Tanya Khovanova's Math Blog

Why are manhole covers round? The manhole covers are round because the manholes are round. Duh! But the cute mathematical answer is that the round shapes are better than many other shapes because a round cover can’t fall into a round hole. If we assume that the hole is the same shape as the cover but slightly smaller, then it is true that circular covers can’t fall into their holes. But there are many other shapes with this property. They are called the shapes of constant width.

Given the width, the shape with the largest area is not surprisingly a circle. The shape with the smallest area and a given constant width is a Reuleaux Triangle. Here is how to draw a Reuleaux triangle. Draw three points that are equidistant from each other at distance d. Then draw three circles of radius d with the centers at given points. The Reuleaux triangle is the intersection of these three circles.

Can we generalize this to 3d? What would be an analogue of a Reuleaux Triangle in 3d? Of course, it is a Reuleaux Tetrahedron: Take four points at the vertices of a regular tetrahedron; take a sphere at each vertex with the radius equal to the edge of the tetrahedron; intersect the four spheres.

Is this a shape of the constant width? Many people mistakenly think that this is the case. Indeed, if you squeeze the Reuleaux tetrahedron between two planes, one of which touches a vertex and another touches the opposite face of the curvy tetrahedron, then the distance between them is equal to d: the radius of the circle. This might give you the impression that this distance is always d. Not so. If you squeeze the Reuleaux tetrahedron between two planes that touch the opposite curvy edges, the distance between these planes will be slightly more than d. To create a shape of constant width you need to shave off the edges a bit.

Theoretically you can shave the same amount off every edge to get to a surface of constant width. But this is not the cool way to do it. The cool way is to shave a bit more but only from one edge of the pair of opposite edges. You can get two different figures this way: one that has three shaved edges forming a triangle, and the other, where three shaved edges share a vertex. These two bodies are called Meissner bodies and they are conjectured to be shapes of the constant width with the smallest volume.

On the picture I have two copies of a pair of Meissner bodies. The two left ones have three edges that share a vertex shaved off. The very left shape gives a top view of this vertex and the solid next to it has its bottom with holes looking forward. The two shapes on the right show the second Meissner body in two different positions.

I recently discovered a TED-Ed video about manhole covers. It falsely claims that the Reuleaux tetrahedron has constant width. I wrote to TED-Ed, to the author, and posted a comment on the discussion page. There was no reaction. They either should remove the video or have an errata page for it. Knowingly keeping a video with an error that is being viewed by thousands of people is irresponsible.

I am running a PRIMES-STEP program for middle school students, where we try to do research in mathematics. In the fall of 2015 we decided to study the following topic in logic.

Suppose there is an island where the following four types of people live:

• Absolute truth-tellers always tell the truth.
• Partial truth-tellers tell the truth with one exception: if they are guilty, they will say that they are innocent.
• Absolute liars always lie.
• Responsible liars lie with one exception: if they are guilty, they will say that they are guilty.

See if you can solve this simple logic puzzle about people on this island.

It is known that exactly one person stole an expensive painting from an apartment. It is also known that only Alice or Bob could have done it. Here are their statements:
Alice: I am guilty. Bob is a truth-teller.
Bob: I am guilty. Alice stole it. Alice is the same type as me.
Who stole the painting and what types are Alice and Bob?

My students and I discovered a lot of interesting things about these four types of people and wrote a paper: Who Is Guilty?. This paper contains 11 cute logic puzzles designed by each of my 11 students.

I envied my students and decided to create two puzzles of my own. You have already solved the one above, so here is another, more difficult, puzzle:

A bank was robbed and a witness said that there was exactly one person who committed the robbery. Three suspects were apprehended. No one else could have participated in the robbery.
Alice: I am innocent. Bob committed the crime. Bob is a truth-teller.
Bob: I am innocent. Alice is guilty. Carol is a different type than me.
Carol: I am innocent. Alice is guilty.
Who robbed the bank and what types are the suspects?

