Tanya Khovanova's Math Blog

I recently posted two trick questions.

First question. What is the answer to this question?

The title to this post was the same as the content except for the question mark: What is the Answer to This Question. The title contained the answer: WHAT.

What I like about trick questions is that sometimes people produce alternative answers that are as good as the intended ones. For this problem I like the following answers:

• If you have a compiler which converts recursion to loops, then infinite loop. Otherwise, stack overflow. (by aphar)
• Chocolate is the answer. It doesn’t matter what the question is. (by Mark James)
• 42. (by Clao)
• This is the Answer to that Question. (by Javifields)

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

The intended answer was FOUR, as four is the only number in the English language for which the number of letters in its name is equal to the number itself. Many people used variations on the theme and supplied the following answers by writing out numbers in non-canonical ways:

• Positive fifteen. (by my AMSA students)
• One plus twelve. (by Michael and MQ)
• Two plus eleven. (by MQ)
• Maybe eleven. (by Michael Albert)
• Certainly sixteen. (by Michael Albert)
• Zero plus one plus two plus three plus …. (by Bob Hearn)

Some people used sentences to express numbers:

• Any number n whose value can be expressed using n letters, for example sixty seven. (by Michael Albert)

Some other people used Roman numerals and digits to express the answer:

• I, II, or III (by my AMSA students)
• 0 (by my AMSA students, Leo, and lvps1000vm)

Many people pointed out that if the puzzle would be asked in other languages, it would produce completely different answers.

But the majority of my AMSA students took a completely different approach:

• 30.

This is because there are thirty letters in the phrase the correct answer to this question.

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.

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

What is the answer to this question?

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?

The puzzle Family Ties was written for the 2013 MIT Mystery Hunt, but it never made it to the hunt. Here’s your chance to solve a puzzle no one has seen before. I wrote the puzzle jointly with Adam Hesterberg. The puzzle is below:

Mathematics professor S. Lee studies genealogy and is interested in the origins of life.

1. Alexei Mikhailovich Ivanov
2. Alexei Petrovich Ivanov
3. Amminadab
4. Anna of Moscow
5. Arador
6. Arahad II
7. Arassuil
8. Arathorn I
9. Arathorn II
10. Aravorn
11. Argonui
12. Asger Thomsen
13. Caecilia Metella Dalmatica
14. Egmont
15. Eldarion
16. Ellesar
17. Endeavour
18. Faustus Cornelius Sulla
19. Henry Frederick
20. Hezron
21. Isaac
22. Ivan the Great
23. Ivan the Terrible
24. Jacob
25. James I and VI
26. James V
27. Jens Knudsen
28. John Francis
29. Joseph Patrick
30. Joseph Patrick
31. Jørgen Jensen
32. Judah
33. Knud Nielsen
34. Margaret Stuart
35. Maria Donata
36. Mary Stuart
37. Matthew Rauch
38. Mikhail Ivanovich Ivanov
39. Niels Møller
40. Ole Pedersen
41. Peder Petersen
42. Peter Jørgensen
43. Petr Alexeyevich Ivanov
44. Pharez
45. Ram
46. Robert Francis
47. Rose Elizabeth
48. Søren Thomsen
49. Thomas Olsen
50. Ursula Gertrud
51. Vasily I of Moscow
52. Vasily II of Moscow
53. Vasili III of Russia
54. Yuri of Uglich
55. Zerah

I already wrote about my favorite problem from the 2015 All-Russian Math Olympiad that involved tanks. My second favorite problem is about coins. I do love almost every coin problem.

A coin collector has 100 coins that look identical. He knows that 30 of the coins are genuine and 70 fake. He also knows that all the genuine coins weigh the same and all the fake coins have different weights, and every fake coin is heavier than a genuine coin. He doesn’t know the exact weights though. He has a balance scale without weights that he can use to compare the weights of two groups with the same number of coins. What is the smallest number of weighings the collector needs to guarantee finding at least one genuine coin?

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?

Professor Bock came to his office at the Math Department of Deys University and discovered that someone had broken in. Luckily he had a lunch scheduled with his friend Detective Radstein. Bock complained to the detective about the break-in and the detective agreed to investigate.

Nothing was stolen from the office. It looked like somebody had just slept on Professor Bock’s couch. The couch was bought recently and it was positioned so that it was impossible to see it through the tiny window of the office door, which had been locked. Interestingly, Professor Bock’s office was the only one with a couch. The detective concluded that the person who broke in knew about the couch and thus was from the Math Department. In addition, the couch was very narrow and couldn’t possibly sleep more than one person.

