28.2.3 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, with these 5 points as vertices, that have an area not greater than $\frac{1}{4}$.
The proof of this problem requires the following common conclusion, which we will use as a lemma: The area of any triangle within a rectangle does not exceed half the area of the rectangle.

In rectangle \(ABCD\), if a triangle formed by any three points has an area not greater than \(\frac{1}{4}\), it is 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 assume it is \(AEOG\), which contains at least two of the given five points (including the boundary, hereinafter the same). Let points \(M\) and \(N\) be in the smaller rectangle \(AEOG\), as shown in Figure a.
(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's assume the known points \(P\) and \(Q\) are both 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 \((P, Q, K)\) 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 \(AFOG\), 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 b, and
\[
S_{AEHCFG}=1-\frac{1}{8}-\frac{1}{8}=\frac{3}{4}.
\]

We will now consider three cases:
(1) If the convex hull of \(M, N, P, Q, R\) is a convex pentagon, let's assume it is \(MNPQR\), as shown in Figure c. 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.
(2) If the convex hull of \(M, N, P, Q, R\) is a convex quadrilateral, let's assume it is \(A_1A_2A_3A_4\), and the other known point is \(A_5\), as shown in Figure d, 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
\[
S_{\triangle A_1A_2A_5} + S_{\triangle A_2A_3A_5} + S_{\triangle A_3A_4A_5} + S_{\triangle A_4A_1A_5} = S_{\text{quadrilateral } A_1A_2A_3A_4} \leq \frac{3}{4},
\]
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.
(3) If the convex hull of \(M, N, P, Q, R\) is a triangle, let's assume it is \(\triangle A_1A_2A_3\), and the other two known points are \(A_4\) and \(A_5\), as shown in Figure e, 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 summary, regardless of the situation, there are at least two good groups among the five known points.
We now 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 f. 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 assume 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 not greater than \(\frac{1}{4}\) is 2.