Tanya Khovanova's Math Blog

You know that the negation of a true statement is a false statement, and the negation of a false statement is a true statement. You also know that you can negate a sentence by preceding it with “It is not true that ….”

Now look at the following statement and its negation, invented by David Bernstein. Which one is true?

• This sentence contains five words.
• It is not true that this sentence contains five words.

How about this pair?

• This sentence contains ten words.
• It is not true that this sentence contains ten words.

by Tanya Khovanova and Alexey Radul

Sleeping Beauty participates in the following experiment. On Sunday she is put to sleep, and a fair coin is flipped. Regardless of the result of the coin flip, she is awakened on Monday and is offered a bet. She may pay $550 in which case she will get $1000 if the coin was tails. If the coin was tails, she is put back to sleep with her memory erased, and awakened on Tuesday and given the same bet again. She knows the protocol. Should she take the bet?

As we discussed in our first essay about Sleeping Beauty, she should take the bet. Indeed, if the coin was heads her loss is $550. But if the coin was tails her gain is $900.

To tell you the truth, when Beauty is offered the bet, she dreams: “It would be nice to know the day of the week. If it were Tuesday, then the coin must have been tails and I would gladly take the winning bet.”

In our next variation of the riddle her dream comes true.

Every time she is awakened she is offered to buy the knowledge of the day of the week. How much should she be willing to pay to know the day of the week?

Sleeping Beauty participates in the following experiment. On Sunday she is put to sleep, and a fair coin is flipped. Regardless of the result of the coin flip, she is awakened on Monday and asked whether she thinks the coin was heads or not. If the coin was tails, however, then she is put back to sleep with her memory erased, and awakened on Tuesday and asked the same question again. She knows the protocol. She is awakened one morning and instead of the expected questions she is offered a bet. She may pay $600 in which case she will get $1000 if the coin was tails. Should she take the bet?

by David Wilson

We know that tripling the triangular number 1 yields the triangular number 3. The figure shows how we can use this fact to conclude that tripling the triangular number 15 yields the triangular number 45.

Using this new fact, can you modify the figure to find even larger examples of tripling triangles?

by Tanya Khovanova and Alexey Radul

This post is inspired by the following problem:

Sleeping Beauty participates in the following experiment. On Sunday she is put to sleep, and a fair coin is flipped. Regardless of the result of the coin flip, she is awakened on Monday and asked whether she thinks the coin was heads or not. If the coin was tails, however, then she is put back to sleep with her memory erased, and awakened on Tuesday and asked the same question again. She knows the protocol. She is awakened one morning: What is her probability that the coin was heads?

Some people argue: asleep or awake, the probability of a fair coin being heads is one half, so her probability should be one half.

Other people, including us, argue that those people didn’t study conditional probability. On the information of the setup to the problem and the information of having awakened, the three situations “Coin was heads and it is Monday”, “Coin was tails and it is Monday”, and “Coin was tails and it is Tuesday” are symmetric and therefore equiprobable; thus the probability that the coin was tails is, on this information, two thirds.

So who is right? We are, of course. A good way to visualize probability judgements is to turn them into bets. Suppose each time Beauty wakes up she is offered the following bet: She pays $600 and gets $1000 if the coin was tails. Should she take it? If her probability of the coin being tails were one half, then obviously not; if her probability of the coin being tails were two thirds, obviously yes. So which is it? Consider the situation from her perspective as of Sunday. She can either always take this bet or always refuse it. If she always refuses, she gets nothing. If she always accepts: If the coin turns up heads, she will be asked the question once and will lose $600. If the coin turns up tails, she will be asked the question twice and will gain $800. So on average she will win, so she should take the bet. By this thought experiment, her probability of tails is clearly not one half.

To make matters more interesting, let’s try another bet. Suppose she is given the above bet just once, in advance, on Sunday. She pays $600, and she gets paid $1000 on Wednesday if the coin was tails. This has nothing to do with sleeping and awakening. If she takes the bet she loses $600 with probability one half and gains $400 otherwise. So she shouldn’t take the bet. Her probability on Sunday that the coin will come up heads is, of course, one half. The point is that just as these two bets are different bets, the sets of information Beauty has on Sunday vs at awakening are different, and lead to different conclusions. On Sunday she knows that the next time she wakes up it will be Monday, but when she then wakes up, she doesn’t know that it’s Monday.

