Tanya Khovanova's Math Blog

I wrote a lot about how during entrance tests for Moscow State University, the examiners were giving Jewish and other undesirable students special (e.g. more difficult) questions during the oral exams. (See, for example, our paper Jewish Problems with Alexey Radul.) Not all examiners agreed to do this. So the administration made sure that there were different exam rooms: brutal rooms with compliant examiners torturing students with difficult questions, and normal rooms with normal examiners testing preapproved students. The administration also had other methods. One of them is the topic of this essay.

The math department of Moscow State University had four entrance exams. The first was a written math test consisting of three trivial problems, a very difficult one, and a brutally challenging one. At the end, I will show you a sample: a trivial problem and a very difficult one from 1976, my entrance year.

What was the point of such vast variation in difficulty, you may ask? There were two reasons.

But first, let me explain some entrance rules. The exam was scored according to the number of solved problems. A score of two or less was a failing score. People with such scores would be disqualified from the next exam. Any applicant with a smidge of mathematical intelligence would be able to solve all three trivial problems. Almost all applicants who qualified for the next test would have the same score of three on the first test, as they wouldn’t be able to solve the last two problems. Thus, mathematical geniuses and people who barely made the cut got the same score.

There was another rule. Officially, people with a gold medal from their high school (roughly equivalent to a valedictorian) could be accepted immediately if they scored 5 on the first exam. So one of the administrative goals was to prevent anyone getting a 5, thus, blocking Jewish applicants from sneaking in after the first exam.

Another goal was to have all vaguely qualified people get the same score. The same goal applied to other exams. After the four admission exams, the passing score, say X, was announced. A few people with a score higher than X were immediately accepted. Then there were hundreds of applicants with a score of X, way more than the quota of people the department was planning to accept. An official rule allowed the math department to pick and choose whoever they wanted from everyone who scored X.

I heard a speech by the famous Russian mathematician, Vladimir Arnold, directed at decent examiners who tested “approved” students at the oral math exam, which was the second admissions exam. His suggestion was brilliant and simple. If the students are good and belong in the department, give them an excellent grade of 5. If not, give them a failing grade of 2. Arnold’s plan boosted the chances of good students doing better than the cutoff passing score X and removed mediocre students from the competition. His idea was not only brilliant and simple but also courageous: he was risking his career by trying to fight the system.

I never experienced the entrance exams firsthand. By ministry order, as a member of the USSR IMO team, I was accepted without taking any exams. I already wrote an essay, A Hole for Jews, about how getting on the IMO team was the only way for Jewish students to get into the Moscow State University, and how the University tried to block them.

But I still looked at the entrance exam problems I would have had to solve to get in. The last two problems scared me. Now I found them again online (in Russian) at: the 1976 entrance test. The trivial problem below is standard and mechanical, while the other problem still looks scary.

Trivial problem. Solve for x:

Solution. We were drilled in school to solve these types of problems, so this one was trivial. First, make a substitution: y = 3^x. This leads to an equation: (2y – 1)(y – 3)/(y² – 2)(y – 1) ≤ 0. From this we get ranges for y: (-∞, -√2], [1/2,1], [√2, 3]. The last step is to take a logarithm.

Very difficult problem. Three spheres are tangent to plane P and to each other. Two of the spheres are the same size. The apex of a circular cone is on P, and the cone’s axis is perpendicular to the plane P. All three spheres are outside the cone and tangent to it. Find the cosine of the angle between the cone’s generatrix and the plane P, if one of the angles of the triangle formed by the intersection points of the spheres and the cone is 150 degrees.

I stumbled upon one of Smullyan’s puzzle on Facebook, in Russian. I couldn’t find the original text, so I just translated it back for my students.

Puzzle. You are on an island where only truth-tellers and liars live. The truth-tellers always tell the truth, and the liars always lie. You meet an islander who sits with you for a long time, then says, “I already said this sentence.” Is he a truth-teller or a liar?

I expected the following solution. If this islander is a truth-teller, then there should have been a time when he said, for the first time, “I already said this sentence.” But this would create a contradiction.

However, my students used this puzzle as an opportunity to teach me some intricacies of the English language. They explained to me the ambiguities of my translation. Here is a shortened and lightly edited quote from one of them:

There are two different linguistic opinions that give different answers to this problem. The first is that the truth of a statement is decided at the moment it starts to be delivered: in this case, when the islander starts saying his statement. With this interpretation, for the statement to be true, he had to have said the sentence before, and for that to be true, he had to have said it even before that, and this continues indefinitely. Clearly, he cannot have been alive forever, so he has to be a liar.
The other opinion is that the verity of a statement is decided at the exact conclusion of its deliverance. Then, when the islander finishes saying his sentence, its truth is judged, and he has at that same instant “already” said the sentence, so he is telling the truth. By this interpretation, the islander is a truth-teller.

