We will show that that line perpendicular(line $x$) or parallel(line $y$) to $L$, passing through the center $O$ of the circle, satysfies our conditions. Let us enumerate our segments with numbers $1,2,...,4n$.For each $i$ let $x_i$ and $y_i$ denote perpendicular projections of the i-th segment to lines $x$ and $y$ respectively. Assume, that our lines $x$ anf $y$ are "bad". That means that no two of segments $x_1,x_2,...,x_{4n}$ have a point in common (and the same for segments $y_1,y_2,...y_{4n}$) Thus we have \[ x_1+x_2+...+x_{4n}<2n \] and \[ y_1+y_2+...+y_{4n}<2n \] as $2n$ is the diameter of our circle. But for each $i$ we obtain \[ (x_i+y_i)^2=(x_i^2+y_i^2)+2x_i y_i\geq 1 \] so $x_i+y_i\geq 1$. Which means that \[ (x_1+x_2+...+x_{4n})+( y_1+y_2+...+y_{4n})\geq 4n \] so we've already obtained a contradiction which means that either $x$ or $y$ are 'good' !!!

jastrzab wrote: But for each $i$ we obtain \[ (x_i+y_i)^2=(x_i^2+y_i^2)+2x_i y_i\geq 1 \] so $x_i+y_i\geq 1$. You can actually use the triangle inequality to show this.

hy i wonder what were the results of the selection test? how many people solved 3 or more problems?

jastrzab wrote: That means that no two of segments $x_{1},x_{2},...,x_{4n}$ have a point in common (and the same for segments $y_{1},y_{2},...y_{4n}$) Thus we have \[x_{1}+x_{2}+...+x_{4n}<2n \] and \[ y_1+y_2+...+y_{4n}<2n Hey, this is not necessary. The projections can be the same but the segments may not be cut by the lines at all.

So what should the solution be?