MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.

share|cite|improve this question
The projection on a convex subset of $R^n$ is a contracting map. – Jean-Marc Schlenker Jul 28 '11 at 16:11
A related result is the "Boxes in boxes" problem at – Richard Stanley Jul 28 '11 at 16:12
Yes, I mean perimeter. Sorry, I read that problem in German, and normally don't do much geometry in English. I fixed it above. Thanks! – Matthias Goergens Jul 28 '11 at 17:36
The elementary problem is wrongly stated. First, it's not clear whether "the circumference of $ABD$" is intended to mean the perimeter of $ABD$ or the circumference of the circumcircle of $ABD$. Either way, the assertion is false: On the cartesian plane, let $A=(−1,0)$, $B=(1,0)$, $D=(0,1)$, and $C=(0,a)$; then $a=2$ is already big enough for a counterexample ($a=100$ would make it obvious). – John Bentin Jul 29 '11 at 11:17
The statement is OK as it is. Anyway, for good measure, I have submitted for approval a general stylistic copy-edit of whe whole question. – John Bentin Feb 24 at 12:00
up vote 8 down vote accepted

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.

share|cite|improve this answer
This also corresponds reasonably well with the following intuitive argument: Stretch a rubber band around $S$. Now remove $S$ but keep $s$ in the same plane, so the rubber band is limited to the complement of $S$. What happens? – Noam D. Elkies Jul 28 '11 at 16:29
Noam, yes, the intuitive argument is what motivated my generalisation. – Matthias Goergens Jul 28 '11 at 17:39
[Of course I meant "the rubber band is limited to the complement of $s$", lower case, not "of $S$".] – Noam D. Elkies Jul 28 '11 at 20:08

Your result, including your proposed generalization and the further generalization to surface area in higher dimensions, is an immediate consequence of Cauchy's surface area formula. This states that, up to a constant depending only on the dimension, the surface area of a convex body is the average of the areas of its 1-codimensional projections. I don't know a good reference in a web page, but see for example Klain and Rota's book.

share|cite|improve this answer
Given your edit (made while I typed my answer), I suspect this is a more sophisticated proof than you're hoping for in the case of triangles. – Mark Meckes Jul 28 '11 at 16:09
Mark, Thanks. I am interested in the general result, but also in a very elementary proof for the case of triangles. – Matthias Goergens Jul 28 '11 at 17:41
@Deane: whoops, thanks for catching that embarrassing typo! – Mark Meckes Jul 28 '11 at 18:19

This also follows immediately from Crofton's formula, which says that, up to a fixed constant factor, the length of a reasonable curve (and polygons are certainly reasonable) is the integral, over straight lines (with respect to a suitable invariant measure), of the number of intersections between the straight line and the curve. Almost every straight line meets the perimeter of a convex polygon either twice or not at all, and shrinking the polygon will only reduce the set of lines that meet it.

share|cite|improve this answer

Motivated by my thoughts about a recent question I want to comment that perhaps this fact is intrinsic to our very definition of arc length. As correctly noted, the result for convex polygons is the triangle inequality and induction (along with some nice reasoning). In the general case we consider the limit of polygonal paths. This works because the definition of arc length (for a convex curve), is the least upper bound of the length of inscribed polygonal paths. My comment is that (part of) the justification for that definition is that we take it as obvious that each inscribed polygonal path should be shorter (or at least no longer) than the curve.

Archimedes in his rectification of the circumference of the circle (the first use of this limiting polygonal path method) gives a careful definition of convexity and takes as a postulate that

If two plane curves C and D with the same endpoints are concave in the same direction, and C is included between D and the straight line joining the endpoints, then the length of C is less than the length D.

There may be more to it than that (in the case that the curve C has endpoints interior to the segment joining the endpoints of D.)

share|cite|improve this answer

For the triangle problem: Euclid I, 21:

If from the ends of one of the sides of a triangle two straight lines are constructed meeting within the triangle, then the sum of the straight lines so constructed is less than the sum of the remaining two sides of the triangle, but the constructed straight lines contain a greater angle than the angle contained by the remaining two sides.

share|cite|improve this answer

For the triangle problem: Let $E$ on $CA$ and $F$ on $CB$ be such that $CEDF$ is a parallelogram. Then the triangle law and the equality of opposite sides of a parallelogram gives $$DA<EA+ED=EA+CF\quad\text{and}\quad DB<FD+FB=CE+FB.$$By addition,$$DA+DB<(CE+EA)+(CF+FB)=CA+CB,$$as required.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.