I have solved too many puzzles in my life. When I see a new one, my solution is always the intended one. But my students invent other ideas from time to time and teach me to think creatively. For example, I gave them this puzzle:

Two boys wish to cross a river, but there is a single boat that can take only one boy at a time. The boat cannot return on its own; there are no ropes or similar tricks; yet both boys manage to cross the river. How?

Here is what my inventive students came up with:

• There was another person on the other side of the river who brought the boat back.
• There was a bridge.
• The boys can swim.
• They just wanted to cross the river and come back, so they did it in turns.

And here is my standard solution: They started on different sides of the river.

I gave a talk about thinking inside and outside the box at the Gathering for Gardner conference. I mentioned this puzzle and the inventiveness of my students. After the conference a guy approached me with another answer which is now my favorite:

• They wait until the river freezes over and walk to the other side.

I want the world to be a better place. My contribution is teaching people to think. People who think make better decisions, whether they want to buy a house or vote for a president.

When I started my blog, I posted a lot of puzzles. I was passionate about not posting solutions. I do want people to think, not just consume puzzles. Regrettably, I feel a great push to post solutions. My readers ask about the answers, because they are accustomed to the other websites providing them.

I remember I once bought a metal brainteaser that needed untangling. The solution wasn’t included. Instead, there was a postcard that I needed to sign and send to get a solution. The text that needed my signature was, “I am an idiot. I can’t solve this puzzle.” I struggled with the puzzle for a while, but there was no way I would have signed such a postcard, so I solved it. In a long run I am glad that the brainteaser didn’t provide a solution.

That was a long time ago. Now I can just go on YouTube, where people post solutions to all possible puzzles.

I am not the only one who tries to encourage people to solve puzzles for themselves. Many Internet puzzle pages do not have solutions on the same page as the puzzle. They have a link to a solution. Although it is easy to access the solution, this separation between the puzzle and its solution is an encouragement to think first.

But the times are changing. My biggest newest disappointment is TED-Ed. They have videos with all my favorite puzzles, where you do not need to click to get to the solution. You need to click to STOP the solution from being fed to you. Their video Can you solve the prisoner hat riddle? uses my favorite hat puzzle. (To my knowledge, this puzzle first appeared at the 23-rd All-Russian Mathematical Olympiad in 1997.) Here is the standard version that I like:

A sultan decides to give 100 of his sages a test. He has the sages stand in line, one behind the other, so that the last person in line can see everyone else. The sultan puts either a black or a white hat on each sage. The sages can only see the colors of the hats on all the people in front of them. Then, in any order they want, each sage guesses the color of the hat on his own head. Each hears all previously made guesses, but other than that, the sages cannot speak. Each person who guesses the color wrong will have his head chopped off. The ones who guess correctly go free. The rules of the test are given to them one day before the test, at which point they have a chance to agree on a strategy that will minimize the number of people who die during this test. What should that strategy be?

The video is beautifully done, but sadly the puzzle is dumbed down in two ways. First, they explicitly say that it is possible for all but one person to guess the color and second, that people should start talking from the back of the line. I remember in the past I would give this puzzle to my students and they would initially estimate that half of the people would die. Their eyes would light up when they realized that it’s possible to save way more than half the people. They have another aha! moment when they discover that the sages should start talking from the back to the front. This way each person sees or hears everyone else before announcing their own color. Finally, my students would think about parity, and voilà, they would solve the puzzle.

In the simplified adapted video, there are no longer any discoveries. There is no joy. People consume the solution, without realizing why this puzzle is beautiful and counterintuitive.

Nowadays, I come to class and give a puzzle, but everyone has already heard the puzzle with its solution. How can I train my students to think?

