We started by working on the following problem from the 2014 Virginia Tech Regional Math Contest:

Suppose we are given a 19×19 chessboard (a table with \(19^2\) squares) and remove the central square. Is it possible to tile the remaining \(19^2 -1 = 360\) squares with 4 × 1 and 1 × 4 rectangles? (So each of the 360 squares is covered by exactly one rectangle.) Justify your answer.

We proved that the answer is no, by adapting the proof of a classical problem: it is not possible to tile an 8×8 chessboard with 2 opposite corners removed using 2×1 and 1×2 tiles.

Then we worked on solutions for three problems from the 2010 Putnam:

- B1: We used the suggestion from last week’s post, and also saw how to do a proof using the Cauchy-Schwarz inequality.
- A2: Using ideas similar to the ones we used in problem 8 of the proof strategies handout, we could prove that the derivative of f is periodic with period 1. Then we took the second derivative of f, which turned out to be zero and thus f must be of the form ax + b for some constants a and b.
- A3: Here we first thought about a similar but simpler problem: \(h(x) = ah'(x) \). Using standard ODE techniques,\(h(x) = Ce^{-x/a}\), so the boundedness condition forces the constant C to be 0 and thus h is zero as well. For the two variable case, fix x and y and take derivatives of \(h(x+at,y+bt) \) with respect to t.

Today’s nuggets of wisdom:

- Once again, invariants made an appearance. Without them, it can be quite difficult to prove that something does not exist!
- Simplification is often a good idea to get you started.

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