A former student runs into an old calculus teacher as she is shuffling home. The student is happy to see her and says, “I recently thought about you and our classes.” “How so?” she perks up. “I was in a bit of a pickle when calculus helped me,” he says. The old lady straightens, and her face begins to glow. “Can you elaborate?”, asks the teacher in anticipation. The student proceeds with his story. “I was walking home in the pouring rain when a gust of wind snatched my hat right off my head. The hat landed in a puddle. This wasn’t just any hat; it was a gift from my dad, so I really wanted to get it back. But I didn’t exactly fancy diving headfirst into the puddle. So, I looked around and saw a piece of wire. I bent it into the shape of an integral and used it to fetch my hat.”

Foams are a recent craze in homology theory. I want to explain what a foam is using a tesseract as an example. Specifically, the 2D skeleton of a tesseract is a foam.

We can view a tesseract as a convex hull of 16 points in 4D space with coordinates that are either 0 or 1. The edges connect two vertices with the same three out of four coordinates. Faces are squares with corners being four vertices that all share two out of four coordinates.

Foam definition. A foam is a finite 2-dimensional CW-complex. Each point’s neighborhood must be homeomorphic to one of the three objects below.

An open disc. Such points are called regular points.

The product of a tripod and an open interval. Such points are called seam points.

The cone over the 1D-skeleton of a tetrahedron. Such points are called singular vertices.

My favorite example of a foam is a tesseract. Or, more precisely, the set of tesseract’s vertices, edges, and faces form a foam.

The regular points are the insides of the tesseract’s faces. Their neighborhoods are obviously open discs.

The seam points are the insides of the tesseract’s edges. Each edge is incident to three faces, and the projection of its neighborhood to a plane perpendicular to the edge is a tripod.

The singular vertices are the tesseract’s vertices. Each vertex is incident to four edges and six faces. We can view this neighborhood as a cone cover of a tetrahedron formed by this vertex’s neighbors.

Some of the coolest foams are tricolorable. A foam is called tricolorable if we can color it using three colors in such a way that each face has its own color, and any three faces that meet at a seam are three distinct colors.

Not surprisingly, I chose a tricolorable foam for our example. Let me prove that the tesseract’s 2D skeleton is tricolorable. We start by coloring the edges in four colors depending on the direction: red, green, blue, or gray, as in the first picture (source: Wikipedia). Each face has two pairs of edges of two different colors. We can color the faces in the following manner: if none of the edges are gray, then the face color is the complementary non-gray color (For example, if the edges are red and blue, the face is green). If the edges are gray and one other color, then the face color matches the non-gray color (For example, if the edges are red and gray, the face is red). I leave it to the reader to prove that this coloring means that each edge is the meeting point of three different face colors.

Proposition. If we remove the regular points of one particular color from a tricolored foam, we will get a closed compact surface containing all the seam points and singular vertices.

In our example, the result is a torus, which you can recognize in the second picture. Here, I use the Schlegel diagram as a model for a tesseract, shown on the right (source: Wikipedia). The exercise for the reader is to explain where the eight green faces of the torus were before they were removed.

The following lemma from the aforementioned paper describes another cool property of a tricolorable foam.

Lemma. If a foam is tricolorable, its 1D skeleton (the graph formed by seams and singular points) is bipartite.

And, surely, I am leaving it up to the reader to check that the tesseract’s 1D skeleton forms a bipartite graph.

A cat is chasing a mouse, and the mouse hides into its little hole. While it’s in there, the mouse hears some barking, “Woof! Woof!” Smart mouse figures a dog scared off the cat, so it peeks out. But guess what? Cat’s still there and catches the mouse, saying, “See, that’s why it pays to know foreign languages!!”

I recently bought a book by Evdokimov, titled Hundred Colors of Math. The book has lovely math puzzles and cute pictures. The book has answers but doesn’t explain them. Also, the English translation is decent but not perfect. For these two reasons, I am not sure I would recommend the book. However, I do like the puzzles, and here is one of them, called Runaway Cell.

Puzzle. The figure depicted in the picture (a 6-by-6 square, in which the top row is moved by one square) was cut along the grid lines into several identical parts which could be put together to form a 6-by-6 square. The parts are allowed to be turned over. What is the minimal possible number of such parts?

I gave a short talk about my favorite math jokes at G4G15. G4G stands for the Gathering for Gardner, my favorite conference. Here is a joke about Heisenberg from my talk.

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Heisenberg gets pulled over on the highway. Cop: “Do you know how fast you were going, sir?” Heisenberg: “No, but I know exactly where I am.”

After my talk, David Albert sent me a sequel to this joke.