Another student had a different brilliant idea. Depending on the islander’s intonation, it is possible that he says, “I already said ‘this sentence’.” In that case, there are no self-referencing sentences, and the islander could be either a truth-teller or a liar.

I consulted my best English consultant: my son, Alexey, and here is his reply. “The basic answer is that neither truth nor semantic meaning are absolute, and edge cases will be judged differently by different observers. A sentence whose truth is time-dependent on the same scale as the duration of uttering the sentence is clearly an edge case. That’s why mathematicians intentionally try to eliminate ambiguity from their communication.”

He suggested the following fix for the puzzle’s translation.

Fixed puzzle. You meet an islander who says, “I have said this sentence before.” Is he a truth-teller or a liar?

Alexey didn’t stop at fixes and suggested the following bonus puzzles.

Bonus puzzle 1. You meet an islander who says, “I will have said this sentence.” Is he a truth-teller or a liar?

Bonus puzzle 2. You meet an islander who says, “I will say this sentence again.” Is he a truth-teller or a liar?

Ukraine is on my mind. Here is a problem for 9-graders from the 1978 Ukrainian Math Olympiad:

Problem. Prove that for every natural number n, the following number is not an integer.

Oskar van Deventer is a designer of beautiful mechanical puzzles. For the recent mini-MOVES gathering at the MoMath, he asked people in his Zoom breakout room to calculate the average age in the room without revealing their actual ages. I know the following solution to this puzzle.

People agree on a large number N that is guaranteed to be greater than the sum of all the ages. The first person, say Alice, thinks of a uniformly random integer R between 0 and N. Alice adds her age to R modulo N and passes the result to the second person, say Bob. Bob adds his age modulo N and passes the result to the third person, and so on. When the result comes back to Alice, she subtracts R modulo N and announces the sum total of all the ages.

During this process, no one gets any information about other people’s ages. But two people can collude to figure out the sum of the ages of the people “sitting” between them.

I gave this problem to my grandson, and he suggested the following procedure. First, people choose two trusted handlers: Alice and Bob. Then, each person splits their age into the sum of two numbers (the splitting should be random and allow one of the numbers to be negative). They then give one number to Alice and another to Bob. Alice and Bob announce the sums of the numbers they receive. After that, the sum total of everyone’s ages is the sum of the two numbers that Alice and Bob announce.

The advantage of this method is that no two people, except Alice and Bob, can collude to get more information. The disadvantage is that if Alice and Bob collude, they would know everyone’s age. Which method would you prefer?

Many years ago, I wrote a blog post about an amusing fact: John Conway put Moscow, the former capital of the USSR, as a coauthor: A Math Paper by Moscow, USSR. Thus, Moscow got an Erdős number 2, thanks to Conway’s Erdős number 1. At that time, my Erdős number was 4, so I wondered if I should try coauthoring a paper with Moscow, my former city of birth, to improve my Erdős number.

This weird idea didn’t materialize. Meanwhile, my Erdős number became 2 after coauthoring a paper with Richard Guy, Conway’s Subprime Fibonacci Sequences. I relaxed and forgot all about my Erdős status. I couldn’t do better anyway, or could I?

I recently heard about a cheater who applied to grad schools. In addition to a bunch of fabricated grades, the cheater submitted an arXiv link to a phony paper. What is fascinating to me is that the cheater put real professors’ names from the university the cheater supposedly attended as coauthors. The professors hadn’t heard of this student and had no clue about the paper. So the cheater added fake coauthors to add legitimacy to their application and boost the perceived value of the cheater’s “research”. As a consequence, the cheater got a fake Erdős number.

I hope that the arXiv withdrew the paper. Cheating is the wrong way to improve one’s Erdős number.

But here is another story. Robert Wayne Thomason named as coauthor his dead friend, Thomas Trobaugh. The paper in question is Higher Algebraic K-Theory of Schemes and of Derived Categories and can be found at https://www.gwern.net/docs/math/1990-thomason.pdf. This paragraph in the paper’s introduction explains the situation.

