Area of Projection of Tetrahedron

Question

this is my first post. Here is a question I found in a handout I am reading on 3D geometry.

A plane passes through the midpoints of two skew lines of a regular tetrahedron. The projection of the tetrahedron to the plane produces a quadrilateral with an angle of $60$. What is the ratio of the area of this quadrilateral to the surface area of the original tetrahedron?

I was able to solve the problem using geogebra to model the diagram, but I am unable to prove it. The answer that I found was about $0.3145$. I am really stuck on how to prove an exact answer.

Please help.

*Here is the link to my geogebra model: https://www.geogebra.org/3d/tjpgzaqb

Answer 1

As we are concerned with a ratio of areas, we can assume that the tetrahedron is inscribed in the cube $(\pm1,\pm1,\pm1)$ with vertices

$$B(1,1,1), \ A(-1,-1,1), \ C(1,-1,-1), \ D(-1,1,-1)$$

We have: $\vec{AB} \perp \vec{CD}$ (resp. the upper edge and lower edge). Their midpoints are $E(0,0,1)$, resp $F(0,0,-1)$ A rotation with angle $\alpha$ gives rise to a projection onto the following isosceles trapezoid:

where the projected points onto the plane are:

$$A_1(-\sqrt{2} \cos \alpha,1), \ B_1(\sqrt{2} \cos \alpha,1), \ C_1(\sqrt{2} \sin \alpha,-1), \ D(-\sqrt{2} \sin \alpha,-1)$$

Indeed, as $\tan \pi/3 = \sqrt{3}$, we must have:

$$\underbrace{\tan \angle C_1D_1A_1}_{\sqrt{3}}=\dfrac{EF}{|D_1F-A_1E|}$$

$$\dfrac{\sqrt{2}}{2}|\cos \alpha - \sin \alpha|=\frac{1}{\sqrt{3}}$$

with a particular solution:

$$\sin(\alpha-\pi/4)=\frac{1}{\sqrt{3}}$$

$$\alpha=\pi/4+\arcsin \frac{1}{\sqrt{3}}$$

From there, it is not difficult to find the ratio of areas.

Indeed, the area of the trapezoid is

$$Area \ = \ half basis \ \times \ height \ = \ 2 \sqrt{2}(\sin(\alpha)+\cos(\alpha))$$

$$ \ Area \ = \ 4\cos(\alpha - \pi/4)=4\cos(\arcsin(1/\sqrt{3}))$$

$$ \ Area \ = \ 4\sqrt{2/3}\tag{1}$$

As the lateral area of the tetrahedron is

$$8 \sqrt{3}\tag{2},$$

the final ratio is the quotient of (2) and (1):

$$\color{red}{\frac{\sqrt{2}}{6}}\approx 0.2357$$

which is below your estimation...

Answer 2

Let's take the vertices of the tetrahedron as being:

$A(1,1,1), B(-1,-1,1), C(1,-1,-1)$ and $D(-1,1,-1)$.

Choose $AB$ and $CD$ as being our skew sides.

The midpoints are given by $M(0,0,1)$ and $N(0,0,-1)$

We want a plane that includes these points. Its normal vector $\bf{n}$ must be perpendicular to $\vec{MN}$

$\begin{bmatrix}{x\\y\\z}\end{bmatrix}.\begin{bmatrix}{0\\0\\2}\end{bmatrix}=0\Rightarrow 2z=0\Rightarrow z=0$

Let $\bf{n}=\begin{bmatrix}{\cos \theta\\ \sin \theta\\ 0}\end{bmatrix}$

Plane is given by $\bf{r}.\begin{bmatrix}{\cos \theta\\ \sin \theta\\0}\end{bmatrix}$$=d$

Since the plane contains $M(0,0,1)$, we know that $d=0$

Plane is given by $\bf{r}.\begin{bmatrix}{\cos \theta\\ \sin \theta\\0}\end{bmatrix}$$=0$

Projections onto this plane can be calculated...

Answer 3

Here's a solution without trigonometry. Let's take a tetrahedron $ABCD$ where the distance $OO'$ between the midpoints of two skew edges is $2$, and the edges have then a length of $2\sqrt2$. The projection is an isosceles trapezoid $A'B'C'D'$, with altitude $OO'=2$ and bases $a=A'B'$, $b=C'D'$.

