Archive for November 2012

Halving Lines

One of the 2012 PRIMES projects, suggested by Professor Jacob Fox, was about bounds on the number of halving lines. I worked on this project with Dai Yang.

Suppose there are n points in a general position on a plane, where n is an even number. A line through two given points is called a halving line if it divides the rest of n−2 points in half. The big question is to estimate the maximum number of halving lines.

Let us first resolve the small question: estimating the minimum number of halving lines. Let’s take one point from the set and start rotating a line through it. By a continuity argument you can immediately see that there should be a halving line through any point. Hence, the number of halving lines is at least n/2. If the point is on the convex hull of the set of points, then it is easy to see that it has exactly one halving line through it. Consequently, if the points are the vertices of a convex n-gon, then there are exactly n/2 halving lines. Thus, the minimum number of halving lines is n/2.

Finding the maximum number of halving lines is much more difficult. Previous works estimated the upper bound by O(n4/3) and the lower bound by O(ne√log n). I think that Professor Fox was attracted to this project because the bounds are very far from each other, and some recent progress was made by elementary methods.

Improving a long-standing bound is not a good starting point for a high school project. So after looking at the project we decided to change it in order to produce a simpler task. We decided to study the underlying graph of the configuration of points.

Suppose there is a configuration of n points on a plane, and we are interested in its halving lines. We associate a graph to this set of points. A vertex in the configuration corresponds to a vertex in the graph. The graph vertices are connected, if the corresponding vertices in the set have a halving line passing through them. So we decided to answer as many questions about the underlying graph as possible.

For example, how long can the longest path in the underlying graph be? As I mentioned, the points on the convex hull have exactly one halving line through them. Hence, we have at least three points of degree 1, making it impossible for a path to have length n. The picture below shows a configuration of eight points with a path of length seven. We generalized this construction to prove that there exists a configuration with a path of length n−1 for any n.

Path

We also proved that:

  • The largest cycle can have length n − 3.
  • The largest clique is of the order O(√n).
  • The degrees of two distinct vertices sum to at most n, if they are connected by an edge, and at most n − 2 otherwise.

After we proved all these theorems, we came back to the upper bound and improved it by a constant factor. Our paper is available at arXiv:1210.4959.

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I am on the Air

Samuel Hansen has an unusual profession: he is a mathematics podcaster. He interviewed me for his Relatively Prime podcast titled 0,1,2,3,…, where we discussed my Number Gossip project. The podcast also includes interviews with Neil Sloane, Michael Shamos, and Alex Bellos.

My previous interview with Samuel is at acmescience.com. There I discuss both math education and gender in math issues.

When I listened to myself, I found it strange that I seemed to have a British accent on top of my Russian accent. Did you notice that too?

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Helpmate

I discovered the following chess puzzle on a Russian blog for puzzle lovers. It is a helpmate-type puzzle. Black cooperates with White in checkmating himself. In this particular puzzle Black starts and helps White to win in one move.

Oops. Something is not quite right. There are not enough pieces on the board. To recover the missing pieces in order to solve the puzzle, you need to retrace your steps. If Black and White go back one move each, they will be able to cooperatively checkmate Black in one move. Find the position one move back and the cooperative checkmate.

Helpmate

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Me and Chess

I am Russian; I know how to play chess. My father taught me when I was three or four. We played a lot and he would always win. I got frustrated with that and one day, when I was five, I didn’t announce my check. On the next move, I grabbed his king and claimed my victory.

He was so angry that he turned red and almost hit me. This frightened me so much that I lost my drive for chess that very moment.

I still understand its beauty and solve a chess problem about once a decade. Look for a cute chess puzzle in my next post.

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Challenge Problems

For my every class I try to prepare a challenge problem to stretch the minds of my students. Here is a problem I took from Adam A. Castello’s website:

There is a ceiling a hundred feet above you that extends for- ever, and hanging from it side-by-side are two golden ropes, each a hundred feet long. You have a knife, and would like to steal as much of the golden ropes as you can. You are able to climb ropes, but not survive falls. How much golden rope can you get away with, and how? Assume you have as many hands as you like.

The next problem I heard from my son Sergei:

You are sitting at the equator and you have three planes. You would like to fly around the equator. Each plane is full of gas and each has enough gas to take you half way around. Planes can transfer gas between themselves mid-air. You have friends, so that you can fly more than one plane at once. How do you fly around the equator?

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Nerdy and Flirtatious Jokes

* * *

Right clicking a file with a mouse will allow you to change it, check it for viruses, or revert to the previous version. I wonder where I can buy a mouse that can do the same thing with my husband.

* * *

— Let’s have sex.
— Sure, but today I want to be the numerator.

* * *

Attention! We want to check that you are not a robot. Please, undress and turn on a web-camera.

* * *

In a drug store:
— I would like input and output cleaners.
— ???
— Toothpaste and toilet paper.

* * *

I used to recount the multiplication tables to delay my ejaculation. Now, each time I see the multiplication tables I get a hard on.

* * *

— Tonight my parents are away. Let’s finally try a forbidden thing.
— Dividing by zero?

* * *

My friend put his mistress in his phone’s contact list as ‘low battery’.

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