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Heisenberg gets pulled over on the highway. Cop: “Do you know how fast you were going, sir?” Heisenberg: “No, but I know exactly where I am.” Cop: “You were going 85 miles per hour”. Heisenberg: “Oh great—now I’m lost!”

Here is another joke from the conference.

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—Did you hear about the mathematician who’s afraid of negative numbers? —He’ll stop at nothing to avoid them.

This joke, too, got an awesome sequel from Jesse Lauzon.

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Did you hear about the mathematician who is afraid of negative numbers? —He’ll stop at nothing to avoid them. —Well, that’s only natural!

Here is the most recent addition to my collection from my friend, Alexander Karabegov.

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A trigonometry professor lost his voice and had to use sine language.

Another cute geometry puzzle was posted on Facebook.

Puzzle. An equilateral triangle in a plane has three vertices with known x-coordinates: a, b, and c. What is the side of the triangle?

I want to describe three different solutions that the readers of the Facebook channel posted. But before doing so, let’s look at the problem’s symmetries. We can immediately say that the answer should be a symmetric function of three variables: |a-b|, |b-c|, and |c-a|. It is possible to coordinate-bash the problem. However, I always prefer geometric solutions. Having said that, if one wants a calculation, using complex numbers might speed things up.

A solution using complex numbers. Suppose c is the origin, then the first vertex corresponds to a complex number a+xi. Then, the second vertex can be found after rotating the first vertex around the origin by 60 degrees. That means it is at (a+xi)exp(±2πi/6). Without loss of generality, we can assume that the second vertex corresponds to (a+xi)(1+i√3)/2. It follows that b = (a−x√3)/2. Thus, x = 2(a/2-b)/√3. And the side length is √(a^{2}+x^{2}) = √(4(a^{2}-ab+b^{2})/3). Adjusting for the choice of the origin, we get that the length is √(2((a-b)^{2}+(b-c)^{2}+(c-a)^{2})/3).

A geometric solution. Draw a line through point A parallel to the x-axis. Denote the intersections of this line with lines x=b and x=c as P and Q, correspondingly. Let R be the midpoint of the side BC. Then, the triangle PQR is equilateral. To prove it, notice that angles ARC and AQC are right, which implies that points ARCB are on the same circle with diameter AC. It follows that the angles RCA and RQA are the same; thus, the angle RQA is 60 degrees. Given that the triangle PQR is isosceles as R has to be on the bisector of PQ, we conclude that the triangle PQR is equilateral. Now, we can calculate the height of PQR and, therefore, the height of ABC, from which the result follows.

A physics solution. Without loss of generality, we can assume that a+b+c=0. Thus, the y-axis passes through the triangle’s centroid. The moment of inertia of the system consisting of the three triangle vertices with respect to the y-axis is a^{2} + b^{2} + c^{2}. Now, we add the symmetry consideration: the inertia ellipse must be invariant under the 60-degree rotation, implying that the ellipse is actually a circle. This means that the inertia moment doesn’t change under any system rotation. Thus, we can assume that one of the vertices lies on the y-axis. In this case, the inertia moment equals L^{2}/2, where L is the length of the triangle’s side. The answer follows.

The homework I give to my students (who are in 6th through 9th grades) often starts with a math joke related to the topic. Once, I decided to let them be the comedians. One of the homework questions was to invent a math joke. Here are some of their creations. Two of my students decided to restrict themselves to the topic we studied that week: sorting algorithms. The algorithm jokes are at the end.

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A binary integer asked if I could help to double its value for a special occasion. I thought it might want a lot of space, but it only needed a bit.

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Everyone envies the circle. It is well-rounded and highly educated: after all, it has 360 degrees.

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Why did Bob start dating a triangle? It was acute one.

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Why is Bob scared of the square root of 2? Because he has irrational fears.

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Are you my multiplicative inverse? Because together, we are one.

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How do you know the number line is the most popular? It has everyone’s number.

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A study from MIT found that the top 100 richest people on Earth all own private jets and yachts. Therefore, if you want to be one of the richest people on Earth, you should first buy a private jet and yacht.

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Why did the geometry student not use a graphing calculator? Because the cos was too high.

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Which sorting algorithm rises above others when done underwater? Bubble sort!

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Which sorting algorithm is the most relaxing? The bubble bath sort.

Puzzle. In front of my dog, Fudge, lies an infinite number of meatballs with a fly sitting on each of them. At each move, Fudge makes two consecutive operations described below.

Eats a meatball and all the flies sitting on it at that time.

Transfers one fly from one meatball to another (there can be as many flies as you want on a meatball).

Fudge wants to eat no more than a million flies. Assuming that flies sit still, prove that Fudge doesn’t have a strategy where each meatball is eaten at some point.