# One-step problems in geometry

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).

• If anyone wonders why I deleted my previous comment, I looked more closely at Anton's file of examples and realized his exercises were at a higher level than I had thought. Dec 8, 2009 at 23:31
• Just a comment and vote up to say nice collection of problems. I have to restrain myself to not go attempt to solve them all and instead study for my topology qual. Dec 9, 2009 at 4:14
• A trick that is useful elsewhere is a method. Jul 21, 2016 at 12:27
• What does "pigtikal" mean? Dec 31, 2020 at 14:09
• I believe that this question, that I posted here some time ago, can be made into a nice problem in the first chapter of your Arxive file...if only you find that one trick which solves it! mathoverflow.net/q/376467/167834 Unfortunately at the moment it's still unsolved. Aug 9, 2021 at 20:11

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.

• Warning: Spoiler ahead! As far as I can tell, this problem is not much different from the problem where n is set to 2. Is that the trick, or is there something else going on that is widely applicable yet specific to geometry? (Or maybe it is different?) Gerhard "Also Likes Zero Step Problems" Paseman, 2016.03.31. Mar 31, 2016 at 19:39
• For $n=2$ there is nothing to do, but how does this help in general case? Mar 31, 2016 at 20:03
• I guess the trick is to realize that you can reduce it to n =2. Gerhard "Should Not Say Anything More" Paseman, 2016.03.31. Apr 1, 2016 at 1:38
• @Gerhard Possibly, but I do not realize it at all. Apr 1, 2016 at 6:09
• Thank you, very good problem --- it is included now. Dec 31, 2020 at 2:37

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!

• Yes, it is a nice problem --- thanks:). I heard nearly the same question from David Berg (UIUC). He was interested in geodesics in rigions of Euclidean space cutted by a graph of Lipschitz function. If I remember right, he has a paper on this subject. Dec 9, 2009 at 18:16
• Shh! Say "geodesic" and you'll give the answer away! Dec 9, 2009 at 18:35
• @Darsh, don't worry, everyone else Frederic asked on the same train journey also failed first time. Dec 13, 2009 at 23:42
• A lasso's loop is meant to contract when pulled tight. Sep 25, 2011 at 19:19
• This ice-mountain problem is from the book "Introduction to Classical Mechanics: With Problems and Solutions" by David Morin (problem 2.10 on p. 45) amazon.com/Introduction-Classical-Mechanics-Problems-Solutions/… Sep 28, 2013 at 16:44

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.

• Nice problem, I solved it :) But I can not see where else such idea can be useful... Dec 14, 2009 at 4:24
• One proof uses intersection of planes with a 4-ball. Arguably this leads to ideas such as hyperbolic geometry in n+1 dimensions to study spheres in n dimensions, as well as the 4-ball definition of linking in 3 dimensions. Dec 14, 2009 at 4:40
• Accepted (my solutions was not that clever). Dec 14, 2009 at 5:00
• Hey Greg, I think I remember hearing this problem mentioned by Oleg Viro when he gave a talk at Davis when I was there. Could this be where Thurston got it? Dec 17, 2009 at 19:08
• Maybe. My recollection is that he cited a discussion between him and Conway. We'd have to ask Bill to find out, I guess. Dec 17, 2009 at 19:29

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.

• There is a 4-sentence solution attributed to Marvin Ortel written up in the first few lines of the paper "Criticality for the Gehring link problem" by Cantarella et al arxiv.org/abs/math.DG/0402212 Is it by chance the one you heard?
– j.c.
Nov 11, 2012 at 16:13
• Yes, it is a good one, thank you. Is the proof linked by jc is the same as the proof of the graduate student you mentioned? Nov 11, 2012 at 18:17
• Yes, this is the same proof:-) But the student was UNDERgraduate:-) I add, that Gehring, when stating the problem, added that he could prove that $length(A)\geq c$, where $c$ is an absolute constant. No one EXPECTED that the solution could be THAT simple. Nov 11, 2012 at 19:57
• The situation reminds me the famous problem of J-J. Sylvester: "Can we have a configuration of finitely many lines in the (real) projective plane such that all intersections are triple." (There is such a configuration in the complex projective plane $C^2$.) In the beginning of XX century this was a famous unsolved problem, sometimes listed next to the "4-colors problem". In 1944 it was solved, and the solution is such that it could be found by a clever high school kid. Also about 2 lines of text, using nothing. Nov 11, 2012 at 20:08
• Just to complete the argument and to make clear where the hypothesis is used: $A$ is entirely outside of the unit ball centered at $O$, hence the projection of $A$ over $A'$ is $1$-Lipschitz. Apr 1 at 18:06

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.

