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
 A: 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.