The first author [Robert Wayne Thomason] must state that his coauthor and close friend, Tom Trobaugh, quite intelligent, singularly original, and inordinately generous, killed himself consequent to endogenous depression. Ninety-four days later, in my dream, Tom’s simulacrum remarked, “The direct limit characterization of perfect complexes shows that they extend, just as one extends a coherent sheaf.” Awaking with a start, I knew this idea had to be wrong, since some perfect complexes have a non-vanishing K₀ obstruction to extension. I had worked on this problem for 3 years, and saw this approach to be hopeless. But Tom’s simulacrum had been so insistent, I knew he wouldn’t let me sleep undisturbed until I had worked out the argument and could point to the gap. This work quickly led to the key results of this paper.

This precedent gives anyone hope that they might achieve an Erdős number 1. You just need to wait for Paul Erdős to come to you in your dreams with a genius idea.

Last year, our junior PRIMES STEP group studied Latin squares. We invented a lot of different types of Latin squares and wrote a paper about it, Fun with Latin Squares. Recall that a Latin square is an n by n table containing numbers 1 through n in every cell, so that every number occurs once in each row and column. In this post, I want to talk about anti-chiece Latin squares.

First, what’s a chiece? A chiece is a portmanteau word made out of two words, chess and piece, and, not surprisingly, it means a chess piece. Given a chiece, an anti-chiece Latin square is a Latin square such that any two cells, where our chiece can move from one cell to the other, according to the rules of chess, can’t contain the same number. Let’s see what this means.

Let’s start with rooks, which move along rows and columns. An anti-rook Latin square can’t have the same numbers repeating in any one row or column. Ha, anti-rook Latin squares are just Latin squares. Anti-bishop and anti-queen Latin squares can’t have the same numbers repeating on any diagonal.

Now, here is a picture of an anti-knight Latin square in which no two identical numbers are a knight’s move apart. This particular Latin square also forms a mini-Sudoku: not only does each row and column, but also each 2 by 2 corner region, contains all distinct numbers.

Consider all instances of some number, say 1, in an anti-chiece Latin square. If the board is n by n, we get n instances of non-attacking chieces. A famous math puzzle asks to place eight non-attacking queens on a standard chessboard. So the instances of any one particular number in an anti-queen Latin square solves the problem of placing n non-attacking queens on an n by n chessboard. Thus, building an anti-queen Latin square is more complicated than solving the non-attacking queens puzzle. The former requires filling the chessboard with n non-overlapping sets of non-attacking queens. The picture below gives an example of an anti-queen 5 by 5 Latin square.

This square has some interesting properties. It can be formed by cycling the first row. It also happens to be one of the chiece Latin squares we study in our paper. A chiece Latin square is a Latin square such that for each number in a cell, there is another cell, a chiece’s move apart, containing the same number. You can check that our anti-queen Latin square is at the same time a knight Latin square.

I wonder, can anyone build an anti-queen Latin square on the standard 8 by 8 chessboard?

Here is the famous barber paradox.

Paradox. The barber shaves all those and only those who don’t shave themselves. Does the barber shave himself?

If he shaves himself, then he doesn’t shave himself. If he doesn’t shave himself, then he shaves himself.

English is not my primary language, and I am fascinated by the variety of verb tenses in English as compared to the Russian language. For example, Russian has one present tense while English has four. I wonder what would happen if we use the other present tenses in this paradox.

Present continuous. The barber shaves all those and only those who aren’t shaving themselves. Does the barber shave himself?

Does this mean that the barber starts shaving himself and then has to stop, and a moment later he has to start again?

Present perfect. The barber shaves all those and only those who haven’t shaved themselves. Does the barber shave himself?

Does this mean that the barber shaves himself every other day?

Present perfect continuous. The barber shaves all those and only those who haven’t been shaving themselves. Does the barber shave himself?

Does this mean the barber shaves himself once in his lifetime and then never again?

Tim Gowers discussed the age of consent on his blog, which I can no longer find. I will talk about his post here based on my old notes and my memory. The age of consent is a legal term to protect young people from being manipulated into agreeing to sex. Having consensual sex with people under the age of consent may be considered statutory rape or child sexual abuse.

Gowers starts with several assumptions.

• Non-triviality: There should exist an age at which a person is qualified to consent to sex and, consequently, have it.
• Simplicity: Whether or not two people are allowed to have sex with each other should depend only on their age in years.
• Monotonicity: If two people are allowed to have sex with each other today, they should be allowed to have sex with each other at all times in the future.

From these assumptions, the following theorem can be deduced.

Theorem. The only possible rule satisfying these assumptions would allow any two people to have sex with each other as long as they both reached some fixed age k.

