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

Home > Just Math
Real and Cyclic Sum (Posted on 2016-08-10) Difficulty: 3 of 5
Determine all possible real solutions to this system of equations:

A+B+C+D = 5, and:
A*B+B*C+C*D+D*A = 4, and:
A*B*C+B*C*D+C*D*A+D*A*B= 3, and:
A*B*C*D= -1

Prove that there are no others.

No Solution Yet Submitted by K Sengupta    
Rating: 4.0000 (1 votes)

Comments: ( Back to comment list | You must be logged in to post comments.)
adding on | Comment 2 of 3 |
Good work, Harry.

You can extend your method to the third equation, then substitute the values you found for (A+C) and (B+D).

AC(B+D)+BD(A+C)=3 becomes 4(AC) + BD=3 using your first values.   Then you can solve that simultaneously with equation 4 to get values for AC and BD.

At that point there's enough information to evaluate A and C.

I don't have the time now for a full listing, but I've found these other values for A:

(1+sqrt(5))/2
(1-sqrt(5))/2
2+sqrt(5)
2-sqrt(5)
(4+sqrt(17))/2
(4-sqrt(17))/2


  Posted by xdog on 2016-08-11 18:20:34
Please log in:
Login:
Password:
Remember me:
Sign up! | Forgot password


Search:
Search body:
Forums (0)
Newest Problems
Random Problem
FAQ | About This Site
Site Statistics
New Comments (3)
Unsolved Problems
Top Rated Problems
This month's top
Most Commented On

Chatterbox:
Copyright © 2002 - 2017 by Animus Pactum Consulting. All rights reserved. Privacy Information