One of my jobs is giving linear algebra recitations at MIT. The most unpleasant aspect of it is dealing with late homework. Students attempt to submit their homework late for lots of different reasons: a sick parent needing help, stress, a performance at Carnegie Hall, a broken printer, flu, and so on. How do I decide which excuse is sufficient, and which is not? I do not want to be a judge! Moreover, my assumption is that people tell the truth. In the case of linear algebra homework, this assumption is unwise. As soon as students discover that I trust everyone, there’s a sharp increase in the number of sick parents and broken printers. So I have the choice of being either a naive idiot or a suspicious cynic. I do not like either role.

Our linear algebra course often adopts a brilliant approach: We announce that late homework is not allowed for any reason. To compensate for emergencies, we drop everyone’s lowest score. That is, we allow the students to skip one homework out of ten. They are free to use this option for oversleeping the 4 pm deadline. If all their printers work properly and they do not get so sick that they have to skip their homework, they can forgo the last homework. Happily, this relieves me from being a judge.

Or does it? Unfortunately, this policy doesn’t completely resolve the problem. Some students continue trying to push their late homework on me, despite the rules. In order to be fair to other students and to follow the rules, I reject all late homework. The students who badger me nonetheless waste my time and drain my emotions. This is very unpleasant.

From the point of view of those students, such behavior makes sense. They have nothing to lose and they might get some points. There is no way to punish a person who tries to hand in homework late and from time to time they stumble onto a soft instructor who accepts the homework against the official rules. Because such behavior is occasionally rewarded, they continue doing it. I believe that wrong behavior shouldn’t be rewarded. As a responsible adult, I think it is my duty to counteract the rewards of this behavior. But I do not know how.

What should I do? Maybe I should…

1. Persuade our course leader, or MIT in general, to introduce a punishment for trying to hand in late homework. We might, for example, subtract points from their final score.
2. Promote the value of honorably following the rules through discussions with the students.
3. Growl at any student who submits their homework late.
4. Explain to them what other people think about them, when they do not follow the rules.

Maybe I should just do number 4 right now and explain what I feel. A student who persists in handing in homework late feels to me that s/he is entitled and is better than everyone else, and shows that s/he doesn’t care about the rules and honor. Again, this wastes my time, puts me in a disagreeable position, and reduces my respect for that student.

Earlier I suggested that students don’t have anything to lose by such behavior, but in fact, they do pay a price, even though they may not understand that.

I already wrote how I build a magic SET hypercube with my students. Every time I do it, I can always come up with a new question for my students. This time I decided to flip over two random cards, as in the picture. My students already know that any two cards can be completed to a set. The goal of this activity is to find the third card in the set without trying to figure out what the flipped-over cards are. Where is the third card in the hypercube?

Sometimes my students figure this out without having an explicit rule. Somehow they intuit it before they know it. But after several tries, they discover the rule. What is the rule?

Another set of questions that I ask my students is related to magic SET squares that are formed by 3 by 3 regions in the hypercube. By definition, each magic SET square has every row, column, and diagonal as a set. But there are four more sets inside a magic SET square. We can call them super and sub-diagonal (anti-diagonal) wrap-arounds. Can you prove that every magic SET square has to have these extra four sets? In addition, can you prove that a magic SET square is always uniquely defined by any three cards that do not form a set, and which are put into places that are not supposed to from a set?

I explained to my AMSA students the idea of meta-solving multiple choice. Sometimes by looking at the choices without knowing what the problem is, it is possible to guess the correct answer. Suppose your choices are 2k, 2, 2/k, 10k, −2k. What is the most probable answer? There are several ideas to consider.

• A problem designer considers different potential mistakes and tries to include the answers corresponding to most common mistakes. That means the answers corresponding to the mistakes are variations of the right answer. Thus, the right answer is a common theme in all the answers.
• A problem designer tries not to allow the students to bypass the solution. So if only one of the choices is negative and the answer is negative, the students do not need to calculate the exact answer, they just need to see that it’s negative. That means the right answer cannot be the odd one out.

