Welcome to our community

Be a part of something great, join today!

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

  • Thread starter
  • Admin
  • #1

anemone

MHB POTW Director
Staff member
Feb 14, 2012
3,601
Given that \(\displaystyle a,\,b\) and \(\displaystyle c\) are real numbers that satisfy the system of equations below:

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

Find \(\displaystyle (a-c)^2+(a^2-c^2)^2\).
 

Opalg

MHB Oldtimer
Staff member
Feb 7, 2012
2,684
Given that \(\displaystyle a,\,b\) and \(\displaystyle c\) are real numbers that satisfy the system of equations below:

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

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

 
  • Thread starter
  • Admin
  • #3

anemone

MHB POTW Director
Staff member
Feb 14, 2012
3,601
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)
 

topsquark

Well-known member
MHB Math Helper
Aug 30, 2012
1,121
This took me much longer than it should have done!
$$\begin{aligned}(a-c)^2+(a^2-c^2)^2 &= \bigl((a-b) + (b-c)\bigr)^2 + \bigl((a^2-b^2) + (b^2-c^2)\bigr)^2 \\ &= (a-b)^2 + 2(a-b)(b-c) + (b-c)^2 + (a^2-b^2)^2 + 2(a^2-b^2)(b^2-c^2) + (b^2-c^2)^2 \\ &= (a-b)^2 + (a^2-b^2)^2 + (b-c)^2 + (b^2-c^2)^2 + 2(a-b)(b-c) + 2(a^2-b^2)(b^2-c^2) \\ &= 85 + 75 + 2(a-b)(b-c)\bigl(1 + (a+b)(b+c)\bigr) \\ &= 160 + 2(a-b)(b-c)(1 - 1) = 160 \end{aligned}$$

Brilliant!

-Dan