All about flooble | fun stuff | Get a free chatterbox | Free JavaScript | Avatars    
perplexus dot info

Home > Just Math
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    
No Rating

Comments: ( Back to comment list | You must be logged in to post comments.)
spoiler | Comment 1 of 2
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.

Then it's just a matter of listing possible values for A,B,C, seeing if the remaining digits fit, and hoping I didn't miss any.

I find these 6 solutions, two of which use the same values of A,B,C highlighted below.


  Posted by xdog on 2018-01-18 08:47:20
Please log in:
Remember me:
Sign up! | Forgot password

Search body:
Forums (0)
Newest Problems
Random Problem
FAQ | About This Site
Site Statistics
New Comments (2)
Unsolved Problems
Top Rated Problems
This month's top
Most Commented On

Copyright © 2002 - 2018 by Animus Pactum Consulting. All rights reserved. Privacy Information