An abstract animal wanders on a unit-radius sphere. The length of its steps is 1.99 (this is the length of the segment defined by the two endpoints). In its journey, it cannot take a step immediately back in the direction it came from. - What is the minimum number of steps required for it to return to where it started?
I. solution. Let's denote the abstract animal by $A$, its starting point by $P$, the center of the sphere by $O$, the endpoint of its next step by $Q$, $R$, and $S$. Since the length of one step is less than the diameter of the sphere, $A$ always has somewhere to step. According to our assumption, $R$ cannot be the same as $P$, and we will show that $S$ cannot be the same either. Otherwise, the sides of the triangle $PQR$ would all be 1.99, which is greater than the side of the largest equilateral triangle that can be inscribed in a unit sphere, $\sqrt{3}$. Therefore, $A$ must take at least four steps, and as we will see, if it is a little careful, it can return to $P$ in four steps.
![](https://cdn.mathpix.com/cropped/2024_05_02_27386365082d3c90a723g-1.jpg?height=612&width=1152&top_left_y=434&top_left_x=473)

If $A$ wants to return to $P$ in four steps, its first step can still be arbitrary. In its second step, it only needs to ensure that $R$ is not in the plane $OPQ$. For example, rotate $P$ by $90^{\circ}$ around the axis $OQ$, then the resulting point $R$ will have $QR = QP$, and the planes $OQP$ and $OQR$ will be perpendicular, into which $A$ can step from $R$. Then reflect $A$ across the plane $OPR$ to $Q$, and let $S$ be the resulting reflection. Since $Q$ is not in the plane $OPR$ (since $R$ is not in the plane $OPQ$), $S$ will be different from each of the points $P$, $Q$, and $R$, and because $RS = RQ$, $A$ can step from $R$ to $S$, and because $SP = QP$, it can return to $P$ from there. This proves that the shortest path of $A$ consists of four steps.

II. solution. In the first step, $A$ can step onto the circle $k_1$, whose points are 1.99 units away from $P$, and which is cut out of the sphere by a plane perpendicular to the diameter of the sphere passing through $P$. If $A$ wants to return to $P$ in four steps, then in the third step, it must step onto a point of this circle again.

In the first step, $A$ can step into any point $Q$ of $k_1$. The points reachable in the second step are the points of the circle $k_2$, which is congruent to $k_1$ and passes through $P$, and whose plane is perpendicular to the diameter of the sphere passing through $Q$.

Similarly, from a point $R$ on $k_2$ that is different from $P$, $A$ can step onto a circle $k_3$ that is congruent to $k_1$ but different from it, and which passes through $Q$. The point $S$ must lie on both $k_1$ and $k_3$ and must be different from $Q$. Such a point does not exist only if $k_1$ and $k_3$ are tangent at $Q$; this, however, follows if $R$ is the antipodal point of $P$ on $k_2$.

We have thus found that if $A$ steps in the second step onto a point $R$ of the circle $k_2$ that is different from $P$ and its antipodal point, then there is always a uniquely determined point $S$ on the sphere, i.e., the intersection point of $k_1$ and $k_3$ different from $Q$, onto which it can step in the third step, and from which it can return to $P$ in the fourth step.