Projecting the figure on a plane perpendicular to $OO'$ (figure below) one realises that triangles $DD'O$ and $BB'O'$ are equal, hence $OD'^2+O'B'^2=O'B^2$, that is: $$ a^2+b^2=8. $$ Moreover, from $\angle O'B'D'=60°$ it follows $O'B'-OD'=OO'/\sqrt3$, that is: $$ a-b={4\over\sqrt3}. $$ Combining the above equations one finds: $$ a={2\over\sqrt3}(\sqrt2+1), \quad b={2\over\sqrt3}(\sqrt2-1) $$ and from there the area of the trapezoid can be found, leading to a ratio $$ {\text{area of trapezoid}\over\text{surface area of tetrahedron}} ={\sqrt2\over6}. $$

Answer 4

As we are concerned with ratio of areas, we can assume that the tetrahedron is inscribed in the cube $(\pm1,\pm1,\pm1)$ with vertices

$$A(1,1,1), \ B(-1,-1,1), \ C(1,0,-1), \ D(0,1,-1)$$

We have: $\vec{AB} \perp \vec{CD}$ (resp. the upper edge and lower edge). Their midpoints are $E(0,0,1)$, resp $F(0,0,-1)$ A rotation with angle $\alpha$ gives rise to a projection onto an isosceles trapezoid with these values:

Answer 5

Similar to the solution by @tomi, nice coordinates for the tetrahedron can be chosen as

$A(1,1,1), B(1, -1, -1), C(-1, 1, -1), D(-1, -1, 1) $

Its side length is $2 \sqrt{2}$

Two opposite edges are $AB$ and $CD$, their respective midpoints are

$M = (1, 0, 0) $ and $N = (-1, 0, 0) $

Hence, the line joining $M$ and $N$ is along the $x$-axis. A plane containing $MN$ must have a unit normal vector perpendicular to $MN$, thus its most general form is

$n = (0, \cos t, \sin t) = (0, c, s) $

where $c = \cos t, s = \sin t $ for some $t$ that we will find.

This is where I depart from @tomi's work.

We can let projection plane pass through the origin without affecting the projected polygon. In addition, define unit vectors $u_1$ and $u_2$ such that the triple $[u_1, u_2, n] = R$ form an orthonormal basis (i.e. matrix is a rotation matrix). Two such vectors can be easily picked as $u_1 = [1, 0, 0]^T $ and $u_2 = [0, s, -c]^T $.

Now the projection of the four vertices $A,B,C,D$ on to the $x'y'$ plane is obtained by premultiplying the coordinates by $R^T$, which is given by

$R^T = \begin{bmatrix} 1 && 0 && 0 \\ 0 && s && - c \\ 0 && c && s \end{bmatrix} $

The $2D$ coordinates of the projection in coordinate system $R$ are

$A' = (1, s - c) $

$B' = (1, c - s) $

$C' = (-1, s + c ) $

$D' = (-1, -s - c )$

Note that $A'$ and $B'$ are mirror images about the $x'$ axis, so are $C'$ and $D'$.

The angle we're concerned with is the angle between $A'B'$ and $A'C'$ which we want to be $60^\circ$

$A'B' = 2(0, c-s ) $ and $A'C' = 2(-1, c) $

Hence the angle $\phi $ between them satisfies

$\cos \phi = \dfrac{1}{2} = \dfrac{(A'B') \cdot (A'C') }{\| A'B' \| \|A'C'\| } = \dfrac{ c (c - s) }{ | c - s| \sqrt{1 + c^2 } } = \text{sgn}(c - s) \dfrac{c }{\sqrt{1+c^2}}$

Assuming $\text{sgn}(c-s)$ to be $+1$ to simplify the equation, we end up with

$\sqrt{1+c^2} = 2 c \Rightarrow c = \dfrac{1}{\sqrt{3}} \Rightarrow s = \pm \sqrt{\dfrac{2}{3}}$

Since we want $\text{sgn}(c - s) $ to be $+1$ we have to take the sine to be negative.

Now the area of the projected trapezoid resulting is

$A = (1 - (-1)) \dfrac{1}{2} (\|A'B'\| + \|C'D'\|) = 2(c - s) + 2(-s - c) = -4 s = 4 \sqrt{\dfrac{2}{3}} $

The area of the tetrahedron is $A_T = 4 \cdot \dfrac{\sqrt{3}}{4} \cdot (2 \sqrt{2})^2 = 8\sqrt{3} $

Hence the ratio of areas is $\dfrac{ \sqrt{2} }{6} $