Tanya Khovanova's Math Blog

I just discovered a Russian Internet Linguistics Olympiad. Even though most linguistics problems are not translatable, this time we are lucky. My favorite problem from this Olympiad is related to chemical elements — their names in Russian have the same logical structure as in English. Keep in mind, the problem doesn’t assume any knowledge of chemistry. Here is the problem:

The formulae for chemical elements and their names are given below in mixed order:
C₃H₈, C₄H₆, C₃H₄, C₄H₈, C₇H₁₄, C₂H₂;
Heptene, Butine, Propane, Butene, Ethine, Propine.

1. Match the formulae with their names. Explain your solution.
2. Write the names of the elements with the following formulae: C₂H₄, C₂H₆, C₇H₁₂.
3. Write the formulae for the following elements: Propene, Butane.

I found a new Russian Olympiad for high schools related to universities. I translated my favorite problems from last year’s final round. These are the math problems:

8th grade. In a certain family everyone likes their coffee with milk. At breakfast everyone had a full cup of coffee. Given that Alex consumed a quarter of all consumed milk and one sixth of all coffee, how many people are there in the family?

8th grade. How many negative roots does the equation x⁴ − 5x³ − 4x² − 7x + 4 = 0 have?

10th grade. Find a real-valued function f(x) that satisfies the following inequalities for any real x and y: f(x) ≤ x and f(x+y) ≤ f(x) + f(y).

I liked the physics problems even more:

8th grade. Winnie-the-Pooh weighs 1 kg. He hangs in the air with density 1.2kg/m³ next to a bee hive. He is holding a rope connected to a balloon. Estimate the smallest possible diameter of the balloon, assuming that this happens on Earth.

10th grade. Two containers shaped like vertical cylinders are connected by a pipe underneath them. Their heights are the same and they are on the same level. The cross-sectional area of the right container is twice bigger than the left’s. The containers are partially filled with water of room temperature. Someone put ice into both containers: three times more ice into the right one than into the left one. After that, the containers are closed hermetically. How will the water level will change after the ice melts completely:

• The levels will not change.
• The level on the left will be higher than on the right.
• The level on the left will be lower than on the right.
• The answer depends on the initial volume of water in the containers.

At the 2011 IMO, Lisa Sauermann received yet another gold medal. Now she tops the Hall of Fame of the IMO with four gold medals and one silver medal.

In addition, in 2011 she achieved the absolute best individual result and was the only person with a perfect score. In previous years, there were several girls who tied for first place, but she is the first girl ever to have an absolute rank of 1.

I told you so. In my 2009 essay Is There Hope for a Female Fields Medalist?, I predicted that a girl will soon become an absolute champion of the IMO.

In that essay I draw a parallel between the absolute champion of IMO and a Fields medalist. Indeed, we get one of each per year. Lisa Sauermann is the best math problem solver in her year. Will she grow up to receive a Fields medal? I am not so sure: the medal is still unfriendly to women. Lisa Sauermann is the best math problem solver ever. Will she grow up to be the best mathematician of our century? I wonder.

I met Leon Vaserstein at a party. What do you think I do at parties? I bug people for their favorite problems, of course. The first riddle Leon gave me is a variation on a famous problem I had already written about. Here’s his version:

The hypotenuse of a right triangle is 10 inches, and one of the altitudes is 6 inches. What is the area?

When Leon told me that he had designed some problems for the Soviet Olympiads, naturally I wanted to hear his favorite:

A closed polygonal chain has its vertices on the vertices of a square grid and all the segments are the same length. Prove that the number of segments is even.

The Moscow Math Olympiad has a different set of problems for every grade. Students need to write a proof for every problem. These are the 8th grade problems from this year’s Olympiad:

Problem 1. There were 6 seemingly identical balls lying at the vertices of the hexagon ABCDEF: at A — with a mass of 1 gram, at B — with a mass of 2 grams, …, at F — with a mass of 6 grams. A hacker switched two balls that were at opposite vertices of the hexagon. There is a balance scale that allows you to say in which pan the weight of the balls is greater. How can you decide which pair of balls was switched, using the scale just once?

Problem 2. Peter was born in the 19th century, while his brother Paul was born in the 20th. Once the brothers met at a party celebrating both birthdays. Peter said, “My age is equal to the sum of the digits of my birth year.” “Mine too,” replied Paul. By how many years is Paul younger than Peter?

