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.
 A: Assuming $a,b,c>0$, then as you noted the equalities are just cosine laws for $3$ triangles which form a larger triangle with sides $2,\sqrt3,\sqrt7$, because the angles add up to $2\pi$. That triangle is right, because $4+3=7$, so we can find the lengths analytically if we draw it like this:

Here we set $A=(0,0)$, $B=\left(0,\sqrt3\right)$ and $C=(2,0)$. The angle $APB$ is right, so $P$ lies on the circle with the center $S_1=\left(0,\frac{\sqrt3}2\right)$ and radius $\frac{\sqrt3}2$. The size of the angle $APC$ is $\frac{2\pi}3$, so $ACX$ is an equilateral triangle and $P$ lies on the circle with the center $S_2=\left(1,-\frac{\sqrt3}3\right)$ and radius $\frac{2\sqrt3}3$.
So if $P=(x,y)$, then
\begin{alignat*}{5}x^2+\left(y-\tfrac{\sqrt3}2\right)^2\ &=\left(\tfrac{\sqrt3}2\right)^2&&=\tfrac34&\ \Longleftrightarrow\ &&0\ &=x^2+y^2-\sqrt3\ y\\
(x-1)^2+\left(y+\tfrac{\sqrt3}3\right)^2\ &=\left(\tfrac{2\sqrt3}3\right)^2&&=\tfrac43&\ \Longleftrightarrow\ &&2x\ &=x^2+y^2+\tfrac{2\sqrt3}3y\end{alignat*}
So $x=\frac12\left(\tfrac{2\sqrt3}3+\sqrt3\right)y=\tfrac{5\sqrt3}6y$ and $0=\left(\tfrac{5\sqrt3}6y\right)^2+y^2-\sqrt3\ y\Longleftrightarrow0=y(111y-36\sqrt3)$, so $y=\frac{12\sqrt3}{37}$ and $x=\frac{30}{37}$, because we want the nonzero solution.
Therefore $a=\sqrt{\left(\frac{30}{37}\right)^2+\left(\frac{12\sqrt3}{37}\right)^2}=\dfrac6{\sqrt{37}}$, $b=\sqrt{\left(\frac{30}{37}\right)^2+\left(\sqrt3-\frac{12\sqrt3}{37}\right)^2}=\dfrac{5\sqrt3}{\sqrt{37}}$ and $c=\sqrt{\left(2-\frac{30}{37}\right)^2+\left(\frac{12\sqrt3}{37}\right)^2}=\dfrac8{\sqrt{37}}$.

If you want all real solutions, then it can be solved geometrically too. Just consider different signs before $ac$ and $bc$. For example $-ac=-2ac\cos\frac\pi3$ and $\sqrt3\,bc=-2bc\cos\frac{5\pi}6$. We see that $\frac\pi2+\frac\pi3=\frac{5\pi}6$, so we can draw the following picture:

The triangle $ACX$ is again equilateral and it can be solved in a similar fashion. This gives $a=\dfrac6{\sqrt{13}}, b=\dfrac{\sqrt3}{\sqrt{13}}, c=\dfrac8{\sqrt{13}}$ and if we negate $a$ we get a solution to the original problem.
Similar configurations can be made for the other $2$ possibilities, but there are no solutions, so these two we found together with their negations are all $4$ real solutions.
A: 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.

