Simultaneous Equations Challenge

[anemone]

Solve the following system in real numbers:

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

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

\(\displaystyle c^2=ab\)

 

[Ackbach]

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!

 

[Mathwebster]

My solution

Spoiler

a = b = c = 1.

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

Spoiler

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.

 

[anemone]

My solution

Spoiler

a = b = c = 1.

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

Spoiler

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:

Spoiler

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

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

 

[CellCoree]

To solve this system of equations, we can use the substitution method. First, we will solve for c in the third equation by taking the square root of both sides:

c = √(ab)

Next, we can substitute this value of c into the first equation to get:

a^2+b^2=2√(ab)

We can then rearrange this equation to solve for b:

b^2 - 2√(ab) + a^2 = 0

Using the quadratic formula, we can solve for b:

b = (√(2a) ± √(2a - a^2)) / 2

Now, we can substitute this value of b into the second equation to get:

1+a^2=2a√(2a - a^2)

Again, we can rearrange this equation to solve for a:

a^4 - 4a^3 + 4a^2 - 1 = 0

Using the quadratic formula again, we can solve for a:

a = (2 ± √(4 - 4(-1))) / 2

a = (2 ± 2√2) / 2

a = 1 ± √2

Now that we have a value for a, we can plug it back into the equations to solve for b and c. We can also use the values of a and c to find the value of b using the first equation. Finally, we can use these values to verify that they satisfy all three equations.

Therefore, the solutions to this system of equations in real numbers are:

a = 1 + √2, b = 1 + √2, c = √(2 + 2√2).