Problem 3. Does there exist a hexagon which can be divided into four congruent triangles by a single line?

Problem 4. Every straight segment of a non-self-intersecting path contains an odd number of sides of cells of a 100 by 100 square grid. Any two consecutive segments are perpendicular to each other. Can the path pass through all the grid vertices inside and on the border of the square?

Problem 5. Denote the midpoints of the non-parallel sides AB and CD of the trapezoid ABCD by M and N respectively. The perpendicular from the point M to the diagonal AC and the perpendicular from the point N to the diagonal BD intersect at the point P. Prove that PA = PD.

Problem 6. Each cell in a square table contains a number. The sum of the two greatest numbers in each row is a, and the sum of the two greatest numbers in each column is b. Prove that a = b.

by Tanya Khovanova and Alex Ryba

The following problem appeared at the Gillis Math Olympiad organized by the Weizmann Institute:

A foreign government consists of 12 ministers. Each minister has 5 friends and 6 enemies amongst the ministers. Each committee needs 3 ministers. A committee is considered legitimate if all of its members are friends or all of its members are enemies. How many legitimate committees can be formed?

Surprisingly, this problem implies that the answer doesn’t depend on how exactly enemies and friends are distributed. This meta thought lets us calculate the answer by choosing an example. Imagine that the government is divided into two factions of six people. Within a faction people are friends, but members of two different factions dislike each other. Legitimate committees can only be formed by choosing all three members from the same faction. The answer is 40.

We would like to show that actually the answer to the problem doesn’t depend on the particular configuration of friendships and enmities. For this, we will count illegitimate committees. Every illegitimate committee has exactly two people that have one enemy and one friend in the committee. Let’s count all the committees from the point of view of these “mixed” people. Each person participates in exactly 5*6 committees as a mixed person. Multiply by 12 (the number of people), divide by 2 (each committee is counted twice) and you get the total 180. This gives an answer of 40 for the number of legitimate committees without using a particular example.

What interests us is the fact that the number of illegitimate, as well as legitimate, committees is completely defined by the degree distribution of friends. For any set of people and who are either friends or enemies with each other, the number of illegitimate committees can be calculated from the degree distribution of friends in the same way as we did above.

Any graph can be thought of as representing friendships of people, where edges connect friends. This cute puzzle tells us that the sum of the number of 3-cliques and 3-anti-cliques depends only on the degree distribution of the graph.

As a non mathematical comment, the above rule for legitimate committees is not a bad idea. In such a committee there is no reason for two people to gang up on the third one. Besides, if at some point in time all pairs of friends switch to enemies and vice versa, the committees will still be legitimate.

by Pavel Etingof and Tanya Khovanova

We worked for several years with RSI where we supervised summer math research projects by high school students. Now, we’ve started an additional program at MIT’s math department called PRIMES, where local high school students do math research during the academic year. In this essay we would like to discuss what makes a good math research project for a high school student.

A doable project. The project should not be believed to be extremely difficult to yield at least results. It is very discouraging for an aspiring mathematician not to produce anything during their first project.

An accessible beginning. The student should be able to start doing something original soon after the start of the project. After all, they don’t come to us for coursework, but for research.

Flexibility. It is extremely important to offer them a project that is adjustable; it should go in many directions with many different potential kinds of results. Since we do not know the strength of incoming students in advance, it is useful to have in mind both easier and harder versions of the project.

Motivation. It is important for the project to be well motivated, which means related to other things that have been studied and known to be interesting, to research of other people, etc. Students get more excited when they see that other people are excited about their results.

A computer component. This is not a must for a good project. But modern mathematics involves a lot of computation and young students are better at it than many older professors. Such a project gives young students the opportunity to tackle something more senior people are interested in but might not have enough computer skills to solve. In addition, through computer experiments students get exposed to abstract notions (groups, rings, Lie algebras, representations, etc.) in a more “hands-on” way than when taking standard courses, and as a result are less scared of them.

A learning component. It is always good when a project exposes students to more advanced notions.

The student should like their project. This is very difficult to accomplish when projects are chosen in advance before we meet the students. However, we try to match them to great projects by using the descriptions they give of their interests on their applications. It goes without saying that mentors should like their project too.

Having stated the desired properties of a good project, let us move on to giving examples: bad projects and good projects. We start with a bad one:

Prove that the largest power of 2 that doesn’t contain 0 is 2⁸⁶.

