Task A-3.4. (8 points)

If for the lengths $a, b, c$ of the sides of a triangle it holds that $(a+b+c)(a+b-c)=3 a b$, determine the angle opposite side $c$.
## Solution.

The given expression can be written in the form $(a+b)^{2}-c^{2}=3 a b$, or $a^{2}+b^{2}-c^{2}=a b$.

According to the cosine rule, $\cos \gamma=\frac{a^{2}+b^{2}-c^{2}}{2 a b}=\frac{a b}{2 a b}=\frac{1}{2}$, so $\gamma=60^{\circ}$.

For 4 points: correct up to the result $\cos \gamma=\frac{1}{2}$.

For 8 points: correct result and procedure.