# Given distances (shortest paths) between four cities, how to show that they cannot be in the same plane?

In the example below we are given distances between four cities. The author of the book says that these distances "suffice to prove that the world is not flat".

• Do I understand this correctly that this just means: Checking that we cannot place four points on plane with these distances? Or am I supposed to interpret this claim differently?
• How can this be shown? I.e., how can I use the distances below to show that it is not possible to find the points on the plane with these distances?

If I understand the problem correctly, I am supposed to find some property which is fulfilled for distance between the vertices of any quadrilateral in a plane. And then show that the given distances do not fulfill this property.

The following excerpt is taken from Einstein Gravity in a Nutshell by Zee, page 66

The logic of differential geometry

Differential geometry, as developed by Gauss and Riemann, tells us that given the metric, we can calculate the curvature. The logic goes as follows. The metric tells you the distance between two nearby points. Integrating, you can obtain the distance along any curve joining two points, not necessarily nearby. Find the curve with the shortest distance. By definition, this curve is the "straight line" between these two points. Once you know how to find the "straight line" between any two points, you can test all of Euclid’s theorems to see whether our space is flat. For example, as described in the prologue, the mite geometers could now draw a small circle around any point, measure its circumference, and see if it is equal to $2\pi$ times the radius. (See appendix 1.) Thus, the metric can tell us about curvature.

Take an everyday example: given an airline table of distances, you can deduce that the world is curved without ever going outside. If I tell you the three distances between Paris, Berlin, and Barcelona, you can draw a triangle on a flat piece of paper with the three cities at the vertices. But now if I also give you the distances between Rome and each of these three cities, you would find that you can’t extend the triangle to a planar quadrangle (figure 1). So the distances between four points suffice to prove that the world is not flat. But the metric tells you the distances between an infinite number of points.

I have looked up distances between these four cities on WolframAlpha:

$$\begin{array}{|c|c|c|c|c|} \hline & \text{Bar} & \text{Ber} & \text{Rom} & \text{Par} \\\hline \text{Bar} & 0 & 1498 & 861.8 & 829.2\\\hline \text{Ber} & & 0 & 1184 & 878.7 \\\hline \text{Rom} & & & 0 & 1109 \\\hline \text{Par} & & & & 0 \\\hline \end{array}$$

• The main reason why I am posting this question is that another user asked the question here and here. This user repeatedly asked how should they improve the question so that it is not put on hold/closed. This is my attempt to show how this question can look with an appropriate context. Of course, I cannot be sure whether this is what that user had in mind. (Especially since they asked for a proof using graph theory.) – Martin Sleziak Jun 8 '16 at 6:48
• You can easily use the standard formula for the volume of a tetrahedron given its edges. But that requires the straightline distances. Are you using great-circle distances? – almagest Jun 8 '16 at 6:56
• BTW the other question has been reopened in the meantime. So the question is whether they are duplicates. (As I said, I am not sure - especially because of the fact that the OP says that they have the question from graph theory textbook. Although after my comment there, they also added (differential-geometry) tag.) – Martin Sleziak Jun 8 '16 at 7:04
• @HKLee I have included quote from a book which inspired this question. I might have misunderstood the question, but I do not think that the author means distances in space. The distances are supposed to be distances on the manifold on which the points are placed (=lengths of shortest paths). – Martin Sleziak Jun 8 '16 at 8:39
• You can use the methodology in this answer to tell whether the world is flat (basically what's in Gerry's answer) and if it is non-flat, estimate the "radius" the world is lying on. – achille hui Jun 8 '16 at 8:53

Take Rome, use cosine rule to find three angles it makes with 3 different triangles, and then check whether $$\cos(a+b)=\cos{a}\cos{b}-\sin{a}\sin{b}$$ is satisfied $\angle a=$ barcelona-rome-paris $\angle b=$ paris-rome-berlin and $\angle a+b=$ barcelona-rome-berlin

• can someone explain the downvote? – avz2611 Jun 8 '16 at 7:20
• Barcelona-Rome-Paris: Sides are $829.2$, $861.8$, $1109$, which gives angle in Rome $47.76^\circ$. Paris-Rome-Berlin: Sides are $1109$, $1184$, $878.7$, which gives angle in Rome $44.92^\circ$. (Sum of these two angles is $92.68^\circ$.) Barcelona-Rome-Berlin: Sides are $861.8$, $1184$, $1489$, which gives angle in Rome $92.04^\circ$. – Martin Sleziak Jun 8 '16 at 9:59
• @avz2611Maybe because the answer was completed by Martin's comment. – Crowley Jun 8 '16 at 13:19
• @Crowley The downvote came before my comment. BTW I have noticed that at some point there were 3 answers, each with one downvote and also the question had one downvote. So it is possible that somebody simply downvoted all posts here (at that time). – Martin Sleziak Jun 8 '16 at 15:50

I also posted this to the other question.

