Tanya Khovanova's Math Blog

Although I cut most carbs from my diet, I still wasn’t losing weight. I started my weight-loss journey three years ago when I was 245 pounds. My initial excitement helped me get to 220, but then I started slowly gaining it back to 235 pounds. A couple of friends of mine who lost a lot of weight explained to me that I didn’t actually cut all the carbs from my diet. I eliminated pizza, potatoes, spaghetti, sugar, cookies, soda, and so on. But there are many carbs hidden in some very natural and healthy foods. For example, a banana has 27 grams of carbs.

I love bananas; I can live on bananas alone. But my friends convinced me to cut the bananas, too. I agreed, but in that week I gained five more pounds. I think that every time I remove something from my diet, I end up becoming more relaxed with other food.

Clearly, carbs-cutting doesn’t work. I am back to 240 pounds. Time for plan B.

The first Moscow Math Olympiad was conducted in 1935. Today, eighty years later, I decided to check it out. Most of the problems look standard, but some of the stereometry problems look too complicated. I found four problems that I really like: all of them are geometry problems.

Problem 1. The lengths of the sides of a triangle form an arithmetic progression. Prove that the radius of the inscribed circle is one third of one of the heights of the triangle.

Problem 2. A median, bisector, and height all originate from the same vertex of a triangle. You are given the three points that are the intersection points of the aforementioned median, bisector, and height with the circumscribed circle. Construct the triangle.

Problem 3. Find the set of points P on the surface of a cube such that the main diagonal subtends the smallest possible angle if viewed from P. Prove that the main diagonal subtends larger angles if viewed from other points on the surface. [Clarification: the two corners the main diagonal passes through don’t count.]

Problem 4. Given three parallel straight lines, construct a square such that three of its vertices belong to these lines.

Each of these problems has a powerful idea that solves it. You can try and solve these problems, but if you want help, the ideas are presented below as hints in a scrambled order.

• Hint. Rotate by 90 degrees.
• Hint. Consider a circumscribed sphere.
• Hint. The line connecting the intersection point of the bisector with the circle and the circle’s center is parallel to the height.
• Hint. Use Heron’s formula.

As a child I felt that society expected me to grow up and be a mother. And only to be a mother. It didn’t make any sense to me, because if all that people do is reproduce, how will progress be made? So in my teens I decided that in addition to having children, I need to do something else, something to make the world a better place.

For many years I wasn’t too successful in my plan. My children were my priority. I believed that for the first three years they needed a lot of my attention, which they couldn’t necessarily get from a nanny. Later, being a single mother was so overwhelming that I didn’t have time for other missions.

As my children grew, I felt myself imposing my own ambitions onto them. It wasn’t fair to burden my children like that. I was meant to do something with my life other than bring up children. There was no substitute — I had to do it myself.

At that time, I was working at BAE Systems, but I hated my job. Its only purpose was to support my family. So as soon as my youngest son turned 16, I resigned to come back to mathematics. Though mathematics has my full attention now, my children are still my priority. I can afford to do what I love, because it is good for my children too. I am living my own ambitions, and they have the space to live theirs.

I was supposed to write about menopause. I am writing about menopause, just give me a second. I was also raised to believe that the best thing I can do for a man is to give him children. Every time I loved someone I wanted to have a child with that person, or I thought that I was supposed to want to. Now that I can’t conceive a child for any potential future husband, I feel relieved. I can’t be blamed any more for not having children. I am so happy that my current attempt at mathematics will not be interrupted any more.

I am so glad that I live in an age when women’s life expectancy is way past menopause. I might love a man again; I might marry. I am so glad that I have an excuse not to have children. I can do what I want to do. I am free.

* * *

—Mike, here are 10 chocolates. Give half of them to your brother.
—OK. I’ll give him three chocolates.
—You can’t count?
—I can, but he can’t.

* * *

—How is your progress?
—50%.
—Done or left to do?

* * *

—Q: What did Al Gore play on his guitar?
—A: An Algorithm.

* * *

I put my root beer in a square glass. Now it’s just beer.

* * *

—Q: Why shouldn’t you argue with a decimal?
—A: Decimals always have a point.

* * *

—I am cold.
—Go stay in the corner. It’s 90 degrees.

* * *

Arithmetic is the art of counting up to twenty without taking off your shoes.

* * *

I bought a book online “How to implement an Internet scam.” Somehow, though, it’s been a while, and I still haven’t received it.

* * *

Sex is a pathetic thrill for losers who are not able to take a triple integral.

* * *

Paradox: Less money, more need to count it.

How many letters are there in the correct answer to this question?

What is the answer to this question?

In my paper Nim Fractals written with Joshua Xiong we discovered an interesting graph structure on P-positions of impartial combinatorial games. P-positions are vertices of the graph and two vertices are connected if they are consecutive P-positions in an optimal longest game.

A longest game of Nim is played when exactly one token is removed in each turn. So in Nim two P-positions are connected if it is possible to get from one of them to the other by removing two tokens.

In the paper we discussed the evolution graph of Nim with three piles. The graph has the same structure as three branches of the Ulam-Warburton automaton.

For completeness, I would like to describe the evolution graph of the 2-pile Nim. The P-positions in a 2-pile Nim are pairs (n,n), for any integer n. Two positions (n,n) and (m,m) are connected if and only if m and n differ by 1. The first picture represents this graph.

