Tanya Khovanova's Math Blog

Romanian Masters in Mathematics might become the most prestigious math competition for high school students. Romania invites teams from the 20 best countries from the previous year’s International Math Olympiad. This way they guarantee that all the competitors are extremely strong and they can give more difficult problems than the usual IMO problems.

This year they held their second competition and my son, Sergei Bernstein, was invited to join the USA team. Here is one of the problems:

For a finite set X of positive integers, define ∑(X) as the sum of arctan(1/x) for all elements x in X. Given a finite set S, such that ∑(S) < π/2, prove that there exists a finite set T such that S is a subset of T and ∑(T) = π/2.

The official solution involved a tedious greedy algorithm. My son, Sergei, got an extra point for his unique solution, which was very different from the official one. Here’s his solution:

To an integer x we associate a complex number x + i, which in polar coordinates is r(cosθ + i sinθ). Note that θ equals arctan(1/x). That means we can interpret ∑(X) as the angle in the polar form of ∏(x+i) — the product of (x+i) for all x in X.

We are given a set S with elements s_j such that ∏(s_j+i) has a positive real part, and we need to find other elements t_k, such that ∏(s_j+i) multiplied by ∏(t_k+i) has real part 0.

But we know that the ∏(s_j+i) = a + bi, for some integers a and b. I claim that we can always find a positive integer y, so that (a + bi)(y + i) has a smaller real part than a.

Note that ∑(S) < π/2, hence a and b are positive. By an ever so slightly modified version of the division algorithm we can find integers q and r such that b = aq + r, 0 < r ≤ a. Now simply set y equal to q + 1 (which is a positive integer). Then the real part of (a + bi)(y + i) equals ay – b = a – r, and is a non-negative number less than a.

Now we just repeat the process, which obviously has a finite number of steps.

My son’s solution wasn’t complete. The problem talked about sets of numbers and that implies that the numbers should be distinct. I leave it to my readers to finish this solution for distinct numbers.

This story happened in the summer of 1975. I was 16. Before that, I was naive and brainwashed; by the end of the summer I had grown up. All that summer I was stunned and didn’t know what to do as I watched this story unfold.

I was invited to the summer training camp for the International Math Olympiad. There I became friends with a fellow team member Sasha (Alexander) Reznikov. Sasha had dreamed of being a mathematician from childhood. He was gifted and brilliant and when he was in the 7th grade he got noticed by professional mathematicians. They told him that the only way to become a mathematician is to do undergraduate studies at the math department of Moscow State University. He was also told that the math department doesn’t accept Jews. Sasha was Jewish.

But there was a hole in the system. If he could get to the International Math Olympiad, he would be admitted to the place of his choice by an order of the Ministry of Education. As the IMO was conducted during entrance exams for universities, there was this special arrangement for the team members. Besides, the Soviet Math Olympiad wasn’t yet corrupted, so the Olympiads would give him a chance.

Sasha was brilliant, but he had a disadvantage — he was 2 years younger than his classmates because he had skipped grades. Since the IMO is only for high school students, he had to make it to the IMO before he graduated at 15 years old.

He worked very hard and really pushed himself, and he made it on to our team. The photo of two kids is of Sasha and me that summer.

We had a supervisor — Zoya Ivanovna — who was a Ministry liaison. She was to compile the list of team members for the Ministry of those who should be accepted without entrance exams to universities and colleges. The team had eight people, so every year the list consisted of eight students. That year was special, since we actually only had six people on our team who were high school seniors. Two of us, Sergey Finashin and I, were not yet seniors. But Zoya Ivanovna who was a good-hearted lady decided to sneak in eight people and added our two alternates to the list. As our alternates were preparing for the IMO instead of preparing for entrance exams, this was a generous and fair thing to do. Everything was fine and everyone was happy.

Until one day when strange things started to happen. We were invited for a meeting where Zoya Ivanovna told us that there was a problem with Moscow State University. We were told that the math department has a limit of four people who can be accepted without an entrance exam, and we had five students applying. Zoya Ivanovna asked if there was a volunteer who might reconsider. At this point Alexey Muzykantov said that he would volunteer since he was an alternate. Besides, he was always as interested in physics as in math, and would be happy to study in the physics department.

After the meeting I stumbled upon Zoya Ivanovna crying in the ladies room. She told me that she didn’t know what to do. The problem was that out of five people applying for the math department of Moscow State University, three were Jews. Three Jews were too many out of the 400 people annually accepted to the department. Our team coordinator, Valentin Anatolievich Skvortsov, was working at the math department, where he was being pressured. Zoya Ivanovna told me he had been threatened with expulsion from the Communist Party if he didn’t reduce the number of Jews by at least one. Being expelled from the Communist Party was a serious threat at that time and Zoya Ivanovna was eager to help, so she invented this idea about the limit. The idea didn’t work, because Alexey Muzykantov, who removed himself from the list, wasn’t Jewish.

