# Sum of angles | Tetrahedron

Prove that the sum of the six angles subtended at an interior point of a tetrahedron by its six edges is greater than 540°.

Any help on getting me started out here? I am not able to get any idea as to how to begin with this

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In the completely symmetrical case, each angle is $\arccos(-1/3)$, so the sum is about $656.8^\circ$. –  Michael Hardy Sep 3 '11 at 17:49

The angles subtended by the edges at an interior point don't change if we move one of the vertices towards the point. Thus we can transform any counterexample into one where the point is arbitrarily close to a vertex, so we only have to consider what happens in the limit of the interior point approaching some vertex $v$ along some fixed direction $d$. In that case, the limit of the angle subtended by an edge $e$ not incident at $v$ is the vertex angle at $v$ of the face containing $e$ and $v$, and the limit of the angle subtended by an edge $e$ incident at $v$ is $\pi-\angle(d,e)$. So the sum of the subtended angles is the sum of the vertex angles at $v$ plus $3\pi$ minus the sum of the angles between $d$ and the edges incident at $v$. That last sum can always be increased by rotating $d$ towards one of the edges incident at $v$, so we only have to consider the case where $d$ is the direction of one of those edges. In this case the contribution from that edge is $\pi$ and the contributions from the other two edges incident at $v$ are each $\pi$ minus one of the vertex angles. Thus, the overall sum is $3\pi$ plus the third vertex angle, which is greater than $3\pi$. Since we can make the third vertex angle arbitrarily small, this is also the best possible bound.

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My example has 4 points on the surface of the earth. Points $A$ and $B$ are on the equator, one inch on each side of the prime meridian. $C$ and $D$ are at the two ends of the international date line, one inch from the north (respectively south) pole. The sum of the 6 edges is (very close to) $3\pi$. –  GEdgar Sep 3 '11 at 21:31
@GEdgar: I don't understand. What is this an example of? Is it intended as a counterexample? Or as an example of how to get close to $3\pi$? (I presume your interior point is the centre of the Earth?) –  joriki Sep 3 '11 at 21:44

Here's another proof.

Observe that moving a vertex directly toward or away from the interior point doesn't change the subtended angle. Hence we can arrange such that the distance from the interior point to any of the four vertices is 1. In other words, we can assume without loss of generality that the interior point is the centre of a sphere of radius 1, and the vertices all lie on the surface of the sphere.

Now fix three of the vertices, call them $A$, $B$, and $C$. And let us use $D$ to denote the antipode of the remaining vertex (which we call $D'$). Call the centre $O$. Observe that by the assumption that $O$ is an interior point, $D$ lies in the interior of the spherical triangle $ABC$.

Since $O$ is the centre, the angle $AOB$ is equal to the arc-length $|AB|$.

Since $D$ and $D'$ are antipodes, $|DA| + |D'A| = |DB| + |D'B| = |DC| + |D'C| = \pi$. So to prove the claim, it suffices to show that for a spherical triangle $ABC$ and a point $D$ in it interior,

$$|DA| + |DB| + |DC| < |AB| + |BC| + |CA|$$

But that follows from the triangle inequality on metric spaces.

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+1, very nice :-) –  joriki Sep 3 '11 at 22:07