Picture a regular tetrahedron where each vertex has a line through the centroid and a plane normal to it. I need to show that the range of the smallest angles of incidence from an arbitrary point to one of these planes, as measured at the centroid, falls within $-arccos(1/3)/2$ to $+arccos(1/3)/2$.

i.e. the angles are measured at each plane/centroid intersection and the smallest of these never exceeds the above range.

I can model it, just not prove it. Any help is much appreciated.

[edit] An analogous question is what is the minimum angle $\theta$ such that four spherical segments defined by +/- $\theta$ as subtended from the centroid in each of the 4 normals to the surface of a sphere will give 100% coverage of that sphere.

I apologise for the possibly vague form of the question. It's just that I'm an IT guy and not a mathematician.



I'm struggling with a way to present this question such that it is clear to others. The best that I can come up with at the moment is an analogy.

Imagine that I construct a regular tetrahedron inside a larger sphere. On this tetrahedron I place poles that go from the vertex to the centroid. At the centroid I mount a spotlight on, but perpendicular to, each of these poles. These spotlights being free to rotate 360 degrees around the pole. Forgetting for a moment the impossibility of having all four in the same place at one time, my question would be, what is the minimum beam angle that I would need in order to be able to illuminate any part of the inside of the sphere.


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    $\begingroup$ When you talk about the "angle of incidence" from a point X to a plane P, measured at another point Y, what do you mean? Do you mean the angle between YX and YZ, where Z is the nearest point to Y on P? Or something else? $\endgroup$ – Matt May 18 '15 at 9:50
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    $\begingroup$ Is your question equivalent to the following: Consider a regular tetrahedron with vertices $T_1, T_2, T_3, T_4$, centered at the origin $O$. Show that for any point $P$ not at $O$, there is some $i$ for which $\angle POT_i < \arccos(1/3)/2$. $\endgroup$ – Matt May 18 '15 at 10:05
  • $\begingroup$ So, if your tetraedron is centered at the origin and has vertices $$(1,1,1),\quad (1,-1,-1)\quad (-1,1,-1)\quad\text{and}\quad (-1,-1,1),$$ what would be the angle of incidence of the point $P=(1,1,-1)$? $\endgroup$ – san May 18 '15 at 18:52
  • $\begingroup$ @Matt I think that you have the gist of it. Only I'm not measuring the angle to the vertex but to a plane nprmal to it. i.e. parallel to it's opposite face but passing through the centroid. $\endgroup$ – Kevin Acres May 18 '15 at 20:22
  • $\begingroup$ @san, sorry but I'm unsure about your notation for the vertices. I don't know how to relate it to my x,y,z measurements and can't really answer a question that I don't completely understand. $\endgroup$ – Kevin Acres May 19 '15 at 0:40

My computations confirm that the maximal angle is $$ arcsin(1/\sqrt{3})=\frac{1}{2}arccos(1/3). $$ You can always put a coordinate system such that the origin is at the centroid and the vertices of the tetrahedron are $$ (1,1,1),\quad (1,-1,-1),\quad (-1,1,-1)\quad\text{and}\quad (-1,-1,1). $$ This means, they are four corners of a 3D cube of sidelength 2 centered at the origin, with faces parallel to the coordinate planes. We claim (proof below) that the maximal angle is attained at the center of any of the faces. On the other hand the angle between the center of a face and an adjacent corner is $arccos(1/\sqrt{3})$. For example if you take the corner to be $A=(1,1,1)$ and the center of the face $M=(0,0,1)$, then $$ \cos(\sphericalangle MOA)=\frac{M\cdot A}{\|A\|\|M\|}=\frac{1}{\sqrt{3}\cdot 1}. $$ Hence the angle we are searching for is the complementary angle $\pi/2-arccos(1/\sqrt{3})$ which gives $$ arcsin(1/\sqrt{3})=arccos(\sqrt{2/3})=\frac 12 arccos(1/3), $$ as desired.

$\textbf{ Proof of the claim:}$ For any point in the sphere the straight line from the origin to the point intersects the surface of the cube at a unique point. Hence, it suffices to show that for any point on the cube, there exists a corner, such that the angle between them is lower than or equal to $\pi/2$ and greater than or equal to $arccos(1/\sqrt{3})$. This shows that the angle to the orthogonal plane, which is the complementary angle, is $\ge 0$ and at most $arcsin(1/\sqrt{3})$.

For this we cover a face of the cube with four symmetric trapezoids. Each trapezoid is opposed to a corner in the following way:

If $A=(1,1,1)$, $B=(-1,1,1)$, $C=(-1,-1,1)$ and $D=(1,-1,1)$, the trapezoid opposed to $A$ is formed by $B,D,E,F$, where $E=(0,-1,1)$ is the midpoint of the segment $BC$ and $F=(-1,0,1)$ is the midpoint of the segment $CD$. Then for any point $P$ in this trapezoid opposed to $A$ the angle between the point $P$ and $A$ (the central angle $\sphericalangle POA$) is lower than or equal to $\pi/2$ and greater than or equal to $arccos(1/\sqrt{3})$:

For the first inequality it suffices to compute the internal product $P\cdot A\ge 0$, which follows from $$ E\cdot A=(0,-1,1)\cdot (1,1,1)=0\quad\text{and}\quad F\cdot A=(-1,0,1)\cdot (1,1,1)=0. $$ For the second inequality we proceed in two steps. First, if $P$ is not on the segment $BD$, then there is a straight line from $P$ to $A$ that intersects the segment $BD$ at one point, let us call it $Q$. Then we can add angles $$ \sphericalangle POA=\sphericalangle POQ+\sphericalangle QOA, $$ and so $\sphericalangle POA\ge\sphericalangle POQ$. Hence it suffices to prove the second inequality for points $Q$ on the segment $BD$.

If $A=(1,1,1)$, $B=(-1,1,1)$ and $D=(1,-1,1)$, the points on the segment $BD$ are parametrized by $$ Q(t)=(0,0,1)+t(1,-1,0),\quad \text{for $t\in[-1,1]$}. $$ Then $$ cos(\sphericalangle QOA)=\frac{Q(t)\cdot A}{\|Q(t)\|\|A\|}=\frac{1}{\sqrt{3(2t^2+1)}}, $$ which attains its maximum $1/\sqrt{3}$ at $t=0$, i.e., at the midpoint of $BD$, which is the center of the face. Therefore all angles $\sphericalangle POA$ of points in the trapezoid are greater than or equal to $\sphericalangle QOA$ for a point $Q$ on $BD$, and $\sphericalangle QOA\ge arccos(1/\sqrt{3})$, as desired.

  • $\begingroup$ Thanks a lot for the proof. Do you know of any references in literature that I could read up? $\endgroup$ – Kevin Acres May 19 '15 at 20:25
  • $\begingroup$ No, I know no reference for this specific construction. Feel free to ask about any details of the proof. It would also help to view the construction in Mathematica, 3Dplot. $\endgroup$ – san May 19 '15 at 23:43
  • $\begingroup$ I may need to include the proof in my paper. In such an event I need to give proper credit for it. What's the best way forward on that? $\endgroup$ – Kevin Acres May 20 '15 at 0:00
  • $\begingroup$ Look at meta.math.stackexchange.com/questions/4259/… For myself I wouldn't complain if you just cite this question in Math.stackexchange. $\endgroup$ – san May 20 '15 at 0:39

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