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

Given $n$ balls in $\mathbb{R}^d$ with radii $r_1,r_2,\dots,r_n$. Assume that this system of balls can not be separated by a hyperplane (that is, if a hyperplane $H$ does not intersect these balls, they necessarily belong to the same half-space bounded by $H$). Prove that all $n$ balls may be covered by a ball of radius $\sum r_i$.

reference: A. W. Goodman and R. E. Goodman, A circle covering theorem, Amer. Math. Monthly 52 (1945), 494-498.

Answer 2

Here's a cute question which Frederic Bourgeois asked me on a train journey recently. He was asked it by Givental, if my memory serves correctly, but I've no idea where it came from originally. Anyway, the question:

There is a mountain of frictionless ice in the shape of a perfect cone with a circular base. A cowboy is at the bottom and he wants to climb the mountain. So, he throws up his lasso which slips neatly over the top of the cone, he pulls it tight and starts to climb. If the mountain is very steep, with a narrow angle at the top, there is no problem; the lasso grips tight and up he goes. On the other hand if the mountain is very flat, with a very shallow angle at the top, the lasso slips off as soon as the cowboy pulls on it. The question is: what is the critical angle at which the cowboy can no longer climb the ice-mountain?

To solve it, you should think like a geometer and not an engineer. (And yes, it needs just one trick which is certainly applicable elsewhere.)

P.S. When I was asked the question, I failed miserably!

Answer 3

Here is a problem that I learned from W. Thurston. I do not remember whose problem it was originally. Possibly Conway?

Suppose that you have a finite collection of round circles in round $S^3$, not necessarily all of the same radius, such that each pair is linked exactly once. (In particular, no two intersect.) Prove that there is an isotopy in the space of such collections of circles so that afterwards, they are all great circles.

Answer 4

Suppose that we have two simple closed curves in $R^3$ which are linked. And suppose that the distance between these curves is $1$. Prove that the length of each curve is at least $2\pi$.

This problem has an interesting history. It was published in the book by W. Hayman, Research problems in Function theory, where it was attributed to F. Gehring. I solved it in 1977, jointly with Oleg Vinkovski, prepared a paper and gave a seminar talk. After the talk, I was approached by an undergraduate student, who proposed a ridiculously simple solution. Just two lines, using nothing. So I did not submit my paper. Later I've seen several published solutions, but none of them was so simple.

EDIT. Here is this proof (due to Igor Syutrik). Fix a point $M$ on $A$. Then one can find another point $M'$ on $A$ such that the interval $[M,M']$ intersects $B$. Indeed, otherwise we can deform $A$ to $M$ moving straight along these intervals $[M,M']$ and deformation will not cross $B$. Let $O$ be a point on $[M,M']$ that belongs to $B$. Let $A'$ be the central projection of $A$ from $O$ onto the unit sphere around $O$. Then $A'$ passes through two diametrically opposite points of the sphere and thus its length is at least $2\pi$.

EDIT2. Exactly the same proof is published in the paper Criticality for the Gehring link problem, Geometry & Topology 10 (2006) 2055–2115, where it is credited to Marvin Ortel.

EDIT3. Our original solution with Vinkovski also has been rediscovered since then. It can be seen in this file: http://www.math.purdue.edu/~eremenko/dvi/gehring.pdf Thanks to Anton Petrunin for finding this file on my computer:-)

EDIT 4. Recently published updated version of Hayman's problem list, W. Hayman and E. Lingham, Research problems in Function theory, Springer 2019, says that the article by R. Osserman, The isoperimetric inequality, Bull AMS 84 (1978), contains on p. 1226 a survey of published solutions of this problem. It does not contain a solution as simple as Syutrik's solution.

Answer 5

Does Steve Fisk's "trick" in solving the "Art Gallery Problem" of Victor Klee qualify?

Show that for a simple plane polygon with n sides, floor function (n/3) vertex guards are sometimes necessary and always sufficient to "see" all of the interior points of the polygon.

Answer 6

Anton asks for more problems in topology and algebraic geometry. One issue is that the concept of a "trick" is treated differently in these two areas than in differential geometry. In topology, not quite as many ideas are called "tricks"; they are sometimes named after people and co-opted as material, e.g., the Alexander trick and the Whitney trick. In algebraic geometry, tricks are sometimes regarded as suspect; they are sometimes taken as a reason to reorganize definitions to either again co-opt the trick or avoid it outright.

Still, a problem based on the Alexander trick could be at a good level for this problem list.

Problem: Prove that space of tame knots, meaning piecewise-linear embeddings $f:S^1 \to \mathbb{R}^3$, is connected in the $C^0$ topology on functions $f$.

Answer 7

Since Gjergji brings up isoperimetric inequalities, there is a lot of attention in combinatorial geometry devoted to combinatorial versions. For instance, if $f$ is a Boolean function $f$ on $n$ bits, define its "instability" to be the number of ways that $f(\vec{x}) \ne f(\vec{y})$ when $\vec{x}$ and $\vec{y}$ differ in one bit. If half of the values of $f$ are 0 and half are 1, then the theorem is that the most stable choice of $f$ is a function $f(\vec{x}) = x_k$ that only depends on one bit.

