Magic square variation

I have a rather difficult variation of magic squares: In the below image, all numbers from 1 to 24 must be placed in the 24 closed areas, in such a way that all numbers in areas of each circle must sum to 80. Each number must be placed only once.

Some observations: Let's consider two "opposite" circles. These two do not have any areas in common, therefore these must have 2 distinct subsets of 8 numbers each. Similarly, the outer areas to the left and right of these circles, also have 8 distinct areas so this is the 3rd subset of 8 numbers. Since all numbers from 1 to 24 add up to 300, and the 2 above distinct circles must have a sum of 160, the remaining 8 areas (4+4 to the left and right of the distinct circles) must have a sum of 140.

The numbers at the 6 outmost areas of each of the 6 outer circles, appear only once. The numbers at the 6 areas right below them, appear twice. The areas right below these areas appear 3 times each, and the numbers at the inner thin areas that look like flower petals appear 4 times each. So we have x+2y+3z+4w=7*80 since we have 7 circles. Each of x, y, z and w is a sum of 6 distinct numbers.

Finally, the inner circle consists of 12 distinct areas, which means that we have to use the smaller numbers (to get a sum of 80).

Is there any way to continue?

• I think the inner circle has the numbers 1,2,3,4,5,6,7,8,9,10,11 and 14 that sum up to 80. This leaves the rest for the outer circles Feb 7, 2018 at 20:20