435 Given a convex polygon $F$, consider all the figures that are positively homothetic to the convex polygon $F$ and smaller than $F$. Let $n(F)$ be the minimum number of such figures (allowing translation but not rotation) needed to cover the convex polygon $F$. Find the value of $n(F)$.
(1) On the one hand, let the convex polygon $F$ be the parallelogram $ABCD$. It is easy to see that the convex polygon $F$ can be covered by four smaller parallelograms that are similar to the convex polygon $F$ (as shown in Figure 3).

On the other hand, for any smaller parallelogram $F_1$ that is similar to the convex polygon $F$, if $F_1$ contains point $A$, then $F_1$ cannot contain any other vertices of the convex polygon $F$ (as shown in Figure 4). This indicates that the convex polygon $F$ cannot be covered by fewer than four smaller and similar figures.
Therefore, $n(F)=4$.
(2) Let the polygon $F$ not be a parallelogram. Then $n(F) \geqslant 3$.

In fact, if the convex polygon $F$ can be covered by two smaller and similar figures, let's assume these two figures are of equal size; otherwise, the smaller one can be enlarged. Establish a Cartesian coordinate system such that the corresponding vertices of these two figures have the same y-coordinates. Consider the vertices of the convex polygon $F$ with the maximum and minimum y-coordinates, which cannot be covered by these two figures simultaneously. Hence, $n(F) \geqslant 3$.
Next, we prove that three figures are sufficient.
First, we prove a lemma.
Lemma There exists a triangle $T$ that contains the convex polygon $F$, and each side of $T$ contains a side of the convex polygon $F$.

Proof If the convex polygon $F$ is a triangle, then let $T=F$.

If the convex polygon $F$ is not a triangle, then there exist two sides of the convex polygon $F$ that are neither parallel nor intersecting. Extend these two sides until they intersect (as shown in Figure 5), resulting in a polygon $F_1$ that contains the convex polygon $F$, and the number of sides of $F_1$ is less than that of the convex polygon $F$, with each side of $F_1$ containing a side of the convex polygon $F$.

Repeating this process, the final figure $F'$ will inevitably be a parallelogram or a triangle (only these two figures do not have sides that are neither parallel nor intersecting), and $F'$ contains the convex polygon $F$, with each side of $F'$ containing a side of the convex polygon $F$.
If $F'$ is a triangle, then let $T=F'$.
If $F'$ is a parallelogram $ABCD$, since the convex polygon $F$ is not a parallelogram, $F'$ must have at least one vertex that is not a vertex of the convex polygon $F$, let's say point $A$. Let point $P$ be the point on line segment $AD$ closest to point $A$. Then $P$ is a vertex of the convex polygon $F$, and one of its sides lies on line segment $AD$. Let the other side of the convex polygon $F$ from point $P$ intersect line segment $AB$ at point $Q$. Then the triangle $T$ formed by lines $PQ$, $BC$, and $CD$ contains the convex polygon $F$, and each side of $T$ contains a side of the convex polygon $F$ (as shown in Figure 6).
Back to the original problem.
Using the lemma, assume the convex polygon $F$ is contained in $\triangle ABC$, and the sides $A_1A_2$, $B_1B_2$, and $C_1C_2$ of the convex polygon $F$ lie on segments $BC$, $AC$, and $AB$ respectively (as shown in Figure 7).

Choose any point $O$ inside the convex polygon $F$ and internal points $X$, $Y$, and $Z$ on sides $A_1A_2$, $B_1B_2$, and $C_1C_2$ respectively. Then segments $OX$, $OY$, and $OZ$ divide the convex polygon $F$ into three polygons $F_1$, $F_2$, and $F_3$. Assume polygon $F_1$ is inside quadrilateral $AZOY$.

By the selection of points $O$, $Y$, and $Z$, if $0 < k < 1$ and $k$ is infinitely close to 1, then the homothety $\varphi_1$ with center $A$ and ratio $k$ satisfies $\varphi_1$ (convex polygon $F \cup$ quadrilateral $AZOY$) contains quadrilateral $AZOY$.

Note that the part of quadrilateral $AZOY$ outside the convex polygon $F$ becomes a subset of itself after the homothety, so $\varphi_1(F)$ covers $F_1$.

Similarly, there exist two homotheties $\varphi_2$ and $\varphi_3$ with ratios less than 1 such that $\varphi_2(F)$ covers $F_2$ and $\varphi_3(F)$ covers $F_3$.

Thus, the convex polygon $F$ is covered by the union of $\varphi_1(F)$, $\varphi_2(F)$, and $\varphi_3(F)$.
Therefore, $n(F)=3$.