**From New Scientist #2426, 20th December 2003** [link]

If you place a number in each of the twelve circles on this star you can then calculate the “line-total” of each of the six lines by adding up the four numbers along the line. Similarly you can calculate the “triangle-total” of each of the six small triangles by adding up the three numbers around the triangle.

Your task is to place the numbers 1 to 12 in the circles so that each of the line-totals is the same, and there is a triangle-total larger than the line-total, and a triangle-total less than half the line-total.

What number is diametrically opposite 1? And what number is diametrically opposite 2?

[enigma1270]

### Like this:

Like Loading...

This Python program runs in 1.8s.

It could be made more efficient (but more complicated), by considering lines with fewer numbers of blanks at each recursive step (and in particular filling out lines with only one missing number), in a similar style to my solution to Enigma 241. But for a problem of this size it is perhaps not worth it.

Solution:3 is diametrically opposite 1, and 5 is diametrically opposite 2.There are essentially 4 different solutions.

One pair of solutions has triangle sums of 12, 18 and 27 (each pair of diametrically opposed triangles have the same sum):

The other pair of solutions has triangle sums of 12, 21 and 27:

As Jim points out, it is not really worth putting too much effort into tuning a solution, but I couldn’t resist having a go.

If the inner ring c,d,f,g,i,j are defined, the values of the outer ring a,b,e,h,k,l can be deduced (see code beolw for the formulae). Using a similar idea to Jim’s to fix c as the smallest member of the inner ring and avoiding reflections by only considering one permutation of d,f :