Advanced book on Mathematics Olympiad

Combinatorics and Probability 737

⌊

200 π

√

2

2 π

⌋

= 100

√

2 = 141.

(Ukrainian Mathematical Olympiad)

844.The solution is based on the pigeonhole principle. Let us assume that the sum of

lengths of the chords is greater than or equal tokπ. Then the sum of the lengths of the

arcs subtended by these chords is greater thankπ. Add to these arcs their reflections

about the center of the circle. The sum of the lengths of all arcs is greater than 2kπ,

so there exists a point covered by at leastk+1 arcs. The diameter through that point

intersects at leastk+1 chords, contradicting our assumption. Hence the conclusion.

(Kvant(Quantum), proposed by A.T. Kolotov)

845.The center of the desired circle must lie at distance at least 1 from the boundary

of the square. We will be able to find it somewhere inside the square whose sides are

parallel to those of the initial square and at distance 1 from them. The side length of this

smaller square is 36.

The locus of all points that lie at distance less than 1 from a convex polygonal surface

Pis a polygonal surfaceQwith sides parallel to those ofP and whose corners are

rounded. The areas ofPandQare related by

S[Q]=S[P]+(perimeter ofP)× 1 +π.

This is because the circular sectors from the corners ofQcomplete themselves to a disk

of radius 1.

So the locus of the points at distance less than 1 from a polygon of area at mostπ

and perimeter at most 2πis less than or equal toπ+ 2 π+π= 4 π. It follows that the

area of the region of all points that are at distance less than 1 from any of the given 100

polygons is at most 400π. But

400 π≤ 400 · 3. 2 = 40 · 32 = 362 − 42 < 362.

So the set of these points does not cover entirely the interior of the square of side length

36. Pick a point that is not covered; the unit disk centered at that point is disjoint from
    any of the polygons, as desired.
    (M. Pimsner, S. Popa,Probleme de geometrie elementara ̆(Problems in elementary
    geometry), Editura Didactica ̧ ̆si Pedagogica, Bucharest, 1979) ̆
    846.Placendisks of radius 1 with the centers at the givennpoints. The problem can be
    reformulated in terms of these disks as follows.

Alternative problem. Givenn≥ 3 disks in the plane such that any 3 intersect, show
that the intersection of all disks is nontrivial.