## English Issue, December 2002 | ||||

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## Solutions of advanced problems A

**A. 269.** *A circular hole is to be completely covered with two square boards. The sides of the squares are *1 *metre. In what interval may the diameter of the hole vary?*

*Figure 1*

It is easy to see that a hole of radius \(\displaystyle 2-\sqrt2\) (and all smaller holes) can be covered by two unit squares, see *Figure 1.*

We will prove that if a unit square covers at least half of the perimeter of a circle of radius *r*, then \(\displaystyle r\le2-\sqrt2\). This implies that the diameter of the hole cannot exceed

\(\displaystyle 4-2\sqrt2\approx1.17~\rm metre. \)

Let *K* be a circle of radius *r*, and *N* be a unit square, which covers at least the half of *K*. If *r*\(\displaystyle \ge\)1/2, then *N* can be translated such that each side of *N* or its extension has at least one common point with *K* and the covered part of the curve is increasing, see *Figure 2.*

*Figure 2*

So, it can be assumed that *r*>1/2 and (the extension of) each side of *N* has a common point with *K*.

First, assume that a vertex of *N* is inside *K*. Let the vertices of *N* be *A*, *B*, *C*, *D* and suppose that *A* is inside *K*. Denote the intersection of *K* and the half-lines *AB* and *AD* by *P* and *Q*, respectively. (*Figure 3.*)

Because *PAQ*\(\displaystyle \angle\)=90^{o}, the centre of *K* and vertex *A* are on the same side of line *PQ*. This implies that the uncovered arc *PQ* is longer than the half of *K*. This is a contradiction, thus this case is impossible.

We are left to deal with the case when there are no vertices of *N* inside *K* and each side or its extension has a common point with *K*. It is easy to check that these common points cannot be on the extension of the sides of *N*; they must lie on the perimeter of *N*.

Figure 3 | Figure 4 |

Denote by \(\displaystyle \alpha\), \(\displaystyle \beta\), \(\displaystyle \gamma\) and \(\displaystyle \delta\) the angles drawn on *Figure 4.*

The total of the uncovered arcs is at most the half of *K*, so \(\displaystyle \alpha\)+\(\displaystyle \beta\)+\(\displaystyle \gamma\)+\(\displaystyle \delta\)90^{o}. By symmetry, it can be assumed that +45^{o}.

As can be read from the Figure, *r*(cos +cos )=*r*(cos +cos )=1. The concavity of the cosine function yields

and hence .

Thus, the radius of the hole can be at most ; its diameter can be at most metre.