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.

These two puzzles were given to me by Andrey Khesin.

Puzzle. My friend and I are going to play the following game at a casino. Each round, each of us (my friend, the dealer, and I) secretly chooses a black or white stone and drops it in the same bag. Then, the contents of the bag are revealed. If all three stones are the same color, my friend and I win the round. If not, we lose to the dealer. One extra caveat. I have a superpower: as soon as we sit down, I can read the dealer’s mind and learn the dealer’s choices for all future rounds. Unfortunately, at that time, it’s too late for me to give this information to my friend and win all the rounds. The only thing we can do is agree on a strategy before the game.

Design a strategy to win 6 out of 10 rounds.

Design a strategy to win 7 out of 11 rounds.

Is it possible to win 6 out of 9 rounds?

Puzzle. In a crowd of 70 people, one person is a murderer, and another person is a witness to said murder. A detective can invite a group into his office and ask if anyone knows anything. The detective knows that everyone except the witness would say nothing. The witness is a responsible person who is more afraid of the murderer than they desire to fulfill their civic duty. If the witness is in the same group as the murderer, the witness will be silent; otherwise, the witness will point to the murderer. The detective knows this will happen and wants to find the murderer in as few office gatherings as possible. What is the minimum number of times he needs to use his office, and how exactly should the detective proceed?