Archive for the ‘Puzzles’ Category.

## Grigori Perelman’s Puzzle

Have you heard of Grigori Perelman? If you like math, you probably have. He is one of the most renowned mathematicians in the world. I recently got a book on the Leningrad Mathematical Olympiads (scheduled for publication in English in 2025) and found Grigori’s name there. He authored one of the Olympiad problems from 1984. For context, he was born in 1966. Here it is.

Puzzle. You are given ten numbers: one “1” and nine “0”s. You are allowed to replace any two numbers with their arithmetic mean. What is the smallest number that can appear in place of the “1” after a series of such operations?

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## Pass-Fail

Recent Facebook Puzzle from Denis Afrisonov.

Puzzle. 100 students took a test where each was asked the same question: “How many out of 100 students will get a ‘pass’ grade after the test?” Each student must reply with an integer. Immediately after each answer, the teacher announced whether the current student passed or failed based on their answer. After the test, an inspector checks if any student provided a correct answer but was marked as failed. If so, the teacher is dismissed, and all students receive a passing grade. Otherwise, the grades remain unchanged. Can the students devise a strategy beforehand to ensure all of them pass?

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## Another Symmetry Puzzle

I recently posted a symmetry puzzle from Donald Bell. He just sent me another one.

Puzzle. Start with a 30-60-90 triangle (half of an equilateral triangle). Divide it into two 30-60-90 triangles of different sizes by dropping a perpendicular from the right-angled corner to the opposite side. Put the resulting two pieces together to form a symmetrical shape. There are two solutions.

It took me some time to find the second solution. I love this puzzle.

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## Is It Possible?

Usually, I only post puzzles to which I know the solution. However, I don’t know the solution to this exciting geometry question from Facebook, yet. But I like the puzzle so much that I’d rather post it than wait until I find time to think about it.

Puzzle. A centrally-symmetric figure is cut into two equal polygons: A and B. Is it possible that the center of symmetry is in A but not in B?

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## An Alternator Coin Puzzle

I run a program at MIT called PRIMES STEP, where we conduct mathematical research with children in grades 6 through 9. Our first research project was about a funny coin called an alternator. This coin exists only in a mathematician’s mind as it can change weight according to its own will. When you put the alternator on the scale, it can either weigh the same as a real coin or a fake coin (the fake coins are lighter than real ones). The coin strictly alternates how much it weighs each time it is put on the scale. My colleague, Konstantin Knop, recently sent me a fresh alternator puzzle.

Puzzle. There are four identical-looking coins: two real, one fake, and one alternator. How do you find the alternator using a balance scale at most three times?

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## 2024 MIT Mystery Hunt

I am not as excited about the MIT Mystery Hunt as I used to be. So, for this year’s hunt, I didn’t go through all the puzzles but present here only the puzzles that were recommended to me. I start with math, logic, and CS.

Then we have some word puzzles.

Now, the rest.

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## Three L-tetraminoes

Here is a cool puzzle I heard from Tiago Hirth at the last Gathering for Gardner, who in turn heard it from Donald Bell.

Puzzle. You have three L-tetrominoes. Arrange them on a plane without overlaps so that the resulting shape has a line of symmetry.

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## Hundred Colors of Math

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?

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## Find the Side

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 √(a2+x2) = √(4(a2-ab+b2)/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 a2 + b2 + c2. 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 L2/2, where L is the length of the triangle’s side. The answer follows.

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## Fudge Likes Meatballs

Here is an interesting puzzle by Ivan Mitrofanov.

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.

1. Eats a meatball and all the flies sitting on it at that time.
2. 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.

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