Tanya Khovanova's Math Blog

Nikolay Konstantinov, the creator and the organizer of the Tournament of the Towns, discussed some of his favorite tournament problems in a recent Russian interview. He mentioned two beautiful geometry problems by Sergey Markelov that I particularly loved. The first one is from the 2003 tournament.

An ant is sitting on the corner of a brick. A brick means a solid rectangular parallelepiped. The ant has a math degree and knows the shortest way to crawl to any point on the surface of the brick. Is it true that the farthest point from the ant is the opposite corner?

The other one is from 1995.

There are six pine trees on the shore of a circular lake. A treasure is submerged on the bottom of the lake. The directions to the treasure say that you need to divide the pine trees into two groups of three. Each group forms a triangle, and the treasure is at the midpoint between the two triangles’ orthocenters. Unfortunately, the directions do not explain how exactly to divide the trees into the groups. How many times do you need to dive in order to guarantee finding the treasure?

I gave my students a problem from the 2002 AMC 10-A:

Tina randomly selects two distinct numbers from the set {1, 2, 3, 4, 5}, and Sergio randomly selects a number from the set {1, 2, …, 10}. The probability that Sergio’s number is larger than the sum of the two numbers chosen by Tina is: (A) 2/5, (B) 9/20, (C) 1/2, (D) 11/20, (E) 24/25.

Here is a solution that some of my students suggested:

On average Tina gets 6. The probability that Sergio gets more than 6 is 2/5.

This is a flawed solution with the right answer. Time and again I meet a problem at a competition where incorrect reasoning produces the right answer and is much faster, putting students who understand the problem at a disadvantage. This is a design flaw. The designers of multiple-choice problems should anticipate mistaken solutions such as the one above. A good designer would create a problem such that a mistaken solution leads to a wrong answer — one which has been included in the list of choices. Thus, a wrong solution would be punished rather than rewarded.

Readers: here are three challenges. First, to ponder what is the right solution. Second, to change parameters slightly so that the solution above doesn’t work. And lastly, the most interesting challenge is to explain why the solution above yielded the correct result.

I am coaching my AMSA students for math competitions. Recently, I gave them the following problem from the 1964 MAML:

The difference of the squares of two odd numbers is always divisible by:
A) 3, B) 5, C) 6, D) 7, E) 8?

The fastest way to solve this problem is to check an example. If we choose 1 and 3 as two odd numbers, we see that the difference of their squares is 8, so the answer must be E. Unfortunately this solution doesn’t provide any useful insight; it is just a trivial calculation.

If we remove the choices, the problem immediately becomes more interesting. We can again plug in numbers 1 and 3 to see that the answer must be a factor of 8. But to really solve the problem, we need to do some reasoning. Suppose 2k + 1 is an odd integer. Its square can be written in the form 4n(n+1) + 1, from which you can see that every odd square has remainder 1 when divided by 8. A solution like this is a more profitable investment of your time. You understand what is going on. You master a method for solving many problems of this type. As a bonus, if students remember the conclusion, they can solve the competition problem above instantaneously.

This is why when I am teaching I often remove multiple choices from problems. To solve them, rough estimates and plugging numbers are not enough. To solve the problems the students really need to understand them. Frankly, some of the problems remain boring even if we remove the multiple choices, like this one from the 2009 AMC 10.

One can holds 12 ounces of soda. What is the minimum number of cans needed to provide a gallon (128 ounces) of soda?

It’s a shame that many math competitions do not reward deep analysis and big-picture understanding. They emphasize speed and accuracy. In such cases, plugging in numbers and rough estimates are useful skills, as I pointed out in my essay Solving Problems with Choices.

In addition, smart guessing can boost the score, but I already wrote about that, too, in How to Boost Your Guessing Accuracy During Tests and To Guess or Not to Guess?, as well as Metasolving AMC 8.

As the AMC 10 fast approaches, I am bracing myself for the necessity to include multiple choices once again, thereby training my students in mindless arithmetic.

