
Problem 5. Find all real numbers $t$ for which there exist real numbers $x, y, z$ such that

$$
3 x^{2}+3 x z+z^{2}=1,3 y^{2}+3 y z+z^{2}=4, x^{2}-x y+y^{2}=t
$$


Solution. We shall prove that the answer is $t \in\left(\frac{3-\sqrt{5}}{2}, 1\right)$.

Let $x, y, t, \alpha$ satisfy the conditions of the problem. Consider four points $A, B, C, O$ in the plane such that $A O=x, B O=y, C O=$ $\frac{z}{\sqrt{3}}$ and $\Varangle A O B=60^{\circ}, \Varangle B O C=\Varangle C O A=150^{\circ}$. By the cosine theorem it follows that $t>0, A B=\sqrt{t}, B C=\frac{2}{\sqrt{3}}, C A=\frac{1}{\sqrt{3}}$. Since $\Varangle A C B\frac{1}{2}
$$

$$
\frac{(\sqrt{t})^{2}+\left(\frac{1}{\sqrt{3}}\right)^{2}-\left(\frac{2}{\sqrt{3}}\right)^{2}}{2 \sqrt{\alpha} \frac{2}{\sqrt{3}}}=\cos \Varangle B A C>-\frac{\sqrt{3}}{2} .
$$

These inequalities are equivalent to $t0$, i.e. $t \in\left(\frac{3-\sqrt{5}}{2}, 1\right)$.

Conversely, let $t \in\left(\frac{3-\sqrt{5}}{2}, 1\right)$. Then we can construct a triangle $A B C$ such that $A B=\sqrt{t}, B C=\frac{2}{\sqrt{3}}, C A=\frac{1}{\sqrt{3}}$. By (1) and (2) it follows that $\Varangle A C B<60^{\circ}$ and $\Varangle A B C<\Varangle B A C<150^{\circ}$. Let $k_{A}$ be the circle through $B$ and $C$ such that the arc $B C$, lying in the half-plane (with respect to $B C$ ) containing the point $A$, is equal to $60^{\circ}$. Denote by $k_{B}$ the analogous circle trough $A$ and $C$. Then it is easy to see that the second common point $O$ of $k_{A}$ and $k_{B}$ lies in the interior of $\triangle A B C$. (Indeed, assume the contrary. If $O$ lies in $\Varangle A C B$, then $\Varangle A O B=\Varangle A O C+\Varangle B O C=150^{\circ}+150^{\circ}=300^{\circ}$ which is impossible. If $O$ lies in the opposite angle of $\Varangle A C B$, then $\Varangle A O B=\Varangle A O C+\Varangle B O C=30^{\circ}+30^{\circ}=60^{\circ}$ which contardicts to the inequalities $\Varangle A O B<\Varangle A C B<60^{\circ}$. The other cases for the position of $O$ can be rejected in the same manner.) Hence $\Varangle A O B=360^{\circ}-\Varangle A O C-\Varangle B O C=360^{\circ}-150^{\circ}-150^{\circ}=60^{\circ}$. Set $x=A O, y=B O, z=C O \sqrt{3}$. Applying the cosine theorem for triangles $A O B, B O C$ and $C O A$ it follows that $x, y, z, t$ satisfy the conditions of the problem.
