An airplane flies from Oslo to city $X$, which is on the equator in South America, without landing, on the shortest route. The plane flew in a westerly direction from Oslo. Determine the geographical coordinates of $X$! Which city is this? (Geographical coordinates of Oslo: latitude $59^{\circ} 55^{\prime}$ north, longitude $10^{\circ} 43^{\prime}$ east.)
The shortest flight path means that the trajectory of the motion - the line on the ground over which the plane traveled - is the great circle cut (marked) out of the surface by the $O X F$ plane; here $O$ represents the city of Oslo, and $F$ the center of the Earth. Indeed, on the sphere, the distance is measured by the angle between the $F O$ and $F X$ radii, and thus $O X = R \cdot \varphi$, where $\varphi = \angle O F X$, in radians, and $R$ is the radius of the Earth (Figure 1).

![](https://cdn.mathpix.com/cropped/2024_05_02_b5c28cb7f5aafc9c978cg-1.jpg?height=873&width=827&top_left_y=336&top_left_x=627)

The great circle $O F X$ plane $S_{1}$ is determined, in addition to the points $O, F$, by the fact that the plane took off in a westerly direction, i.e., perpendicular to the north-south direction, the meridian connecting Oslo with the North Pole, and thus also perpendicular to the plane $O E E F = S_{2}$ containing this meridian. Therefore, the flight path plane is also perpendicular to $S_{2}$, since the velocity vector (and tangent) at Oslo is in $S_{1}$, and if a plane contains a line perpendicular to another plane, then the two planes are perpendicular to each other.

This means that the projection of $S_{1}$ onto $S_{2}$ is a straight line. Since $F$ is also in $S_{1}$ and its projection onto $S_{2}$ is itself, the projection of the plane's path lies on the diameter of the meridian in $S_{2}$ starting from Oslo (Figure 2).

![](https://cdn.mathpix.com/cropped/2024_05_02_b5c28cb7f5aafc9c978cg-1.jpg?height=618&width=569&top_left_y=1602&top_left_x=759)

Figure 2

The plane $S_{3}$ of the Equator, which contains $X$ and is perpendicular to $S_{2}$, is also perpendicular to the Earth's $E ́ F$ axis and passes through $F$; thus, the projection of the Equator is the diameter of the meridian in Oslo perpendicular to $E ́ F$. Since the projection of $X$ onto $S_{2}$, $X^{\prime}$, is on both $O F$ and the diameter of the Equator, $X^{\prime}$ coincides with $F$, meaning that the Earth radius $X F$ is perpendicular to $S_{2}$.

This implies that the meridian plane $X \dot{E} F (= S_{4})$ of the city $X$ is perpendicular to the meridian plane of Oslo, and is rotated $90^{\circ}$ westward, so the western longitude of $X$ is $|10^{\circ} 43' - 90^{\circ}| = 79^{\circ} 17'$. The point on the Equator at this longitude is near Quito, the capital of Ecuador.

Remarks. 1. The answer did not require the geographical latitude of Oslo, so the same answer would apply if we had mentioned any other point on Oslo's meridian, such as Lübeck, Augsburg, or Cabinda, Angola.

2. If the points $P_{0}, P_{1}, P_{2}, \ldots, P_{n-1}, P_{n}$ are given on a plane, then by the triangle inequality, it can be shown that the length of the $P_{0} P_{n}$ segment cannot be greater than the sum of the lengths of the segments $P_{0} P_{1}, P_{1} P_{2}, \ldots, P_{n-1} P_{n}$. This is why we say that the shortest path between two points is a straight line. Those who think that by connecting the points $P_{i-1}, P_{i}$ with segments, we implicitly used the statement, should complete the statement by saying that the straight line is the shortest among all paths consisting of (and a finite number of) straight segments. And for those who want to extend the statement to other types of paths, they must first define the length of those paths. And what they want to get rid of, the length of curves is usually approximated by broken lines tangent to them.

These are the things to keep in mind for those who are looking for the shortest path between two points on a sphere. We can say that the "segment" defined by two points is the shorter arc of the great circle passing through the two points - if the two points are not antipodal points of the sphere. (If the points are antipodal, we do not define the "segment" passing between them.)

Let's examine whether the planar statement can be extended to this case. The number of points can be reduced step by step to two, if we have already proven the statement for $n=2$. This is the triangle inequality, which can be reduced to the fact that in a triangle, the larger side is opposite the larger angle. This can lead to some confusion, as the sides of the triangles are actually angles, arcs, and the angles of the triangles are the angles between the planes containing the sides, which is not easily visible in space. But perhaps precisely because of this, it can be useful to practice thinking through what can be transferred from the geometry of planar triangles to spherical geometry. Those who do this will find that it is more convenient to generalize the theorems to be proven and show that they can be derived from each other without referring to the specific content of the concepts involved, and using more properties of the geometric elements than those provided by the theorems in question.

The planar triangle inequalities can be shown by reflecting over the angle bisector: if $A C > B C$, then reflecting $B$ over the angle bisector $t$ at $C$, it lands on the segment $A C$.

![](https://cdn.mathpix.com/cropped/2024_05_02_b5c28cb7f5aafc9c978cg-2.jpg?height=675&width=637&top_left_y=1039&top_left_x=728)

Figure 3

If we denote the new position of $B$ by $B^{\prime}$, then we only need to use the fact that the exterior angle at $B^{\prime}$ in the triangle $A B B^{\prime}$ is greater than the interior angle at $A$. In space, $t$ corresponds to the half-plane bisector of the angles between the "sides" $A C$ and $B C$, and reflecting $B$ over this, we indeed get the internal point $B^{\prime}$ on the arc $A C$.

![](https://cdn.mathpix.com/cropped/2024_05_02_b5c28cb7f5aafc9c978cg-3.jpg?height=897&width=832&top_left_y=137&top_left_x=625)

Thus, it remains to decide whether the statement about the exterior angle is true. In the plane, more is true: the exterior angle is equal to the sum of the two non-adjacent interior angles, and the sum of the angles is $180^{\circ}$. This is not true on the sphere at all: for "small", planar-like spherical triangles, the deviation is not large, but we have excluded antipodal pairs so far. It is noticeably the parallel lines that cause the confusion, or rather, the fact that the "segments" we are considering are contained in "lines" that intersect at two points (Figures 3-4).

This is the point where we stop the examination of the question. Perhaps it is more useful to pose new problems related to spherical geometry.