Magic triangular square

Logic Level 3

An $n\times n$ triangular square is an equilateral triangle of side length $n$ filled with $n$ rows of unit equilateral triangles. There are $n^2$ unit equilateral triangles, hence the name.

If we fill a $3\times 3$ triangular square with all the integers from $1$ to $9$ such that the sum of the numbers in each $2\times 2$ sub-triangular square is the same, what is the maximum value that this sum can take?

This triangular square is not magical because the sums of every 2 x 2 triangular square are all different (15, 22 and 20).

The answer is 23.

1 solution

Romain Bouchard
Apr 29, 2018

Let $S$ be the value of the maximum sum.

We have three $2\times 2$ sub-triangular squares. Thus $S = a+b+c+d = b+e+f+g = d+g+h+i$ .

Since $a+b+c+d+e+f+g+h+i = 1+2+3+4+5+6+7+8+9 = 45$ and $a+b+c+d+e+f+g+h+i = 3S-b-d-g$ , then

$S = \frac{45+b+d+g}{3}=15+\frac{b+d+g}{3}$ .

If we can choose $b,d$ and $g$ maximal, then $b+d+g = 7+8+9 = 24$ and $S = 15+8=23$ .

Here is such a configuration :

Magic triangular square

The answer is 23.

1 solution

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