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How do I calculate: The height of a regular tetrahedron, side length 1.

Just to be completely clear, by height I mean if you placed the shape on a table, how high up would the highest point be from the table.

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I'm curious. If you mind telling us, what was the purpose of the interview? What were the other mathematically relevant questions asked? –  user02138 Nov 20 '10 at 15:25
    
It's an old question for and Economics and Management interview at Oxford. –  Patrick Beardmore Nov 20 '10 at 16:43
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3 Answers 3

up vote 8 down vote accepted

The first thing you need to do is to note that the apex of a regular tetrahedron lies directly above the center of the bottom triangular face. Thus, find the length of the segment connecting the center of an equilateral triangle with unit length to a corner, and use the Pythagorean theorem with the length of an edge as the hypotenuse, and the length you previously derived as one leg. The height you need is the other leg of the implied right triangle.


Here's a view of the geometry:

tetrahedron

and here's a view of the bottom face:

triangle

In the second diagram, the face is indicated by dashed lines, and the (isosceles) triangle formed by the center of the triangle and two of the corners is indicated by solid lines.

Knowing that the short sides of the isosceles triangle bisect the 60° angles of the equilateral triangle, we find that the angles of the isosceles triangle are 30°, 30° and 120°.

Using the law of cosines and the knowledge that the longest side of the isosceles triangle has unit length, we have the equation for the length $\ell$ of the short side (the length from the center of the bottom face to the nearest vertex):

$$1=2\ell^2-2\ell^2\cos 120^{\circ}$$

Solving for $\ell$, we find that the length from the center of the bottom face to the nearest vertex is $\frac{1}{\sqrt{3}}$, as indicated here.

From this, the Pythagorean theorem says that the height $h$ (the length from the center of the bottom face) satisfies

$$h^2+\left(\frac{1}{\sqrt{3}}\right)^2=1$$

Solving for $h$ in the above equation, we now find the height to be $\sqrt{\frac23}=\frac{\sqrt{6}}{3}$, as mentioned here.

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It's not homework, it was an interview question I struggled with. I'd love to know the answer. –  Patrick Beardmore Nov 20 '10 at 14:57
    
Good to know. I'll edit with an explicit answer. –  J. M. Nov 20 '10 at 15:00
    
Thank-you very much. P.S: how did you make your lovely 3d shape diagrams? –  Patrick Beardmore Nov 20 '10 at 16:43
    
@Patrick: I used Mathematica 5.2 (yes, it's an old copy :) ) for these diagrams. –  J. M. Nov 20 '10 at 21:33
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Consider the tetrahedron inscribed in the unit cube, with vertices at (0,0,0), (1,1,0), (0,1,1), (1,0,1). Its height is the distance from (0,0,0) to the centre of the opposite face, which is given by the equation $x+y+z = 2$. Thus its height is $\frac{2}{\sqrt 3}$, and since the edges of this tetrahedron have length $\sqrt 2$, the height of a regular tetrahedron with side $x$ is $x \sqrt{\frac{2}{3}}$.

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Note that this way of doing this also gives that the distance from the centers of opposite sides of the tetrahedron is always ${1 \over \sqrt{2}}$ times the side length. –  Zarrax Nov 20 '10 at 15:26
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You can also use trig based on the dihedral angle between two faces of the tetrahedron.

Writing $ABC$ for the base triangle, $O$ for the apex, $K$ for the center of $ABC$ (the foot of the perpendicular dropped from $O$), and $M$ for the midpoint of (for instance) side $BC$, we have a right triangle $OKM$ with right angle at $K$. So,

$$\text{height of tetrahedron} = |OK| = |OM|\sin{M}$$

$OM$ is the height of the (equilateral) face $OBC$, measuring $\frac{\sqrt{3}}{2}s$, where $s$ is the length of a side.

As for the measure of angle $M$ ... Note that this is the dihedral angle between faces $OBC$ and $ABC$; it is also the angle between (congruent) segments $OM$ and $AM$ in triangle $OMA$. We can use the Law of Cosines as follows:

$$\begin{eqnarray} |OA|^2 &=& |OM|^2 + |AM|^2 - 2 |OM||AM|\cos{M} \\ s^2 &=& \left(\frac{\sqrt{3}}{2}s\right)^2 + \left(\frac{\sqrt{3}}{2}s\right)^2 - 2 \left(\frac{\sqrt{3}}{2}s\right)\left(\frac{\sqrt{3}}{2}s\right) \cos{M} \\ s^2 &=& \frac{3}{4} s^2 + \frac{3}{4}s^2 - 2 \frac{3}{4} s^2 \cos{M} \\ 1 &=& \frac{3}{2} - \frac{3}{2} \cos{M} \\ \frac{-1}{2} &=& - \frac{3}{2} \cos{M} \\ \frac{1}{3} &=& \cos{M} \;\;\; (**)\\ \Rightarrow \sqrt{1-\left(\frac{1}{3}\right)^2} = \frac{\sqrt{8}}{3} =\frac{2\sqrt{2}}{3}&=& \sin{M} \end{eqnarray}$$

Therefore,

$$\text{height of tetrahedron} = |OK| = |OM|\sin{M} = \frac{\sqrt{3}}{2} s \cdot \frac{2\sqrt{2}}{3} = \frac{\sqrt{6}}{3}s$$

(**) This cosine is the reason I posted this approach. It's sometimes handy to know (as in this problem); even better, it's easy to remember, because it turns out that it fits a simple pattern (which might be more-likely to impress interviewers):

$$\begin{eqnarray} \cos\left({\text{angle between two sides of a regular triangle}}\right) &=& \frac{1}{2}\\ \cos\left({\text{angle between two faces of a regular tetrahedron}}\right) &=& \frac{1}{3}\\ \cos\left({\text{angle between two facets of a regular n-simplex}}\right) &=& \frac{1}{n} \end{eqnarray}$$

(Who would've suspected, upon first encountering it, that the "$2$" in "$\cos{60^{\circ}}=\frac{1}{2}$" was actually a reference to the dimension of the triangle?)

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Thank-you v. much! What's a simplex? –  Patrick Beardmore Nov 20 '10 at 17:14
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A simplex is the analog of "triangle" in any dimension: the simplest possible shape. It's what you get when you join $(n+1)$ points in $n$-dimensional space. A triangle is a "$2$-simplex" ($3$ points in $2$ dimensions); a tetrahedron is a "$3$-simplex" ($4$ points in $3$ dimensions); going upward in dimensions, one generally just says "$4$-", "$5$-", ..., "$n$-simplex"; going downward, one can say the line segment is a "1-simplex" (determined by $2$ points) and the point is a "0-simplex" ($1$ point!). See en.wikipedia.org/wiki/Simplex and mathworld.wolfram.com/Simplex.html –  Blue Nov 20 '10 at 17:30
    
Very nice! I never knew about this pattern until now, thanks! :D –  J. M. Nov 20 '10 at 21:37
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