## An Orthogonal Quadrangle

Consider a triangle with vertices A, B, and C. Let O be its orthocenter. Let’s connect O to the vertices. We get six lines: three sides of the triangle and three altitudes. These six lines are pair-wise orthogonal: AO ⊥ BC, BO ⊥ AC, and CO ⊥ AB.

It is easy to see that A is the orthocenter of the triangle OBC, and so on: each vertex is the orthocenter of the triangle formed by the other three. We say that these four points form an orthocentric system.

I heard a talk about this structure at the MOVES 2015 conference by Richard Guy. What I loved in his talk was his call to equality and against discrimination. The point O plays the same role as the other three points. It should be counted. Richard Guy suggested calling this system an orthogonal quadrangle. I am all for equality. This is not a triangle, this is a quadrangle!

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## The Battle I am Losing

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?

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## A Frog Puzzle

I stumbled upon a TED-Ed video with a frog puzzle:

You’re stranded in a rainforest, and you’ve eaten a poisonous mushroom. To save your life, you need an antidote excreted by a certain species of frog. Unfortunately, only the female of the species produces the antidote. The male and female frogs occur in equal numbers and look identical. There is no way to distinguish between them except that the male has a distinctive croak. To your left you spot a frog on a tree stump. You hear a croak from a clearing in the opposite direction, where you see two frogs. You can’t tell which one made the sound. You feel yourself starting to lose consciousness, and you realize that you only have time to run in one direction. Which way should you go: to the clearing and lick both frogs or to the tree stump and lick the stump frog?

My first thought was that male frogs croak to attract female frogs. That means the second frog in the clearing is probably an already-attracted female. The fact that the stump frog is not moving means it is male. I was wrong. This puzzle didn’t assume any knowledge of biology. The puzzle assumes that each frog’s gender is independent from other frogs. Thus this puzzle is similar to two-children puzzles that I wrote so much about. I not only blogged about this, but also wrote a paper: Martin Gardner’s Mistake.

As in two-children puzzles, the solution depends on why the frog croaked. It is easy to make a reasonable model here. Suppose the male frog croaks with probability p. Now the puzzle can be solved.

Consider the stump frog before the croaking:

• It is a female with probability 1/2.
• It is a croaking male with probability p/2.
• It is a silent male with probability (1-p)/2.

Consider the two frogs in the clearing before the croaking:

• Both are female with probability 1/4.
• One is a female and another is a croaking male with probability p/2.
• One is a female and another is a silent male with probability (1-p)/2.
• Both are silent males with probability (1-p)2/4.
• Both are croaking males with probability p2/4.
• One is a silent male and another is a croaking male with probability p(1-p)/2.

The probabilities corresponding to our outcome—a non-croaking frog on the stump and one croaking frog in the clearing—are in bold. Given that the stump frog is silent, the probability that it’s a female is 1/(1-p). Given that one clearing frog croaked, the probability that one of them is a female is p/2 divided by p(1-p)/2. The ratio is the same 1/(1-p). It doesn’t matter where you go for the antidote.

The TED-Ed’s puzzle makes the same mistake that is common in the two-children puzzles. I don’t want to repeat their incorrect solution. The TED-Ed’s frog puzzle is wrong.

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## Is My Bank Smart Enough?

When I receive a bank statement, I review all the transactions. The problem is my failing memory. I do not remember when and where I last took money from an ATM, and how much. So I decided to create a pattern. I used to take cash in multiples of 100. Probably most people do that. A better idea is to always take the amount that has a fixed remainder modulo 100, but not 0. For example, let’s say I take one of the following amounts: 40, 140, 240, or 340. Sometimes I need more money, sometimes less. This set of numbers covers all of my potential situations, but my pattern is that all the numbers end in 40. This way if someone else gets access to my account, they will almost surely take a multiple of 100. I will be able to discover a fraud without remembering the details of my last withdrawal.

In addition, when I first started doing this, I was hoping I wouldn’t need to wait until I review my own statement to discover problems. My hope was that if a thief tried to take cash from my account, my bank would notice a change in the pattern and notify me immediately. Now I realize that this was wishful thinking. I doubt that banks are as smart as I am.

