what we have:

• $$12 \mid mn$$

• $$4 \mid m$$

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

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

(ii) If $$12 \mid m$$ and by point $$3$$, $$m \geq 6$$ then by a simple induction we can show that $$m$$ can be represented in the form $$3a + 4b$$. 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 $$3 \times 4$$ rectangles is more predictable than the second kind of pairing. Then all we do is that we split up the board into $$a$$ strips of $$3 \times m$$ and $$b$$ strips of $$4 \times m$$ which both can be tiled. □

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

Note by Anqi Li
4 years, 7 months ago

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