• Sounds great, I'll try to solve it first... Dec 9, 2009 at 4:28

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 $f$.

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.

• I did not solve it --- any hints? Dec 18, 2009 at 14:32
• The domain of $f$ is the set of vertices of an $n$-cube. Make a correspondence between these vertices and orthant subsets of the sphere in $n-1$ dimensions. Dec 18, 2009 at 15:28
• This is a special case of the 'edge isopermetric inequality' on the Boolean cube [Harper '64, "Optimal assignments of numbers to vertices"], which is proved by induction on n. It also follows from the 'Poincare inequality' for the Boolean cube, which can be proven by elementary Fourier analysis (or again by induction). Jan 1, 2020 at 12:43
• I would say the spherical isoperimetry is a way much more hard result than Harper. Dec 31, 2020 at 8:05

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 $\mathbb{R}^6$.

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 $B_7=B_1$.

• Nice problem, but it is more for school students; the target is graduate students... Mar 15, 2010 at 18:46
• I have some statistic... I think it is a hard problem even for professionals. You have good graduate students! (Vladimir Igorevich could say it is a problem for a kindergarten.) Mar 16, 2010 at 2:33

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 $p$.

• (1) I do not know a proof of first problem which use Brun--Minkowski :( Can you give a hint? (2) Second problems looks nice --- I have to think a bit :)... Thank you Dec 14, 2009 at 4:36
• A possible hint: Let Y be the complement of $X_{\delta}$. Now form $X'$ as all $ax$ with $x\in X,a\in [0,1]$, similarly for $Y'$. What can be said about $X',Y'$ and $\frac{X'+Y'}{2}$ as subsets of $\mathbb{R}^{n+1}$? Dec 17, 2009 at 7:36
• Do you know, what is the source of this proof? Dec 31, 2020 at 0:29
• @AntonPetrunin: It's in Matousek's book "Lectures on discrete geometry" (chapter 14). There I see that it is attributed to a paper by J. Arias-de-Reyna, K. Ball, R. Villa called "Concentration of the distance in finite dimensional normed spaces". Dec 31, 2020 at 0:44

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,...)

• I like the problem, usually I sign it by author or give a ref where a nice solution can be found (or at least a hint). Any suggestions? Dec 19, 2009 at 13:38
• I do not have a good reference other than that Wikipedia has a page called the Harnack curve theorem. I learned the problem in graduate school from a paper, but I do not remember which paper. As for a hint: A smooth complex curve of degree $d$ has genus $g = (d^2-3d+2)/2$. This looks sophisticated to prove, but it is not very hard. It may look sophisticated to use that to prove Harnack's bound $g+1$, but that's not very hard either. Dec 19, 2009 at 16:03

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.

• Not bad --- I solved it :). BTW do you know that a similar problem is open: find a point on outer curve which has two tangent segments from this point to the inner curve has equal size (at least if outer curve is not convex). Dec 29, 2009 at 4:33
• I did not know that problem and could not solve it (even if outer curve is convex). I can comment it: Each point x of inner curve defines "left" and "right" segment of a chord tangent at x and corresponding values l(x) and r(x). Consider images L and R of functions l and r correspondingly. If L \subset R or R \subset L then for any diffeomorphism g of the inner curve there exists x such that L(g(x))=R(x). It seems that, in general, it is not true that L \subset R either R \subset L. Jan 1, 2010 at 18:01
• I can prove your problem for a case when l (or r) is a constant function. Another variant (generalization) of the initial problem, unknown to me, is the following. Consider (locally) affine coordinate x on the inner curve, which is an angle of a tangent line with a fixed line. This coordinate is defined uniquely up to a sum with constant. Then, for any angle a an equation l(x)=r(x+a) is solvable. Jan 1, 2010 at 18:14
• Petya, this statement which you can do is everything known about the problem (if I remember right my conversation with Sergei Tabachnikov). Jan 30, 2010 at 19:38

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).

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.

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.

