2. The medians $t_{a}$ and $t_{b}$ in triangle $A B C$ form angles with side $A B$ whose sum is $60^{\circ}$. Calculate the area of triangle $A B C$, if $t_{a} t_{b}=\sqrt{3}$.
Solution. Since the medians with side $AB$ form angles whose sum is $60^{\circ}$, it follows that

$$
\begin{aligned}
& \Varangle A T B=120^{\circ}, \measuredangle A_{1} T B_{1}=120^{\circ}, \\
& \measuredangle A_{1} T B=60^{\circ} \text { and } \measuredangle A T B_{1}=60^{\circ} .
\end{aligned}
$$

Let $\overline{A T}=n, \overline{A_{1} T}=m, \overline{B T}=s$ and $\overline{B_{1} T}=t$. Now

$$
\begin{aligned}
& \sqrt{3}=t_{a} t_{b}=(m+n)(s+t)=m t+m s+n t+n s \quad A \\
& P_{A B A_{1} B_{1}}=\frac{1}{2} n s \sin 120^{\circ}+\frac{1}{2} s m \sin 60^{\circ}+\frac{1}{2} m t \sin 120^{\circ}+\frac{1}{2} n t \sin 60^{\circ} \\
&=\frac{\sqrt{3}}{4}(n s+s m+m t+n t)=\frac{3}{4} .
\end{aligned}
$$

![](https://cdn.mathpix.com/cropped/2024_06_05_e95f2dbd17df8adc2c7cg-05.jpg?height=312&width=430&top_left_y=1692&top_left_x=1049)

Since $A_{1} B_{1}$ is the midline in $\triangle A B C$, it follows that triangles $A B C$ and $A_{1} B_{1} C$ are similar with a similarity coefficient of 2. Therefore, $P_{A B C}=4 P_{A_{1} B_{1} C}$, which means $P_{A_{1} B_{1} C}=\frac{1}{3} P_{A B A_{1} B_{1}}=\frac{1}{4}$. Thus, $P_{A B C}=P_{A B A_{1} B_{1}}+P_{A_{1} B_{1} C}=1$.