5 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, formed by these 5 points as vertices, whose area is not greater than $\frac{1}{4}$. (Supplied by Leng Gangsong)
The proof of this problem requires the following commonly used conclusion, which we will use as a lemma: The area of any triangle within a rectangle is no more than half the area of the rectangle.

In rectangle $ABCD$, if a triangle formed by any three points has an area no greater 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, and the intersection of line segments $EF$ and $GH$ be $O$. The line segments $EF$ and $GH$ divide rectangle $ABCD$ into four smaller rectangles. Therefore, there must be a smaller rectangle, let's say $AEOG$, which contains at least two of the given five points (including the boundaries, as below). Let points $M$ and $N$ be in the smaller rectangle $AEOG$, as shown in Figure 5.
(1) If rectangle $OHCF$ contains no more than 1 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 points $X$, there are at least two good groups.
(2) If rectangle $OHCF$ contains at least 2 known points, let's say points $P$ and $Q$ are both in rectangle $OHCF$, and consider the remaining known point $R$. If $R$ is in rectangle $OFDG$, then the triplet $(M, N, R)$ is in rectangle $AEFD$, and the triplet $(P, Q, R)$ is in rectangle $GHCD$, thus they are both good groups, so there are at least two good groups. Similarly, if point $R$ is in rectangle $EBHO$, there are also at least two good groups. If point $R$ is in rectangle $OHCF$ or rectangle $AEOG$, let's assume point $R$ is in rectangle $OHCF$. We will examine the convex hull of the five points $M, N, P, Q, R$, which must be within the convex hexagon $AEHCFG$, as shown in Figure 6, and
$$
S_{AEHCFG}=1-\frac{1}{8}-\frac{1}{8}=\frac{3}{4}.
$$

We will now consider three cases:
i) If the convex hull of $M, N, P, Q, R$ is a convex pentagon, let's say it is $MNPQR$, as shown in Figure 7. In this case,
$$
S_{\triangle MQR} + S_{\triangle MNQ} + S_{\triangle NPQ} \leq \frac{3}{4},
$$
thus at least one of $(M, Q, R)$, $(M, N, Q)$, and $(N, P, Q)$ is a good group. Since $(P, Q, R)$ is in rectangle $OHCF$, it is certainly 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, let's say it is $A_1A_2A_3A_4$, and the other known point is $A_5$, as shown in Figure 8, 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
$$
S_{\triangle A_1A_2A_5} + S_{\triangle A_2A_3A_5} + S_{\triangle A_3A_4A_5} + S_{\triangle A_4A_1A_5} = S_{A_1A_2A_3A_4} \leq \frac{3}{4},
$$
thus at least two of $(A_1, A_2, A_5)$, $(A_2, A_3, A_5)$, $(A_3, A_4, A_5)$, and $(A_1, A_4, A_5)$ are good groups.
iii) If the convex hull of $M, N, P, Q, R$ is a triangle, let's say it is $\triangle A_1A_2A_3$, and the other two known points are $A_4$ and $A_5$, as shown in Figure 9, where $A_i \in \{M, N, P, Q, R\} (i=1,2,3,4,5)$. Connect $A_4A_i (i=1,2,3)$, then
$$
S_{\triangle A_1A_2A_4} + S_{\triangle A_2A_3A_4} + S_{\triangle A_3A_1A_4} = S_{\triangle A_1A_2A_3} \leq \frac{3}{4},
$$
thus at least one of $(A_1, A_2, A_4)$, $(A_2, A_3, A_4)$, and $(A_1, A_3, A_4)$ is a good group. Similarly, $A_5$ also forms a good group with two of $A_1, A_2, A_3$, so there are at least two good groups in this case as well.

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 $AN:NB = AM:MD = 2:3$, as shown in Figure 10. 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 $M$ and $N$, let's 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
$$
S_{\triangle MBC} = \frac{1}{2}, \quad S_{\triangle MCD} > S_{\triangle BCD} = \frac{1}{4},
$$
thus $(M, B, C)$ 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, and $S_{\triangle MNB} = S_{\triangle MND} = \frac{1}{5} < \frac{1}{4}$, thus there are exactly two good groups $(M, N, B)$ and $(M, N, D)$.
Therefore, the minimum number of triangles with an area no greater than $\frac{1}{4}$ is 2.