One-step problems in geometry

Question

I'm collecting advanced exercises in geometry. Ideally, each exercise should be solved by one trick and this trick should be useful elsewhere (say it gives an essential idea in some theory).

If you have a problem like this please post it here.

Remarks:

• I have been collecting such problems for many years. The current collection is at arXiv; the paper version is available at amazon.

• At the moment, I have just a few problems in topology and in geometric group theory and only one in algebraic geometry.

• Thank you all for nice problems --- I decided to add bounty once in a while and choose the best problem (among new or old).

Answer 1

Let $\Sigma$ be a surface. A loop $\gamma\colon S^1\to \Sigma$ is called a piecewise injective $n$-gon if is it is a concatenation of $n$ injective paths. A constant loop is by convention a $0$-gon. Let $g \in \pi_1(\Sigma)$ and define $P(g)\in \mathbb Z$ to be the smallest integer such that $g$ is represented by piecewise injective $n$-gon.

Question: What is the supremum of $P(g)$?

Answer 2

This is admittedly not very hard even without getting the trick, but it's super-easy with the trick. Hopefully it isn't too easy to be interesting (or even amusing):

Problem. Let $v_1,\ldots,v_n$ be vectors in $\mathbb{R}^m$, and let $V$ be the $m\times n$ matrix whose columns are $v_1,\ldots,v_n$. Show that the $n$-dimensional volume of the $n$-dimensional parallelepiped in $\mathbb{R}^m$ determined by $v_1,\ldots,v_n$ is $\sqrt{\det(V^TV)}$.

We should probably take as given that the determinant of a square matrix is the signed volume of the parallelepiped determined by its columns (or by its rows); I would consider justifying that to be a separate problem (for which I don't know any simple tricks).

Answer 3

Another problem which is solved in 'one step' (if one assumes proving the existence of a Delaunay triangulation of a finite point set in the Euclidean plane as 'one step') is Thue's Theorem on optimal circle (disk) packing.

(I know that there was already an answer involving Delaunay triangulation, but I still wanted to mention this one).

Answer 4

Von Neumann's law for two-dimensional cell growth

A soap froth or polycrystalline slab is modeled by a two-dimensional network of piecewise smooth curves, joined at vertices at internal angles of $2\pi/3$ (see figure). Any given point of a curve moves toward its center of curvature with a speed that is proportional to the curvature.

Prove that cells with six neighbors have a constant area (although their shape can change). Cells with more than six neighbors grow and those with fewer shrink.
Deduce that the average number of vertices per cell is 6.

Solution: By integrating the curvature along the perimeter of a cell one finds that the area $A_i$ of cell number $i$ varies in time according to Von Neumann's law [1,2] $$\frac{d}{dt}A_i(t)=k(n_i-6),$$ with $k$ a constant and $n_i$ the number of vertices of the cell. Since $\sum_i A_i$ must be time independent, it follows that the cells in the network have 6 vertices on average, in agreement with Euler's polyhedron formula.

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• [1] J. von Neumann, in Metal Interfaces (American Society for Testing Materials, Cleveland, 1952), p. 108.

• [2] W.W. Mullins, Two-Dimensional Motion of Idealized Grain Boundaries

Answer 5

I learned this one from my advisor: the Borromean rings are not realized by round circles.

Really this is two steps, since one needs to know that the Borromean rings are nontrivial (not the unlink), but the geometric part is one step.

Answer 6

(1) Prove: Every simple polygon may be triangulated (partitioned into triangles) via diagonals, vertex-to-vertex segments that are strictly interior (except at their endpoints). [This is a precursor to Joe Malkevitch's post.]

(2) Can every polyhedron be partitioned into tetrahedra via diagonals?

     

Answer 7

I like the following problem:

Let $S$ be a finite collection of circles in the plane such that the area of their union is $1$. Show that you can pick disjoint circles whose area is at least $1/9$.

The solution idea is to do a greedy selection: First pick the circle of maximum radius. Then take away any of the other circles that are contained in three times this radius and repeat.

This greedy idea is a bit 'algorithmic' and the many similar variants are used for a wide variety of approximation algorithms.