Parallelogram 32 • Level 3 • 11 Apr 2024Olympic mascots

All you have to do is a subtraction: 2012 - 1850 = 162.

The triangle in the centre of the diagram is equilateral since each of its sides is equal in length to the side of one of the squares. The sum of the angles at a point is 360°, so x=360−90+90+60=120.

The two lines of reflection symmetry of the grid are shown in the figure on the right as broken lines.

We see from this that if the grid with some of the small squares shaded grey has both these lines of reflection symmetry, then all the four squares labelled P must be the same colour. Similarly, all the four squares labelled Q must be the same colour, the two squares labelled R must be the same colour, and the two squares labelled S must be the same colour.

It follows that there are only three ways in which Mathias can shade exactly four of the small squares of the grid grey so that the result has two lines of reflection symmetry. These are:

(i) shade grey all the squares labelled P, and no others,
(ii) shade grey all the squares labelled Q, and no others,
and (iii) shade grey all the squares labelled R and all those labelled S, and no others.

The three different grids that Mathias could produce are shown in the figure below.

There are only two 1-pip dominoes among the six Dominic has. These must therefore be adjacent. Likewise for the two 2-pips, the two 6-pips and the two 4-pips. So the ring must include the three adjacent dominoes:

and also the three adjacent dominoes:

The ring may now be completed two ways, i.e., by connecting top-left 5-pip to bottom-left 5-pip and the top-right 5-pip to bottom-right 5-pip … or by connecting top-left 5-pip to bottom-right 5-pip and the top-right 5-pip to bottom-left 5-pip.

We therefore see that Dominic can create two different rings of six dominoes, as shown below.

We now see that there are just 2 dominoes that the [1:5] domino cannot be adjacent to, namely the domino [2:5] and, of course, the domino [4:6].

We first consider what happens if one vertex of the tetrahedral die remains fixed on the grid and the die rolls about this fixed point. In doing this the die covers in succession six faces making up a hexagon. The faces adjacent to the fixed vertex are each face down twice in this hexagon. If the die is rolled about any of the edges on the perimeter of the hexagon, the remaining face is then face down. This is shown in the diagram on the right for the case where the faces with 1, 2 and 3 pips are adjacent to the fixed vertex, and the remaining face has 4 pips on it.

The pattern shown above can be extended to the whole of the given triangular grid, as shown in the diagram on the right. (Indeed, this pattern could be extended to a triangular grid filling the entire plane.)

For any triangle in the grid, the face then face down together with the two preceding faces form part of a hexagon, and determine the orientation of the die. (Here, by the orientation of the die, we mean the positions of all four of its faces.) The die can roll to this triangle across any of its edges. However, the pattern implies that the same face is face down in all these cases. In particular, the face with 1 pip is face down when the die reaches the shaded triangle.