The project satisfies only one requirement: it contains a computer component. Otherwise, it doesn’t have an accessible beginning. It is not very flexible: if the student succeeds, the long-standing conjecture will be proven; if s/he doesn’t, there is not much value in intermediate results. The question is not very interesting. The only motivation is that it has been open for a long time. Also, there is not much to learn. Though, almost any theoretical question can be made flexible. We can start with the question above and change its direction to make it more promising and enticing.

Another bad example is a project where the research happens after the programs are written. This is bad because it is difficult to estimate the programming abilities of incoming students. It doesn’t have an accessible beginning and there is no flexibility until the programming part is finished. If the student can’t finish the programming quickly, s/he will not have time to look at the results and produce conjectures. For example, almost any project in studying social networks may fall into this category:

Study an acquaintance graph for some epic movies or fiction, for example Star Wars or The Lord of the Rings. In this graph people are vertices and two people are connected by an edge if they know each other. The project is to compare properties of such graphs to known properties of other social networks.

Though the networks in movies are much smaller than other networks that people study, the amount of programming might be substantial. This project can be a good project for a person with a flexible time frame or a person who is sure in advance that there will be enough time for him/her to look at the data.

Now on to an example of a good project. Lynnelle Ye and her mentor, Tirasan Khandhawit, chose to analyze the game of Chomp on graphs during RSI 2009.

Given a graph, on each turn a player can remove an edge or a vertex together with all adjacent edges. The player who doesn’t have a move loses. This game was previously solved for complete graphs and forest graphs, so the project was to analyze the game for other types of graphs.

It is clear how to analyze the game for any particular new graph. So that could be a starting point providing an accessible beginning. After that the next step could be to analyze other interesting sets of graphs. The flexibility is guaranteed by the fact that there are many sets of graphs that can be used. In addition, the project entails learning some graph theory and game theory. And the project has a computational component.

Lynnelle Ye successfully implemented this project and provided a complete analysis of complete n-partite graphs for arbitrary n and all bipartite graphs. She also gave partial results for odd-cycle pseudotrees. The paper is available at the arxiv. Not surprisingly, Lynelle got fourth place in the Intel Science Talent Search and second place in the Siemens Competition.

33 horsemen are riding in the same direction along a circular road. Their speeds are constant and pairwise distinct. There is a single point on the road where the horsemen can pass one another. Can they ride in this fashion for an arbitrarily long time?

The puzzle appeared at the International Tournament of the Towns and at the Moscow Olympiad. Both competitions were held on the same day, which incidentally fell on Pi Day 2010. Just saying: at the Tournament the puzzle was for senior level competitors; at the Moscow Olympiad it was for 8th graders.

Warning: If you want to solve it yourself first, pause now, because here is the solution I propose.

First, consider two horsemen who meet at that single point. The faster horseman passes the slower one and gallops ahead and the slower one canters along. The next meeting point should be at the same place in the circle. Suppose the slower horseman rides n full circles before the next meeting, then the second horseman could not have passed the first in between, so he has to ride n+1 full circles. That means their speeds should have a ratio of (n+1)/n for an integer n. And vice versa, if their speeds have such a ratio, they will meet at the same location on the circle each time. That means that to solve the problem, we need to find 33 different speeds with such ratios.

As all speed ratios are rational numbers, we can scale speeds so that they are relatively prime integers. The condition that two integers have a ratio (n+1)/n is equivalent to the statement that two integers are divisible by their difference. So the equivalent request to the problem is to find a set of 33 positive integers (or prove non-existence), such that every two integers in the set are divisible by their difference.

Let’s look at examples with a small number of horsemen. For two riders we can use speeds 1 and 2. For three riders, speeds 2, 3 and 4.

Now the induction step. Suppose that we found positive integer speeds for k horsemen. We can add one more horseman with zero speed who quietly stays at the special point and everyone else passes him. The difference condition is satisfied. We just need to tweak the set of speeds so that the lazy horseman starts moving.

We can see that if we add the least common multiple to every integer in a set of integers such that every two numbers in a pair are divisible by their difference, then the condition stays satisfied. So by induction we can find 33 horsemen. Thus, the answer to the problem is: Yes they can.