One day I should try an experiment. I should go to an ATM I never use and withdraw 200 dollars. I wonder if my bank would notice.

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## Late Homework

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.

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## An Asteroid Walk

I found the following puzzle on a facebook page of Konstantin Knop (in Russian):

Seven astronauts landed on a small spherical asteroid. They wanted to explore it and walked in different directions starting from the same location. All used the same walking algorithm: walk x kilometers forward, turn 90 degrees left and walk another x kilometers, turn 90 degree left again and walk the last x kilometers. The value of x was different for different astronauts and was one of 30, 40, 50, 60, 70, 80, and 90.
All but one astronaut finished in the same location. What was the value of x for the astronaut who finished alone? What is the size of the asteroid?

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## More Jokes

I’ve been collecting math jokes for many years. I thought I’ve seen them all. No. Inventive people continue to create them. I was recently sent a link to a math joke website that features many jokes that are new to me. Here are my favorites:

* * *

With massive loss of generality, let $n=5$.

* * *

How do you prove a cotheorem? Using rollaries.

* * *

$0\to A\to B \to C \to 0$. Exactly.

* * *

Let $\varepsilon\to 0$. There goes the neighborhood!

* * *

Take a positive integer $N$. No, wait, $N$ is too big; take a positive integer $k$.

* * *

Calculus has its limits.

* * *

There is a fine line between a numerator and a denominator.

* * *

There’s a marked difference between a ruler and a straightedge.

* * *

Suppose there is no empty set. Then consider the set of all empty sets.

* * *

Q: Why is it an insult to call someone “abelian”?
A: It means they only have a 1-dimensional character, and are self-centered.

* * *

Q: What’s a polar bear?
A: A rectangular bear after a coordinate transform.

* * *

A logician rides an elevator. The door opens and someone asks:
—”Are you going up or down?”
—”Yes.”

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## Mikhail Khovanov at Kvant

Russians cherish the issues of Kvant, a famous Soviet monthly magazine for high-school students devoted to math and physics. At its peak its circulation was about 300,000, which is unparalleled for a children’s science journal. I still have my old childhood issues somewhere in my basement. But one issue is very special: it has a prominent place in my office. I didn’t receive it by subscription; I received it as a gift from my brother Misha.

Strictly speaking Misha is not my brother, but rather a half-brother. His full name is Mikhail Khovanov and he is a math professor at Columbia. The signed issue he gave me contains his math problem published in the journal. He invented this problem when he was a 10th grader. Here it is:

In a convex n-gon (n > 4), no three diagonals are concurrent (intersect at the same point). What is the maximum number of the diagonals that can be drawn inside this polygon so that all the parts they divide into are triangles?

He designed other problems while he was in high school. All of them are geometrical in nature. The journal is available online, and a separate document with all the math problems is also available (in Russian). His problems are M1038, M1103, M1108 (above), M1119, M1153, M1204. I like his other problems too. M1153 (below) is the shortest problem on his list: as usual I am guided by my laziness.

What’s the greatest number of turns that a rook’s Hamiltonian cycle through every cell on an 8 by 8 chessboard can contain?

I wanted to accompany this post with a picture of my brother at the age he was when he invented these problems—about 16. Unfortunately, I don’t have a quality picture from that period. I do have a picture that is slightly off: by about ten years.

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## The Ford Circles Game

What are the Ford circles? A picture is worth a thousand words, so here is a picture.

We draw a circle for any rational number p/q between 0 and 1 inclusive. We assume that p/q is the representation of the number in the lowest terms. Then the center of the circle is located at (p/q,1/q) and the radius is 1/q. The number inside a circle is q.

Here’s the game. We start with any circle on the picture, except for the two largest circles corresponding to integers 0 and 1. In one move we can switch to a larger circle that touches our circle. The person who ends up at the two largest circles corresponding to integers 0 or 1 loses. Equivalently, the person who ends in the central circle marked “2” wins.

There are other ways to describe moves in this game in terms of rational numbers corresponding to circles, that is, the x-coordinates of their centers. Two circles corresponding to numbers p/q and r/s touch each other iff one of the following equivalent statements is true:

• The cross-determinant of two numbers p/q and r/s that is defined as |ps-qr| equals to 1;
• p/q and r/s are neighbors in some Farey sequence;
• One of the numbers is the parent of the other in the Stern-Brocot tree.