I am starting yet another part-time job as the Head Mentor at PRIMES, a new MIT research program for high schoolers. I am very excited about this program, for it will be valuable not only to kids who want to become researchers, but also to kids who just want to see what research is like. Kids who want to learn to think in a new way will also find it highly useful.

PRIMES is in many ways similar to RSI, which it augments and complements. There are also a lot of differences. Keep in mind that I am only comparing PRIMES to the math part of RSI, with which I was working as a coordinator for two years. I do not know how RSI handles other sciences.

Different time scale. RSI lasts six weeks; PRIMES will take about a year. I already wrote about some peoples’ skepticism towards RSI in my piece called “Fast Food Research?.” PRIMES creates a more natural pace for research.

Choices. Because of the time schedule at RSI, students get their project as soon as they start. Students who realize by the end of the second week that they do not like their project are at a disadvantage: if they do not change their project, they’re stuck with something that does not inspire them or is too difficult, and if they do change their project, they won’t have enough time to do a great job. At PRIMES students will have time to talk to the mentors before starting their project, so that they can participate in choosing their project. Depending on how it goes later, they’ll have time to try several different directions. I believe that the best research comes from the heart: students who have the time and opportunity to shape their choices will be more invested in their project.

Application process. At RSI, The Center for Excellence in Education reviews the applications. Even though they usually do a superb job at sending us great students, I believe it would be an advantage if mentors were able to influence the review process, for they might find even better matches to their projects. At PRIMES, the mentors will have this opportunity to review the applications.

Geography. RSI accepts students from all over the US and from some other countries. PRIMES can only accept local students — those who live close enough to visit MIT once a week for four months. Because of this restriction, PRIMES is recruiting from a smaller pool of students than RSI. But for local students it means that it will be easier to get accepted to PRIMES than to RSI.

Coaching. At RSI, students get a lot of coaching. I think that every student is in close contact with four adults. Two of them are from the math department — mentor and coordinator (that’s me!) — and two tutors from CEE. PRIMES will have less coaching. A student will have a mentor and me, the head mentor. In addition, mentors might arrange for students to talk to the professors who originated their projects.

Immersion. RSI students are physically present. They are housed at MIT with the expectation that they completely devote their time to their research. Students at PRIMES will be integrating their research into the rest of their lives and their commitments. That will require good organizational skills and a lot of self-discipline. RSI students have discipline imposed on them by their situation — which may be an advantage to them.

Olympiads. While they are at RSI, students can’t go to IMO or other summer activities. This is why many strong Olympiad students choose not to go to RSI, or they turn down an RSI acceptance if in the meantime they have gotten on to an Olympic team. At PRIMES you can do both. It is possible to go to an Olympiad, in addition to writing a paper.

Grade. RSI students have to be juniors. There are no grade limitations for PRIMES. Thus, it is possible to go to PRIMES in one’s senior year. In this case, it may be too late to use PRIMES on college applications, but it is perfectly fine for the sake of research itself. Or it might be possible to go to PRIMES as a sophomore, and then apply for RSI the next year. This will strengthen the student’s application for RSI.

RSI is well-established and has proven itself. PRIMES is new and hopefully will offer young mathematicians additional opportunities to try research.

I think that the American system of education creates a lot of pressure for teachers to drill their students for standardized tests and multiple choice questions. This blocks creative thinking. Every program like PRIMES is very good for unleashing students’ creativity and contributing to the development of the future thinkers of American society.

From the 1966 Moscow Math Olympiad:

Prove that you can choose six weights from a set of weights weighing 1, 2, …, 26 grams such that any two subsets of the six have different total weights. Prove that you can’t choose seven weights with this property.

Let us define the sequence a(n) to be the largest size of a subset of the set of weights weighing 1, 2, …, n grams such that any subset of it is uniquely determined by its total weight. I hope that you agree with me that a(1) = 1, a(2) = 2, a(3) = 2, a(4) = 3, and a(5) = 3. The next few terms are more difficult to calculate, but if I am not mistaken, a(6) = 3 and a(7) = 4. Can you compute more terms of this sequence?

