Sunday, April 20, 2008

Election time

In Italy, there are just three parties: L (the left), R (the right), and N (north league). Elections are done, L and R get each 49\% of votes, N just 2\%. Since an agreement cannot be found, two tie-break rounds are done, where the two biggest parties are matched, and at the second round the winner is matched against the other party. Prove that there is a strategy for N's electors to make their party win the tie-break rounds, knowing that:
• L's electors, if they cannot vote for L they will vote for R, because they hate N.
• R's electors, if they cannot vote for R they will vote for N, because they hate L.

Wednesday, April 16, 2008

Minimal values

Let $f:\mathbb{R}\rightarrow\mathbb{R}$ be a function (any function). Let's call
$$M := \{ x \in \mathbb{R} : \exists\epsilon_x>0: f(x) \leq f(y)\;whenever\;|y-x| < \epsilon_x\}$$
the set of points where $f$ has a (non-strict) local minimum. Let $V=f(M)$ be the set of the local minimal values. Can $V$ have positive Lebesgue measure? (note that $M$ surely can, just take as $f$ a constant function, and every point is a local minimum).

Friday, April 4, 2008

An elettric circuit

There exists an electric circuit that has an interruption. The -proper- length of the interruption is $L_0$. A tract of electric wire runs along the circuit with speed $v$. The -proper- length of the wire is $L_0$ too. In circuit reference system wire length is $L<L_0$, so the circuit is never closed. In wire reference system interruption length is $L<L_0$, so at some instant the circuit is closed. How can you explain the apparent contradiction?
Update(22/04/2008): Solution posted! (show solution)