Let me explain the last bullet. Given two rational numbers in their lowest terms a/b and c/d, we generate their mediant as: (a+c)/(b+d). We call the two numbers a/b and c/d the parents of the mediant (a+c)/(b+d). The Stern-Brocot tree starts with two parents 0/1 and 1/1. Then their mediant is inserted between them to create a row: 0/1, 1/2, and 1/1. Then all possible mediants of two consecutive numbers are inserted in a given row to get a new row. The process repeats ad infinitum. The famous theorem states that any rational number between 0 and 1 will appear in the process.

What I like about this game is a simple and beautiful description of P-positions (These are the positions you want to end your move at in order to win.) P-positions are numbers with even denominators in their lowest terms.

In the picture above P-positions are blue, while other positions are red. All circles touched by blue are red. And if we look at the larger neighbors of every red circle, one of them is blue and one is red.

Let’s prove that the numbers with even denominators satisfy the conditions for P-positions. First, two circles corresponding to numbers with even denominators can’t touch each other. Indeed, the cross-determinant of two such fractions is divisible by 2. Second, each red circle has to touch one blue and one red circle with larger radii. Indeed, the circles with larger radii touching a given circle are exactly the parents of the circle. If the mediant has an odd denominator, then one of the parents must have an even denominator and the other an odd denominator.

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## Weighted Mediants

Let us start with regular (non-weighted) mediants. Suppose we have two fractions a/b and c/d, the mediant of these two numbers is the wrong way a child might sum these two fractions: (a+c)/(b+d). There is nothing wrong with this childish way of summing, it is just not a sum of two numbers, but rather an operation the result of which is called a mediant. One must be careful: the result doesn’t depend on the initial numbers chosen, but on their representations. The way to deal with this is to assume that a/b, c/d, and (a+c)/(b+d) are in their lowest terms.

The numerical value of the mediant is always in between given numbers. This is probably why it is called a mediant.

Suppose you start with two rational numbers 0/1 and 1/1, and insert a mediant between them. If you continue doing it recursively, you can reach any rational number between 0 and 1. This is a well-known property of the mediants. For example, after three rounds of inserting mediants into the initial sequence (0/1, 1/1), we get the sequence: 0/1, 1/4, 1/3, 2/5, 1/2, 3/5, 2/3, 3/4, 1/1.

The mediants are easy to generalize if we assign weights to initial fractions. Suppose the first fraction has weight m and the second n, then their weighted mediant is (am+cn)/(bm+dn). The simplest generalization happens when the total weight, m+n is 3. In this case, given two rational numbers a/b and c/d, the left mediant is (2a+c)/(2b+d) and the right one is (a+2c)/(b+2d). Obviously the inequality property still holds. If a/b ≤ c/d, then a/b ≤ (2a+c)/(2b+d) ≤ (a+2c)/(b+2d) ≤ c/d.

James Propp suggested the following question for our PRIMES project. Suppose the starting numbers are 0/1 and 1/1. If we recursively insert two weighted mediants in order between two numbers we will get a lot of numbers. For example, after two rounds of inserting weighted mediants into the initial sequence (0/1, 1/1), we get the sequence: 0/1, 1/5, 2/7, 1/3, 4/9, 5/9, 2/3, 5/7, 4/5, 1/1. It is easy to see that new fractions must have an odd denominator. Thus unlike the standard case, not every fraction will appear. The question is: will every rational number between 0 and 1, which in reduced form has an odd denominator appear?

I started working on this project with Dhroova Aiylam when he was a high-school junior. We didn’t finish this project during the PRIMES program. But Dhroova finished another project I already wrote about: he analyzed the case of the standard mediants with any starting points. He showed that any rational number in between the starting points appears.

Dhroova became an undergraduate student at MIT and we decided to come back to the initial PRIMES project of weighted mediants. In our paper Stern-Brocot Trees from Weighted Mediants we prove that indeed every fraction with an odd denominator between 0 and 1 appears during the recursive procedure. We also analyzed what happens if we start with any two rational numbers.

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