49. For which $a>1$ does there exist a convex polyhedron, the ratio of the areas of any two of its faces being greater than $a$?
- Answer: for all $a \in(1,2)$. We will use the following lemma. Let the faces of a convex polyhedron have areas $S_{1}, S_{2}, \ldots, S_{n}$. Then the inequality holds: $S_{n} \leqslant S_{1}+S_{2}+\ldots+S_{n-1}$. (Analogous to the triangle inequality for polyhedra).

Proof of the lemma: project the polyhedron onto the plane of the face with area $S_{n}$. Then the projections of the other faces will obviously cover the face $S_{n}$ (due to the convexity of the polyhedron). From this, it follows that $S_{n}$ does not exceed the sum of the areas of the projections of the other faces. But the area of the projection of each face does not exceed the area of the face itself (since it is obtained by multiplying by the cosine of the angle between the faces), which leads to the required inequality.

Now we will show that for $a \geqslant 2$ such a polyhedron cannot exist. Let the areas of the faces of this polyhedron be $S_{1}a>a_{0}$. First, note that there exists a natural number $n$ such that $\frac{1}{a}+$ $\frac{1}{a^{2}}+\ldots+\frac{1}{a^{n-1}}>1$. Indeed, the sum of the geometric progression on the left side of this inequality is $\frac{a^{n-1}-1}{a^{n}-a^{n-1}}$. The condition that this fraction is greater than 1 is equivalent to $2 a^{n-1}>a^{n}+1$. Rewriting the last inequality as $2>a+\frac{1}{a^{n-1}}$, it is clear that for $a \geqslant 2$, this inequality cannot be satisfied for any natural $n$.

For $a \in (1,2)$, we can construct a polyhedron with the required properties. Let the areas of the faces be $a_{0}, \frac{a_{0}}{a}, \frac{a_{0}}{a^{2}}, \ldots, \frac{a_{0}}{a^{n-1}}$. We will construct the polyhedron inductively. Suppose we have already constructed a polyhedron with faces of areas $a_{0}, \frac{a_{0}}{a}, \frac{a_{0}}{a^{2}}, \ldots, \frac{a_{0}}{a^{k}}$. We will add a new face with area $\frac{a_{0}}{a^{k+1}}$. To do this, we need to choose the height $h$ of the new face such that the inequality

$$
h>\sqrt{\frac{a_{0}^{2} h_{k+1}^{2}-a^{2} h_{k}^{2}}{a^{2}-a_{0}^{2}}}
$$

is satisfied. By choosing an appropriate $h$, we can ensure that all such inequalities are satisfied for all $k$. It remains to choose $h$ such that the area $\frac{1}{2} \cdot \frac{\sqrt{h^{2}+h_{n-1}^{2}}}{a^{n-1}}$ is greater than $a_{0}$ (i.e., greater than the area of the base multiplied by $a_{0}$). Clearly, this can also be achieved.

(A. Shapovalov)