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Given the following information about the tetrahedron pictured below, can the radius of the sphere that inscribes the tetrahedron be calculated?

Given:

Complete information about triangle BCD (side lengths and angles)

The angles formed by BAC, BAD, and CAD

Tetrahedron with limited information

Some physical experimentation with rings representing the circles that inscribe faces BAC, BAD, and CAD (the radii of which can be calculated from the given information) held together on one side at fixed distances (as would be perscibed by the side lengths of face BCD) has shown me that using only the given information I am capable of constructing a unique sphere, though I have yet to find a way to do this mathematically.

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  • $\begingroup$ If $BD = CD$ and $\angle BAD=\angle CAD$, then there are two points $D1$ and $D2$ on line $AD$ that will work for $D$, and they won't give the same sphere. $\endgroup$ – Michael May 30 '18 at 13:27
  • $\begingroup$ Hmm, it looks like you're right, but in the two cases the circle that inscribes face BAD would be two different sizes, and I was under the impression that knowing one side length and the opposing angle allowed you to calculate the radius of the inscribing circle of a triangle using the law of sines. Where does my intuition break down? $\endgroup$ – DaggerOfMesogrecia May 30 '18 at 13:35
  • $\begingroup$ Looking more closely at the equivalent in two dimensions, even though very different triangles can be constructed given the same angle and opposite side length, the radius of the inscribing circle is always the same (though not the location). Can you be certain that this doesn't hold for tetrahedra? $\endgroup$ – DaggerOfMesogrecia May 30 '18 at 13:46
  • $\begingroup$ You might want to compare notes with shadow_wxh: see math.stackexchange.com/q/2758699/265466. $\endgroup$ – amd May 31 '18 at 6:28
  • $\begingroup$ I'll have to reach out to him. Thanks again for all the help. $\endgroup$ – DaggerOfMesogrecia May 31 '18 at 14:08
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The inscribed circles on the faces of the tetrahedron don’t have a particularly simple relationship to its inscribed sphere. A way to view one of these circles is as the intersection of an elliptical cone with the face. This cone is tangent to the other three faces and to the inscribed sphere. I don’t believe that the axes of these cones pass through the sphere or circle centers, so it doesn’t seem to me that examining them really gets you any closer to finding the inscribed sphere.

Based on your comments it seems that you’re really interested in finding $A$ by somehow using the inspheres of a bunch of tetrahedrons that share this vertex. All of the ways that I can imagine to compute the insphere either require knowing $A$ or computing things like the face normals that are tantamount to knowing $A$. I don’t really see how working with inspheres would lead to anything more tractable than a direct computation.

Unfortunately, the information you have about the tetrahedron doesn’t lend itself to a straightforward solution: You have some of the side lengths and angles, but not the ones that could be used in the usual side-angle-angle formulas. Michael gives a direct approach to this in his answer, essentially turning the problem into one of finding the intersection of three quadrics. This is likely to have multiple solutions, but it seems to me that you should be able to disambiguate by considering the other tetrahedra in your larger problem.

Bringing either projective geometry or spherical trigonometry to bear on the problem might be useful. For example, given the three known angles at $A$, one can easily construct three rays from it with those angles between them. The problem of placing $A$ is then equivalent to finding a plane such that the pairwise distances between its intersections with those rays match the known edge lengths. Using spherical trigonometry, the dihedral angles of the faces meeting at $A$ can be computed, and it should be possible to develop constraints on the unknown angles and relationships among them. (There is, for instance, a tetrahedral counterpart of the Law of Sines that can relate adjacent angles at a vertex.)

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  • $\begingroup$ I do have an idea for how to use the incircles and related tangent cones, but it also involves knowing the face normals. If it pans out, I’ll update this answer. $\endgroup$ – amd May 30 '18 at 20:29
  • $\begingroup$ Thank you! I will take a look at this counterpart of the Law of Sines and see if I can work it out from there. $\endgroup$ – DaggerOfMesogrecia May 30 '18 at 20:31
  • $\begingroup$ @DaggerOfMesogrecia One way or another, I’m pretty sure that you eventually end up having to intersect three spheres to find $A$. $\endgroup$ – amd May 30 '18 at 20:52
  • $\begingroup$ The question just comes out to be what spheres will those be? This question was part of a larger approach where I would have multiple tetrahedra, that all share the unknown point. I was hoping to be able to calculate the radius of each sphere from the information given above and use three of them to find A. $\endgroup$ – DaggerOfMesogrecia May 30 '18 at 20:56
  • $\begingroup$ @DaggerOfMesogrecia Hmm... All of the ways that I can imagine of computing the insphere are equivalent to computing $A$. Might as well do that directly. $\endgroup$ – amd May 30 '18 at 21:03
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Set up the angles at $A$, with angles $\alpha,\beta$ and $\gamma$. We want to know $AB,AC$ and $AD$ so that $BC=d,CD=b$ and $BD=c$ are the right lengths. So we want to find $AB=x,AC=y$ and $AD=z$ for which $$x^2-2xy\cos\alpha+y^2=d^2\\ y^2-2yz\cos\beta+z^2=b^2\\ z^2-2xz\cos\gamma+x^2=c^2$$ I think three degree-2 equations will, in general, have eight ($=2^3$) solutions.

