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This is my first math stackexchange question, so if it is a duplicate, I will remove it. My question is as follows:

Say I have a point at the origin, in 3D space. Now from this origin point, I have 3 rays going out, where the 3 are not coplanar. I would like to find a plane where the three points where these 3 rays meet the plane form a specific nondegenerate triangle, specified by its 3 angles.

Anyone know how I would go about this? If it is not possible for any triangle, what about a right or equilateral triangle, or a square?

Vincent
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SupaMaggie70 b
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1 Answers1

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Let us somewhat invert the issue by working on a triangle installed in the $xy$ plane and finding the position of the summit $M(x,y,z)$ from which the rays are "emitted".

Without loss of generality, one can assume that the triangle has coordinates:

$$A(-a,0,0),\ \ B(a,0,0), \ \ C(b,c,0)$$

Let $R_1, \ R_2, \ R_3$ be the three rays with resp. cosine of angles :

$$c_{1,2}=\cos(R_1,R_2), \ \ c_{2,3}=\cos(R_2,R_3), \ \ c_{3,1}=\cos(R_3,R_1)\tag{1}$$

Let us consider the particular case, where, say, it is rays $(D_1,D_2)$ that "see" side $AB$. Let us equate the two ways to express the following dot product

$$\vec{MA}.\vec{MB}=c_{1,2} \|\vec{MA}\| \|\vec{MB}\|$$

or more exactly its square :

$$((-a-x)(a-x)+y^2+z^2)^2=c_{1,2}^2 ((-a-x)^2+y^2+z^2)((a-x)^2+y^2+z^2)\tag{2}$$

With words: quartic surface with equation (2), known under the name "spindle torus" (more or less apple-shaped) is therefore the set of points in $\mathbb{R^3}$ from which one "sees" line segment $AB$ under a given angle.

What we have done for this pair of rays $(R_1,R_2)$ and this side $AB$ can be done of course in the same way for the two other pairs of rays and sides.

If these 3 spindle torii intersect in a common point (a single point if we impose $z>0$ ; otherwise symmetry will give another point), this point is summit $M$, hence a solution exist for the given issue.

Such a solution exists if the angles mentionned in (1) are all less than a certain threshold (is it $\pi/2$ or less ? the question remains open).

Here is a graphical representation :

enter image description here

Fig. 1: A Matlab representation of the intersection of two spindle torii associated with sides $AB$ and $BC$. These surfaces look like half-apples with the little seeds' region around line segments $AB$ or $AC$ which are excluded areas. The third spindle torus hasn't been represented because it would have obscured the interpretation.

Jean Marie
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    Jesus christ thank you so much. The image actually makes a lot of sense. I appreciate you for putting this much effort into the problem! – SupaMaggie70 b Dec 28 '22 at 23:04
  • Important: I was remembering a similar question (to which I had given a solution) ; I have retrieve it : https://math.stackexchange.com/q/4407068/305862 – Jean Marie Dec 29 '22 at 00:17
  • this looks like exactly my problem! Should I mark mine as a duplicate or anything, or is it fine? – SupaMaggie70 b Dec 29 '22 at 00:44
  • Ok so it's impossible to find one point for a triangle that has an angle >90 degrees? – SupaMaggie70 b Dec 29 '22 at 02:00
  • About marking your question as a duplicate : as the other question hasn't been asked exactly with the same formulation, leave this task to other people, in particular moderators, to appreciate if it's really the case. 2) It looks like $\pi/2$ is a threshold, but we must have a thorough analysis to ascertain it.
    • – Jean Marie Dec 29 '22 at 08:07
  • ok, that's interesting. I appreciate all of the help. I am just surprised that obtuse triangles don't work and that right triangles might not, granted I don't understand a lot of what was said yet. – SupaMaggie70 b Dec 29 '22 at 16:31
  • In fact, I am not certain at all that $\pi/2$ is this limit. It might be $2\pi/3$. I am going to reconsider what I have written about the limit. – Jean Marie Dec 29 '22 at 16:37
  • oh I thought you meant it was that or lower, I misunderstood sorry. – SupaMaggie70 b Dec 29 '22 at 16:42
  • To represent it as a 2D problem, transform the line such that it is between (-1,0) and (1,0). Say (-1,0) is A, the origin is B, and the point you are searching for is C. By trigonometry, BC=AB/tan(angle C). Thus BC=tan/C=1/tan(angle we want over two, as we sliced original triangle in half). Since any tangent value greater than pi/2 will wrap around into negatives, the highest limit would be pi radians. I might've done something wrong, or missed something that comes with multiple points or 3D space. I appreciate your help! – SupaMaggie70 b Dec 31 '22 at 04:52
  • Your line of reasoning looks right : indeed the "allowed angles" should be between $0$ and $\pi$. Besides, I am going "in a near future" to make an edit to my answer. – Jean Marie Dec 31 '22 at 08:23
  • Another similar question : https://math.stackexchange.com/q/1110809/305862 – Jean Marie Dec 31 '22 at 08:26