Estonian Mathematical Olympiad

Let the sizes of the angles of the pentagon be denoted in decreasing order as $\alpha \ge \beta \ge \gamma \ge \delta \ge \epsilon$. The sum of all the interior angles is $(5-2)\cdot 180^{\circ }$, in other words $\alpha +\beta +\gamma +\delta +\epsilon =540^{\circ }$.

The angle that equals the sum of the other four is greater than the other four. Therefore $\alpha =\beta +\gamma +\delta +\epsilon =\frac{540^{\circ }}{2}=270^{\circ }$.

The angle which equals the sum of some other three cannot be equal to $\alpha$, because then $\epsilon =0^{\circ }$. As it must be greater than the other three angles, $\beta =\gamma +\delta +\epsilon =\frac{270^{\circ }}{2}=135^{\circ }$.

Analogously, the angle that is the sum of some other two angles can be equal to neither $\alpha$ nor $\beta$, therefore $\gamma =\delta +\epsilon =\frac{135^{\circ }}{2}=67.5^{\circ }$.

Finally, the pentagon cannot have more angles of size $270^{\circ }$, $135^{\circ }$ or $67.5^{\circ }$, therefore $\delta =\epsilon =\frac{67.5^{\circ }}{2}=33.75^{\circ }$.