23. Given a positive number $c$. In space, 99 points are marked in such a way that for each of the marked points, the distances to the two nearest marked points differ by at least a factor of $c$. What is the largest $c$ for which this is possible?
23. Answer: $c=\frac{1+\sqrt{5}}{2}$.

First, let's prove that $c \leqslant \frac{1+\sqrt{5}}{2}$.

From the condition, it follows that for any marked point, the distance to any other marked point (except the nearest one) exceeds the distance to the nearest one by at least a factor of $c$.

Suppose there are three such marked points $A, B$, and $C$, such that the nearest point to $A$ is point $B$, and the nearest point to $B$ is point $C \neq A$. Let $A B=x$. Since $A B$ is the shortest distance from point $A$, then $A C \geqslant c x$. Since $B C$ is the shortest distance from point $B$, then $B C \leqslant \frac{x}{c}$. Then, by the triangle inequality,

$$
x + \frac{x}{c} \geqslant A B + B C \geqslant A C \geqslant c x
$$

from which $c^{2} - c - 1 \leqslant 0$. This means that the number $c$ lies on the number line between the roots of the quadratic polynomial $t^{2} - t - 1$. In particular, $c \leqslant \frac{1+\sqrt{5}}{2}$, as required.

If such a triplet of points does not exist, then all points can be divided into pairs (in each pair, the points are nearest to each other). But this is impossible, as the number of points is odd!

It remains to provide an example where $c=\frac{1+\sqrt{5}}{2}$. The first part of the solution helps us here: in inequality $(*)$, equality was achieved when point $B$ lies on the segment between $A$ and $C$ and, for example, $A B=1$, $B C=\frac{1}{c}$, $A C=1+\frac{1}{c}=c$. Mark 33 such triplets of points $A_{1}, B_{1}, C_{1}, \ldots, A_{33}, B_{33}, C_{33}$,

$$
\overrightarrow{A_{1} B_{1} C_{1}} \quad \overrightarrow{A_{2} B_{2} C_{2}} \quad \overrightarrow{A_{3} B_{3} C_{3}}
$$

and place them far enough from each other so that all distances between points from different triplets are greater than $c$. Then, for each point $A_{i}$, the nearest points will be $B_{i}$ and $C_{i}$, and $A_{i} C_{i}: A_{i} B_{i}=c: 1=c$; for each point $B_{i}$, the nearest points will be $C_{i}$ and $A_{i}$, and $B_{i} A_{i}: B_{i} C_{i}=1:(1 / c)=c$; and, finally, for each point $C_{i}$, the nearest points will be $B_{i}$ and $C_{i}$, and $C_{i} A_{i}: C_{i} B_{i}=c:(1 / c)=c^{2}>c$. The requirement of the problem is satisfied.