I entered it into Maple, which gave me a degree-8 polynomial for $x$. Here is how you might do it:

  • Subtract equation (2) from equation (1). The result is linear in $y$.
  • Multiply equation (1) by $(2x\cos\alpha-2z\cos\beta)^2$. That lets you substitute $y$ altogether, leaving $P(x,z)=0$, a degree-4 polynomial in $x$ and $z$.
  • Regard equation (3) as a quadratic in $z$. It has two solutions $z_1(x)$ and $z_2(x)$. The only things I need from this is $z_1(x)+z_2(x)=2x\cos\gamma$ and $z_1(x)z_2(x)=c^2-x^2$.
  • $P(x,z_1)P(x,z_2)=0$ is a degree-8 polynomial in $x,z_1$ and $z_2$, that is symmetric in $z_1$ and $z_2$. So it can be written in terms of $x,z_1+z_2$ and $z_1z_2$. Now it can be written as a degree-8 polynomial in just $x$.
  • The original equations have a symmetry, that $(x,y,z)$ can be replaced by $(-x,-y,-z)$. Therefore, the degree-8 polynomial is even, and is a degree-4 polynomial in $x^2$.

EDIT 2:
Here is a formula if all six edges are known, and the three angles mentioned above.
Let R be the circumcentre. It is the intersection of the planes that are the perpendicular bisectors of the edges. Any three of these planes will do so long as all four vertices are involved.
Put $A$ at the origin. The equation of the plane bisecting $AB$ is $$2\vec{R}.\vec x=\vec x.\vec x$$ Let $M$ be the matrix whose rows are $\vec x,\vec y,\vec z$, and $V$ the column vector whose entries are $x^2,y^2,z^2$. Then $2M\vec R=V$, and $\vec R=\frac12M^{-1}V$. Dot product $\vec R$ with itself to get $R^2$, the square of the circumradius $$R^2 = \frac14V^T(M^{-1})^TM^{-1}V= \frac14V^T(MM^T)^{-1}V\\ =\frac14 (x^2,y^2,z^2) \left(\begin{array}{ccc} x^2&\vec x.\vec y&\vec x.\vec z\\ \vec y.\vec x&y^2&\vec y.\vec z\\ \vec z.\vec x&\vec z.\vec y&z^2\end{array}\right)^{-1}\left(\begin{array}{c}x^2\\y^2\\z^2\end{array}\right)$$ Recall that $\vec x.\vec y=xy\cos\alpha,\vec y.\vec z=\cos\beta,\vec z.\vec x=\cos\gamma$, and I think it works out to $$\frac{x^2\sin^2\beta+y^2\sin^2\gamma+z^2\sin^2\alpha+2xy(\cos\beta\cos\gamma-\cos\alpha)+2xz(\cos\alpha\cos\beta-\cos\gamma)+2yz(\cos\alpha\cos\gamma-\cos\beta)} {4(1+2\cos\alpha\cos\beta\cos\gamma-\cos^2\alpha-\cos^2\beta-\cos^2\gamma)}$$ That can be written just in terms of $b,c,d,x,y,z$ by using my first three equations. I don't know whether it can be written just in terms of $b,c,d,\alpha,\beta,\gamma$, which was your original question.

EDIT 3:
The denominator is $$16\sin(\frac{\alpha+\beta+\gamma}2)\sin(\frac{\alpha+\beta-\gamma}2)\sin(\frac{\alpha-\beta+\gamma}2)\sin(\frac{-\alpha+\beta+\gamma}2)$$ while the numerator is $$\frac{(xb+yc+zd)(xb-yc+zd)(xb+yc-zd)(-xb+yc+zd)}{b^2c^2c^2}$$

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  • $\begingroup$ Ah well, I suppose this was just some wishful thinking on my part that this would be easy to generalize into 3 dimensions. $\endgroup$ – DaggerOfMesogrecia May 30 '18 at 13:56
  • $\begingroup$ I still can't help but feel that there is some geometric relationship being ignored that may help reduce this to solvability. If you can agree that the radii of the three circles that inscribe faces BAD, CAD, and BAC are set in stone, try placing three circles together such that they meet to form the triangle BCD. It seems to me that when bringing them up until they meet at one point that there is only one sphere that can be created. $\endgroup$ – DaggerOfMesogrecia May 30 '18 at 14:05
  • $\begingroup$ Take a look at this, and perhaps you'll see what I'm saying. imgur.com/a/AdPvHfb $\endgroup$ – DaggerOfMesogrecia May 30 '18 at 14:09

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