Methods of Proof 347......
P
P
P
iPjM
j+ (^1) i+ 1
Figure 52
60.LetAiAi+ 1 be the longest side of the polygon (or one of them if more such sides
exist). Perpendicular to it and at the endpointsAi andAi+ 1 take the linesLandL′,
respectively. We argue on the configuration from Figure 53.
If all other vertices of the polygon lie to the right ofL′, thenAi− 1 Ai >AiAi+ 1 ,
because the distance fromAito a point in the half-plane determined byL′and opposite
toAiis greater than the distance fromAitoL′. This contradicts the maximality, so it
cannot happen. The same argument shows than no vertex lies to the left ofL. So there
exists a vertex that either lies on one ofLandL′, or is between them. That vertex projects
onto the (closed) sideAiAi+ 1 , and the problem is solved.
......
LL
Ai Ai+ 1
Figure 53Remark.It is possible that no vertex projects in the interior of a side, as is the case with
rectangles or with the regular hexagon.
(M. Pimsner, S. Popa,Probleme de geometrie elementara ̆(Problems in elementary
geometry), Editura Didactica ̧ ̆si Pedagogica, Bucharest, 1979) ̆
61.First solution: Consider the oriented graph of roads and cities. By hypothesis, the
graph has no cycles. Define a partial order of the cities, saying thatA<Bif one can
travel fromAtoB. A partial order on a finite set has maximal and minimal elements. In
a maximal city all roads enter, and from a minimal city all roads exit.
Second solution: Pick an itinerary that travels through a maximal number of cities (more
than one such itinerary may exist). No roads enter the starting point of the itinerary, while
no roads exit the endpoint.
(Kvant(Quantum))