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How do I calculate the height of a regular tetrahedron having 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
Here there are good explanation for this: – Herman Jaramillo Nov 28 '15 at 23:20
up vote 11 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:


and here's a view of the bottom face:


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


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

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

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:

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


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

(**) 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
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 and – 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
btw, $ \cos\left({\text{angle between two faces of a regular tetrahedron}}\right) = -\frac{1}{3} $ – Narasimham Mar 16 '15 at 0:07
@Narasimham: The angle between two faces of a regular tetrahedron is acute, so its cosine must be positive. – Blue Mar 16 '15 at 0:45

I'd like to offer a slightly simpler approach to part of the 1st answer above. We know that the equilateral triangle of the base of the tetrahedron has sides of 1, 1, and 1, and we know we can split that in half, creating two right triangles having sides of hypotenuse=1, base=1/2, and perpendicular=√3/2, along with angles of 30, 60, and 90 degrees.

Now consider the 2nd diagram of the 1st answer, which shows a solid-line triangle having angles of 30, 30, and 120 degrees. That triangle could be divided in half, creating two right triangles having angles of 30, 60, and 90 degrees. If we consider the base of each triangle to be its shortest side, then the perpendicular of either one of those triangles has a length of 1/2. We can now use the power of ratios to compute the other two sides:

(1/2):(√3/2):(1) --triangle 1: half of tetrahedral face, angles of 30, 60 & 90 degrees.

( ):( 1/2):( ) --triangle 2: has unknown base & hypotenuse, but is proportionate to triangle 1.

We really only need to compute the hypotenuse of triangle 2, because that is the desired distance from the corner to the center of the tetrahedron's base:

Multiply triangle-1-hypotenuse by triangle-2-perpendicular; divide by triangle-1-perpendicular.

[(1/2)(1)]/(√3/2) = (1/2)(2/√3) = 1/√3, as the 1st answer also computes in a more complicated way.

For the sake of completeness, since in any 30-60-90-degree triangle the base is simply half the length of the hypotenuse, the second unknown is 1/(2√3) or √3/6 (although it could also have been figured by using ratios, as above). The reader is invited to verify that the square of (√3/6) plus the square of 1/2 equals the square of (1/√3).

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The normal height ($H_{n}$) of any regular tetrahedron having edge length $a$ is equal to the sum of radii of its inscribed & circumscribed spheres which is given as follows $$H_{n}=\frac{a}{2\sqrt{6}}+\frac{a}{2}\sqrt{\frac{3}{2}}=\frac{4a}{2\sqrt{6}}=a\sqrt{\frac{2}{3}}$$ Hence, the normal height ($H_{n}$) of regular tetrahedron with edge length $a$ is generalized by the formula $$\bbox[4pt, border: 1px solid blue;] {H_{n}=a\sqrt{\frac{2}{3}}}$$ As per given value of edge length $a=1$ in the question, the normal height of tetrahedron is $\sqrt{\frac{2}{3}}$

Note: for derivation & detailed explanation, kindly go through HCR's Formula for Regular n-Polyhedrons

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