• Is it your problem? Jan 2, 2021 at 22:43
• No, the friend I quoted proposed this to me, but I don't know where this comes from. If you are interested in, I can ask him :) Jan 2, 2021 at 23:46
• Could you give clues to your solutions? I have one which is likely different from yours: choose a line that does not pass thru any intersection point of the balls --- it remains to show that a line cannot be presented as a union of disjoint closed intervals (which is easy). Jan 3, 2021 at 1:10
• Can I ask why do you think such line exist? Such covering is not even ensured to be countable a priori, but maybe there is an easy argument I am not seeing. About my solution: notice that in such a covering all opens are contractile, pairwise intersections are whether empty or a point, and threewise intersections are empty. Thus $\mathbb{R}^n$ would be homotopy equivalent to the nerve of the covering, which is a graph with vertices indexed by balls and connected if they intersect each other. Simple homotopical observations on $\mathbb{R}^n$ will make you conclude such graph does not exist :) Jan 3, 2021 at 1:41
• Thank you. Every ball has nonzero area, so it is a countable covering. Since there is at most one point of intersection for any pair of balls, the set of intersection points is countable as well. Jan 3, 2021 at 1:50

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.

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

• It is included now. Dec 31, 2020 at 0:30
• BTW, did you see this question: mathoverflow.net/q/360066 Jan 2, 2021 at 22:24

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).

• This is a nice problem! I was wondering if this is true for an arbitrary convex polytope: Can we always find one facet which can be covered by folding its neighbors? Dec 18, 2020 at 19:11
• Happy new year! Yes, I also mean by a congruent copy. What is an example when a projection can not be covered by the projected body? Jan 1, 2021 at 8:42
• @FedorPetrov it was a conjecture of Zalgaller, solved by Kovalev; a short solution was given by Kós and Törőcsik in "Convex disks can cover their shadow”. Discrete Comput. Geom. 5.6 (1990), pp. 529–531. Jan 2, 2021 at 2:44
• @AntonPetrunin but that's in dimension 2, in higher dimensions is it wrong or unknown? Jan 2, 2021 at 15:55
• @FedorPetrov it is false in higher dimensions. For example take the regular octahedron with vertices $\pm e_i$ and project it on the plane $x_1+x_2+x_3=0$ to get a regular hexagon. You can show the hexagon cannot fit inside the octahedron. Jan 2, 2021 at 19:09

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.

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.

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).

• Was it an illustration for the probabilistic method? Nov 4, 2018 at 5:18
• I don't know what was intended. I just saw it written in some problem list. Nov 4, 2018 at 5:20
• An easier version is same thing on a flat torus. Nov 4, 2018 at 5:21
• @AntonPetrunin here the probability is called area: for any of our $n$ points consider a disk of area $1/n$ centered in it, there exists an uncovered point $x$. The disk centered in $x$ does not contain our points. Dec 31, 2020 at 7:30

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.

• I am not smart enough to come up with this solution but I have a very naive way to attack this: for every $u\in\mathbb C$, let $S_u$ denote the set of complex numbers $z$ of modulus $1$ such that $\lvert 2u-z\rvert=1$. Then the cardinality of $S_u$ is at most $2$ if $u\neq0$, and $S_{-u}$ is symmetric to $S_u$ with respect to $0$. The result then follows. An obvious geometric interpretation of $S_u$: it is the intersection of the unit circle and the line passing through $u$ and perpendicular to $u-0$.
– Z. M
Apr 8 at 20:21

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

• Yes, I deside to include one problem on Delauney triangulation (it is 6.6 "inscribed triangulation"). Mar 16, 2010 at 1:22
• I mean another proof. There is very nice proof through stereographic projection on sphere Mar 16, 2010 at 2:42
• Nice proof, I did not know it before :) Mar 16, 2010 at 18:06

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...

• This is really cute. Apr 8 at 15:23

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 $P(g)$?

• Is it $\infty$? Jul 7, 2017 at 21:11
• What is your next (immediate and intuitive) guess if I tell you that it is not $\infty$? Jul 9, 2017 at 14:37
• 2? good problem, thank you very much again. Jul 9, 2017 at 16:16

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.

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).

(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?

• Is every polyhedra a simplicial set $X$ with $X_0$ equal its vertices? Jan 2, 2021 at 16:02
• @AndreaMarino: Yes, every triangulated polyhedron surface. Apr 1 at 23:39

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.