The Wythoff’s game is more interesting. There are two piles of tokens. In one move a player can take any number of tokens from one pile or the same number of tokens from both piles.

The P-positions (n,m) such that n ≤ m start as: (0,0), (1,2), (3,5), (4,7), (6,10) and so on. They can be enumerated using φ: the golden ratio. Namely, n_k = ⌊kφ⌋ and m_k = ⌊kφ²⌋ = n_k + k, where k ≥ 0.

In a longest Wythoff’s game the difference between the coordinates decreases by 1. That is, it takes a maximum of 2k steps to end an optimal game starting from position (n_k,m_k). The picture shows the evolution graph.

The interesting part of the picture is the crossover between two “lines”. From positions with large coordinates like (6,10) with a difference of 4 you can get to only one position with a difference of 3: (4,7) and not (7,4). But from position (3,5) with a difference of 2 you can get to both positions with a difference of 1: (1,2) and (2,1).

I enjoy doing research in recreational mathematics. The biggest problem is that the probability that something I’ve done has already been done before is very high. This is partially because of the nature of this area of research. Trying to find new questions that are not buried deep in the abstract means someone else could have stumbled upon the same question. In addition, I like going in different directions: geometry, number theory, probability, combinatorics, and so on. That means I am not an expert in any of the areas. This too increases my probability of doing something that has already been done before.

This is quite unpleasant: to discover that I reinvented the wheel. And I keep reinvented different wheels on a regular basis. When I complain, my mathematician friends give me the same advice over and over:

Find your own niche!

I’ve been trying to understand what they mean, and finally I got it:

Do something no one else is interested in using methods no one else can understand.

Finding a topic that is not interesting almost guarantees that other people didn’t do it before. Developing new complicated methods helps limit the number of followers and prevents the niche from becoming crowded.

Now I know why I see so many boring math papers I can’t understand.

To hell with my own niche!

Let me start with a real story that happened a long time ago when I worked with my co-author on a math project. When I told him the idea I came up with, he dismissed it. The next day he came to me very excited with his new idea. He repeated the idea I had proposed the day before.

I said, “I suggested this exact idea to you yesterday.” He didn’t believe me. Luckily for our relationship, I knew him very well. He was not a person who lied. I remembered how preoccupied he was when I had laid out the idea to him. Now I think that he didn’t pay attention to my idea at the time, but it was lodged somewhere in his subconscious and surfaced later. I am sure that he truly believed that the idea was his. (Unfortunately, he didn’t believe me. But this is another story.)

This story made me think about the nature of collaboration and how people can’t really separate ownership of the results of a joint project. Let me tell you another story, one that I do not remember happening, although it could have.

I was working on a paper with my co-author on Sharelatex. One day an idea for a lemma came to me that we hadn’t discussed before. My co-author lives in a different time zone, so instead of discussing the idea with him, I just added Lemma 3 and its proof to our paper. I was truly proud of my own contribution.

The next day I came up with Lemma 4 while driving back home. It was another completely new direction for our research. When I came home to my laptop, I discovered my Lemma 4 neatly written and proven by my co-author. Is this lemma his? It can’t be completely his. It was a natural extension of what we discussed; he just got to a computer before me.

Did you notice in the last story that the situation is symmetric, but my hypothetical feelings are not? In a true collaboration ideas are bounced off each other. Very often people come up with the same idea at the same time, but you can’t ascribe ownership to the first person who speaks. Because of many stories like this, I made a rule for myself: never discuss in public who did what in a joint paper. Just always say “we.”

Unfortunately I keep forgetting to teach my PRIMES/RSI students that. The question usually doesn’t come up until it is too late, when one of the students in a group project during a presentation says, “I proved Theorem 5.” Oops!

Round 1 of Who Wants to Be a Mathematician had the following math problem:

Bob and Jane have three children. Given that one child is their daughter Mary, what is the probability that Bob and Jane have at least two daughters?

In all such problems we usually make some simplifying assumptions. In this case we assume that gender is binary, the probability of a child being a boy is 1/2, and that identical twins do not exist.

In addition to that, every probability problem needs to specify the distribution of events over which the probability is calculated. This problem doesn’t specify. This is a mistake and a source of confusion. In most problems like this, the assumption is that something is chosen at random. In this type of problem there are two possibilities: a family is chosen at random or a child is chosen at random. And as usual, different choices produce different answers.

The puzzle above is not well-defined, even though this is from a contest run by the American Mathematical Society!

Here are two well-defined versions corresponding to two choices in randomization:

Bob and Jane is a couple picked randomly from couples with three children and at least one daughter. What is the probability that Bob and Jane have at least two daughters?

Mary is a girl picked randomly from a pool of children from families with three children. What is the probability that Mary’s family has at least two daughters?

Now, if you don’t mind, I’m going to throw in my own two cents, that is to say, my own two puzzles.

Harvard researchers study the influence of identical twins on other siblings. For this study they invited random couples with three children, where two of the children are identical twins.

1. Bob and Jane is a couple picked randomly from couples in the study with at least one daughter. What is the probability that Bob and Jane have at least two daughters?
2. Mary is a girl picked randomly from a pool of children participating in the study. What is the probability that Mary’s family has at least two daughters?