Consider a sphere with 3 diameters that are pairwise perpendicular. Divide each of them into 4 equal parts. Draw a plane perpendicular to the corresponding diameter through each division point. How many parts does the surface of the sphere get divided into by the 9 circles cut out by the 9 planes? - How many parts does the surface of the sphere get divided into if the 3 diameters are divided into 6 equal parts each and the same procedure is followed?
Three planes perpendicular to the same diameter of a sphere are parallel to each other, the middle one of which cuts a great circle from the sphere, the other two diameters of the sphere are also diameters of this great circle. The 3 great circles cut the sphere's surface into 8 congruent and similarly shaped spherical triangles due to the obvious symmetry, so it is enough to determine the number of parts of one such spherical triangle created, and its 8-fold gives the total number of parts. The 1st figure shows the spherical triangle $ABC$ and instead of the spherical arcs lying in it, we can examine the (perpendicular) projection of the figure onto one of the great circle planes, e.g., $OAB$, because this way the projections of the spherical surface parts do not overlap.

![](https://cdn.mathpix.com/cropped/2024_05_02_6c28e1c418d2b73dbd00g-1.jpg?height=718&width=824&top_left_y=430&top_left_x=629)

1st figure

![](https://cdn.mathpix.com/cropped/2024_05_02_6c28e1c418d2b73dbd00g-1.jpg?height=651&width=615&top_left_y=1298&top_left_x=733)

2nd figure

The projection of the spherical triangle is a quarter circle $OAB$ of the respective great circle, its two bounding radii are the projections of the other two great circles $AC$, $BC$ (2nd figure); the quarter arcs of the spherical radii $OA$, and $OB$ at the midpoints $A_1$, and $B_1$ of the planes cutting through the sphere give the projections of the quarter arcs of the spherical circles falling into the triangle $ABC$ as half-chords perpendicular to $OA$ at $A_1$, and to $OB$ at $B_1$ in the circular segment; finally, the projection of the quarter arc of the circle cut from the sphere by the plane through the midpoint $C_1$ of $OC$ is a quarter circle $i$ of equal radius around $O$. The radius of $i$ is equal to the length of the said half-chords (the $\sqrt{3}/2$ times the radius of the sphere). The intersection point of the half-chords is closer to $O$ than this radius, so $i$ intersects the half-chords one by one. Thus, 7 plane parts are formed in the circular segment, so according to the above, the 9 circles divide the sphere's surface into $8 \cdot 7 = 56$ parts.

In the case of the three diameters of the sphere being divided into 6 equal parts each, proceeding similarly, we get 2 half-chords parallel to the radii $OA$, $OB$, and two quarter circles of equal radius to the half-chords around $O$ (3rd figure). According to the construction, the smaller quarter circle passes through the 2 intersection points of the half-chords where a longer and a shorter half-chord intersect, while the larger quarter circle passes through the intersection point $P'$ of the two shorter half-chords. The distance of this from $O$ is $OP' = OA' \sqrt{2} = 2r \sqrt{2}/3$, where $r$ is the radius of the sphere.

![](https://cdn.mathpix.com/cropped/2024_05_02_6c28e1c418d2b73dbd00g-2.jpg?height=677&width=634&top_left_y=125&top_left_x=732)

3rd figure

Let $P$ be the point of the spherical triangle $ABC$ where the perpendicular to the plane $OAB$ through $P'$ intersects the spherical triangle. Thus, from the right triangle $OP'P$,

$$
P'P = \sqrt{r^2 - OP'^2} = \sqrt{r^2 - \frac{8r^2}{9}} = \frac{r}{3}
$$

so $P$ is indeed on that circular section parallel to the plane $OAB$ which is cut from the sphere by the plane perpendicular to $OC$ through its first third point. This shows that the other two intersection points found are also projections of spherical surface points through which 3 circular sections pass. Indeed, the distances of $P$ from the planes of the three great circles are $2r/3$, $2r/3$, $r/3$, and the distances of the other two intersection points are given by swapping these (the distance of the intersection point of the half-chords from $O$ is $r\sqrt{5}/3$, so the height of the spherical surface point above the plane $OAB$ is $2r/3$).

According to this, the number of plane parts of the circular segment shown in the 3rd figure is 16, and the 15 circles resulting from the points dividing the 3 diameters into 6 equal parts each divide the sphere's surface into $8 \cdot 16 = 128$ parts.

Remark. The spherical surface parts corresponding to the 1, 3, and 7, as well as 2, 6, and 4 labeled plane parts in the 2nd figure are congruent in threes, as the figure of the spherical circles has a "three-beat" (120°) rotational symmetry, whose axis goes through $O$, and also through the center of the spherical surface part represented by the "," labeled plane part in the 2nd figure. - Similarly, the spherical parts corresponding to the 1, 5, 13, as well as 3, 11, 16, and 7, 9, 14 labeled parts in the 3rd figure are congruent in threes, and the spherical surface parts corresponding to the 2, 4, 6, 12, 10, and 15 labeled parts are congruent; 3 of these go into each other by rotation, and one member of each pair goes into the other by a suitable reflection, possibly followed by a rotation; the center of the spherical surface part corresponding to the 8 labeled part goes through the 3-beat rotation axis, as well as the planes of the mentioned reflections.