After several days, Sasha Reznikov’s mother appeared at the summer program. She told me that she was being pushed to persuade Sasha not to go to Moscow State University.

I asked Zoya Ivanovna why she chose Sasha. She told me that out of the three Jews, one was from Moscow, so she didn’t consider him, and Sasha was much younger than the third student and, besides, he had health problems. So she tried to convince Sasha’s mother that Sasha would be better off in his home town Kiev than in Moscow.

Sasha went to Kiev University. The system had had a hole through which two Jews passed that year, so even though Sasha had made astonishing efforts, he hit a wall. He was crushed.

Later he tried to transfer to Moscow State University, but was ridiculed, humiliated and denied. Eventually Sasha moved to Israel and got his PhD in mathematics. He died in 2003 by, according to rumors, suicide.

Computational Linguistics Olympiads started in Moscow in 1962. Finally in 2007 the US caught up and now we have the NACLO — North American Computational Linguistics Olympiad.

The problems from past Soviet Olympiads are hard to find, so here I present a translation from Russian of a sample problem from the Moscow Linguistics Olympiad website:

You are given sentences in Niuean language with their translations into English:

1. To lele e manu. — The bird will fly.
2. Kua fano e tama. — The boy is walking.
3. Kua koukou a koe. — You are swimming.
4. Kua fano a ia. — He is walking.
5. Ne kitia he tama a Sione. — The boy saw John.
6. Kua kitia e koe a Pule. — You are seeing Pule.
7. To kitia e Sione a ia. — John will see him.
8. Ne liti e ia e kulï. — He left the dog.
9. Kua kai he kulï e manu. — The dog is eating the bird.

Translate the following sentences into Niuean:

1. John swam.
2. You will eat the dog.
3. Pule is leaving you.
4. The bird will see the boy.
5. The dog is flying.

I ran an experiment. I copied multiple choices from the 2007 AMC 8 into a file and asked my son Sergei to try to guess the answers, looking only at the choices. I allowed him to keep several choices. The score I assigned depended on the number of leftover choices. If the leftover choices didn’t contain the right answer, the score for the problem was zero. Otherwise, it was scaled according to the number of choices he left. For example, if he had four choices left and the right answer was among them he got 1/4 of a point. Here are the choices:

1. 9, 10, 11, 12, 13.
2. 2/5, 1/2, 5/4, 5/3, 5/2.
3. 2, 5, 7, 10, 12.
4. 12, 15, 18, 30, 36.
5. 24, 25, 26, 27, 28.
6. 7, 17, 34, 41, 80.
7. 25, 26, 29, 33, 36.
8. 3, 4.5, 6, 9, 18.
9. 1, 2, 3, 4, cannot be determined.
10. 13, 20, 24, 28, 30.
11. Choose picture: I, II, III, IV, cannot be determined.
12. 1:1, 6:5, 3:2, 2:1, 3:1.
13. 503, 1006, 1504, 1507, 1510.
14. 5, 8, 13, 14, 18.
15. a+c < b, ab < c, a+b < c, ac < b, b/c = a.
16. Choose picture: 1, 2, 3, 4, 5.
17. 25, 35, 40, 45, 50.
18. 2, 5, 6, 8, 10.
19. 2, 64, 79, 96, 131.
20. 48, 50, 53, 54, 60.
21. 2/7, 3/8, 1/2, 4/7, 5/8.
22. 2, 4.5, 5, 6.2, 7.
23. 4, 6, 8, 10, 12.
24. 1/4, 1/3, 1/2, 2/3, 3/4.
25. 17/36, 35/72, 1/2, 37/72, 19/36.

It is clear that if you keep all choices, your score will be 5, which is the expected score for AMC if you are randomly guessing the answers. Sergei’s total score was 7.77, which is noticeably better than the expected 5.

There were two questions where Sergei felt that he knew the answer exactly. First was question number two with choices: 2/5, 1/2, 5/4, 5/3, 5/2. All but one of the choices has a 5 in it, so 1/2 must be wrong. Numbers 2/5 and 5/2 are inverses of each other, so if organizers expect you to make a mistake by inverting the right answer, then one of these choices must be the right answer. But 5/4 and 5/3 are better suited as a miscalculation of 5/2, than of 2/5. His choice was 5/2, and it was correct. The second question for which he was sure of the answer was question 19, with his answer 79. I still do not have a clue why.

Sergei’s result wasn’t too much better than just guessing. That means that AMC 8 organizers do a reasonably good job of hiding the real answer. You can try it at home and see if you can improve on Sergei’s result. I will publish the right answers as a comment to this essay in a week or so.

China Girls Math Olympiad is becoming an international math Olympiad for girls. When I first heard about this competition I felt very sad. I need to explain myself here.

