<s>[INST] How find this system $a^2+b^2=3,a^2+c^2+ac=4,b^2+c^2+\sqrt{3}bc=7$ Find the  this system real solution
$$\begin{cases}
a^2+b^2=3\\
a^2+c^2+ac=4\\
b^2+c^2+\sqrt{3}bc=7
\end{cases}$$
I think that  one can use Geometry to solve this system. Maybe there exist an algebraic method.
$$a^2+b^2=\sqrt{3}^2$$
$$a^2+c^2-2ac\cos{(120°)}=2^2$$
$$b^2+c^2-2bc\cos{150°}=\sqrt{7}$$
and note $$150°+120°+90°=360°$$
This is Mathematics competition of Zhejiang province today.
 [/INST] Here is an algebraic solution that yields all the real answers.
Let us introduce the complex numbers:
 $$x=\frac{-b+i a}{\sqrt{3}},\quad y=\frac{2a+c}{4}-i\frac{\sqrt{3}}{4}c.$$
The first two equations are equivalent to the statement: $\vert x\vert=\vert y\vert=1$, and the third equation tells us that
$$\vert \sqrt{3} x-2i y\vert^2=\left\vert b+\frac{\sqrt{3}+i}{2} c\right\vert^2
=b^2+c^2+\sqrt{3}bc=7$$
On the other hand, since $\vert x\vert=\vert y\vert=1$ we see that
$\vert \sqrt{3} x-2i y\vert^2=3+4+4\sqrt{3}\,\Re(ix\bar{y})$. Thus, we have
$\Re(ix\bar{y})=0$ that is $x\bar{y}$ is a real number of modulus $1$, or $y=\pm x$.


*

*If $y=x$, we conclude comparing real parts, and imaginary parts that
$$b=-\frac{1}{2\sqrt{3}}a,\qquad c=-\frac{4}{3}a,$$
and replacing in the first equation we get $b=\epsilon\sqrt{3/13}$ with $\epsilon\in\{+1,-1\}$, and this yields the couple of solutions:
$$
a=-\epsilon\frac{6}{\sqrt{13}},~b=\epsilon\frac{\sqrt{3}}{\sqrt{13}},~
c=\epsilon\frac{ 8}{\sqrt{13}},\quad\hbox{with }\epsilon\in\{+1,-1\}$$

*If $y=-x$, we conclude similarly that
$$b=\frac{5}{2\sqrt{3}}a,\qquad c=\frac{4}{3}a,$$
and Replacing in the first equation we obtain the second couple of solutions:
$$
a= \epsilon\frac{6}{\sqrt{37}},~b=\epsilon\frac{5\sqrt{3}}{\sqrt{37}},~
c=\epsilon\frac{ 8}{\sqrt{37}},\quad\hbox{with }\epsilon\in\{+1,-1\}$$
and we are done, with four solutions in total.

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