Philippine Mathematical Olympiad

Solution:

Denote by $(XYZ)$ the area of $\triangle XYZ$. Let $w = (APF) = (BPD) = (CPE)$, $x = (BPF)$, $y = (CPD)$, and $z = (APE)$.

Having the same altitude, we get
$$\frac{BD}{DC} = \frac{(BAD)}{(CAD)} = \frac{2w + x}{w + y + z}$$
and
$$\frac{BD}{DC} = \frac{(BPD)}{(CPD)} = \frac{w}{y}$$
which implies

$$w y + x y = w^2 + w z$$
Similarly, we also get
$$w z + y z = w^2 + w x \quad \text{and} \quad w x + x z = w^2 + w y$$
Combining equations (1) and (2) gives
$$x y + y z + x z = 3w^2$$
On the other hand, by Ceva's Theorem, we have
$$\frac{AF}{FB} \cdot \frac{BD}{DC} \cdot \frac{CE}{EA} = \frac{(APF)}{(BPF)} \cdot \frac{(BPD)}{(CPD)} \cdot \frac{(CPE)}{(APE)} = \frac{w}{x} \cdot \frac{w}{y} \cdot \frac{w}{z} = 1$$
or
$$w^3 = x y z$$
Applying equation (5) to equation (3) gives
$$\frac{w}{z} + \frac{w}{x} + \frac{w}{y} = 3$$
Equations (4) and (6) assert that the geometric mean and the arithmetic mean of the positive numbers $\frac{w}{x}$, $\frac{w}{y}$, and $\frac{w}{z}$ are equal. By the equality condition of the AM-GM Inequality, it follows that
$$\frac{w}{x} = \frac{w}{y} = \frac{w}{z} = 1 \quad \text{or} \quad w = x = y = z$$
Therefore, we conclude that $AF = FB$, $BD = DC$, and $CE = EA$, which means that $P$ is the centroid of $\triangle ABC$.