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Zoom and Wallpaper

Recently I’ve been using zoom frequently, both for my classes and for socializing with friends and family. Zoom has a feature called virtual background, where a participant in a Zoom meeting can use an image or a video as her background. It’s common to use scenic views for a background to give the illusion that you’re on vacation in exotic locations or to use family photos to suggest being with your loved ones.

For the more mathematically inclined, backgrounds of abstract designs might be more appealing. Since I’ve been working on symmetry designs for one of my classes, I decided to start using them as backgrounds. Here’s as example of me with one of the designs as a background:

The designs are all examples of patterns whose symmetry groups are one of the 17 wallpaper groups. Here’s a collection of some of my patterns:

If any of them catch your fancy you can download the collection here.

Coronavirus

For any of you who have followed my blog you may recall that I have been tracking the number of cases of COVID-19 in Santa Clara County. When I started do this, there was no other tracking site that I knew about. Shortly thereafter, the Santa Clara County Department of Public Health started publishing the numbers and displaying graphs of the data on their web site. Over time the presentation of the data became more and more informative and complete. So I am terminating my data displays and suggest that those interested follow the information at the Santa Clara County Departmnet of Public Health.

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More on the Desargues Configuration

One of the remarkable properties of the Desargues configuration is its symmetry. Recall that Desargues Theorem states that if two triangles are in perspective from a point j, the center of perspectivity, then the three points of intersection of the corresponding sides of the triangles are collinear. This is illustrated in the configuration on the left below. What is surprising is that any point of the configuration can be chosen as the center of perspectivity, and there will be triangles in the configuration in perspective from that point illustrating Desargues Theorem! For example, if the original point d is chosen as the center j, the relabelling at the right shows another instance of Desargues Theorem.

One way to see why this is so uses yet another interesting property of the Desargues configuration. In the figure to the left below, the points are labeled with pairs of digits from 1 to 5 such that the points on each line are the three pairs in a triplet of digits from 1 to 5. For example, the points labeled with pairs from the triplet 123, namely 12, 13, 23, are on a line. Since this is true for any of the ten triplets, permuting the digits produces another labeling with the same properties.

The triangles in perspective from 12 are 13, 14, 15 and 23, 24, 25, and the collinear points of intersection of corresponding sides are 34, 35, 45. Swapping the digits 1 and 3 produces the labeling on the right, the same configuration illustrating Desargues Theorem but with a different center of perspectivity. With a suitable permutation any of the points can be labeled 12, the center of perspectivity.

Coronavirus Update

The latest estimate I have for the Basic Reproduction Number R0 and a projection for the number of cases are:

Based on this estimate, we are very close the inflection point where the R0 falls below 1, and less than a month away from where the total cases curve flattens.

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Desargues Configuration

In my previous blog I posted a problem that I used in the San Jose State Math Circle. In the problem we assume that a committee consists of members of the board of directors of a corporation and is subject to the following conditions:

(1) Each director belongs to the same number of committees.
(2) Each committee has the same number of members.
(3) Two directors are on at most one committee together.

The problem statement is:

There are 10 directors on the board of Starlight
Corporation.

a. The directors would like to form five committees.
If each director is on two committees, then how many
directors must belong to each committee? Show how the
committees may be formed, or prove that it is not possible.

b. After a few years, the company realizes that it needs
more committees and wants to increase the number of
committees to 10, while reducing the size of each committee
to three. Assuming this is possible, how many committees
must each director belong to? Show how the committees may
be formed, or prove that it is not possible.

We can represent a committee graphically by using points to represent the directors and lines to represent the committees, where a director is on a committee if his point is on the committee’s line. (It is not necessary for the lines to be “straight”, but for these two problems all solutions except one for part b. can be drawn with straight lines in the Euclidean plane.)

In part a. each director is on two committees means that the number of pairs (d,c) of a director d and a committee c that d sits on is 20: for each of the 10 directors d, there are two choice for c. On the other hand there are five choices for the committee c, so there must be four directors on each committee. Thus a graphical representation has 10 points and five lines such that each point is on two lines and each line has four points. One drawing of this graph is the pentagram, a five-pointed star:

In this graph, we have 10 points, the directors, and five lines, the committees. Each point is on two lines and each line is on four points.

