5 In a rectangle $A B C D$ with an area of 1 (including the boundary) there are 5 points, no three of which are collinear. Find the minimum number of triangles, formed by these 5 points as vertices, that have an area not greater than $\frac{1}{4}$.
The proof of this problem requires the use of the following common 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 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, and the intersection of line segments \(EF\) and \(CH\) be \(O\). Line segments \(EF\) and \(GH\) divide rectangle \(ABCD\) into four smaller rectangles. Therefore, there must be a smaller rectangle, let's assume it is \(AEOG\), which contains at least 2 of the given 5 points (including the boundaries, hereinafter the same), let points \(M\) and \(N\) be in the smaller rectangle \(AEOG\), as shown in Figure 1.
(1) If there is no more than 1 known point in rectangle \(OHCF\), 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 there are at least 2 known points in rectangle \(OHCF\), let the known points \(P\) and \(Q\) both be in rectangle \(OHCF\), and consider the remaining last 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 point \(R\) be in rectangle \(OHCF\), we will examine the convex hull of the 5 points \(M, N, P, Q, R\), which must be within the convex hexagon \(AEHCFG\), as shown in Figure 2, and
\[
S_{\text{AEHCFG}} = 1 - \frac{1}{8} - \frac{1}{8} = \frac{3}{4}
\]

We will now consider three cases:
i If the convex hull of \(M, N, P, Q, R\) is a convex pentagon, let it be \(MNPQR\), as shown in Figure 3, at this time
\[
S_{\triangle MQR} + S_{\triangle MNQ} + S_{\triangle NPQ} \leq \frac{3}{4}
\]

Thus, 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 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 it be \(A_1A_2A_3A_4\), and the other known point be \(A_5\), as shown in Figure 4, 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 \((A_1, A_2, A_5)\), \((A_2, A_3, A_5)\), \((A_3, A_4, A_5)\), \((A_1, A_4, A_5)\) are good groups.
iii If the convex hull of \(M, N, P, Q, R\) is a triangle, let it be \(\triangle A_1A_2A_3\), and the other two known points be \(A_4, A_5\), as shown in Figure 5, 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 \((A_1, A_2, A_4)\), \((A_2, A_3, A_4)\), \((A_1, A_3, A_4)\) 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 5 known points.

Next, we provide an example to show that the number of good groups can be only 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 6. Then, among the 5 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 \(M, N\), 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
\[
S_{\triangle MBC} = \frac{1}{2}, \quad S_{\triangle MCD} > S_{\triangle ECD} = \frac{1}{4}
\]

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 more than \(\frac{1}{4}\) is 2.