Let’s see what can be said about upper and lower bounds for a(n). If we take weights that are different powers of two, we are guaranteed that any subset is uniquely determined by the total weight. Thus a(n) ≥ log₂n. On the other hand, the total weight of a subset has to be a number between 1 and the total weight of all the coins, n(n+1)/2. That means that our set can have no more than n(n+1)/2 subsets. Thus a(n) ≤ log₂(n(n+1)/2).

Returning back to the original problem we see that 5 ≤ a(26) ≤ 8. So to solve the original problem you need to find a more interesting set than powers of two and a more interesting counting argument.

The International Math Olympiad started in Eastern Europe in 1959. Romania was the first host country. The Olympiad grew and only in 1976 did it move outside the Eastern bloc. The competition was held in Austria.

I was on the Soviet team in 1975 and 1976, so I was able to compare competitions held in Eastern vs. Western countries. Of course, the Austrian Olympiad was much better supported financially, but today I want to write about the differences in how our team was prepped.

Before our travel to Austria the Soviet team members were gathered in a room with strangers in suits for a chat. I assumed that we were talking to the KGB. They gave us a series of instructions. For example, they told us not to leave the campus during the competition, to always walk in groups, and to avoid talking to kids from countries that are enemies of the USSR. They warned us that they would be watching, and I was scared to death.

Now that I am older and wiser, I understand that their goal was to frighten us. Our team traveled with adult supervisors, who were trusted by the KGB. But for several days during the grading period of the competition, our supervisors were not allowed to see us. So the KGB wanted us to be too afraid to be very adventurous when we were left on our own.

In addition, the KGB had a Jewish problem. In general, Jews were not allowed to go abroad. I had many Jewish friends who qualified for the pre-IMO math camp where the team was chosen, but who were not able to get on the IMO because of delays with their travel documents. Some local bureaucrats were eager to impress the KGB and therefore held up visas for Jewish students, preventing them from being on the team. But the team selection process itself wasn’t yet corrupt in 1976. So every year despite the efforts of the system, some young Jewish mathematicians would end up on the team.

Before 1976, the Olympiad was in the Eastern bloc, so the KGB wasn’t quite so concerned about having Jewish members on the team. But Austria was not only a Western country, it was also the transition point for Jewish refugees from the Soviet Union. The speed with which the IMO moved their competition to a Western country was much faster than the Soviet bureaucratic machine could build a mechanism for completely preventing Jews from joining the team.

One very strong candidate, Yura Pass, didn’t get his documents, but two other Jewish boys made it on to the team that was going to Austria. They were joking that they would be the only Soviet Jews who would go to Austria and actually come back. They did come back, only to go forward later: both are now math professors working in the US.

Because we had Jewish members on our team, it gave the KGB a special extra reason to scare us. But the biggest pressure was to win. We were told that 1976 was the most important year for the Soviet team to be the best. We were told that capitalist countries spread rumors that the judges in Eastern bloc countries favored the Soviet team and that the relative success of the Soviet team throughout the years had not been fully deserved. Now that the competition was in Austria, the suits told us, the enemies of the USSR were hoping for the downfall of the Soviet team. Our task was to prove once and for all that the Soviet students were the best at math, and that the rumors were unfounded. We had to win the team competition not only to prove ourselves, but also to clear the name of the Soviet team for all the previous years.

We did have a very strong team. The USSR came out first with 250 points, followed by the UK with 214 points and the USA with 188 points. Out of nine gold medals, we took four.

We could have gotten one more gold medal if Yura Pass had been allowed on the team. Yura was crushed by the machine’s treatment of Jews and soon afterwards quit mathematics.

My coauthor Konstantin Knop publishes cute math problems in his blog (in Russian). Recently he posted a coin weighing problem that was given at the 2010 Euler math Olympiad in Russia to eighth graders. The author of the problem is Alexander Shapovalov.

