# Professor Halfbrain's first cutting theorem

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

Halfbrain's first 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 longest side of $A$ equals the length of the longest side of $B$.

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

Pick a point $P$ on the original convex polygon, and then place the point $Q$ on $P$ and move it continuously clockwise along the perimeter of the polygon until it comes back to $P$. As $Q$ is moving, keep track of the clockwise distance along the perimeter from $P$ to $Q$. By the intermediate value theorem, at one point that distance was equal to half the perimeter. Thus for any point on the perimeter, there is a unique "opposite" point that splits the perimeter in half.
Now move $P$ and $Q$ along the perimeter of the polygon together. As they are moving, they cut the original polygon into $A$ and $B$, and we consider the longest edge of each of the two polygons. Clearly the length of the longest edge is continuous because it's unchanging when it is an edge of the original polygon and changing continuously when it is $PQ$. We move $P$ and $Q$ until they switch places, so $A$ and $B$ also switched places, so the lengths of the longest edges of $A$ and $B$ also swapped.
Thus by the intermediate value theorem there was a $PQ$ that split the original polygon into $A$ and $B$ with equal perimeters and equal longest edges.
• One thing I don't understand about this proof. $A$ and $B$ have switched places, but what if "the longest edge of $A$" is the same length as "the longest edge of $B$"? Mightn't it be possible that, even though the length of the longest edge is continuous, it might not ever be $PQ$? – Snowbody Sep 27 '15 at 13:04
• @Snowbody, Yes, it is possible for the equal longest edge not to be $PQ$, but that still fits the requirements. $A$ and $B$ just have to have equal longest sides, whether it is shared (as in $PQ$) or not. In fact, if the longest side had to be $PQ$, the claim would in fact be false. An equilateral triangle would be a counterexample. – Ben Frankel Sep 27 '15 at 16:16