Now I would like to apply that procedure from the solution to calculate what kind of speeds we get. If we start with one rider with the speed of 1, we add the second rider with speed 0, then we add 1 to both speeds, getting the solution for two riders: 1 and 2. Now that we have a solution for two riders, we add a third rider with speed 0 then add 2 to every speed, getting the solution for three horsemen: 2, 3 and 4. So the procedure gave us the solutions we already knew for two and three horsemen.

If we continue this, we’ll get speeds 12, 14, 15 and 16 for four riders. Similarly, 1680, 1692, 1694, 1695, and 1696 for five riders.

We get two interesting new sequences out of this. The sequence of the fastest rider’s speed for n horsemen is: 1, 2, 4, 16, 1696. And the sequence of the least common multiples for n−1 riders — which is the same as the lowest speed among n riders — is: 1, 1, 2, 12, 1680, 343319185440.

The solution above provides very large numbers. It is possible to find much smaller solutions. For example for four riders the speeds 6, 8, 9 and 12 will do. For five riders: 40, 45, 48, 50 and 60.

I wonder if my readers can help me calculate the minimal sequences of the fastest and slowest speeds. That is, to find examples where the integer speed for the fastest (slowest) horseman is the smallest possible.

My blog is getting more famous. Now I don’t need to look around for nice problems, for my readers often send them to me. In response to my blog about him, Sergey Markelov’s Best, Markelov sent me more of his problems. Here is a cute tetrahedron problem that he designed:

Six segments are such that you can make a triangle out of any three of them. Is it true that you can build a tetrahedron out of all six of them?

Another reader, Alexander Shen, sent me a different tetrahedron problem from a competition after reading my post on Problem Design for Multiple Choice Questions:

Imagine the union of a pyramid based on a square whose faces are equilateral triangles and a regular tetrahedron that is glued to one of these faces. How many faces will this figure have?

Shen wrote that the right answer to this problem had been rumored to have a negative correlation with the result of the entire test.

Janet Mertz encouraged me to find IMO girls and compare their careers to that of their teammates. I had always wanted to learn more about the legendary Lida Goncharova — who in 1962 was the first girl to win an IMO gold medal. So I located her, and after an interview, wrote about her. Only 14 years later, in 1976, did the next girl get a gold medal. That was me. I was ranked overall second and had 39 points out of 40.

As I did in the article about Lida, I would like to compare my math career to that of my teammates.

I got my PhD in 1988 and moved to the US in 1990. My postdoc at MIT in 1993 was followed by a postdoc at Bar-Ilan University. In 1996 I got a non-paying visiting position at Princeton University. In 1998 I gave up academia and moved to industry, accepting an offer from Bellcore. There were many reasons for that change: family, financial, geographical, medical and so on.

On the practical level, I had had two children and raising them was my first priority. But there was also a psychological element to this change: my low self-esteem. I believed that I wasn’t good enough and wouldn’t stand a chance of finding a job in academia. Looking back, I have no regrets about putting my kids first, but I do regret that I wasn’t confident enough in my abilities to persist.

I continued working in industry until I resigned in January 2008, due to my feeling that I wasn’t doing what I was meant to do: mathematics. Besides, my children were grown, giving me the freedom to leave a job I did not like and return to the work I love. Now I am a struggling freelance mathematician affiliated with MIT. Although my math blog is quite popular and I have been publishing research papers, I am not sure that I will ever be able to find an academic job because of my non-traditional curriculum vitae.

The year 1976 was very successful for the Soviet team. Out of nine gold medals our team took four. My result was the best for our team with 39 points followed by Sergey Finashin and Alexander Goncharov with 37 points and by Nikita Netsvetaev with 34 points.

Alexander Goncharov became a full professor at Brown University in 1999 and now is a full professor at Yale University. His research is in Arithmetic Algebraic Geometry, Teichmuller Theory and Integral Geometry. He has received multiple awards including the 1992 European Math Society prize. Sergey Finashin is very active in the fields of Low Dimensional Topology and Topology of Real Algebraic Varieties. He became a full professor at Middle East Technical University in Ankara, Turkey in 1998. Nikita Netsvetaev is an expert in Differential Topology. He is a professor at Saint Petersburg State University and the Head of the High Geometry Department.

Comparing my story to that of Lida, I already see a pattern emerging. Now I’m curious to hear the stories of other gold-winning women. I believe that the next gold girl, in 1984, was Karin Gröger from the German Democratic Republic. I haven’t yet managed to find her, so can my readers help?