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Magic triangles (Posted on 2018-01-18) Difficulty: 3 of 5
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?

No Solution Yet Submitted by Ady TZIDON    
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Comments: ( Back to comment list | You must be logged in to post comments.)
Some Thoughts re: spoiler.....SQUARES AS WELL Comment 2 of 2 |
(In reply to spoiler by xdog)

Did you (or anyone) notice that 837249516 is a very  special case"?


Compare the sums of the squares  of the numbers
 on all three sides...



  Posted by Ady TZIDON on 2018-01-20 17:15:18
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