System of $3$ nonlinear equations

Question

Find positive solutions to the following:

\begin{align*} x^2+y^2+xy=&1\\ y^2+z^2+yz=&3\\ z^2+x^2+xz=&4 \end{align*}

I simplified and got $x+y+z=\sqrt{7}$ and $x^2+y^2+xy=1$. How do I continue?

Answer 1

Through the cosine theorem, the problem can be stated in the following way:

In a triangle $ABC$ with side lengths $1,\sqrt{3},2$, what are the distances of the Fermat point $F$ from the vertices of $ABC$?

Well, such a triangle is a right triangle, since $1^2+\sqrt{3}^2=2^2$, and $FA+FB+FC$ is the length of the Steiner net of $ABC$. It follows that:

$$ x+y+z = \sqrt{1^2+\sqrt{3}^2-2\sqrt{3}\cos 150^\circ}=\sqrt{7}$$

To find $x,y,z$, i.e. the lenghts of $FA,FB,FC$, it is enough to embed this construction in the plane by setting, for instance, $A=(0,0), B=(1,0), C=(0,2)$, then finding the coordinates of $F$ by intersecting a couple of lines. As an alternative, since the trilinear coordinates of $F$ are given by $$ F = \left[\frac{1}{\sin(A+\pi/3)};\frac{1}{\sin(B+\pi/3)};\frac{1}{\sin(C+\pi/3)}\right]$$ with the previous assumptions we have $$ F=\left[1;\frac{\sqrt{5}}{2+\sqrt{3}};\frac{\sqrt{5}}{1+2\sqrt{3}}\right]$$ and the tripolar coordinates of $F$ are $[1,2,4]$.

On the other hand, once we get $x+y+z=\sqrt{7}$ our job is done, since the original system of equations implies: $$\left\{\begin{array}{rcr}(x-z)(x+y+z)&=&-2\\(y-x)(x+y+z)&=&-1\\(z-y)(x+y+z)&=&3 \end{array}\right. $$ or: $$\left\{\begin{array}{rcr}z-x&=&\frac{2}{\sqrt{7}}\\x-y&=&\frac{1}{\sqrt{7}}\\z-y&=&\frac{3}{\sqrt{7}} \end{array}\right. $$ so that $(x,y,z)=\left(\tau,\tau-\frac{1}{\sqrt{7}},\tau+\frac{2}{\sqrt{7}}\right)$.

By solving the original system with respect to $\tau$, we get $\tau=\frac{2}{\sqrt{7}}$. The other solutions (non-geometrical, associated with negative lengths) can be derived in a similar way.

Answer 2

All positive real solutions are given by $$ (x,y,z)=\frac{1}{\sqrt{7}}(2,1,4). $$ I obtained the solutions by finding one linear equation (note that $x+y+z= 7$ does not hold), namely $2y + z - 3x=0$. Then replacing $z=3x-2y$ yields two quadratic equations, which are easy to solve with the formula for quadratic equations. This gives all solutions, namely $$ (x,y,z)=\pm\frac{1}{\sqrt{7}}(2,1,4), \pm (0,-1,2). $$