## [Soln] 2013 AIME I Problem 4

2013 AIME I 4. In the array of 13 squares shown below, 8 squares are coloured red, and the remaining 5 squares are coloured blue. If one of all possible such colourings is chosen at random, the probability that the chosen coloured array appears the same when rotated $90^\circ$ around the central square is $\frac{1}{n}$, where $n$ is a positive integer. Find $n$.

The fraction we are trying to compute is

$\displaystyle\frac{\text{No. of colourings which look the same when rotated }90^\circ}{\text{Total no. of colourings}}.$

The denominator is easy to calculate: it’s just how many ways to select 5 out of 13 squares to colour blue, i.e. $\displaystyle\binom{13}{5} = 1287$.

To count the number of colourings which look the same when rotated $90^\circ$, let’s label all the squares. It is then a simple matter of rotating the picture and equating the values in each square!

For the original to look exactly like the rotated version, we must have

$\begin{cases} B = E = H = K, \\ C = F = I = L, \\ D = G = J = M. \end{cases}$

It’s also clear that as long as the 3 “equations” above hold, the rotated version will indeed look like the original. (It should be noted that the question did not specify whether the rotation is clockwise or anti-clockwise. A moment’s thought will show that the equations that must hold are the same in either case.)

Now, how can we choose the 5 blue squares so that the “equations” above hold? The equations essentially partition the squares on the outside into 3 groups of 4, and the squares in each group must have the same colour. Hence, exactly one of the groups must be coloured blue, and the square in the middle must be coloured blue.

What choices do we have in colouring this way? The only choice we have is choosing which of the 3 groups should be blue. Hence, the numerator is 3.

Summing up, we have

\begin{aligned} \displaystyle\frac{1}{n} &= \frac{3}{1287} = \frac{1}{429}, \\ n &= \fbox{429}. \end{aligned}

Done!