Triominoes

We have received many solutions that have started to scratch the surface of this 
problem.

The best example of this comes from Elizabeth Daniels, aged 14, from Stamford High 
School:

I read through the question first and I then decided to try a few examples to see whether 
a pattern would emerge. I first used a square with the measurements 2 by 2 (21 by 21). 

 

After that, I then tried a square with measurements 4 by 4 (22 by 22). I tried different 
ways to fit everything together, but I worked it out by putting a triomino in each corner 
leaving a central square where another triomino could be added and one square being 
the red one, could be left.  On the diagram the triominoes are shown with corresponding 
numbers.

 


When I came to try a square with measurements of 8 by 8 (23 by 23) then I realised that 
four, 4 by 4 squares would fit into the 8 by 8 square. I then did the 4 by 4 square again, 
this time so as the red square was in the corner which was going to be in the centre of 
the 8 by 8 square. Therefore in the centre of the square there would be a square of four 
reds and so three of those could be covered with a triomino and the remaining one 
would be the red square.

 


I then realised that the 2 by 2 square had 4 squares, the 4 by 4 square had 16 squares 
and the 8 by 8 square had 64 squares. If 1 was subtracted from these numbers and the 
number that was left was divisible by three, then it was likely that the triominoes would 
fit into the square.  For example 82 is 64 and if 1 is subtracted then you get 63 which 
divided by 3 is 21. This therefore shows that 21 triominoes are needed to fill the whole 
square.

I then thought about how the square would be constructed if I put two or more 
triominoes together like below:

 

I then tried to fit the above shape into the 8 by 8 square. I managed to do it as shown 
below. I then realised that as long as there is a 2 by 2 square left in the corner then it 
will be possible to fit all the triominoes in because it will leave only 1 square for the 
red and the other three for the remaining triomino.

          By Elizabeth Daniels Aged 14 (Year 10 Stamford 
High School)


To crack this tough nut there will need to be some follow up work:

a)	Can you show that all chessboards of size 2^k by 2^k will have a total number 
of squares which is 1 more than a multiple of 3?

b)	Elizabeth used her solution for the 4 by 4 square to complete the 8 by 8 square. 
Can the solution for the 8 by 8 square be used for the 16 by 16 square, and so 
on? Having seen solutions for the 2 by 2, 4 by 4 and 8 by 8 squares, we now 
want to know that we can find a solution for all 2^k by 2^k squares. There are 
an infinite number of squares so it is impossible to show all the solutions – 
instead we need to know that there exists a method for solving all 2^k by 2^k 
squares.

c)	Elizabeth has placed the red square in one position in each of her examples. Is it 
possible to find a solution for all possible positions of the red square? For 
example, in the 4 by 4 square the red square could be in any of 16 different 
positions; is it possible to cover the remaining squares on every occasion? Can 
we place the red square in all possible positions and still cover any sized 
chessboard?