On the theme of one-step proofs in geometry that depend on another theorem, there is a one-step proof of this fact using the standard spherical isoperimetric inequality.

Answer 8

I believe this is a problem given in J. Hirsch's Differential Topology. This may be much simpler than the ones posted here already. But for what it's worth, here it is.

Show that given a collection of spheres the product manifold embeds into an Euclidean space of one dimension higher, viz., for instance $S^2 \times S^3$ embeds in $\mathbb{R}^6$.

Answer 9

I like the following problem. It has a very short solution based on a (very) useful trick (idea).

We fix three pairwise tangent (at distinct points) spheres $A_1$, $A_2$ and $A_3$ in 3-space. Let us construct a sequence of spheres: first sphere $B_1$ is a (generic) sphere which is tangent to each $A_i$. $B_2$ is a sphere which is tangent to $B_1$ and to each $A_i$ (in fact there are two such spheres, we chose one of them as a $B_2$). $B_3\ne B_1$ is a sphere which is tangent to $B_2$ and to each $A_i$ (such a sphere is unique), etc: $B_{n+1}\ne B_{n-1}$ is a sphere which is tangent to $B_n$ and to each $A_i$.

The problem is to show that $B_7=B_1$.

Answer 10

Measure concentration for the sphere $S^n$, is a nice problem if the use of a theorem (Brun-Minkowski inequality) counts as a trick.

Problem: Let $\mu$ be the normalized measure in the unit sphere $S^n$. Let $X$ be a measurable subset such that $\mu(X)=\frac12$, and $X_{\delta}$ be the set of all $x\in S^n$ for which there is $x'\in X$ such that $||x-x'||=\delta$. Prove that $\mu(X_{\delta})\geq 1-2e^{-\frac{n\delta^2}{4}}$. (So increasing $X$ by just a little gives almost the entire sphere.)

On a similar line stands the "Classical Isoperimetric Inequality". (Among all bodies of the same volume, the ball has the least surface area.)

Another fact that comes to mind (with the trick being a compactness argument) is Bunt-Motzkin theorem:

Problem: Show that a simple polytope $P\subset \mathbb{R}^n$ is convex iff for every point $p$ not in $P$, there exists a unique point in $P$ that's closest to $p$.

Answer 11

Since you asked for a problem in algebraic geometry, here is a popular result whose proof in modern terms is very short. It could be called a one-step problem:

Harnack's inequality on curves: Prove that a smooth algebraic curve of degree $d$ in $\mathbb{R}P^2$ consists of at most $(d^2-3d+4)/2$ circles. (1,1,2,4,7,11,...)

Answer 12

Problem: Consider two closed smooth strictly convex planar curves, one inside another. Show that there is a chord of the outer curve, which is tangent to the inner curve and divided by the point of tangency into equal parts.

Answer 13

Every planar closed curve contains all four vertices of a rectangle (here is an entertaining video with a solution due to Vaughan: https://www.youtube.com/watch?v=AmgkSdhK4K8).

Answer 14

Let us equip ${\bf M}_2({\mathbb R})$, a $4$-dimensional space, with the standard operator norm. Consider the unit sphere $S$, which is homeomorphic to $S^3$. It contains ${\bf O}_2$, which is the disjoint union of two circles, ${\bf O}_2^+$ (the rotations) and ${\bf O}_2^-$ (the symmetries).

Show that ${\bf O}_2^+$ and ${\bf O}_2^-$ are linked.

Answer 15

Can you cover $\mathbb{R}^n$ with closed balls so that they intersect only at the boundary? By closed disk I mean a set of the form $$ B_r(x) =\{y: \ d(y, x) \le r\} $$

I think this is a cool problem, because you can spend some time trying to craft your own "fractal cover"; after a while you will get the feeling that a tiny bit of the space will always get left but in a mysterious way.

A friend of mine solved this with measure theoretic ideas, but it can be done in a tricky way using homotopy colimits, which are definitely useful in algebraic topology.

Answer 16

Greg Kuperberg's comment on Alexander tricks reminded me of a nice one due to Tom Goodwillie.

Let $K_n$ be the space of $C^k$-smooth embeddings of $\mathbb R$ into $\mathbb R^n$ in the $C^k$-topology $k>1$, where the embeddings are required to be 'long' in the sense that $f(t)=(t,0,\cdots,0)$ for $t \notin [-1,1]$. Let $Imm_n$ be the corresponding space of long immersions $\mathbb R \to \mathbb R^n$.

Then the inclusion map $K_n \to Imm_n$ is null-homotopic. It's a one-line proof provided you know the Smale-Hirsch theorem. Or if you want to remove Smale-Hirsch, replace $Imm_n$ by $\Omega S^{n-1}$ and let the map $K_n \to \Omega S^{n-1}$ be the normalized velocity vector.

