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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

suomynonA
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Jeppe Stig Nielsen
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    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. –  Jan 06 '17 at 18:48
  • @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? – Jeppe Stig Nielsen Jan 06 '17 at 19:28
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    When Mecca is on the equator the L of the first component meets the ┐ of the second component and you get a +. –  Jan 06 '17 at 19:34
  • @Rahul You are right. I wonder if this is a well-known problem which has a name or a reference. – Jeppe Stig Nielsen Jan 06 '17 at 20:26
  • That's a good question; I don't know. You should edit the question to ask that. –  Jan 06 '17 at 21:40
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    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 ${(21.4^{\circ}N, (39.8 - t)^{\circ} E);|;t\in (0, 180]}$? – ChocolateAndCheese Jan 06 '17 at 22:02
  • @ChocolateAndCheese True about the North Pole. However consider a small circle of radius 1 centimeter around the North Pole; there will be a point on that small circle where following the circle tangent leads directly to Mecca. If you are at that tangent point, you face east when you face Mecca. The reason why your set will not work, is because a parallel (or "line" of latitude) such as $21.4^\circ$N is not "straight" (think of the 1 centimeter circle around the pole again). If you are located on the set you propose, you will have to face somewhat northerly to look directly towards Mecca. – Jeppe Stig Nielsen Jan 06 '17 at 22:20
  • @JeppeStigNielsen Ah yeah, makes sense, thanks! – ChocolateAndCheese Jan 06 '17 at 22:35
  • Rotate lat-log grid lines around a Mecca-Earth center global axis until North Pole falls on Mecca. Facing direction should avoid antipodal point in the Pacific. – Narasimham Jan 06 '17 at 23:05
  • @Narasimham I do not think I understand how, when rotating the grid lines around that axis, the North Pole of the grid can ever fall on Mecca. What does this rotation of the grid show; can you elaborate? You are right there is a pair of antipodal points where the correct facing is not defined. The solution set "meets" (but arguable does not include) these two points. – Jeppe Stig Nielsen Jan 06 '17 at 23:46
  • I am confused too and the comments have not helped clarified my confusion. If by "properly oriented towards Mecca" you mean looking at the direction of a great circle (as you seem to imply in the comments) then there is no point in the globe that satisfies your requirements. Looking east means going along latitude lines, so this would work only if Mecca was on the equator. It would be helpful if you were much more rigorous with our definitions. – Thanassis Jan 19 '17 at 04:01
  • Your example of a latitude circle 1cm away from the north pole also got me confused. First of all, there is no single line tangent to a circle in 3D space. If we are talking about our globe there is a whole plane that is tangent to a point you select. If you want to further restrict the selection to the plane of the circle then yes, you get a single line tangent, but this faces east/west and there is no way you can reach Mecca on that direction. All you can do by going east/west is travel on that latitude line. Maybe I misunderstood something completely about the problem. – Thanassis Jan 19 '17 at 04:39
  • @Thanassis Traveling along that latitude line is like traveling along the circumference of a coin. That is not going "straight ahead". If you go "straight ahead", i.e. if you follow a geodesic curve on the sphere, you will eventually cross the equator. – Jeppe Stig Nielsen Jan 19 '17 at 07:44
  • Yes, I understand the notion of geodesics (aka great circles). Note that looking/going east is not looking/going along a geodesic, unless we are on the equator. Since Mecca is not on the equator there are no points on the globe to satisfy the conditions of your question. Can you respond to this? Can you also elaborate the example of the latitude circle 1cm away from the north pole? Can you reply to the remarks I left on the previous comment? What kind of tangent are you talking about? – Thanassis Jan 19 '17 at 08:24
  • @Thanassis Looking and going can be interpreted differently here. You are absolutely right that going east and keeping adjusting your direction (turning) along the way so that you are still going east is ___not___ going straight ahead. But you are not supposed to go to Mecca during prayer. You are just assumed to face Mecca. Rotate where you stand, without changing location. Those are the conditions of my question. Facing Mecca is interpreted as having the orientation you would ___start out___ from if you were to go in a straight line to Mecca. I do not know how I can explain it better. – Jeppe Stig Nielsen Jan 19 '17 at 08:49
  • I think I understand what you are saying. These can indeed be different things, hadn't thought of that before. I believe I now understand the idea of the tangent you are talking about. Before I was thinking of straight lines in 3D space, whereas I should be thinking of geodesics (straight lines on a sphere) tangent to other curves on the sphere. Not sure how to define these rigorously but I'll think about the problem more in this new light. Thanks. – Thanassis Jan 19 '17 at 10:00

1 Answers1

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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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    $x^2+y^2=cxz$ (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. – Jeppe Stig Nielsen Jan 06 '17 at 23:19
  • Wow, free bounty! Thanks! –  Jan 23 '17 at 14:00
  • 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. – Jeppe Stig Nielsen Jan 23 '17 at 15:40