MHB System of Equation Challenge (a+b)(b+c)=-1

  • Thread starter Thread starter anemone
  • Start date Start date
  • Tags Tags
    Challenge System
AI Thread Summary
The discussion revolves around solving a system of equations involving real numbers a, b, and c, specifically focusing on the expression (a-c)² + (a²-c²)². The equations provided include (a+b)(b+c) = -1, (a-b)² + (a²-b²)² = 85, and (b-c)² + (b²-c²)² = 75. Through algebraic manipulation, it is concluded that (a-c)² + (a²-c²)² equals 160. The thread highlights the complexity of the problem and expresses appreciation for the insights shared by a participant named Opalg. The solution is confirmed as correct and is noted to be an unsolved problem on the AOPS forum.
anemone
Gold Member
MHB
POTW Director
Messages
3,851
Reaction score
115
Given that $$a,\,b$$ and $$c$$ are real numbers that satisfy the system of equations below:

$$(a+b)(b+c)=-1\\(a-b)^2+(a^2-b^2)^2=85\\(b-c)^2+(b^2-c^2)^2=75$$

Find $$(a-c)^2+(a^2-c^2)^2$$.
 
Mathematics news on Phys.org
anemone said:
Given that $$a,\,b$$ and $$c$$ are real numbers that satisfy the system of equations below:

$$(a+b)(b+c)=-1\\(a-b)^2+(a^2-b^2)^2=85\\(b-c)^2+(b^2-c^2)^2=75$$

Find $$(a-c)^2+(a^2-c^2)^2$$.
This took me much longer than it should have done!
[sp]$$\begin{aligned}(a-c)^2+(a^2-c^2)^2 &= \bigl((a-b) + (b-c)\bigr)^2 + \bigl((a^2-b^2) + (b^2-c^2)\bigr)^2 \\ &= (a-b)^2 + 2(a-b)(b-c) + (b-c)^2 + (a^2-b^2)^2 + 2(a^2-b^2)(b^2-c^2) + (b^2-c^2)^2 \\ &= (a-b)^2 + (a^2-b^2)^2 + (b-c)^2 + (b^2-c^2)^2 + 2(a-b)(b-c) + 2(a^2-b^2)(b^2-c^2) \\ &= 85 + 75 + 2(a-b)(b-c)\bigl(1 + (a+b)(b+c)\bigr) \\ &= 160 + 2(a-b)(b-c)(1 - 1) = 160 \end{aligned}$$

[/sp]
 
Awesome, as always, Opalg!(Yes)

This actually is an unsolved problem at AOPS forum, and I will share the link of this thread at their site now, and thanks again Opalg for your great insight to attack the problem!(Cool)
 
Opalg said:
This took me much longer than it should have done!
[sp]$$\begin{aligned}(a-c)^2+(a^2-c^2)^2 &= \bigl((a-b) + (b-c)\bigr)^2 + \bigl((a^2-b^2) + (b^2-c^2)\bigr)^2 \\ &= (a-b)^2 + 2(a-b)(b-c) + (b-c)^2 + (a^2-b^2)^2 + 2(a^2-b^2)(b^2-c^2) + (b^2-c^2)^2 \\ &= (a-b)^2 + (a^2-b^2)^2 + (b-c)^2 + (b^2-c^2)^2 + 2(a-b)(b-c) + 2(a^2-b^2)(b^2-c^2) \\ &= 85 + 75 + 2(a-b)(b-c)\bigl(1 + (a+b)(b+c)\bigr) \\ &= 160 + 2(a-b)(b-c)(1 - 1) = 160 \end{aligned}$$

[/sp]
Brilliant!

-Dan
 
Suppose ,instead of the usual x,y coordinate system with an I basis vector along the x -axis and a corresponding j basis vector along the y-axis we instead have a different pair of basis vectors ,call them e and f along their respective axes. I have seen that this is an important subject in maths My question is what physical applications does such a model apply to? I am asking here because I have devoted quite a lot of time in the past to understanding convectors and the dual...
Insights auto threads is broken atm, so I'm manually creating these for new Insight articles. In Dirac’s Principles of Quantum Mechanics published in 1930 he introduced a “convenient notation” he referred to as a “delta function” which he treated as a continuum analog to the discrete Kronecker delta. The Kronecker delta is simply the indexed components of the identity operator in matrix algebra Source: https://www.physicsforums.com/insights/what-exactly-is-diracs-delta-function/ by...

Similar threads

Replies
10
Views
1K
Replies
2
Views
1K
Replies
2
Views
1K
Replies
19
Views
3K
Replies
3
Views
1K
Replies
2
Views
1K
Replies
1
Views
1K
Replies
3
Views
987
Back
Top