Height of an irregular tetrahedron with an equilateral base and lateral faces making angles $60^\circ$, $60^\circ$, $80^\circ$ with that base

Question

An irregular tetrahedron has a base that is an equilateral triangle of side length $10$. The lateral faces make angles of $60^\circ, 60^\circ$ and $80^\circ$ with the base. Find the height of the tetrahedron.

So, one way I thought I could solve this problem is using coordinate geometry, specifically attaching a reference frame to the base, and writing the equations of the three planes that represent the three lateral faces, and then solving the linear system for the apex coordinates.

Answer 1

Given a tetrahedron with base $ABC$ and apex $D$, let

• $E$ be the orthogonal projection of $D$ onto the plane holding $ABC$.
• $h = |DE|$ will be the height of tetrahedron.
• $\theta_A / \theta_B / \theta_C$ be the angle between faces $DBC$ / $DCA$ / $DAB$ and base $ABC$.
• $\ell_A / \ell_B / \ell_C$ be the distance of $E$ to edges $BC$ / $CA$ / $AB$.

As long as all $\theta_A, \theta_B, \theta_C < 90^\circ$, $E$ lies inside $ABC$. Furthermore, we have

• $\ell_A = h \cot\theta_A$, $\ell_B = h \cot \theta_B$ and $\ell_C = h\cot\theta_C$
• $|BC|\ell_A + |CA|\ell_B + |BC|\ell_C = 2\verb/Area/(ABC)$

For the tetrahedron at hand, we have $|AB| = |BC| = |CA| = 10$ and $\verb/Area/(ABC ) = \frac{\sqrt{3}}{4}(10)^2$.
This leads to

$$\ell_A + \ell_B + \ell_C = 5\sqrt{3}$$

and as a result, $$\begin{align} h = \frac{\ell_A + \ell_B + \ell_C}{\cot\theta_A + \cot\theta_B + \cot\theta_C} &= \frac{5\sqrt{3}}{2\cot(60^\circ) + \cot(80^\circ)} = \frac{15}{2 + \sqrt{3}\cot(80^\circ)}\\ &\sim 6.506442514261543\end{align}$$

Answer 2

I believe this is a good idea. First write down the equations of the three sides of the triangle, then find the equations of the straight line for each side such as:

1. The line is perpendical to the side
2. The lines intercepts the side
3. The line has an angle of $\pi/3$ with the side (use trigonometry)

Then you will find the plane with the normal vector $\vec{n}$ (with $\vec{n} = \vec{d}_{line} \times \vec{d}_{side}$) intercepting the only point belonging to both side and line, and finally calculate the interception of the three planes. The final steps should not be a problem (you may use Pythagore or some geometry formula's). Sorry for my poor English, I hope it will help.

Answer 3

Let $ABC$ denote the equilateral triangle where the two dihedral angles with values 60° taken along $AB$ and $AC$. Let $H$ be the midpoint of $BC$ and $D$ be the apex of the tetrahedron. Vertical plane $HAD$ is clearly a plane of symmetry of this tetrahedron.

Let $F$ be the foot of the altitude dropped from $D$ onto the horizontal plane; $F$ belongs clearly to line $AH$. We have to find $f:=\text{length}(DF)$.

Let $d:=\text{length}(FH)$.

First, we have:

$$\frac{d}{h}=c \ \ \text{where} \ \ c:=\cot(80°)\approx 0.1763269 \tag{1}$$

It is known (see for example page 3 of this document) that the barycentric coordinates of $F$ with respect to base triangle $ABC$ are proportional to the products of the different sidelengths of the base triangle by the cotangent of their associated dihedral angles:

$$(\text{length(AB)} \cot(60°), \ \ \ \text{length(AC)} \cot(60°), \ \ \ \text{length(BC)} \cot(80°)),$$

themselves proportional to:

$$(\frac{1}{\sqrt{3}},\frac{1}{\sqrt{3}},c) \tag{2}$$

The "true" barycentrical coordinates being obtained by dividing expressions (2) by their sum: $c+\frac{2}{\sqrt{3}}$.

Using the classical areal interpretation of these "true" barycentric coordinates and taking into account the fact that the area of triangle $ABC$ is $25 \sqrt{3}$, one can identify 2 different ways to express the ratio $\frac{area(HBC)}{area(ABC)}$, i.e.,

$$\frac{\tfrac12 d \times 10}{25 \sqrt{3}}=\frac{c}{c+\frac{2}{\sqrt{3}}}\tag{3}$$

From (3), one deduces

$$d=\dfrac{5 \sqrt{3}c}{c+\frac{2}{\sqrt{3}}}$$

Then, from (1), we get the value of

$$h=\frac{d}{c} \approx 6.5064425...$$