Methods of Proof 349black sequence 2l 1 <l 1 +l 2 <···< 2 lm. By maximal we mean that these sequences
cannot be extended any further. Without loss of generality, we may assume thatkn<lm.
We look at all white even numbers between 2kn+1 and some arbitrary 2x; letWbe
their number. If for one of these white even numbers 2kthe numberk+knwere white as
well, then the sequence of whites could be extended, contradicting maximality. Hence
k+knmust be black. Therefore, the numberbof blacks between 2kn+1 andx+knis
at leastW.
Similarly, ifBis the number of black evens between 2lm+1 and 2x, the numberw
of whites between 2lm+1 andx+lmis at leastB. We haveB+W≥x−lm, the latter
being the number of even integers between 2lm+1 and 2x, whileb+w≤x−kn, since
x−knis the number of integers between 2kn+1 andx+kn. Subtracting, we obtain
0 ≤(b−W)+(w−B)≤lm−kn,and this inequality holds for allx. This means that asxvaries there is an upper bound
forb−Wandw−B. Hence there can be only a finite number of black squares that
cannot be written askn+kfor some white 2kand there can only be a finite number of
white squares which cannot be written aslm+lfor some black 2l. Consequently, from a
point onward all white squares are of the formlm+lfor some black 2land from a point
onward all black squares are of the formkn+kfor some white 2k.
We see that forksufficiently large,kis black if and only if 2k− 2 knis white, while
kis white if and only if 2k− 2 lmis black. In particular, for each suchk,2k− 2 knand
2 k− 2 lmhave the same color, opposite to the color ofk. Soifweletlm−kn=a>0,
then from some point onward 2xand 2x+ 2 aare of the same color. The arithmetic
sequence 2x+ 2 na,n≥0, is thus monochromatic. It is not hard to see that it also
satisfies the condition from the statement, a contradiction. Hence our assumption was
false, and sequences with the desired property do exist.
(communicated by A. Negu ̧t)
66.We begin with an observation that will play an essential role in the solution.
Given a triangleXY Z,if∠XY Z≤ π 3 , then either the triangle is equilateral or else
max{YX,YZ}>XZ, and if∠XY Z≥π 3 , then either the triangle is equilateral or else
min{YX,YZ}<XZ.
Choose verticesAandBthat minimize the distance between vertices. IfCis a vertex
such that∠ACB=π 3 , then max{CA, CB}≤AB, so by our observation the triangle
ABCis equilateral. So there exists an equilateral triangleABCformed by vertices of
the polygon and whose side length is the minimal distance between two vertices of the
polygon. By a similar argument there exists a triangleA 1 B 1 C 1 formed by vertices whose
side length is the maximal distance between two vertices of the polygon. We will prove
that the two triangles are congruent.
The linesAB, BC, CAdivide the plane into seven open regions. Denote byRAthe
region distinct from the interior ofABCand bounded by sideBC, plus the boundaries