Among 100 coins exactly 4 are fake. All genuine coins weigh the same; all fake coins, too. A fake coin is lighter than a genuine coin. How would we find at least one genuine coin using two weighings on a balance scale?

It is conceivable that your two weighings may find more than one genuine coin. The more difficult question that Konstantin and his commentators discuss is the maximum number of genuine coins you can guarantee to identify in two weighings. Konstantin and the others propose 14 as the answer, but do not have a proof yet.

I wonder if one of you can find a bigger number than Konstantin or alternatively a proof that indeed 14 is the largest possible.

You might ask, considering the title of this piece, why I think that coins are scary. No, I am not afraid of coins. It scares me that this problem was given to eighth graders in Russia, because I cannot imagine that it would be given to kids that age in the USA.

By the way, ten eighth grade students in Russia solved this problem during the competition.

A week ago I chatted with my son Sergei about memorable math problems. He mentioned problem 5 from USAMO 2007. The problem can be reduced to the following:

Prove that (x⁷ + 1)/(x + 1) is composite for x = 7⁷ⁿ, where n is a non-negative integer.

Perhaps Sergei remembered this problem because as far as he knew, he was the only one in that competition to solve it. That made me curious as to how he solved it. His solution is available as solution 2 at the Art of Problem Solving website. His solution seemed mysterious and impossible to invent on the spot. I became even more curious to understand the thought process underlying his solution.

Here is his recollection:

We need to factor x⁶ − x⁵ + x⁴ − x³ + x² − x + 1. If such factoring existed it would have been known. Therefore, we need to somehow use the fact that x = 7⁷ⁿ. What is the simplest way to factor? We should try to represent the polynomial in question as the difference of squares. Luckily, x is an odd power of 7. We can make it a square if we multiply or divide it by 7 or another odd power of 7. So with a supply of squares on one side, we need to find a match for one of them to build the difference.

Let us simplify the problem and see what happens for (y³ + 1)/(y + 1) for y = 3³ⁿ, when n = 1. In other words we want to represent 703 as a difference of squares. This can be done: 703 = 28² − 9². Now let us see how we can express 28² and 9² through y which in this case is equal to 27. The first term is (y + 1)², and the second is 3y.

Now let’s go back to 7 and x, and check whether (x + 1)⁶ − (x⁶ − x⁵ + x⁴ − x³ + x² − x + 1) is 7x. Oops, no. The difference is 7x⁵ + 14x⁴ + 21x³ + 14x² +7x. On the plus side, it is divisible by 7x which we know is a square. The leftover factor is x⁴ + 2x³ + 3x² +2x + 1, which is a square of x² +x + 1.

The problem is solved, but the mystery remains. The problem can’t be generalized to numbers other than 3 and 7.

I decided to take a closer look at the Putnam Competition. I analyzed the results of the three top contenders for the best Putnam teams: Harvard, MIT, Princeton. I looked at the annual number of Putnam Fellows from each of these three schools starting from 1994.

As you may notice MIT couldn’t even generate a Putnam Fellow until 2000, but starting from 2003 MIT consistently had more Putnam Fellows than Harvard or Princeton.

Richard Stanley, the coach of the MIT team, kindly sent me the statistics for the most recent competition, held in 2009.

Furthermore, MIT had 40% in the top 5, 33% in the top 15, 32% in the top 25, and 35% in the top 81. For comparison, in the top 81, MIT had 28 winners — more than the next three schools together: Caltech 11, Harvard 9, Princeton 7.

No comment.

When the Abel Prize was announced in 2001, I got very excited and started wondering who would be the first person to get it. I asked my friends and colleagues who they thought was the greatest mathematician alive. I got the same answer from every person I asked: Alexander Grothendieck. Well, Alexander Grothendieck is not the easiest kind of person to give a prize to, since he rejected the mathematical community and lives in seclusion.

