IOAA 2019, Теория — Задача 10. South → East → North

1.

The trivial solution of the problem is the North Pole ($N$). Starting from here we travel 6378 km to the South ($\widehat{NA}$) along a meridian (e.g. Greenwich meridian), then 6378 km to the East along a parallel ($\widehat{AB}$), and finally 6378 km to the North ($\widehat{BN}$) (see figure ($a$), where the black dashed lines are the Equator and the Greenwich meridian).

For the general solution, mark the path to be taken in one direction with $s$ and determine what it may be!

Let us denote the longitude and latitude of the point $A$ by $\lambda_A$ and $\varphi_A$, and the point $B$ by $\lambda_B$ and $\varphi_B$.

The triangle $ANB$ is a spherical one only if the points $A$ and $B$ are on the Equator (see figure ($b$)), i.e. $\varphi_A=\varphi_B=0^{\circ}$ and $s=R\pi/2$, because the arc $\widehat{AB}$ is lying along a great circle only in this case, otherwise it is a part of a parallel. The point $A$ and $B$ can be below the Equator as well (figure ($c$)).

Next, we measure the angles in radians.

The central angle of the arc $\widehat{NA}$ on the meridian fitting to the point $A$ ($O$ is the centre of the sphere):

Thus the length $s$ of the arc $\widehat{NA}$:

The radius of the latitude circle fitting to the points $A$ and $B$:

The central angle of the arc $\widehat{AB}$ is $\lambda_B-\lambda_A$, thus the length of the arc $\widehat{AB}$:

From the Eq. (10.2) and Eq. (10.4):

Substitute the expression of $\varphi$ from the first equation into the second equation:

For a given arc length of $s$ ($s<R\pi$) the latitude $\varphi$ of the points $A$ and $B$, and the difference between their longitudes we can calculate from the following equations:

***** Supplement. It is just for the sake of completeness, not necessary for the full marks.

It is well known, that

thus $\lambda_B-\lambda_A=1$ in case of $s\to 0$.

Therefore the difference in longitude has a lower limit (1), but it has no upper limit. This means, that the distance between the points $A$ and $B$ will increase by increasing $s$, the difference between their longitudes will approximate $2\pi$ (see figure ($d$)), and in case of $\lambda_B-\lambda_A=2\pi$ the points $A$ and $B$ will be the same (see figure ($e$)). With further increasing of $s$ the difference $\lambda_B-\lambda_A$ will be larger than $2\pi$ (see figure ($f$)), thus we have to stop at that point on the circle where the length of the path along it reaches the value of $s$, and then turn to the North.

The arc length $s_{\mathrm{lim}}$ corresponding to the "limit case" of $\lambda_B-\lambda_A=2\pi$ we can derive from the numerical solution of the following transcendent equation:

Thus find the root of the function

i.e. the value $s_{\mathrm{lim}}$, where $f(s_{\mathrm{lim}})=0$. We can solve it by iteration using e.g. the Newton–Raphson method:

The first derivative of the function $f(s)$:

We know that the parallel in question runs south from the Equator, thus a good initial value for the iteration could be $s_0=R\pi/2+\Delta s$, where $\Delta s=1000$ km.

The result of the iteration and the corresponding latitude with three decimal places:

$\lambda_B-\lambda_A$ is strictly increasing function of $s$ and

thus for any values of $0<s<R\pi$ can be found an appropriate angle $\lambda_B-\lambda_A$ (see the figure below).

***** End of supplement.

In conclusion, the solution in this case is the North Pole ($N$). [2 p]

For the value of $s=6378$ km the solution is an ($a$)-type curve. The coordinates of the turning points are:

[3 p]

2.

The non-trivial solutions are similar to the ($e$)-type ("lasso") solution of the previous part, only the starting point is not the North Pole, but its location depends on the given value of arc length $s$. The circle of the lasso will run south from the ($e$)-type "limit circle", because the circumference of parallels running north from it are greater than the north-south arc.

The procedure is the following.

Knowing $s$, we have to find the parallel around the South Pole ($S$), whose circumference is an integer multiple of $s$. [2 p]

If the corresponding latitude is $\varphi_D$, then:

[5 p]

If the latitude of the sought starting point $C$ is $\varphi_C$, then the length $s$ of the arc to be taken along the meridian:

From the Eqs. (10.10) and (10.11):

Thus:

[2 p]

and

[2 p]

The negative sign denotes that the parallel is in the southern hemisphere.

So if the starting point is not the North Pole ($N$), there are infinitely many solutions.

For the value of $s=6378$ km the first 5 ($k=1,\ldots,5$) value for $\varphi_D$ in degrees with 3 decimal places:

[2 p]

The first 5 ($k=1,\ldots,5$) value for $\varphi_C$ in degrees with 3 decimal places:

[2 p]

In conclusion, starting to the South from a point of the parallel marked with a dashed red line ($C$), we get to a point on a parallel marked with a continuous red line ($D$). On this parallel we travel a full circle to the East, then we get back to the point $D$, then turn to North and travel back to starting point $C$.

For the further values of $k$ the procedure is very similar, but we have to perform two full circles along the continuous green parallel, three along the blue, four along the gold, five along the brown and so on.

Such kind of solution exists for any values of $0<s<s_{\mathrm{lim}}$.