Selected Problems from Open Contests

By finding the common denominator on the left hand side, transform the equation to $(a+2b+c)(a+b+c)=3(a+b)(b+c)$. Expanding the brackets and simplifying gives $b^2=a^2+c^2-ac$. Comparing the latter with the cosine law $b^2=a^2+c^2-2ac\cos \beta$, we see that the equality holds if and only if $\cos \beta =\frac{1}{2}$, i.e., $\beta =60^{\circ }$.