Example 6 In a rectangle $A B C D$ with an area of 1 (including the boundary) there are 5 points, among which no three points are collinear. Find the minimum number of triangles, with these 5 points as vertices, whose area is not greater than $\frac{1}{4}$.
In rectangle $ABCD$, if the area of a triangle formed by any three points is no more than $\frac{1}{4}$, then these three points are called a "good triplet" or simply a "good group."

Let the midpoints of $AB$, $CD$, $BC$, and $AD$ be $E$, $F$, $H$, and $G$ respectively. The intersection of line segments $EF$ and $GH$ is denoted as $O$. Line segments $EF$ and $GH$ divide rectangle $ABCD$ into four smaller rectangles. Therefore, there must be a smaller rectangle, say $AEOG$, which contains at least two of the given five points (including the boundaries, hereinafter the same). Let $M$ and $N$ be in the smaller rectangle $AEOG$ (Figure 6-31).

(1) If rectangle $OHCF$ contains no more than one of the known points, consider any known point $X$ not in rectangle $OHCF$ and different from $M$ and $N$. It is easy to see that the triplet $(M, N, X)$ is either in rectangle $ABHG$ or in rectangle $AEFD$. By the lemma, $(M, N, X)$ is a good group. Since there are at least two such $X$, there are at least two good groups.

(2) If rectangle $OHCF$ contains at least two known points, say $P$ and $Q$ are in rectangle $OHCF$, consider the remaining known point $R$. If $R$ is in rectangle $OFDG$, then the triplet $(P, Q, R)$ is in rectangle $GHCD$, and the triplet $(M, N, R)$ is in rectangle $AEFD$, so they are both good groups, thus there are at least two good groups. Similarly, if $R$ is in rectangle $EBHO$, there are also at least two good groups. If point $R$ is in rectangle $OHCF$ or rectangle $AEOG$, without loss of generality, assume $R$ is in rectangle $OHCF$. We consider the convex hull of the five points $M, N, P, Q, R$. This convex hull must be within the convex hexagon $AEHCFD$ (Figure 6-32), and
$$
S_{AEHCFD} = 1 - \frac{1}{8} - \frac{1}{8} = \frac{3}{4}.
$$

We now consider three cases:
(i) If the convex hull of $M, N, P, Q, R$ is a convex pentagon, say $MNPQR$ (Figure 6-33), then
$$
S_{\triangle MQR} + S_{\triangle MQP} + S_{\triangle NQP} \leq \frac{3}{4},
$$
so at least one of $(M, Q, R)$, $(M, N, Q)$, $(N, P, Q)$ is a good group. Since $(P, Q, R)$ is in rectangle $OHCF$, it is also a good group, so there are at least two good groups.
(ii) If the convex hull of $M, N, P, Q, R$ is a convex quadrilateral, say $A_1A_2A_3A_4$, and the other known point $A_5$ is inside $A_1A_2A_3A_4$ (Figure 6-34), where $A_i \in \{M, N, P, Q, R\} (i=1,2,3,4,5)$, connect $A_5A_i (i=1,2,3,4)$, then
$$
S_{\triangle A_1A_2A_5} + S_{\triangle A_2A_3A_5} + S_{\triangle A_3A_4A_5} + S_{\triangle A_4A_1A_5} \leq \frac{3}{4},
$$
so at least two of $(A_1, A_2, A_5)$, $(A_2, A_3, A_5)$, $(A_3, A_4, A_5)$, $(A_4, A_1, A_5)$ are good groups.
(iii) If the convex hull of $M, N, P, Q, R$ is a triangle, say $\triangle A_1A_2A_3$, and the other two known points $A_4, A_5$ are inside $\triangle A_1A_2A_3$, then
$$
S_{\triangle A_1A_2A_4} + S_{\triangle A_2A_3A_4} + S_{\triangle A_3A_1A_4} = S_{\triangle A_1A_2A_3} \leq \frac{3}{4},
$$
so at least one of $(A_1, A_2, A_4)$, $(A_2, A_3, A_4)$, $(A_3, A_1, A_4)$ is a good group. Similarly, at least one of $(A_1, A_2, A_5)$, $(A_2, A_3, A_5)$, $(A_3, A_1, A_5)$ is a good group, so there are at least two good groups.
In summary, regardless of the situation, there are at least two good groups among the five known points.

Next, we provide an example to show that the number of good groups can be exactly two. In rectangle $ABCD$, take a point $M$ on side $AD$ and a point $N$ on side $AB$ such that $AN:NB = AM:MD = 2:3$ (Figure 6-35). Then, among the five points $M, N, B, C, D$, there are exactly two good groups. In fact, $(B, C, D)$ is clearly not a good group. If a triplet contains exactly one of $M$ or $N$, say it contains point $M$, and let the midpoint of $AD$ be $E$, then $S_{\triangle MBD} > S_{\triangle EBD} = \frac{1}{4}$, so $(M, B, D)$ is not a good group, and $S_{\triangle MBC} = \frac{1}{2}$, $S_{\triangle MCD} > S_{\triangle ECD} = \frac{1}{4}$, so $(M, B, C)$ and $(M, C, D)$ are not good groups. If a triplet contains both $M$ and $N$, then
$$
\begin{aligned}
S_{\triangle MNC} & = 1 - S_{\triangle NBC} - S_{\triangle MCD} - S_{\triangle AMN} \\
& = 1 - \frac{3}{5} S_{\triangle ABC} - \frac{3}{5} S_{\triangle ACD} - \frac{4}{25} S_{\triangle ABD} \\
& = 1 - \frac{3}{10} - \frac{3}{10} - \frac{2}{25} = \frac{8}{25} > \frac{1}{4}.
\end{aligned}
$$
So $(M, N, C)$ is not a good group, but $S_{\triangle MNB} = S_{\triangle MND} = \frac{3}{5} S_{\triangle MMB} = \frac{3}{5} \times \frac{2}{5} S_{\triangle ABD} = \frac{3}{25} < \frac{1}{4}$, so there are exactly two good groups $(M, N, B)$ and $(M, N, D)$.
Therefore, the minimum number of triangles with an area no more than $\frac{1}{4}$ is 2.