We came to think of this problem:

Ali is a good Muslim who happens to travel a lot. On one occasion when Ali is praying, properly oriented towards Mecca, he notices that he is also facing exactly east.

Where can Ali be?

The geographical coordinates of Mecca are $21.4^\circ\text{N}$ and $39.8^\circ\text{E}$ (you may switch to a coordinate system using Mecca's longitude as the zero meridian). You may assume the Earth is a perfect sphere, and Ali is on its surface.

We have realized that the solution consists of two "components". One is a curve with Mecca (where the correct orientation is not well-defined) as the southern endpoint and the North Pole (where east is "all directions") as the northern endpoint. The other component is obtained from the first by rigidly moving the curve along the surface of the Earth such that it connects the antipodal point of Mecca with the South Pole.

We do not know if each "component" is an arc of a small circle.

Instead of solving the problem ourselves (we are lazy), we thought some people would come up with some nice solutions on this forum. It would be interesting with both formulas and visual representations (such as globes with the solution curve plotted on them).

Extending the problem: What if, in the problem text, you change the direction "east" to "northeast"; you would get a new curve? Or "east-northeast" etc.? This gives a whole family of curves which could all be plotted on a globe.

Also, does anyone know if this is a well-known problem that has its own name or reference?

Wikipedia link: Qibla

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    $\begingroup$ This is half the locus of points where the geodesic to the North Pole and the geodesic to Mecca are perpendicular (the other half is the points where facing Mecca = facing west). So if Mecca was very close to the north pole the locus would indeed be nearly semicircular. But it is not true in general; for example, if Mecca was very close to the equator, the locus would instead be an L-shape. $\endgroup$ – user856 Jan 6 '17 at 18:48
  • $\begingroup$ @Rahul Very interesting. It feels as if in the extreme case where Mecca is located on the equator we get a shape of the symbol $\perp$ rather than an L? $\endgroup$ – Jeppe Stig Nielsen Jan 6 '17 at 19:28
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    $\begingroup$ When Mecca is on the equator the L of the first component meets the ┐ of the second component and you get a +. $\endgroup$ – user856 Jan 6 '17 at 19:34
  • $\begingroup$ @Rahul You are right. I wonder if this is a well-known problem which has a name or a reference. $\endgroup$ – Jeppe Stig Nielsen Jan 6 '17 at 20:26
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    $\begingroup$ I'm trying to understand this but I'm confused. If I'm at the North pole, then facing Mecca is facing south, right? Why isn't the answer the set of coordinates $\{(21.4^{\circ}N, (39.8 - t)^{\circ} E)\;|\;t\in (0, 180]\}$? $\endgroup$ – ChocolateAndCheese Jan 6 '17 at 22:02

I will give a simple solution using coordinates. This does not answer the question of "Is this a well-known problem that has its own name or reference?"

We have a sphere with two distinguished points: the North Pole, $N$, and Mecca, $M$. At any third point $A$, facing east means facing in the direction perpendicular to the geodesic from $A$ to $N$. Facing $M$ means facing in the direction of the geodesic $AM$. Therefore, we seek the locus of points $A$ such that the geodesics $AN$ and $AM$ are perpendicular.

Take the Earth to be a unit sphere centered at the origin $O$, with the coordinates of the points being $N=(0,0,1)$, $M=(\cos\theta,0,\sin\theta)$, and $A=(x,y,z)$. The angle between the great circles $AN$ and $AM$ is equal to the angle between the normals of the planes $AON$ and $AOM$ containing them. This gives the condition $(A\times N)\cdot(A\times M)=0$, which simplifies to $$(x^2+y^2)\sin\theta=xz\cos\theta.$$

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    $\begingroup$ $x^2+y^2=cxz$ (where $c$ is the cotangent to Mecca's latitude) appears to be a kind of infinite (double) cone with apex in $O$. Its intersection with our unit sphere gives the two closed curves (the locus facing west instead of east is also included in your formula if I forget you said $y<0$). It makes sense I think. Also on the sphere it is $1-z^2=cxz$ which describes a kind of hyperbola in the $xz$-plane. Anyway, the solution is the intersection set of two second-degree polynomials. $\endgroup$ – Jeppe Stig Nielsen Jan 6 '17 at 23:19
  • $\begingroup$ Wow, free bounty! Thanks! $\endgroup$ – user856 Jan 23 '17 at 14:00
  • $\begingroup$ You are very welcome. I was hoping for a lot of new reactions in this thread, but almost nothing happened during the bounty period. In that case your response is the best response. $\endgroup$ – Jeppe Stig Nielsen Jan 23 '17 at 15:40

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