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I discovered empirically what to me is an amazing lemma concerning face angles of a tetrahedron. Let $\triangle abc$ be a triangle in the $xy$-plane, and $d$ the apex of a tetrahedron with positive $z$ coordinate. The lemma is this:

Lemma. The locus of points $d$ for which the sum of the nonbase tetrahedron face angles incident to $b$ sum to $\pi$, $$\angle dba + \angle dbc = \pi$$ is a vertical (parallel to $z$) bisector plane which meets the $xy$-plane in a line $L$ that has the property that the angle of incidence $\beta$ between $ab$ and $L$ is equal to the angle of reflection $\beta$ between $bc$ and $L$.

If I express this relationship in terms of ArcCos( )'s of the relevant angles, it all works out algebraically/trigonometrically. So I have a "proof" in this (limited) sense. But surely for such a simple angle of incidence = angle of reflection relationship there is a concise geometrical explanation—maybe involving reflecting light rays...?

Amidst a much longer proof, this was at one point a critical lemma, but now I have circumvented its need (I think?!). Nevertheless, it would be illuminating to see a more revealing proof. Thanks for ideas and/or insights!

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@Joseph: Take $a'$ obtained by reflecting $a$ in the point $b$ (central symmetry) and dilating with center $b$ so that $|ab'|=|cb|$. Then you are asking for $d$ so that $\angle dba'=\angle dbc$ (since $a, b, a', d$ are coplanar. Since the triangles $\Delta dba'$ and $\Delta dbc$ are now congruent (two sides and the angle between them), you are asking for $|da'|=|dc|$, i.e. $d$ is on the bisector plane of $a', c$, which is exactly your claim. Of course, this is very similar to the arguments of both fedja's and Gjergji's. – Misha Apr 20 '12 at 11:58
up vote 2 down vote accepted

Reflect $a$ across $L$ in the $xy$-plane to $a'$, so $\angle a'bc$ is a straight angle, and hence $\angle dba' + \angle dba = \pi$. If $d$ is in the bisector plane then $\angle dba' = \angle dba$, if $d$ is on the same side of the bisector plane as $a$ then $\angle dba' > \angle dba$, and if $d$ is on the opposite side then $\angle dba' < \angle dba$.

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Nice! $\mbox{ }$ – Joseph O'Rourke Apr 20 '12 at 15:47

In other words, you want the set of points $d$ such that if you rotate the plane $bcd$ so that it makes one plane with $abd$, then $ab$ and $bc$ make one line. Since you know that line, the question becomes "given two rays emanating from one point, find all possible axes of rotation containing the intersection point such that one ray can be rotated to the other one. Clearly, the distance from each point on the axis of rotation to both rays must be the same and clearly that sameness depends only on the projection of the axis point to the plane spanned by the rays. Should I say more?

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@fedja: Aside from your unclosed quotation marks :-), I think you nailed it, although the relationships between the various parameters is a bit complex... – Joseph O'Rourke Apr 20 '12 at 1:59

This is still in terms of trigonometry, but it's pretty quick. Let $\vec{u}$ and $\vec{v}$ be the unit vectors in the direction of $\vec{bc}$ and $\vec{ba}$, respectively. We have $$\angle dba + \angle dbc = \pi \iff \cos(\angle dba)+\cos(\angle dbc)=0\iff \vec{u}\cdot \vec{bd}+\vec{v}\cdot\vec{bd}=0$$ $$\iff (\vec{u}+\vec{v})\cdot \vec{bd}=0$$ But the line $\vec{u}+\vec{v}$ is the internal angle bisector of $\angle abc$ so this is precisely the plane orthogonal to that passing through $b$.

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@Gjergji: Succinct and clear! Thanks!! – Joseph O'Rourke Apr 20 '12 at 1:58

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