The two diagonals $p$ and $q$ of a plane quadrilateral and the four side lengths $a$, $b$, $c$, $d$ are related by the Cayley-Menger determinant: $$\det\pmatrix{0&a^2&p^2&d^2&1\cr a^2&0&b^2&q^2&1\cr p^2&b^2&0&c^2&1\cr d^2&q^2&c^2&0&1\cr1&1&1&1&0\cr}=0$$ See https://en.wikipedia.org/wiki/Quadrilateral#Properties_of_the_diagonals_in_some_quadrilaterals

So, if you don't get zero, your points are not in a plane.

• For $a=861.8$, $b=1184$, $c=878.2$ $d=829.2$, $p=1498$, $q=1109$ WolframAlpha indeed returns non-zero determinant. (And the value is large enough so that the non-zero value does not come from rounding errors.) – Martin Sleziak Jun 8 '16 at 8:07
• Is the value large enough so that the non-zero value does not come from differences in elevation? For reference, the elevations of Barcelona, Berlin, Paris and Rome are $12$, $34$, $35$ and $37$ meters. – Servaes Jun 8 '16 at 12:18
• It's perhaps worth emphasizing that the nonzero value of the Cayley-Menger determinant has itself a geometric interpretation: It equals $288V^2$ where $V$ is the volume of the tetrahedron formed by the four points. (This vanishes, of course, precisely when the four points lie in the same plane.) – Semiclassical Jun 8 '16 at 16:49

You have the lengths of all the sides of your triangles, so you should be able to obtain all the corresponding angles. Now, pick three triangles, say: Paris-Barcelona-Berlin, Paris-Barcelona-Rome, Barcelona-Rome-Berlin.

In each of these triangles, the sum of the angles will be 180°. This should be obvious; of course, any three points define a plane. We need to check whether having these four cities on the same plane is consistent with the angles (and the distances) we measured.

One way to go about it is checking whether the angle Paris-Barcelona-Berlin from the first triangle plus the angle Berlin-Barcelona-Rome from the third triangle is equal to the angle Paris-Barcelona-Rome from the second triangle.

A few calculations should convince you that it is not the case: the world is not flat. In particular, we find greater angles than expected (and their sum add up to more than 180°), which is indicative of positive curvature (spherical geometry).

A simple geometric approach. Take suppose Rome to be the point $(0,0,0)$ . Take Paris as $(1109,0,0)$ on $X$ axis. Given 2 points plot Berlin on the $XY$ plane(2 distances being given). Say you get $(X_1,Y_1,0)$. Now you are given three points , Draw three spheres centering each of the three points and radius of each sphere being the distance of Barcelona from each of them. So you get the coordinates of Barcelona. From the given data it will show that the perpendicular distance of each of Barcelona from the plane of the other three cities(namely the $Z$ component) will not be $0$

• there may be to possible locations for berlin , two circles can intersect at two positions, similarly two spheres intersect forming circles , so this won't give the answer imo, i could be wrong though – avz2611 Jun 8 '16 at 7:15
• @avz2611 Does it matter? Where ever the 2 circles intersect I will get a point . And hence define a palne with 3 points. I AM NOT CONCERNED WITH THEIR ACCURATE POSITIONS. JUST WITH THE PLANE THEY ARE FORMING.. – Qwerty Jun 8 '16 at 7:20

We need to consider only a single spherical triangle between any 3 cities and compute angles $\alpha, \beta, \gamma$ opposite them using spherical trigonometry.

To convince that earth is indeed not flat, the spherical excess should be checked to be positive.Or,

$$\alpha + \beta + \gamma -\pi > 0.$$

• But if I have only one triangle and the distance fulfills triangle inequality, then such triangle can be drawn in a plane. (I mean, we are not given in advance that the triangles are on a sphere. We want to check whether it is possible to have such distances in a plane.) – Martin Sleziak Jun 8 '16 at 8:13

I would set up a 4-hedron:
Barcelona-Berlin[a=1498] skew to Paris-Rome[f=1109];
Barcelona-Paris[b=829.2] skew to Rome-Berlin[e=1184];
Barcelona-Rome[c=861.8] skew to Berlin-Paris[d=878.7]
$A=a^2,\,B=b^2,\,C=c^2,\,D=d^2,\,E=e^2,\,F=f^2$
$\text{volume}=\frac{\sqrt{AF(-A+B+C+D+E-F)+BE(A-B+C+D-E+F)+CD(A+B-C-D+E+F)-ABD-AEC-BCF-DFE}}{12}$

If its volume does not vanish, then the cities are not coplanar.

Of course, it should be borne in mind that all of the tetrahedral solutions, including this one, are predicated on the supposition that the intercity distances supplied by Wolfram Alpha are straight lines, when they are actually more akin to geodesics.