Years later I told this story to my friend Ingrid Daubechies. She pointed out to me that my spontaneous poll was extremely biased. Indeed, I was asking only Russian mathematicians living abroad who belonged to “Gelfand’s school.” Even so, the unanimity of those responses continues to amaze me.

Now several years have passed and it does not seem that Alexander Grothendieck will be awarded the Prize. Sadly, my advisor Israel Gelfand died without getting the Prize either. I am sure I am biased with respect to Gelfand. He was extremely famous in Soviet Russia, although less well-known outside, which may have affected the decision of the Abel’s committee.

I decided to assign some non-subjective numbers to the fame of Gelfand and Grothendieck. On Pi Day, March 14, 2010, I checked the number of Google hits for these two men. All the Google hits in the rest of this essay were obtained on the same day, using only the full names inside quotation marks.

• Alexander Grothendieck — 95,600
• Israel Gelfand — 47,900

Google hits do not give us a scientific measurement. If the name is very common, the results will be inflated because they will include hits on other people. On the other hand, if a person has different spellings of their name, the results may be diminished. Also, people who worked in countries with a different alphabet are at a big disadvantage. I tried the Google hits for the complete Russian spelling of Gelfand: “Израиль Моисеевич Гельфанд” and got an impressive 137,000.

Now I want to compare these numbers to the Abel Prize winners’ hits. Here we have another problem. As soon as a person gets a prize, s/he becomes more famous and the number of hits increases. It would be interesting to collect the hits before the prize winner is announced and then to compare that number to the results after the prize announcement and see how much it increases. For this endeavor, the researcher needs to know who the winner is in advance or to collect the data for all the likely candidates.

• Jean-Pierre Serre — 63,400
• Michael Atiyah — 34,200
• Isadore Singer — 44,300
• Peter Lax — 118,000
• Lennart Carleson — 47,500
• Srinivasa Varadhan — 15,800
• John Thompson — 1,610,000
• Jacques Tits — 90,900
• Mikhail Gromov — 61,900

John Thompson is way beyond everyone else’s range because he shares his name with a famous basketball coach. But my point is that Gelfand and Grothendieck could have been perfect additions to this list.

I have this fun book at home written by Clifford Pickover and titled Wonders of Numbers: Adventures in Mathematics, Mind, and Meaning. It was published before the first Abel Prize was awarded. Chapter 38 of this book is called “A Ranking of the 10 Most Influential Mathematicians Alive Today.” The chapter is based on surveys and interviews with mathematicians.

The most puzzling thing about this list is that there is no overlap with the Abel Prize winners. Here is the list with the corresponding Google hits.

1. Andrew Wiles — 64,900
2. Donald Coxeter — 25,200
3. Roger Penrose — 214,000
4. Edward Witten — 45,700
5. William Thurston — 96,000
6. Stephen Smale — 151,000
7. Robert Langlands — 48,700
8. Michael Freedman — 46,200
9. John Conway — 203,000
10. Alexander Grothendieck — 95,600

Since there are other great mathematicians with a lot of hits, I started trying random names. In the list below, I didn’t include mathematicians who had someone else appear on the first results page of my search. For example, there exists a film director named Richard Stanley. So here are my relatively “clean” results.

• Martin Gardner — 292,000
• Ingrid Daubechies — 76,900
• Timothy Gowers — 90,500
• Persi Diaconis — 84,700
• Michael Sipser — 103,000
• James Harris Simons — 107,000
• Elliott Lieb — 86,100

If prizes were awarded by hits, even when the search is polluted by other people with the same name, then the first five to receive them would have been:

1. John Thompson — 1,610,000
2. Martin Gardner — 292,000
3. Roger Penrose — 214,000
4. John Conway — 203,000
5. Stephen Smale — 151,000

If we had included other languages, then Gelfand might have made the top five with his 48,000 English-language hits plus 137,000 Russian hits.

This may not be the most scientific way to select the greatest living mathematician. That’s why I’m asking you to tell me, in the comments section, who you would vote for.