Here is a puzzle I found in *Mitteilungen der DMV* (roughly, "Letters of the German Society of Mathematicians"), issue 19/2011. It was posed by Alfred Schreiber in "Wie man Hasen fangt" (How to catch rabbits), and he claims that less than 5% of his subjects could solve it in under 1 hour. He tested it on students of mathematics and professors of mathematics, computer science, and engineering.


See if you have more luck. The problem is deceptively simple:


Suppose that you have a triangle $ABC$ and a point $D$ inside the triangle. Prove that the perimeter of $ABC$ is larger than the perimeter of $ABD$.


I am currently working on a generalization: Given two convex shapes $S$ and $T$, where $T$ totally encloses $S$. Prove that the perimeter of $S$ is no bigger than the perimeter of $T$.


Alternatively, for a shapes with straight edges: Prove that the perimeter of the convex hull of a set of points increases monotonically (but not strictly monotonically) as points are added to the set.


Please try to find an elementary proof for the special case of the triangle.


**Edit:** Thanks for all the nice answers. By now, I have found a really elementary proof on my own that just uses the triangle inequality twice.



Circumference of Convex Shapes
That's a standard inequality, though perhaps not well enough known for the usual "take it to stackexchange" comment. Denote by $\partial \Sigma$ the boundary of any set $\Sigma$ in the plane. In the case that $s$ is a polygon we can use induction on the number $k$ of edges of $\partial s$ not contained in the $\partial S$. If $k=0$ we're done. If $k>0$, choose an edge $e$ of $s$ not in $\partial S$, and let $H$ be the closed half-plane such that $H \supset s$ and $\partial H \supset e$. Then $S' := S \cap H$ is a convex set containing $s$ whose boundary is shorter than $\partial S$ because we've replaced part of $\partial S$ with the line segment joining the same two points. Moreover $\partial s$ has $k-1$ edges not contained in $\partial S'$. This completes the induction step and the proof.


That argument applies more generally when $\partial s \setminus \partial S$ is polygonal. If it's not, we can reduce to that case via a limiting argument, replacing each component of $\partial s \setminus \partial S$ by an arbitrarily close polygonal path.


