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?

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.

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?

There are a lot of puzzles where you need to guess something asking only yes-or-no questions. In this puzzle, there are not two but three possible answers.

Puzzle. Mike thought of one of three numbers: 1, 2, or 3. He is allowed to answer “Yes”, “No”, or “I don’t know”. Can Pete guess the number in one question?

Yes, he can. This problem was in one of my homeworks, and my students had a lot of ideas. Here is the first list were ideas are similar to each other.

I am thinking of an odd number. Is my number divisible by your number?

If I were to choose 1 or 2, would your number be bigger than mine?

If I were to pick a number from the set {1,2,3} that is different from yours, would my number be greater than yours?

If I have a machine that takes numbers and does nothing to them except have a 50 percent chance of changing a two to a one. Would your number, after going through the machine, be one?

If I were to choose a number between 1.5 and 2.5, would my number be greater than yours?

If your number is x and I flip a fair coin x times, will there be at least two times when I flip the same thing?

I am thinking of a comparison operation that is either “greater” or “greater or equal”. Does your number compare in this way to two?

One student was straightforward.

Mike, please, do me a favor by responding ‘yes’ to this question if you are thinking about 1, ‘no’ if you are thinking about 2, and ‘I don’t know’ if you are thinking about 3?

One student used a famous unsolved problem: It is not known whether an odd perfect number exists.

Is every perfect number divisible by your number?

Then, I gave this to my grandchildren, and they decided to answer in a form of a puzzle. Payback time.

I’m thinking of a number too, and I don’t know whether it’s double yours. Is the sum of our numbers prime?

I rarely post physics puzzles, but this one is too good to pass on.

Puzzle. A wireframe icosahedron is assembled so that each of its edges has a resistance of 1. What is the total resistance between opposite vertices of the icosahedron?

While we are at it, another interesting question would be the following.

Puzzle. A wireframe cube is assembled so that each of its edges has a resistance of 1. What is the total resistance between opposite vertices of the cube?

And this reminds me of a question I heard when I was preparing for an IMO many years ago.

Puzzle. A wireframe infinite square grid is assembled so that each of its edges has a resistance of 1. What is the total resistance between two neighboring vertices?

Here is the homework problem I gave to my PRIMES STEP students.

Puzzle. A man called his wife from the office to say that he would be home at around eight o’clock. He got in at two minutes past eight. His wife was extremely angry at this lateness. Why?

The expected answer is that she thought he would be home at 8 in the evening, while he arrived at 8 in the morning. However, my students had more ideas.

For example, one student extended the time frame.

The man was one year late.

Another student found the words “got in” ambiguous.

He didn’t get into his house two minutes past eight. He got into his car.

A student realized that the puzzle never directly stated why she got angry.

The wife already got angry when he said he would be home around eight, as she needed him home earlier.

The students found alternative meanings to “called his wife from the office” and “minutes.”

He had an office wife whom he called. But the wife at home was a different wife, and she was angry.

Puzzle. There are 100 cards with integers from 1 to 100. You have three possible scenarios: you pick 18, 19, or 20 cards at random. For each scenario, you need to estimate the probability that the sum of the cards is even. You do not need to do the exact calculation; you just need to say whether the probability is less than, equal to, or more than 1/2.

I like including warm-up puzzles with every homework.

Puzzle. Farmer Giles has four sheep. One day, he notices that they are all standing the same distance away from each other. How can this be so?

The expected answer: The configuration is impossible in 2D. So, one of the sheep is on a hill or in a pit.

Some students thought big: The sheep could be placed at different locations around the Earth, forming a really big tetrahedron. In this case, we need to explain what it means for the farmer to “notice”, but this minor issue could be resolved in many ways.

Some of the students questioned the meaning of the word distance. They argued that if sheep are all touching each other, they are the same distance 0 from each other. One way this could happen is if the tails of the four sheep were entangled.

I loved coin puzzles, but after several research projects related to those shiny discs, I got tired of them. The fatigue was temporary, as confirmed by the following Facebook puzzle that reignited my interest.

Puzzle. There are 30 coins in a circle that look the same. However, 20 of them are fake, and the rest are real. Fake coins weigh the same, and real coins weigh the same but heavier than fake ones. You need to find as many fake coins as possible using a balance scale once, given that the fake coins are positioned consecutively. What is your strategy?