60.4. Let $A B C D$ be a square. Let $M$ be the midpoint of $\overline{D C}, N$ the midpoint of $\overline{A C}$, and $P$ the intersection of $\overline{B M}$ and $\overline{A C}$. What is the ratio of the area of $\triangle M N P$ to that of the square $A B C D$ ?
60.4. $1: 24$
Refer to Figure 8. Notice that $\triangle M N P \sim \triangle B C P$, so that
$$
\frac{N P}{P C}=\frac{M N}{B C}=\frac{1}{2} \quad \text { and } \quad \frac{M P}{B P}=\frac{M N}{B C}=\frac{1}{2} .
$$

Recall that the ratio of the areas of two triangles of equal altitudes is equal to the ratio of the corresponding bases. With the notation $(Z)$ to mean the area of polygon $Z$.
$$
(M N P)=\frac{1}{2}(M P C)=\frac{1}{4}(B P C)=\frac{1}{4}\left[\frac{2}{3}(M B C)\right]=\frac{1}{6}\left[\frac{1}{4}(A B C D)\right],
$$
and so $(M N P):(A B C D)=1: 24$.