# Perimeters of Convex Polygons, One within the Other

### Problem

Let P(X) denote the perimeter of polygon X.

Given two convex polygons S and T, S ⊂ T, S ≠ T. Prove that

### Hint

At first site the statement seems obvious, although in general it touches on fundamental and deep properties of geometric shapes. But the problem is not about arbitrary plane regions - it's about convex **polygons**, and this could be utilized to our advantage. More specifically, the statement yields to an elegant proof by induction. What you want to do is find a discrete parameter that would make the inductive step all but obvious.

### Solution

We are given that S ⊂ T, but it does not mean that the two polygons may not share a side line. For example, in the diagram below side A'B' of S lies on side AB of T (and, therefore, A'B' is not longer than AB.) The same is true concerning sides C'D' and CD of S and T respectively.

So S may or may not have some sides on the boundary of T. Let's agree to call sides of S that *do not lie on any side of T free*. The induction will be on the number of free sides of S.

If S has no free sides then each of its sides lies on a side of T and is, therefore, not longer than the latter. It follows that in this case

Now, for an inductive step, introduce statement X(k): "For two polygons

So let this be the case that S has k+1 free sides. Since

That line will split T into two, say, T' and T''. As we know, convex polygon are the intersections of half planes and so always lie on one side from their side lines, implying that S may belong to only one of T' or T''. Let it be T':

### Acknowledgment

This is an example from a Russian book *Simplest Examples of Mathematical Proofs* by V. A. Uspensky (MCCME, 2009.)

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