There are many solutions for part b., but I think the most interesting one arises from a drawing of Desargues Theorem:

Let △abc and △def be triangles and j = ad · be · cf
(△abc and △def are in perspective from j). If
g = ab · de, h = bc · ef, and i = ac · df, then g, h, i
are collinear.

Here the dot between lines represents the point of intersection of the lines. A drawing for Desargues Theorem looks like:

As you can see it has 10 points, the directors, and 10 lines, the committees, and each directors is on three committees and each committee has three directors.

The Desargues configuration has some remarkable properties which I will discuss in my next blog but now it is time to update the

Coronavirus in the Santa Clara County

In my last post I tried estimating the Basic Reproduction Number R0 for COVID-19 in Santa Clara County based on the daily new cases count and its relationship to new cases five days previously. As of today, the graph of this estimate of R0 looks like:

As you can see assuming a linear decline in the R0 estimate, the number of daily new cases should start to trend down in about one week and the total number of cases will start to level off in the first week of May:

I’m cautiously optimistic!

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Zoom!

I did my first class over zoom last Friday. For the San Jose State Math Circle I talked about configurations which I introduced as a problem of forming various committees from the board of directors of a corporation subject to the following conditions:

(1) Each director belongs to the same number of committees.
(2) Each committee has the same number of members.
(3) Two directors are on at most one committee together.

Here’s a sample problem from the math circle:

There are 10 directors on the board of Starlight
Corporation.

a. The directors would like to form five committees.
If each director is on two committees, then how many
directors must belong to each committee? Show how the
committees may be formed, or prove that it is not possible.

b. After a few years, the company realizes that it needs
more committees and wants to increase the number of
committees to 10, while reducing the size of each committee
to three. Assuming this is possible, how many committees
must each director belong to? Show how the committees may
be formed, or prove that it is not possible.

The directors and committees are the points and lines of a (combinatorial) configuration. In some case the configuration can be drawn in the plane using points and lines; if that is the case, then the configuration is realized as a geometric configuration. Both problems above have a realization as a geometric configuration (the configuration in part a. is unique, but part b. has 10 different solutions).

And more zoom ahead: I’ll be resuming several of my classes over zoom starting tomorrow.

Coronavirus Update

In this blog I have been graphing the growth of COVID-19 cases in Santa Clara county. Since I lack any background in mathematical epidemiology, and my statistics background, is in general, not very deep, I decided to do some reading in modeling epidemics, so I downloaded Modeling Paradigms and Analysis of Disease Transmission Models, Gumel & Lenhart (Eds.).

From reading just a portion of the introductory material I am changing the variable I track use to make projections. Instead of using the number of new cases as a percentage of the total cases, I will use the Basic Reproduction Number R0. I estimate R0 as Nt/Nt-5 where Nt is the number of new cases on day t. I use 5 because it is believed to be approximately the median incubation period of COVID-19; the idea being that those newly infected 5 days ago infect those detected today.

So here is the graph of the estimate of R0:

As you can see R0 is in a decreasing trend. Assuming the trend is linear, on April 26, R0 will fall below 1, a critical threshold. At that point the number of new cases per day will start to diminish!

Based on that assumption, here is a projection for new cases for next few weeks:

We can see in this graph an inflection point at April 26. It still appears that we are more than a month away from a peak in the number of cases…

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Circle and Points Problem and Coronavirus Update

Circle and Points Problem

Last post I posed this problem:

Place n points on a circle and draw all the line segments that connect any two of the points. The points are placed such that no three of the segments intersect at a single interior point. Into how many regions do the segments divide the circles interior?

If you start to tabulate the answers for small values of n, you find:

n12345678
Regions1248????

It looks like the solution to this problem is that the number of regions follows the pattern of the powers of 2.

If you draw the figure for 5 points, you will find as expected the number of regions is 16. But when you draw the figure for 6 points:

— no matter where you place 6 points around the circle, the number of regions is 31, not 32!

This problem demonstrates that finding a pattern is not as simple as examining the first 3 or 4 or 5 cases. Ultimately, it will be necessary to prove that any pattern you find is in fact the solution to the problem.

Coronavirus Update

I’m continuing to track the number of cases in Santa Clara county. The latest graph looks like:

As you can see if the rate of change of the daily case count continues its trend, the number of cases will start to level off in two weeks. That would be great news!