Prove that a convex polygon cannot have three sides, each of which is greater than the longest diagonal.

Question

Consider two of the sides which do not have a common endpoint-say sides AB and CD. Then, on the one hand, AC + BC < AB+ CD (since AC and BD are diagonals). On the other hand, if AC and BD intersect at point O, then OA+OB> AB and OC + OD > CD. Adding these inequalities we find that AB+CD < AC + BD, which is a contradiction. This the the answer to this question. However, I can't understand it properly. Can you re-engineer this solution for the sake of simplicity?

Answer 1

There are two facts about polygons that must be understood to follow the start of this proof:

1. In a simple (i.e., non-intersecting) polygon of four or more sides, any set of three sides must have at least two that have no endpoints in common - otherwise the three sides would form a triangle, which would mean one of the endpoints has to be an intersection of the polygon with itself (since there is a fourth side, at least one of these vertices must have another side eminating from it).
2. Convex polygons are simple. An intersection of sides would allow interior points being connected by a line segment passing through the exterior.

The theorem vacuously holds for three-sided polygons. They have no diagonals, so their sides cannot be longer than any diagonal. So it only needs to be proven for convex polygons of four or more sides.

Assume the theorem is false, and a polygon has three sides that are longer than any of its diagonals. At least two of those sides cannot share any endpoint. Label them $\overline{AB}$ and $\overline{CD}$. Now either the segments $\overline{AC}$ and $\overline{BD}$ intersect, or the segments $\overline{AD}$ and $\overline{BC}$ intersect. Swap the labels as necessary so that $\overline{AC}$ and $\overline{BD}$ intersect. Neither $\overline{AC}$ nor $\overline{BD}$ can be a side. For if $\overline{AC}$ is a side, then it is crossed by the segment $\overline{BD}$ which means that segment contains points outside the polygon, in contradiction to convexity. Thus $\overline{AC}$ and $\overline{BD}$ must be diagonals. Let $O$ be the point of intersection:

By assumption, both $\overline{AB}$ and $\overline{CD}$ are longer than any diagonal. So $$AB + CD > AC + BD$$ By the triangle inequality, $$AO + BO > AB\\CO + DO > CD$$ so $$AO + CO + BO + DO > AB + CD$$

But as $O$ is between $A$ and $C$, $AO + CO = AC$ and similarly $BO+DO = BD$. Therefore $$AC + BD > AB + CD$$ contradicting the earlier inequality. Thus the assumption that the polygon has three sides longer than any diagonal must be false.

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An alternate approach would have been to first prove a lemma that for any two disjoint segments $\overline{AB}$ and $\overline{CD}$ with $\overline{AC}$ and $\overline{BD}$ intersecting, that $\overline{AB}$ and $\overline{CD}$ cannot both be longer than $\overline{AC}$ and $\overline{BD}$. This proof is the latter part of the proof above, and does not involve contradiction.

Then the proof of the full theorem could be applied with a much smaller scope for the contradiction, just in selecting two non-adjacent sides both of whom must be longer than their connecting diagonals.