For many years I felt very proud that math Olympiads do not separate the genders. Most Olympic sports, like running or swimming, have separate competitions for men and women. I felt that joint competitions for math demonstrated the spirit of equality in our math community. I felt that insofar as gender didn’t matter, mathematics was more democratic than other sports.

At the same time I do understand how people might assume for the following reasons that a math competition among only girls would be useful:

• They promote the idea of math to girls.
• They can help girls who are into math to feel less lonely.
• They generate additional resources and training for girls.
• They might be less stressful for some girls, than mixed math competitions.
• They help promote the image of female mathematicians to society
• They provide further opportunities for girls to earn prizes and improve their resumes.

See also the article: First US Team to Compete in the China Girls Mathematical Olympiad.

On the other hand, this development scares me. If we have a separate girls Olympiad, will that soon lead us to have two Olympiads, one for boys and one for girls? Two separate Olympiads would be a defeat for women mathematicians. Or, maybe I shouldn’t be scared. The percentage of girls at the most prestigious mathematics competition, the International Mathematical Olympiad, is so small that it can be viewed as virtually boys-only.

Mathematics is becoming similar to chess. There is a World Chess Championship where both men and women are allowed to compete, and there is a separate Women’s World Chess Championship. The interesting part is that Judith Polgar, by far the strongest female chess player in history, never competed in the Women’s World Chess Championship. I suspect that I understand Judith. She probably feels that women-only competitions diminish her, or that chess is about chess, not about gender. In any case, I hope that one day the separate girls Olympiad will not be needed.

I am interested in a career in mathematics. How hard is it to be a woman mathematician?

Let us look at some numbers from the American Mathematical Society Survey Reports for the year 2005:

• There were 40% of women among graduating math majors.
• There were 30% of women among Math PhDs granted.
• There were 11% of women among full-time tenured or tenure-track positions.

You can’t just say that women do not like math — 40% of those choosing math as a major is quite a large number, after all.

On the other hand, the downward trend of these percentages is striking. If women’s opportunities and abilities are the same as men’s, these percentages should grow with every age step, since, as we know, the percentage of women in the population increases with age due to men dying earlier.

But the numbers go down and very fast. There are many potential explanations for this, but today we’re going to look at one of them:

Women have less ability for high-level mathematics.

Was Larry Summers right when in his speech that cost him his Harvard presidency he compared math ability to height and to the propensity for criminality, and suggested that the distribution, especially standard deviation, of math ability differs for men and women?

To answer this question, I wanted to find some other data that correlates gender with math abilities. I took the results of the American Mathematical Competitions (AMC 12) for the year 2008. Among 120,000 students who participated, 43% were females. Here are some results:

• Among students scoring 72 points or higher there were 40% of girls.
• Among students scoring 98 points or higher there were 30% of girls.
• Among students scoring 134.5 points or higher there were 11% of girls.

This picture is similar to that of the academic career: the closer you climb to the top, the smaller percentage of girls you see there. Of course, winning a competition is very different from getting tenure. People who win competitions are smart and competitive — smart and competitive enough to go for money, rather than academia. On the other hand, people who are interested in mathematics often are not interested in anything else. Why would they waste their time in competitions when the Riemann Hypothesis is still waiting to be solved?

But still, both achieving tenure and winning math competitions represent mathematical ability in some sense. If Larry Summers was right and the distribution of math ability is different among males and females, then by looking around you at the percentage of females at your level, you should be able to assess how close you are to the top of the math field.

I propose the following math career predictor: Take your results in AMC 12. If among kids who did better than you, the percentage of girls is more than 11%, you do not have a chance at tenure. If the percentage of girls is more than 30%, do not waste your time working on a math PhD. If the percentage of girls is more than 40% maybe math majoring is not for you.

I hate my math career predictor. I hate it not only because it has so many flaws that it might just deserve the Ig Nobel Prize, but because it doesn’t take people’s effort into account. You really have to work very hard to be a math professor, whether you were a winner or a loser in math competitions.

You might ask why I created a math career predictor that is so flawed. My mathematician friends, those who are more honest than polite, tell me that I have no chance at getting back to academia. On the other hand, I had the second best result at the 1976 IMO, which means I have the ability. My predictor may be my only hope.

A puzzle from the 1977 USSR math Olympiad can be solved naturally with linear algebra:

Seven dwarfs are sitting at a round table. Each dwarf has a cup partially filled with milk. Each dwarf in turn divides all his milk evenly between the six other cups. After the seventh dwarf has done this, every cup happens to contain the initial amount of milk. What was the initial distribution of milk?

Can you use linear algebra to intelligently solve this puzzle?

Here is a math puzzle for kids from a nice collection of ThinkFun Visual Brainstorms:

Wendy has cats. All but two of them are Siamese, all but two of them are Persian, and all but two of them are Maine Coon. How many cats does Wendy have altogether?

This puzzle has two answers: the expected answer and an unexpected answer. Can you find both?