Similarly, there's a nice one-line proof that the inclusion map $K_n \to K_{n+1}$ is null-homotopic. The original idea is ancient but this formulation (as far as I know) is due to me. You don't need any theorems for this, it's a construction for which you can write down the null-homotopy using simple functions.

Answer 17

I think the question Is it possible to capture a sphere in a knot? is an excellent one-step problem.

Answer 18

The Ham Sandwich theorem

Answer 19

All side facets of a convex pyramid fall onto its base facet. Prove that the fallen faces cover the base.

Boring clarifications:

Consider a $d$-dimensional convex pyramid $P={\rm conv}(\{S\}\cup F)$, where $F$ is a $(d-1)$-dimensional convex polytope in $\mathbb{R}^d$ (a base facet of $P$), and $S\in \mathbb{R}^d$ is a point which does not belong to a base plane $\alpha$ of $F$ (an apex of $P$).

For any facet $F_i$ of $F$ consider the point $S_i\in \alpha$ such that

(i) $S_i$ and $F$ are on one side of the $(d-2)$-dimensional plane $\beta$ of $F_i$, and

(ii) $S_iA=PA$ for any point $A\in \beta$.

Then the polytopes ${\rm conv}(\{S_i\}\cup F_i)$ (fallen facets) cover $F$.

This was proposed by Igor Pak (link with spoilers) to All-Russian olympiad in 2012 (in dimension 3).

Answer 20

Show that $\mathbb{R}^3$ can be partitioned into disjoint unit circles. The proof involves the Axiom of Choice. I don't know the status of this problem if you don't assume the Axiom of Choice. See Komjáth and Totik, Problem and Theorems in Classical Set Theory, Chapter 13, Problem 13.

Answer 21

The Tarski Plank Problem:

Unit disc can be covered by $n$-rectangles $1\times n^{-1}$. Prove that it cannot be covered by a smaller number of such rectangles.

Solutions. Take the unit sphere of the same center as the disc. Let $\pi$ be the orthogonal projection on the plane containing the disc. Taking $\pi^{-1}$ of a strip (assuming not too much of it is "outside") we get a "ring" on the sphere. The point is that the area of that ring depends only on its width which is $n^{-1}$ and not on its position. This is the Archimedes Hat-Box Theorem. Since we cover sphere by sets of equal areas, they cannot overlap and the problem follows. $\Box$

Of course, this solution is too short, only an idea.

Answer 22

Given $n$ points on the sphere of area 1, show that there exists an open spherical disk of area $1/n$ which does not contain any of these points.

(I learned this in the 1970s, with credit to Gelfand).

Answer 23

This is not really an advanced problem and only indirectly related to geometry, but I instantly though of it because of its short solution and trick, which is really nice for the introduction to complex numbers.

Let $z_1,z_2,z_3,z_4\in\mathbb{C}$ be points with $|z_1|=|z_2|=|z_3|=|z_4|$ and $z_1+z_2+z_3+z_4=0$, then prove there are two pairs of antipodal points among them.

Consider the polynomial $(z-z_1)(z-z_2)(z-z_3)(z-z_4)$. The cubic coefficient vanishes by proposition, so does the linear one as: \begin{align*} &z_2z_3z_4 +z_1z_3z_4 +z_1z_2z_4 +z_1z_2z_3 =z_1z_2z_3z_4\left( \frac{1}{z_1} +\frac{1}{z_2} +\frac{1}{z_3} +\frac{1}{z_4}\right) \\ =&z_1z_2z_3z_4\left( \frac{z_1^*}{|z_1|^2} +\frac{z_2^*}{|z_2|^2} +\frac{z_3^*}{|z_3|^2} +\frac{z_4^*}{|z_4|^2}\right) =\frac{z_1z_2z_3z_4}{|z_1|^2} (z_1+z_2+z_3+z_4)^*=0. \end{align*} As a result for every root $z$ of the polynomial, $-z$ is also a root.

Answer 24

Every finite point set has Delaunay triangulation (circumsphere of any simplex doesn't contain points from this set).

Answer 25

I believe this is somewhere in Hirsch's book on differential topology, but I am not sure. The question is:

When is $S^m \times S^n$ parallelizable?

If $m$ and $n$ are even, $\chi(S^m \times S^n) = 4$, hence the Euler class is nonzero and the manifold cannot be parallelizable.

If, however, at least one of $m,n$ is odd then one can use that spheres have stably trivial tangent bundles and 'borrow' a line bundle from the other sphere...

Answer 26

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 27

How many lines meet 4 general lines in space?

Solution is to point out that lines meeting three general lines in space form a hyperboloid = quadric.

The fourth line then meets the quadric in two points.

Answer 28

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 29

(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 30

This may not be suitable for several reasons. But anyway, I like Exercise 4.18.3 from A. Beauvilles "Complex Algebraic Surfaces".

Let $S$ be an irreducible surface in $\mathbb{CP}^n$ of degree $d \leq n-2$. Show that $S$ is contained in a hyperplane.

The idea of the solution is definitely used elsewhere in algebraic geometry but outside of it maybe not.