Parting thought: The phenomenon of predictably losing information leads to the phenomenon of predictably changing one’s assessments. Suppose for some reason she decided to take that unprofitable bet on Sunday. When she wakes up during the experiment, should she feel happy or sad? From her perspective during the experiment, the odds of gaining $400 vs losing $600 are two to one, so she should be happy. Given that she knows on Sunday how she will (with complete certainty!) feel about this bet on Monday, should she take it, even given her Sunday self’s assessment that it’s a bad bet?

My e-friend and coauthor, Konstantin Knop, designed the following problem for the 2011 All-Russia Olympiad:

Some cells of a 100 by 100 board have one chip placed on them. We call a cell pretty if it has an even number of neighboring cells with chips. Neighbors are the cells that share a side. Is it possible for exactly one cell to be pretty?

The problem is not easy. Only one person at the Olympiad received full credit for it.

by Gregory Marton

I recently had the following chat with a particular calculator:

• e^(e^(e^(e^e))) = 10^(10^(10^6.219196780089781))
• e^(e^(e^(e^(e^e)))) = 10^(10^(10^(10^6.219196780089781)))
• e^(e^(e^(e^(e^(e^(e^(e^(e^e))))))) = 10^(10^(10^(10^(10^(10^(10^(10^6.219196780089781)))))))

It seems odd to me that putting a few more e’s down the bottom should result in it thinking there were the same number of extra 10s at the bottom. In fact, I’ve never seen a calculator answer in this form at all. I’m especially intrigued that the final power of ten seems to be the same in all three cases, so it can’t even just be estimating. Do you have any thoughts on what screwy counting could be behind these particular answers?

I met Leon Vaserstein at a party. What do you think I do at parties? I bug people for their favorite problems, of course. The first riddle Leon gave me is a variation on a famous problem I had already written about. Here’s his version:

The hypotenuse of a right triangle is 10 inches, and one of the altitudes is 6 inches. What is the area?

When Leon told me that he had designed some problems for the Soviet Olympiads, naturally I wanted to hear his favorite:

A closed polygonal chain has its vertices on the vertices of a square grid and all the segments are the same length. Prove that the number of segments is even.

I wonder what the largest number is that can be represented with one character. Probably 9. How about two characters? Is it 9⁹? What about three or four?

I guess I should define a character. Let’s have two separate cases. In
the first one you can only use keyboard characters. In the second one
you can use any Unicode characters.

I’m awaiting your answers to this.

The Moscow Math Olympiad has a different set of problems for every grade. Students need to write a proof for every problem. These are the 8th grade problems from this year’s Olympiad:

Problem 1. There were 6 seemingly identical balls lying at the vertices of the hexagon ABCDEF: at A — with a mass of 1 gram, at B — with a mass of 2 grams, …, at F — with a mass of 6 grams. A hacker switched two balls that were at opposite vertices of the hexagon. There is a balance scale that allows you to say in which pan the weight of the balls is greater. How can you decide which pair of balls was switched, using the scale just once?

Problem 2. Peter was born in the 19th century, while his brother Paul was born in the 20th. Once the brothers met at a party celebrating both birthdays. Peter said, “My age is equal to the sum of the digits of my birth year.” “Mine too,” replied Paul. By how many years is Paul younger than Peter?

Problem 3. Does there exist a hexagon which can be divided into four congruent triangles by a single line?

Problem 4. Every straight segment of a non-self-intersecting path contains an odd number of sides of cells of a 100 by 100 square grid. Any two consecutive segments are perpendicular to each other. Can the path pass through all the grid vertices inside and on the border of the square?

Problem 5. Denote the midpoints of the non-parallel sides AB and CD of the trapezoid ABCD by M and N respectively. The perpendicular from the point M to the diagonal AC and the perpendicular from the point N to the diagonal BD intersect at the point P. Prove that PA = PD.

Problem 6. Each cell in a square table contains a number. The sum of the two greatest numbers in each row is a, and the sum of the two greatest numbers in each column is b. Prove that a = b.