Both these considerations suggest a rule of thumb: the answers that are odd ones out are probably wrong. In the given example (2k, 2, 2/k, 10k, —2k), the second choice is an odd one out because it’s the only one that does not contain k. The third choice is an odd one out because it’s the only one in which we divide by k. The fourth choice is an odd one out because it’s the only one with 10 instead of 2. The last one is an odd one out because of the minus sign. Thus the most probable answer is 2k.

So I explained these ideas to my students and gave them a quiz, in which I took the 2003 AMC 10A test, but only gave them the choices without the problems. I was hoping they would do better than randomly guessing.

Luckily for me, I have six classes in a row doing the same thing, so I can make adjustments as I go along. Looking at the results of the first two groups of students, I discovered that they were worse than random. What was going on?

I took a closer look, and what do you know? Nobody marked the first or the last choice. The answers are in an increasing order, so the first is the smallest and the last is the largest. So these two numbers are odd ones out, in a way.

It is a good idea to consider 189 as an odd one out in the list 1, 2, 4, 5, 189. In many other cases, the fact that the number is either the smallest or largest is insufficient reason to consider it as the odd one out. For example, there is no reason to consider 1 to be an odd one out in the list 1, 2, 3, 4, 5. And the designers of AMC are good: a lot of problems have an arithmetic progression as a list of choices, where none of the numbers are obviously odd ones out.

To correct the situation of worse than random results, I discussed it with my students in my next classes. Problem designers cannot have a tradition in which the first answer is never the correct answer. If such a tradition existed, and people knew about it, that knowledge would help them guess. So the first answer should be correct approximately five times, which is a fifth of the total number of questions (25).

And we came up with a strategy. Use the odd one out method except for arithmetic progressions. Then add the choices to balance out the total number of the first answers, the second answers, etc.

That method worked. In my next four classes my students were better than random.

My post with Kvantik’s 2012 problems for middle school was a success. So I scanned the 2013 issues and found 7 more problems that I liked. Here are two cute problems I’ve seen before:

Problem 1. In the equation 30 − 33 = 3 move one digit to make it correct.

Problem 2. A patient got two pills for his headache and two pills for his cough. He was supposed to use one of each type of pill today and do the same tomorrow. The pills all looked the same. By mistake, the patient mixed up the pills. How should he use them so that he follows the prescription exactly?

Now logic and information-theoretical problems:

Problem 3. One strange boy always tells the truth on Wednesdays and Fridays and always lies on Tuesday. On other days of the week he can lie or tell the truth. He was asked his name seven days in a row. The first six answers in order are Eugene, Boris, Vasiliy, Vasiliy, Pete, Boris. What was his answer on the seventh day?

Problem 4. Ten people are suspected of a crime. It is known that two of them are guilty and eight are innocent. Suspects are shown to a psychic in groups of three. If there is a guilty person in the group the psychic points him out. If there are two guilty people in the group, the psychic points to one of them. If all of them are innocent, the psychic points at one of the three.

1. How would you find at least one guilty person in four séances?
2. How would you find both criminals in six séances?

Problem 5. There are four balloons: red, blue, green and black. Some of the balloons might be magical. There is also a detector box that can say how many balloons out of the ones put inside are magical. How can you find all the magical balloons using the detector box not more than three times?

I conclude with two miscellaneous problems.

Problem 6. Three runners started their loops at the same time at the same place on the same track. After some time they ended at their starting point together. The fastest runner passed the slowest runner 23 times. Assuming each runner has a constant speed, how many times was one runner passed by another runner in total?

Problem 7. Given a point inside a circular disk, cut the disk into two parts so that you can put them back together into a new disk such that the given point is the center.

Kvant was a very popular science magazine in Soviet Russia. It was targeted to high-school children and I was a subscriber. Recently I discovered that a new magazine appeared in Russia. It is called Kvantik, which means Little Kvant. It is a science magazine for middle-school children. The previous years’ archives are available online in Russian. I looked at 2012, the first publication year, and loved it. Here is the list of the math puzzles that caught my attention.

