Unlike we saw with the 15-sided shape a couple of weeks ago, 5 is prime and so cannot be broken down into products. But since the five piles need to be different, we do still need to find a way of placing a subset of weights such that they balance. One way would be to have 1kg blocks at two adjacent vertices, and then to calculate what the opposite vertex would need to be to balance.

We can call the distance from the middle to each of the vertices ‘d’. Two 1kg weights on adjacent vertices ‘d’ from the middle would be equivalent to 2kg midway between those two points. That midway point is about 0.8*d from the centre. To balance this with a single weight ‘d’ from the centre on the opposite side, the distance multiplied by the weight needs to come to the same value. 0.8d x 2kg = d x 1.6kg. This lies between 1kg and 2kg as the question dictates.

A bit more precise calculation and this balancing block needs to weigh ‘phi’ kg. Phi is the golden ratio, equal to (sqrt(5)+1)/2 or ~1.618.

So, we have an arrangement which balances with three blocks: one at ~1.618kg and two at 1kg. How many of these three-block arrangements do we need to make the five piles all different? We can do it with just three of them. Place 1 heavy block on a vertex, 2 heavy blocks to an adjacent vertex, and the 1kg blocks required to balance them will be in piles of 1, 3 and 2.

So, we have achieved a balance using nine blocks in total: six 1kg blocks and three ~1.618kg blocks. The weight at each vertex is: ~1.618kg, ~3.236kg, 1kg, 3kg, 2kg.

How about now, we see what happens if we place two of the heavy weights on adjacent vertices and calculate what we would need in the single opposite vertex to rebalance the star? Since the pair of weights are scaled up by phi from our original (1,1,phi) subset, the balancing vertex will also be, and will equal to (phi)^2. But because phi is such a special number, (phi)^2 is equal to phi+1, which we can achieve by using one of each of our two weights. Now if we combine one (1,1,phi) subset, and one (phi,phi,phi+1) subset, we can have a different weight on each vertex and balance the star, but this time only using seven weights in total, three of 1kg and four of ~1.618kg. The weight at each vertex is: ~1.618kg, ~3.236kg, 0kg, ~3.618kg, 1kg.