Next Tanya Khovanova

Many years ago at Gelfand’s seminar in Moscow, USSR, someone pointed out a young girl and told me: “This is Natalia Grinberg. In her year in the math Olympiads, she was the best in the country. She is the next you.”

We were never introduced to each other and our paths never crossed until very recently.

Several years ago I became interested in the fate of the girls of the IMO (International Math Olympiad). So, I remembered Natalia and started looking for her. If she was the best in the USSR in her year, she would have been a gold medalist at the IMO. But I couldn’t find her in the records! The only Grinberg I found was Darij Grinberg from Germany who went to the IMO three times (2004, 2005, and 2006) and won two silver medals and one gold.

That was clearly not Natalia. I started doubting my memory and forgot about the whole story. Later I met Darij at MIT and someone told me that he was Natalia’s son.

I was really excited when I received an email from Natalia commenting on one of my blog posts. We immediately connected, and I asked her about past events.

Natalia participated in the All-Soviet Math Olympiads three times. In 1979 as an 8th grader she won a silver medal, and in 1980 and 1981 she won gold. That indeed was by far the best result in her year. So she was invited to join the IMO team.

That year the IMO was being held in the USA, which made Soviet authorities very nervous. At the very last moment four members of the team did not get permission to travel abroad. Natalia was one of them. The picture below, which Natalia sent to me, was taken during the Soviet training camp before the Olympiad. These four students were not allowed to travel to the IMO: Natalia Grinberg, Taras Malanyuk, Misha Epiktetov, and Lenya Lapshin.

1981 IMO training camp

Because of the authorities’ paranoia, the Soviet team wasn’t full-sized. The team originally contained eight people, but as they rejected four, only six traveled to the USA, including two alternates.

I have written before how at that time the only way for a Jewish student to get to study mathematics at Moscow State University was to get to the IMO. I wrote a story about my friend Sasha Reznikov who trained himself to get to the IMO, but because of some official machinations, still was not accepted at MSU.

Natalia’s story surprised me in another way. She didn’t get to the IMO, but she was accepted at MSU. It appears that she was accepted at MSU as a member of the IMO team, because that decision was made before her travel documents were rejected.

Natalia became a rare exception to the rule that the only way for a Jewish person to attend MSU was to participate in the IMO. It was a crack in the system. They had to block visas at the last moment, so that people wouldn’t have time to make a fuss and do something about it. Natalia slipped through the crack and got to study at the best university in the Soviet Union.

Unfortunately, the world lost another gold IMO girl. Three Soviet team members won gold medals that year. Natalia, being better then all of them, would have also won the gold medal.

Share:facebooktwittergoogle_plusredditpinterestlinkedinmail

6 Comments

  1. George R.:

    Good morning Tanya!
    If I may ask a somewhat indiscreet question (?) does Prof. Gelfand has any relation to the world famous chess player Boris Gelfand from Bellarussia? (now representing Israel at international chess competitions)

  2. Tanya Khovanova:

    George R,

    Israel Gelfand lived in Moscow for many years, so the chess player can’t be a close relative. Otherwise, I do not know.

  3. Faibsz:

    Hello,

    in the mid-1970s Moscow Volodya Grinberg was a star of every maths competition I went to.
    He didn’t make it to Moscow University because of anti-Semitism, but is now a full maths professor at UCLA

    Do you know if he is related to Natalia?

    Regards

    Faibsz

  4. Tanya Khovanova:

    Faibsz,

    I do not know who you are talking about.

  5. alex:

    I was interested in the history of IMO and learned by accident that in 2007 the gold took a girl. From the interview with her I also learned that there was a particularly hard problem that year. It was the problem no.3 and it was solved by four participants only including that girl. I took a look at the problem and it appeared to me that the solution was obvious and I wrote it down in the next 5 minutes. I never participated in math olympiads nor any other olympiads and my background is not related to math. I suspect that there is an error in the solution that I am missing so I will present the solution below so that someone might check it. If it doesn’t contain an error then I would say that the problem was fairly obvious.

    Problem

    In a mathematical competition some competitors are friends. Friendship
    is always mutual. Call a group of competitors a clique if each two of them are friends. (In
    particular, any group of fewer than two competitors is a clique.) The number of members
    of a clique is called its size.
    Given that, in this competition, the largest size of a clique is even, prove that the
    competitors can be arranged in two rooms such that the largest size of a clique contained
    in one room is the same as the largest size of a clique contained in the other room.

    Solution

    Let the largest clique C have the size 2n. We can divide this clique in two parts and send one part to room A and another part to room B. Let’s call these cliques CA and CB. The rest of the participants will go to room A. In room A there might be a clique V of the size s>n. There are participants in clique V that do not belong to clique CB (If it’s not the case then send the rest of the participants to room B and apply the same reasoning. This subset of the participants can’t belong to both CA and CB). Send one participant from room A belonging to clique L to room B. The size of L decreased by one. The size of the largest clique in B remains unchanged. Proceed till the size of L will become n+1. Send one more participant to B. The size of the largest clique in B will not exceed n since the size of L was less or equal to 2n. Therefore in both rooms the largest clique will have the size n. QED.

  6. alex:

    correction: By “L” I mean “V”.