The first three problems are well known, but I still like them.

Problem 1. There are 6 glasses on the table in a row. The first three are empty, and the last three are filled with water. How can you make it so that the empty and full glasses alternate, if you are allowed to touch only one of the glasses? (You can’t push one glass with another.)

Problem 2. If it is raining at midnight, with what probability will there be sunshine in 144 hours?

Problem 3. How can you fill a cylindrical pan exactly half-full of water?

I like logic puzzles, and the next two seem especially cute. I like the Parrot character who repeats the previous answer: very appropriate.

Problem 4. The Jackal always lies; the Lion always tells the truth. The Parrot repeats the previous answer—unless he is the first to answer, in which case he babbles randomly. The Giraffe replies truthfully, but to the previous question directed to him—his first answer he chooses randomly.
The Wise Hedgehog in the fog stumbled upon the Jackal, the Lion, the Parrot, and the Giraffe, although the fog prevented him from seeing them clearly. He decided to figure out the order in which they were standing. After he asked everyone in order, “Are you the Jackal?” he was only able to figure out where the Giraffe was. After that he asked everyone, “Are you the Giraffe?” in the same order, and figured out where the Jackal was. But he still didn’t have the full picture. He started the next round of questions, asking everyone, “Are you the Parrot?” After the first one answered “Yes”, the Hedgehog understood the order. What is the order?

Problem 5. There are 12 cards with the statements “There is exactly one false statement to the left of me,” “There are exactly two false statements to the left of me.” …, “There are 12 false statements to the left of me.” Pete put the cards in a row from left to right in some order. What is the largest number of statements that might be true?

The next three problems are a mixture of puzzles.

Problem 6. Olga Smirnov has exactly one brother, Mikhail, and one sister, Sveta. How many children are there in the Smirnov family?

Problem 7. Every next digit of number N is strictly greater than the previous one. What is the sum of the digits of 9N?

Problem 8. Nine gnomes stood in the cells of a three-by-three square. The gnomes who were in neighboring cells greeted each other. Then they re-arranged themselves in the square, and greeted each other again. They did this one more time. Prove that there is at least one pair of gnomes who didn’t get a chance to greet each other.

I was afraid of my advisor Israel Gelfand. He used to place unrealistic demands on me. After each seminar he would ask his students to prove by the next week any open problems mentioned by the speaker. So I got used to ignoring his requests.

He also had an idea that it is good to learn mathematics through problem solving. So he asked different mathematicians to compile a list of math problems that are important for undergraduate students to think through and solve by themselves. I still have several lists of these problems.

Here I would like to post the list by Andrei Zelevinsky. This is my favorite list, partially because it is the shortest one. Andrei was a combinatorialist, and it is surprising that the problems he chose are not combinatorics problems at all. This list was compiled many years ago, but I think it is still useful, just keep in mind that by calculating, he meant calculating by hand.

Problem 1. Let G be a finite group of order |G|. Let H be its subgroup, such that the index (G:H) is the smallest prime factor of |G|. Prove that H is a normal subgroup.

Problem 2. Consider a procedure: Given a polygon in a plane, the next polygon is formed by the centers of its edges. Prove that if we start with a polygon and perform the procedure infinitely many times, the resulting polygon will converge to a point. In the next variation, instead of using the centers of edges to construct the next polygon, use the centers of gravity of k consecutive vertices.

Problem 3. Find numbers a_n such that 1 + 1/2 + 1/3 + … + 1/k = ln k + γ + a₁/k + … + a_n/kⁿ + …

Problem 4. Let x₁ not equal to zero, and x_k = sin x_k-1. Find the asymptotic behavior of x_k.

Problem 5. Calculate the integral from 0 to 1 of x^−x over x with the precision 0.001.

I regret that I ignored Gelfand’s request and didn’t even try to solve these problems back then.

I didn’t have any photo of Andrei, so his widow, Galina, sent me one. This is how I remember him.