There is a problem with this type of rule. Suppose k is 18. If two people who are slightly younger than 18 have consensual sex, they can’t both be predators. These are two children with raging hormones. There is no reason to punish anyone. Now imagine that one of the partners turns 18. Society would still consider this a Romeo-and-Juliet case and would tend not to punish such a partner. Now imagine a child younger than 18 having sex with a partner over 40. The older partner has no raging hormones, knows what they are doing, and probably knows how to manipulate little children into having sex. So, it might be desirable to have a rule that differentiates between these two cases. The rule would take into account the difference in ages while forgiving younger offenders and still punishing predators.

Consider the most common type of law to resolve this issue: Anyone older than 18 can have sex, and, in addition, a person who is not older than 20 can have sex with someone between the ages of 16 and 18. This law doesn’t satisfy monotonicity. It could be that one day the older partner is not yet 20, and the next day, oops, they have a birthday. So, as a birthday gift, they are not allowed to have sex with each other anymore.

Here is a simple idea to resolve the issue by having the law focus on the age gap instead of the age of the older partner. We can have an adjusted law: Anyone older than 18 can have sex, and, in addition, a person can have sex with someone between the ages of 16 and 18 as long as the age gap is not more than four years. This rule doesn’t satisfy the simplicity assumption above, but it is simple enough. It is close in spirit to the previous rule and satisfies monotonicity. The problem with this rule is continuity.

• Continuity: If the age gap between couple A is only slightly larger than the age gap between couple B, then couple A should not have to wait significantly longer to be allowed to have sex.

According to the adjusted rule, the couple with the age gap of four years and one day might have to wait two years longer to have sex than the couple with the age gap of four years. This seems unfair.

Tim Gowers suggests dropping the simplicity rule. We can use days rather than years. For example, the rule might be that if one person in a couple is under 18, but at least 16, and has age x, then the other partner has to be not more than age y, where for example, y − x = 4 + (x − 16)/2. So when one partner turns 16, their partner has to be not older than 20. When one partner is 16 and two months, the other cannot be older than 20 and three months. With the younger partner getting older, the allowable age gap is increasing slowly. By the time the younger partner is a day from turning 18, their partner can be almost five years older.

It might be complicated for two people to calculate if they are allowed to have sex according to this formula. But Gowers’ big idea was that apps and websites could do this easily: two people plug in their birthdays and know whether they are allowed to have sex.

Do you know what an isosceles tetrahedron is? I didn’t until recently. An isosceles tetrahedron has all of its faces congruent. Equivalently, all pairs of opposite edges are of equal length. Such tetrahedra are also called disphenoids. Here are some cute statements about them.

• If the four faces of a tetrahedron all have the same perimeter, then the tetrahedron is a disphenoid.
• If the four faces of a tetrahedron all have the same area, then the tetrahedron is a disphenoid.

Disphenoids have three nontrivial symmetries: 180-degree rotations around three lines connecting the midpoints of opposite edges.

Is it possible to have a tetrahedron with fewer symmetries than disphenoids? Yes, it is. Dan Klain just published a fascinating paper about such tetrahedra: Tetrahedra with Congruent Face Pairs. The results are so elegant and simple that I was surprised that they were new. I got curious and started to google aggressively. I found an official name for this kind of tetrahedron: phyllic disphenoid, but no theorems about them. Their name is quite unappealing: no wonder people didn’t want to study them much. Obviously, Dan wasn’t as aggressive at googling, so he didn’t find this official name and called them reversible tetrahedra. But my favorite name for them is the name Dan thought of but didn’t use in his paper: bi-isosceles tetrahedra.

Here is the picture of a bi-isosceles tetrahedron from Dan’s paper. It has two faces with sides a, b, and c, and two faces with sides a, b, and d. The edges of this tetrahedron are: a, a, b, b, c, and d. It has one nontrivial symmetry: a 180-degree rotation around the line connecting the midpoints of the unequal opposite edges c and d. The picture emphasizes this symmetry. The figure in the picture is a projection of a bi-isosceles tetrahedron onto a plane, such that the line of symmetry is projected onto the point of symmetry of the projection. The two cute statements above, about disphenoids, can be generalized to the case of bi-isosceles tetrahedra.

• If the four faces of a tetrahedron can be split into two pairs with the same perimeter, then the tetrahedron is bi-isosceles.
• If the four faces of a tetrahedron can be split into two pairs with the same area, then the tetrahedron is bi-isosceles.

The first property is trivial, while the second one is proven in Dan’s paper. Dan also shows how to calculate the volume of a bi-isosceles tetrahedron:

Here is a problem from the 1978 Kyiv Olympiad for 7 graders.

Is it possible to place seven points on a plane so that among any three of them, two will be at distance 1 from each other?