Home > Just Math
Real and Cyclic Sum (Posted on 20160810) 

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)


Possible Solution

 Comment 1 of 3

Writing
the first two equations as (A + C) + (B +
D) = 5
and (A + C)(B + D) = 4
and eliminating B + D gives a quadratic equation in (A + C):
(A + C)^{2} – 5(A + C)
+ 4 = 0
with solution (A + C, B + D) = (1,
4) or
(4, 1).
First consider A + C = 1, B +
D = 4.
Sub. C = 1 – A and D = 4 – B into the 3rd and 4th equations:
The 3rd equation AB(C + D) + CD(A + B) = 3
becomes AB(5 – A – B) + (1 – A)(4 – B)(A + B) = 3
and simplifies to (B – 2)^{2} =
4A – 4A^{2} + 1 (1)
The 4th equation AB(1 – A)(4 – B) = 1
becomes (B – 2)^{2}
= 4 + 1/(A(1 – A)) (2)
Using (1) and (2): 4A – 4A^{2}
+ 1 = 4 + 1/(A(1 – A))
which simplifies to 4A^{4}
– 8A^{3} + 7A^{2} – 3A  1 = 0
Thus (A^{2}
– A + 1)(4A^{2} – 4A + 1) = 0
The first factor yields no real roots, but the second factor gives
the two real roots:
A = (1 + sqrt2)/2
and (1 – sqrt2)/2
For both of these, substitution in (1) gives B = 2 and C and D can
then be found from earlier work to give the two following answers:
{A, B, C, D} = {(1 + sqrt2)/2, 2, (1 –
sqrt2)/2, 2}
or {A, B, C, D} = {(1  sqrt2)/2, 2, (1 +
sqrt2)/2, 2}
Second consideration of A + C = 4, B + D = 1 gives two further
answers (otherwise found by cycling A>B, B>A, C>D, D>C):
{A, B, C, D} = {2, (1 +
sqrt2)/2, 2, (1 – sqrt2)/2}
{A, B, C, D} = {2, (1 – sqrt2)/2, 2, (1 + sqrt2)/2}
All possibilities considered and all real roots retained but I don’t
know a simple way of checking that no other real solutions exist.

Posted by Harry
on 20160811 16:31:53 


Please log in:
Forums (0)
Newest Problems
Random Problem
FAQ 
About This Site
Site Statistics
New Comments (1)
Unsolved Problems
Top Rated Problems
This month's top
Most Commented On
Chatterbox:
