Irish Mathematical Olympiad

First solution. Let $Q$ be the point inside $AC$ such that $MQ$ is parallel to $BC$. Then, because $\triangle ABC$ and $\triangle AMQ$ are homothetic, $|AN| = |NQ| = |QC|$ and, if $(ABC)$ denotes the area of the triangle $ABC$ etc., we also have
$$\frac{(AMQ)}{(ABC)} = \frac{|AM|^2}{|AB|^2} = \frac{4}{9}$$

On the other hand, the line $AP$ intersects $MQ$ at the midpoint $E$ of $MQ$, again due to the homothety with centre $A$ which takes $\triangle AMQ$ to $\triangle ABC$. Thus $D$ is the centroid of $\triangle AMQ$ and so
$$\frac{(ADN)}{(AMN)} = \frac{|DN|}{|MN|} = \frac{1}{3}, \quad \text{while} \quad \frac{(AMN)}{(AMQ)} = \frac{|AN|}{|AQ|} = \frac{1}{2}$$
Multiplication of these three identities yields
$$\frac{(ADN)}{(ABC)} = \frac{2}{27}.$$

Second solution. We note that
$$\frac{(AMD)}{(ABD)} = \frac{|AM|}{|AB|} = \frac{2}{3}, \quad \text{while} \quad \frac{(ADN)}{(ADC)} = \frac{|AN|}{|AC|} = \frac{1}{3}.$$
On the other hand, $(ABP) = (APC)$ and $(DBP) = (DPC)$, because $|BP| = |PC|$ and the corresponding triangles have identical altitudes. Taking differences, this implies $(ABD) = (ADC)$ and we obtain $(AMD) = 2(ADN)$. From
$$(AMD) + (ADN) = (AMN) = \frac{|AM| \cdot |AN|}{|AB| \cdot |AC|} (ABC) = \frac{2}{9},$$
$$\text{we now obtain } 2(ADN) + (ADN) = \frac{2}{9}, \text{ hence } (ADN) = \frac{2}{27}.$$