Given three poins $v_1,v_2,v_3$ on the sphere $S^2$, it is obvious that one can construct the circle with radius $|v_1v_2|$ around the point $v_3$, as seen in the following cartoon image:
In this image, you see how one can measure the distance between $v_1$ and $v_2$ with a compass, and then construct a circle with that radius. (Alternatively, one can construct the sphere around $v_3$ with that radius and then intersect it with the unit sphere)
Now my question is: When I stereographically project this sphere to $\mathbb{R}^2$, (how) does this construction translate to a standard construction with compass and straigtedge?
In this 2D setting, you are given:
- A circle $e$ that is the equator of the stereographic projection.
- Three arbitrary points $v_1,v_2,v_3$
This question came up when I was trying to answer a different question on this website.
EDIT: I think the construction described in this answer can be used in this case. But then I still need to know the following constructions:
1: Given two points A and B, construct the great circle through these points. In 3D, you would take the intersection of the plane spanned by A, B and O (the origin) with the unint sphere, but in 2D this becomes a circle.
2: Given two points A and B, construct the circle with center A through B. In 3D this is trivial, but in 2D A is not in the center of the circle anymore.
I have never done these type of things, does anyone have tips for these two constructions?
Best Answer
This is probably not nearly the nicest way to do it, but it should at least serve as an existence proof for a compass-and-straightedge construction.
As it turns out, I have already presented an outline of a compass-and-straightedge construction of the stereographic projection of a great circle through two arbitrary points in an answer to another question. (In that answer, the point $P''$ corresponds to either of the points $v_1$ or $v_2$ in the stereographic projected image, and $Q''$ is a third point on the projected great circle, which is sufficient to construct that circle. In that answer I used a stereographic projection onto a plane tangent to the sphere, but the method could easily be adapted to a projection plane through the center of the sphere.)
But I suggest that we instead begin by finding the the angular distance between two arbitrary points on the sphere given their stereographic projections. For this I will assume that the projection plane of the stereographic projection passes through the center of the sphere so that the circle $e$ is simply the intersection of the sphere with the projection plane.
Given the points $v_1$ and $v_2$ on the projection plane and the equatorial circle $e$ on the projection plane, we find $C$, the center of $e$, and construct the line through $C$ and $v_1$. This line intersects $e$ at two points; let $e_1$ be the intersection on the same side of $C$ as $v_1$.
Let $P$ and $Q$ be the points of intersection of circle $e$ and a line through $C$ perpendicular to line $Ce_1$.
Now let $v_1'$ be the intersection of line $Pv_1$ with the circle $e$ that is not $P$. The angle $\phi_1 = \angle QOv_1'$ is then the co-latitude of the preimage of $v_1$ on the original sphere. In fact, we have now constructed a figure congruent to a geometric figure projecting the original point from the sphere to $v_1,$ except that we have constructed this figure on the projection plane rather than in the plane where it originally occurred, which was perpendicular to the projection plane.
In a similar fashion, we can construct $\phi_2,$ the co-latitude of the preimage of $v_2$ on the original sphere.
And of course the angle $\lambda = \angle v_1Cv_2$ is the angle between the meridians of the original sphere passing through the preimages of $v_1$ and $v_2.$
We now have enough information about the spherical coordinates of the two points in order to apply the spherical law of cosines: $$ \cos(\alpha) = \cos(\phi_1)\cos(\phi_2) + \sin(\phi_1)\sin(\phi_2)\cos(\lambda), $$ where $\alpha$ is the central angle between the two points on the sphere (the preimages of $v_1$ and $v_2$).
We can construct right triangles with angles $\phi_1,$ $\phi_2,$ and $\lambda$ in order to obtain sets of segments in the ratios $\sin(\phi_1) : 1,$ $\sin(\phi_2) : 1,$ $\pm\cos(\phi_1) : 1,$ $\pm\cos(\phi_2) : 1,$ $\pm\cos(\lambda) : 1$ (using the $-$ sign for a cosine of any obtuse angle). Then we can use compass-and-straightedge techniques for finding a pair of segments in the ratio $xy : 1$ from segments in the ratios $x : 1$ and $y : 1$ and for adding or subtracting lengths in order to obtain a pair of segments in the ratio $\pm\cos(\alpha) : 1.$ By constructing a right triangle using the longer of these segments as the hypotenuse and the shorter as a leg, we can construct the angle $\alpha.$
We finish as follows. We construct a line through $C$ and $v_3,$ construct the line perpendicular to $Cv_3$ through $C,$ and identify one of the intersections of that new line with the circle $e$ as the point $P.$ We find $v_3'$, the intersection (other than $P$) of the line $Pv_3$ with the circle $e,$ and construct two points $v_4'$ and $v_5'$ on the circle $e$ such that $\angle v_3'Cv_4' = \angle v_3'Cv_5' = \alpha$ as shown in the figure below.
The points $v_4'$ and $v_5'$ represent the two points of the desired circle around $v_3'$ on the same meridian as $v_3'$ on the sphere (although as in the other construction, we have actually constructed a congruent figure in the projection plane rather than reconstructing the "actual" figure in the plane of the meridian). We project these two points to the points $v_4$ and $v_5$ on the line $Cv_3.$ Since that line is the projection of the meridian through the preimage of $v_3,$ the points $v_4$ and $v_5$ are the two ends of a diameter of the projected circle. Construct a circle with diameter $v_4v_5$; that is the projected circle.