This very entry was written first and foremost as a simple test.  I wanted to determine how easily math-jargon could be incorporated into and displayed in wordpress posts.  Spoiler alert: its really easy.  For starters, check out this formula:

n(A) = 2\Big{\lceil} 2\sqrt{A} \Big{\rceil} + 4

Not only is this my first math blog, today was also my first day working as a research mentor at St. Mary’s 2012 REU program.  (Although our program began yesterday, we did not start working on problems until today.)  I, along with Manny Daring, Isabel Guadarrama, Samantha Sprague, and Carrie Winterer, will be studying various Lego Isoperimetric Problems for the next few weeks.

Tentatively (and somewhat inexplicably) deemed the “Whale formula,” the expression above is the conjectured solution to the following geometric question: What is the smallest number of bricks one needs to enclose a region with area A?  (I realize there are a number of concepts to clarify and expound upon here, like defining “enclose” and “area,” but I will leave it to the students to supply missing details in their upcoming blog posts and write-ups, some of which will be available here.)

After stumbling into our classroom over 90 minutes late, I discovered that Manny, Isabel, Samantha, and Carrie had already made much progress on this question.  In fact, after spending a few minutes chatting with them, the “Whale conjecture” emerged.

Later in the day the group outlined a proof verifying the conjecture for certain areas (specifically, the conjecture appears to be true when A is a perfect square or a “nearly” perfect square), and all it seems to require is some ingenuity, a few known results, and a bit of calculus.

Next up we’ll take a crack at all of the other possible area values (with an eye towards using induction).  And after that, well… its not so clear what we’ll do.  We’ll have no shortage of ideas and questions to tackle.  What happens when we alter the types of bricks under consideration?  What happens when we price the bricks being used?  What happens if we force our brick arrangement to enclose two separate regions?

I don’t know!  (Do you?)

I do know that I’m ending this entry with a toast, raising my glass of Tripel Karmeliet to all the other firsts to come this summer.  Hear, hear.


3 thoughts on “First

  1. What if they’re triangular bricks on a triangular grid? Hexagonal bricks on a hexagonal grid? Penrose tiles on a Penrose “grid”?

  2. These sound like excellent follow up questions, and each one counts twice since the classical isoperimetric problem and its “converse” (the iso-area problem, I suppose) yield different answers in these discrete settings. What about using variously sized circles (restricted only to be tangent to one another)?

  3. Actually, here is one that I think might be quite interesting: Suppose you restrict yourself to an already blocked off m by n region of the grid. You then have the option of creating extra “curves” to further block off a region. The goal is to create a region R so that its area A and perimeter P have the largest ratio A/P. This is different from the classical isoperimetric goal of maximizing the scale-invariant ratio A/(P^2), and of course there isn’t a maximum without restricting the size. Note that adding no curves gives a region with A=mn and P=2m+2n. If I figure out how to say math better, I’ll comment again to explain why this has interesting ties to image analysis.

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