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 R_{0} 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 R_{0} looks like:

As you can see assuming a linear decline in the R_{0} 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!