Imagine a triangle ABC with 3 distinct non-zero digits assigned to each of its vertices, say A=1 B=4 C=7.
Now place the six remaining digits, 2 per side, trying to get a sum of 20.
If we place 6 and 9 between A and B the sum on AB will be 1+6+9+4=20
as requested, but the sequel will in no way create a sum of 20 per side.
The above example was given just to illustrate the process.

What are the triplets (A,B,C) that allow fulfilling the required task?
How many distinct generic solutions (define) are there?

The sum of the numbers on all three sides is 60.  This sum counts A,B,C twice and the remaining digits once.  Since the digits 1-9 sum to 45, A+B+C=15.

Posted by xdog on 2018-01-18 08:47:20