A conjecture about an angle on a solid body

Question

I've come accross this problem as a problem of hyperbolic geometry. I transformed it to get a tamed problem of Euclidean geometry. In the original problem the point $N$ will be in the infinity so that $AN$ and $BN$ will be parallel in the hyperbolic sense. But for the time being it would be helpful to learn the answer to my problem within the framework of euclidean (absolute) geometry.

Please consider the following figure:

Let the plane $AA'N$ be perpendicular to the plan $ABN$ and the dihedral angle of the planes $ABN$ and $BNA'$ be less than $\frac{\pi}2$. The planes $ANA'$ and $ABN$ intersect in the line $A'N$. Finally, let $AB$ be perpendiculat to the plane $AA'N$.

Now, drop a $\color{red}{\text{perpendicular}}$ from $A$ to the plane $A'BN$. This $\color{red}{\text{perpendicular}}$ meets $BNA'$ in $\color{red} F$. Join $B$ and $\color{red} {F}$ with a straight and construct $\color{red} {F'}$.

My conjecture is that $\color{red} F\color{red} {F'}$ is perpendicular to $NA'$. Please help me prove or disprove my conjecture.

Answer 1

An analytic proof. Without loss of generality, we can take $A$ as the origin of a coordinate system and $A'$, $B$, $N$ along the axes: $$ A=(0,0,0);\quad A'=(0,0,a);\quad B=(b,0,0);\quad N=(0,n,0). $$ A vector $\vec v$ perpendicular to plane $A'BN$ ca be readily found as $$ \vec v=(N-B)\times(A'-B)=(an,ab,bn), $$ so that plane $A'BN$ has equation $anx+aby+bnz=abn$. Substituting here $F=t\vec v$ one finds $t=abn/(a^2n^2+a^2b^2+b^2n^2)$.

To show that $BF'$ and $A'N$ are perpendicular, it suffices to check that the dot product of vectors $F-B$ and $N-A'$ vanishes. We have indeed: $$ (F-B)\cdot(N-A')=(ant-b,abt,bnt)\cdot(0,n,-a)=0. $$

EDIT.

A synthetic proof. Let $F'$ be the perpendicular from $F$ to $A'N$: I will show that line $FF'$ passes through $B$.

By the theorem of three perpendiculars, plane $AFF'$ is perpendicular to $A'N$, and thus it is also perpendicular to plane $AA'N$. Line $AB$ is perpendicular to $AA'N$ too: it follows that $AB$ lies on plane $AFF'$.

Point $B$ belongs then to the intersection of planes $A'BN$ and $AFF'$, which is line $FF'$, as we wanted to show.

EDIT 2.

A more direct synthetic approach might be helpful. Let, as before, $F$ be the perpendicular from $A$ to plane $A'NB$ and $F'$ the perpendicular from $F$ to line $A'N$: I will show that line $FF'$ passes through $B$.

Construct points $P$ and $Q$ on line $A'N$, on either side of $F'$, such that $PF'=QF'$. Triangles $PFF'$ and $QFF'$ are equal, whence $PF=QF$. Triangles $PAF$ and $QAF$ are then also equal, implying that $PA=QA$. It follows that triangles $PAB$ and $QAB$ are equal, whence $PB=QB$. Point $B$ belongs then the the vertical bisector of $PQ$, that is to line $FF'$.

Answer 2

For the hyperbolic result, with finite $N$, let's examine what happens in plane $A^\prime B N$ ... although I'm going to use different notation to match my personal mental model; for specificity, I'm making these substitutions: $$A \to O \qquad N \to A \;\text{(finite)} \qquad A^\prime \to B \qquad B \to C \qquad F \to P \qquad F \to Q $$

Let $Q$ lie on $\overleftrightarrow{AB}$, and let $P$ lie on $\overleftrightarrow{CQ}$, and let $\angle AQC$ have measure $\theta$.

Consider the unmotivated quantity $$m := \left|\;\cosh{CA}\cosh{PB}-\cosh{CB}\cosh{PA}\;\right| \tag{1}$$ Using the hyperbolic Law of Cosines for Sides, we have $$\begin{align} \cosh CA &= \cosh{QA}\cosh{QC} - \sinh{QA}\sinh{QC} \cos\theta \\ \cosh CB &= \cosh{QB}\cosh{QC} \pm \sinh{QB}\sinh{QC} \cos\theta \\ \cosh PA &= \cosh{QA}\cosh{QP} - \sinh{QA}\sinh{QP} \cos\theta \\ \cosh PB &= \cosh{QB}\cosh{QP} \pm \sinh{QB}\sinh{QP} \cos\theta \end{align}$$ where "$\pm$" is "$+$" if $A$ and $B$ are on opposite sides of $Q$, and "$-$" if $A$ and $B$ are on the same side of $Q$. After a little algebraic manipulation, we find $$\begin{align} m &= \left|\cos\theta (\sinh{QA}\cosh{QB} \pm \cosh{QA}\sinh{QB}) (\sinh{QP}\cosh{QC} - \cosh{QP} \sinh{QC})\right| \\[4pt] &= \left|\;\cos\theta \sinh\left(QA\pm QB\right)\sinh\left(QP-QC\right)\;\right| \\[4pt] &= \left|\;\cos\theta\;\sinh AB \;\sinh PC \;\right| \end{align}$$

Reasonably declaring $A\neq B$ and $P\neq C$, have $m = 0$ if and only if $\cos\theta = 0$. That is,

$$\overleftrightarrow{AB}\perp \overleftrightarrow{CP} \quad\iff\quad \cosh{CA}\cosh{PB}=\cosh{CB}\cosh{PA} \tag{2}$$

Moving into three dimensions, let $\overline{OP}$ be perpendicular to plane $ABC$.

The hyperbolic Pythagorean Theorem, applied to right triangles $\triangle OPA$ and $\triangle OPB$, gives $$\cosh OA = \cosh{OP}\cosh{PA} \qquad\qquad \cosh OB = \cosh OP \cosh PB $$ This allows us to re-write condition $(2)$ as

$$\overleftrightarrow{AB}\perp \overleftrightarrow{CP} \quad\iff\quad \cosh{OA}\cosh{BC}=\cosh{OB}\cosh{CA} \tag{3}$$

We have yet to apply specific conditions from your setup. Of course, $(2)$ isn't at all helpful if $A$ (your $N$) is a point at infinity. That aside, we have $\overline{OC} \perp \triangle OBA$, so that $\angle OCB = \angle OCA = \pi/2$; therefore, again by hyperbolic Pythagoras, $$\cosh OA = \cosh OC \cosh CA \qquad\qquad \cosh OB = \cosh OC \cosh BC$$ whereupon the equality in $(3)$ is satisfied, so that the perpendicularity condition holds for all finite $A$. $\square$

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For $A$ at infinity, one should derive a condition that doesn't rely directly on lengths of edges through $A$. An approach using the hyperbolic Law of Cosines for Angles would seem appropriate, but I haven't found a clean argument.

FYI: Here are some references about various metric properties of hyperbolic tetrahedra:

• "Elementary Formulas for a Hyperbolic Tetrahedron" (PDF) by A. D. Mednykh and M. G. Pashkevich

• "Hedronometric Formulas for a Hyperbolic Tetrahedron" (PDF) by myself.