Investigating crimes at the Math Department of Deys University was very easy. This was due to a fact discovered by Detective Radstein on a previous case: every member of the department was one of three types:

• A dodger who always tells the truth and answers the question directly, with one exception. If such a person is guilty and asked to confirm that, then s/he remains truthful, but dodges the question.
• A liar who’s every statement is a lie. A liar might dodge or not dodge the question.
• A pathological truth-teller who always answers the question directly and truthfully.

Thus solving a crime at the department is very easy. Detective Radstein just needs to ask every person two questions, “How much is 2+2?” and “Are you guilty?” Only a guilty dodger would answer “4” and dodge. Only a guilty truth-teller would answer “4” and “yes”. Only a guilty liar would answer something other than “4” and “no”.

Detective Radstein decided to enjoy himself by making this investigation more challenging. He asked everyone only one question “Are you guilty of breaking-in?” These are the first three replies:

—Albert: I did it.
—Bill: Albert didn’t do it.
—Connie: Albert did it.

It was amusing how many people knew where Albert slept, but these answers were enough for Detective Radstein to figure out the culprit. After this issue was resolved he asked Professor Bock, “While I am here, what other problems do you have at the department?”

“Well,” said the professor, “In fact, I would be grateful if you could help us resolve two more issues. This semester our expenditure for tea increased three times. It is clear someone is stealing tea. It’s also clear that this could only be someone who started working at the department this semester. There are two people who fit this description: David and Eve. So Detective Radstein asked each of them, “Are you stealing the tea?” He got the following answers:

—David: Eve steals the tea.
—Eve: Only one person steals the tea.

Having solved the Math Department’s tea problem Detective Radstein then asked Professor Bock, “What else?” “Well,” said Professor Bock, “one more thing. We have a lot of blackboards around the department. Every day a famous three-letter Russian curse word appears on the blackboards. The handwriting is always the same, so it is one person. Luckily the word starts with XY, so most people assume that this is some math formula. Anyway, I want to look into the eyes of the person who does this. I left the department late yesterday; only Fedor, Grisha and Harry were here. The blackboards were clean. This morning when I opened up the Math Department, the vulgar swearword had been written on the blackboards. I don’t suspect someone of sneaking into the department at night to scribble on our blackboards. I am certain it is one of the three people I mentioned.”

Once again Detective Radstein asked the suspects whether they did it.

—Fedor: Grisha did it.
—Grisha: Fedor did it.
—Harry: I do not speak Russian.

And again Detective Radstein solved the case. He was surprised that everyone in the department knew what everyone else was doing. Only his friend Professor Bock seemed clueless.

If you were the detective, would you be able to help Professor Bock?

One evening Detective Radstein visited Professor Bock. He was hanging out in the kitchen and overheard a conversation between Professor Bock and his wife. It appeared that Mrs. Bock had discovered that all the cutlets she had prepared for the next day were missing. She asked her husband:

—Did you eat the cutlets?
—I ate soup, the professor replied.
—Oh well, the children were probably very hungry.

Detective Radstein smiled to himself. Many mathematicians have trouble making false statements. Some of them adjust to social situations by learning how to lie while formally telling the truth. Radstein calls them dodgers. Fortunately, dodgers only dodge a question when they are threatened. It was obvious that Professor Bock was such a dodger. He implied that he didn’t eat the cutlets, because he had soup for lunch. The detective was ready to bet $10,000 that, in truth, the soup was just an appetizer to Professor Bock’s meal of cutlets.

Detective Radstein talked to his friend Professor Bock about this obsession mathematicians have to make true statements. Professor Bock agreed that many mathematicians are like that. In fact, all the faculty members in his department are either dodgers or pathological truth-tellers. Ironically, all other staff members at Professor Bock’s math department are liars.

A pathological truth-teller is another term that Detective Radstein uses to describe people who tell the truth no matter what. They never dodge. They answer a question exactly, often disregarding the context and purpose of the question. For example, when someone enters an elevator and asks a pathological truth-teller, “Is this elevator going up or down?” the answer s/he gets is “Yes.”

One day Professor Bock asked Detective Radstein for help, following a series of laptop thefts at his department. It was clear that the thefts were committed by someone working at the department and that the criminal acted alone.

Detective Radstein decided that the easiest starting point would be to ask everyone the same question: Did you steal the laptops? If a pathological truth-teller is the perpetrator, s/he would admit to the crime. A dodger would evade the question, but only if they are guilty. A liar is flexible: s/he might either answer the question with a lie or dodge with a lie. These are the first three answers the detective got from members of the department:

—Alice: No, I didn’t steal the laptops.
—Bob: Alice stole the laptops.
—Clara: Alice didn’t steal the laptops.

Who stole the laptops?