Anyone up with a solution for this nice, unsolved problem?

Here's a sketch: This is the Jar of Fun. Right now it only has 29 vertices, though we can make a similar version with 101 vertices by inserting pairs of the curvy pac-man looking ( )s in the side 'arm's. So suppose we have 101 beetles such that, if we assign each to a vertex on the jar of fun, two beetles are friends if and only if they're connected by a non-red line. It's not too hard to see that if three beetles form an equilateral triangle in the Jar, any arrangement involving them must have them forming an equilateral triangle again. Even better, if four beetles $ABCD$ have $ABC$ and $BCD$ both being equilateral triangles in the Jar, any arrangement involving them must have them forming a rhombus $A'B'C'D'$ that's congruent to $ABCD$ in that specific order. From there, it shouldn't be too hard to show that any arrangement of all the beetles must reform the Jar, which isn't a valid arrangement since the red edge creates a contradiction. Now if we remove any beetle, then try to arrange the other 100, we can arrange them into something very similar to the Jar, but using the freedom created by the removal of the beetle to 'swing' one of the 'arms' of the Jar either towards or away from the other, changing the distance of the red edge. This is just a sketch so I'm not going to elaborate, hopefully it's intuitive enough.