6. In a rectangle $A B C D$ (including the boundary) with an area of 1, 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 $\frac{1}{4}$.
6. The minimum number of triangles with an area not greater than $\frac{1}{4}$ is 2.

To prove this, we will use a common conclusion, which we will state as a lemma: The area of any triangle within a rectangle is not greater than half the area of the rectangle.

In rectangle $ABCD$, if a triangle formed by any three points has an area not greater than $\frac{1}{4}$, we call these three points 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 $O$. Line segments $EF$ and $GH$ divide rectangle $ABCD$ into four smaller rectangles.

There must be a smaller rectangle, let's assume it is $AEOG$ (including the boundary, as below), which contains at least two of the given 5 points. Let points $M$ and $N$ be in the smaller rectangle $AEOG$, as shown in Figure (1).
(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 the known points $P$ and $Q$ be 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 point $R$ be in rectangle $OHCF$. We will examine the convex hull of the 5 known points $M, N, P, Q, R$, which must be within the convex hexagon $AEHCFG$, as shown in Figure (2).
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 it be $MNPQR$, as shown in Figure (3).

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 it be $A_1A_2A_3A_4$, and the other known point be $A_5$.

As shown in Figure (4), where $A_i \in \{M, N, P, Q, R\} (i=1,2,3,4,5)$.
Connecting $A_5A_i (i=1,2,3,4)$, we have $S_{\triangle A_1A_2A_5} + S_{\triangle A_2A_3A_5} + S_{\triangle A_3A_4A_5} + S_{\triangle A_4A_1A_5} = S_{\triangle A_1A_2A_3A_4} \leq \frac{3}{4}$.
Thus, at least two of $(A_1, A_2, A_5)$, $(A_2, A_3, A_5)$, $(A_3, A_4, A_5)$, and $(A_4, A_1, A_5)$ 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 $A_i \in \{M, N, P, Q, R\} (i=1,2,3,4,5)$, connecting $A_4A_i (i=1,2,3)$, we have
$$
S_{\triangle A_1A_2A_4} + S_{\triangle A_2A_3A_4} + S_{\triangle A_3A_1A_4} = S_{\triangle A_1A_2A_3} \leq \frac{3}{4}.
$$

Thus, at least one of $(A_1, A_2, A_4)$, $(A_2, A_3, A_4)$, and $(A_3, A_1, A_4)$ 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.
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 exactly two.
On the side $AD$ of rectangle $ABCD$, take a point $M$, and on the side $AB$, take a point $N$, 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 $M$ 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
$$
S_{\triangle MBC} = \frac{1}{2}, \quad S_{\triangle BCD} > S_{\triangle BCD} = \frac{1}{4}.
$$

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.
However, $S_{\triangle MNB} = S_{\triangle MND} = \frac{1}{5} < \frac{1}{4}$, 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.