# Angle between lines joining tetrahedron center to vertices

What are the angles formed at the center of a tetrahedron if you draw lines to the vertices?

I'm trying to make these:

I need to know what angles to bend the metal.

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I've taken the liberty of adding more information to the title, so that the question is clear at first glance. – Rahul Aug 24 '11 at 4:44

The required tetrahedral angle is $\arccos\left(-\frac13\right)\approx109.5^\circ$. You can use the law of cosines to show this... or more transparently, you can exploit the fact that a tetrahedron is easily embedded inside a cube:

I suppose now's as good a time as any to post the synthetic proof.

One can use the Pythagorean theorem to show that a square with unit edge length has a diagonal of length $\sqrt 2$. The Pythagorean theorem can be used again to show that a right triangle with leg lengths $1$ and $\sqrt 2$ will have a hypotenuse of length $\sqrt 3$ (corresponding to the triangle formed by an edge, a face diagonal, and a cube diagonal). We know that the diagonals of a rectangle bisect each other; this can be used to show that the diagonals of a cube bisect each other. From this, we find that the side lengths of the (isosceles!) triangle formed by two half-diagonals of the cube (corresponding to two of the arms of your caltrops) and a face diagonal are $\frac{\sqrt 3}{2}$, $\frac{\sqrt 3}{2}$, and $\sqrt{2}$. From the law of cosines, we have

$$2=\frac34+\frac34-2\frac34\cos\theta$$

where $\theta$ is the obtuse angle whose measure we are seeking. Algebraic manipulation yields $\cos\,\theta=-\frac13$.

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Anecdotal addendum: that 109.5° bit is stuck in my head, as that fact seems to be drummed into every chemistry student (tetrahedra do pop up a lot in chemistry). – J. M. Aug 11 '11 at 15:09

(Assuming the tetrahedron is supposed to be regular) Take the tetrahedron with vertices $(1,1,1), (1,-1,-1), (-1,1,-1), (-1,-1,1)$, which has centre at the origin, and use the dot product formula:

$a\cdot b = |a| |b|\cos\theta$

which gives $\cos\theta=-\frac13$

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This is essentially J.M.'s second observation, realised with a convenient cube. – Mark Bennet Aug 11 '11 at 5:14
Using coordinates indeed makes for a shorter derivation. :) The "synthetic" route would rely on repeated use of the law of cosines (which is of course equivalent to your use of the dot product) and the Pythagorean theorem... – J. M. Aug 11 '11 at 5:19

One way is to write the vertices as vectors $a,b,c,d$ with norm $\|\cdot\|=1$. Then $a+b+c+d=0$. But

$$0=\|a+b+c+d\|^2=4+2{4 \choose 2}\cos\theta,$$

so $\theta = \arccos(-1/3)$.

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+1 A neat argument! Even though it relies on the existence of the obvious symmetries. – Jyrki Lahtonen Aug 23 '11 at 8:15
@Jyrki: Even though? Because! :-) – joriki Aug 24 '11 at 7:18
@joriki: :-) :-) – Jyrki Lahtonen Aug 24 '11 at 7:37

If the tetrahedron is regular, we can use statical equilibrium of four equal isotropic forces with angle $\theta$ between any two of them. Referencing with reference to any one force,

$$F \cos \theta + .. + .. + F = 0$$

$$\cos \theta = -\frac13$$

To generalize for all $i$ direction forces on a particular Z direction the vector dot product sum can be used:

$$\Sigma F_i . Z =0$$

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