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 $P_1P_4$ and $P_3P_5$ meet at $Q$. Since $Q \in P_3P_5$, $\text{area } P_1P_2Q \le \max(\text{area } P_1P_2P_5, \text{area } P_1P_2P_3) < \alpha$, so $\text{area } P_1P_2P_4 = 1 - \text{area } P_1P_4P_5 - \text{area } P_2P_3P_4 > 1 - 2\alpha$. Thus
$$\frac{P_1Q}{P_1P_4} = \frac{\text{area } P_1P_2Q}{\text{area } P_1P_2P_4} < \frac{\alpha}{1 - 2\alpha}$$
We also have $\frac{P_1Q}{P_4Q} = \frac{\text{area } P_1P_3P_5}{\text{area } P_3P_4P_5}$. Since $\text{area } P_3P_4P_5 < \alpha$ and $\text{area } P_1P_3P_5 > 1 - 2\alpha$,
$$\frac{P_1Q}{P_4Q} > \frac{1 - 2\alpha}{\alpha} \iff \frac{P_1Q}{P_1P_4} > \frac{1 - 2\alpha}{1 - \alpha}$$
Therefore
$$\frac{1-2\alpha}{1-\alpha} < \frac{P_1Q}{P_1P_4} < \frac{\alpha}{1-2\alpha} \implies 5\alpha^2 - 5\alpha + 1 < 0 \iff \frac{5-\sqrt{5}}{10} < \alpha < \frac{5+\sqrt{5}}{10},$$
contradiction.

The proof of the other inequality is analogous.