Problem 2038

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

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

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

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

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