# Professor Halfbrain's second cutting theorem

Professor Halfbrain has recently made several fascinating discoveries on cutting convex polygons in the plane.

Halfbrain's second cutting theorem: Every convex polygon can be cut (by a perfectly straight cut) into two polygons $A$ and $B$ so that $A$ and $B$ have the same perimeter and so that the length of the shortest side of $A$ equals the length of the shortest side of $B$.

Question: Is this second theorem indeed true, or has the professor once again made one of his notorious mathematical blunders?

1. A convex polygon has $n$ pairwise distinct vertices. A side of the polygon is the straight line segment connecting two consecutive vertices along the convex hull.
2. For the professor's first cutting theorem, see Professor Halfbrain's first cutting theorem.

• This theorem is true due to the same proof as in the first cutting theorem, with "longest" replaced by "shortest". – Ben Frankel Sep 27 '15 at 16:20
• @Ben Frankel: This would be extremely boring... Do you have a proof for your claim? – Gamow Sep 27 '15 at 16:42
• Do you really have a proof for your claim "Clearly the length of the /shortest/ edge is continuous"? – Gamow Sep 27 '15 at 17:01
• You're right, I don't. The shortest edge is min(E1, ..., En, PQ) but if the shortest edge is say E1 and P is moving along it, it will shrink until it's no longer one of the edges. Or, when Q reaches a vertex, a new edge is introduced of length ~0 which becomes the new min. – Ben Frankel Sep 27 '15 at 17:09
• If it turns out Halfbrain was wrong, he should try restricting the theorem to even-sided polygons. I know it's true then. – Lopsy Sep 27 '15 at 17:55

The professor made a mistake, this theorem is false.

Consider a regular pentagon. The point opposite of a vertex is the midpoint of the opposite edge, and the point opposite of a point on an edge is somewhere on an edge adjacent to the opposite vertex. In both cases the cutting edge is longer than the sidelength of the original polygon and so it cannot be the shortest edge of either $A$ or $B$.

If the cutting edge is between two sides of the original polygon, it must be, as pictured, like the green line and not the red line. Thus one of $A$ or $B$ will have both of the short edges and it cannot satisfy the conditions.

Now consider the pentagon below. It is close enough to a regular pentagon that our considerations about the regular pentagon hold. So we are left with 5 cases to check, one for each vertex.

From A: On one side we have 1 + 0.99 = 1.99, and on the other side 0.96 + 0.97 = 1.93. Thus we can't split the opposite edge 0.98 right in half like we need to for the shortest edges to be equal, as then the perimeters of $A$ and $B$ wouldn't be equal.

From B: Same reasoning, 1.97 = 0.99 + 0.98 > 1.96 = 1 + 0.96 = 1.96.

From C: Same reasoning, 1 + 0.99 = 1.99 > 1.95 = 0.98 + 0.97.

From D: Same reasoning, 0.99 + 0.98 = 1.97 > 1.95 = 0.97 + 0.96.

From E: Same reasoning, 1 + 0.96 = 1.96 > 1.95 = 0.98 + 0.97.

Thus the above convex polygon is a counterexample.

(note that if it's not actually close enough to a regular pentagon--which I doubt--we could add more 9's to the edges as in 0.9996, 0.9997, etc.)

• A simpler counterexample would be an almost-equilateral triangle. I skipped to pentagon because it's more difficult to prove for a triangle that the cutting edge cannot be the shortest edge. – Ben Frankel Sep 27 '15 at 18:20