Brazilian Math Olympiad

Let's prove that there exists a triangle $P_j{P}_{j+1}{P}_{j+2}$ with area less than or equal to $\alpha =\frac{5-\sqrt{5}}{10}$. Suppose that all triangles $P_j{P}_{j+1}{P}_{j+2}$ have area greater than $\alpha$.



Let diagonals $P_1{P}_4$ and $P_3{P}_5$ meet at $Q$. Since $Q\in P_3{P}_5$, $\text{area }{P}_1{P}_{2}Q\le \max (\text{area }{P}_1{P}_2{P}_5,\text{area }{P}_1{P}_2{P}_3)<\alpha$, so $\text{area }{P}_1{P}_2{P}_4=1-\text{area }{P}_1{P}_4{P}_5-\text{area }{P}_2{P}_3{P}_4>1-2\alpha$. Thus $$\frac{P_{1}Q}{P_1{P}_4}=\frac{\text{area }{P}_1{P}_{2}Q}{\text{area }{P}_1{P}_2{P}_4}<\frac{\alpha }{1-2\alpha }$$ We also have $\frac{P_{1}Q}{P_{4}Q}=\frac{\text{area }{P}_1{P}_3{P}_5}{\text{area }{P}_3{P}_4{P}_5}$. Since $\text{area }{P}_3{P}_4{P}_5<\alpha$ and $\text{area }{P}_1{P}_3{P}_5>1-2\alpha$, $$\frac{P_{1}Q}{P_{4}Q}>\frac{1-2\alpha }{\alpha }⟺\frac{P_{1}Q}{P_1{P}_4}>\frac{1-2\alpha }{1-\alpha }$$ Therefore $$\frac{1-2\alpha }{1-\alpha }<\frac{P_{1}Q}{P_1{P}_4}<\frac{\alpha }{1-2\alpha }⟹5{\alpha }^2-5\alpha +1<0⟺\frac{5-\sqrt{5}}{10}<\alpha <\frac{5+\sqrt{5}}{10},$$ contradiction.

The proof of the other inequality is analogous.