Welcome to our community

Be a part of something great, join today!

Simultaneous Equations Challenge

  • Thread starter
  • Admin
  • #1

anemone

MHB POTW Director
Staff member
Feb 14, 2012
3,684
Solve the following system in real numbers:

\(\displaystyle a^2+b^2=2c\)

\(\displaystyle 1+a^2=2ac\)

\(\displaystyle c^2=ab\)
 

Ackbach

Indicium Physicus
Staff member
Jan 26, 2012
4,192
Solve the following system in real numbers:

\(\displaystyle a^2+b^2=2c\)

\(\displaystyle 1+a^2=2ac\)

\(\displaystyle c^2=ab\)
I find it fascinating that the first equation requires $c\ge 0$. Then the second equation in conjunction with the first requires $a>0$, and incidentally tightens the $c$ inequality to $c>0$. Then the third equation, in conjunction with $c>0$ and $a>0$ requires $b>0$. Dominos!
 

Jester

Well-known member
MHB Math Helper
Jan 26, 2012
183
My solution
a = b = c = 1.

I have a proof that this is the only solution but I'm looking for something a little more elegant.


If we add the first two equations and then 2 times the third we end up with

$(a-b)^2+(a-c)^2+(c-1)^2 = 0$

then $a = b = c$ and $c = 1$ giving our result.
 
Last edited:
  • Thread starter
  • Admin
  • #4

anemone

MHB POTW Director
Staff member
Feb 14, 2012
3,684
My solution
a = b = c = 1.

I have a proof that this is the only solution but I'm looking for something a little more elegant.


If we add the first two equations and then 2 times the third we end up with

$(a-b)^2+(a-c)^2+(c-1)^2 = 0$

then $a = b = c$ and $c = 1$ giving our result.
Thank you Jester for participating and your solution is so much neater and shorter than mine! Well done!:)

My solution:


From \(\displaystyle a^2+b^2=2c\) and \(\displaystyle c^2=ab\), we have

\(\displaystyle a^2+\frac{c^4}{a^2}=2c\;\rightarrow\;\;a^4-2a^2c+c^4=0\)(*)

And from \(\displaystyle a^2+1=2ac\) we square both sides and get

\(\displaystyle a^4+2a^2+1-4a^2c^2=0\)(**)

Now, subtracting the equation (*) from (**) gives

\(\displaystyle 2a^2+1-2a^2c-c^4=0\)

\(\displaystyle 2a^2(1-c)+(1-c^4)=0\)

\(\displaystyle 2a^2(1-c)+(1-c)(1+c+c^2+c^3)=0\)

\(\displaystyle (1-c)(2a^2+1+c+c^2+c^3)=0\)

Since a, b, and c >0, \(\displaystyle 2a^2+1+c+c^2+c^3 \ne 0\) and thus it must be \(\displaystyle 1-c=0\;\rightarrow\;\;c=1\).

Back substituting \(\displaystyle c=1\) to the equation \(\displaystyle a^4-2a^2c+c^4=0\) and gives

\(\displaystyle a^4-2a^2+1=0\)

\(\displaystyle a^2-1=0\)

\(\displaystyle a^2=\pm1\) Since \(\displaystyle a>0\), we conclude that \(\displaystyle a=1\).

And from \(\displaystyle c=ab\), we have

\(\displaystyle 1=1(b)\;\rightarrow b=1\).