350 Methods of Proof
of this region except for the verticesBandC. DefineRBandRCanalogously. These
regions are illustrated in Figure 54. Because the given polygon is convex, each ofA 1 ,
B 1 , andC 1 lies in one of these regions or coincides with one ofA,B, andC.
RCAB
RRAB
CFigure 54If two ofA 1 ,B 1 ,C 1 , sayA 1 andB 1 , are in the same regionRX, then∠A 1 XB 1 <
π
3. Hence max{XA^1 ,XB^1 }>A^1 B^1 , contradicting the maximality of the lengthA^1 B^1.
Therefore, no two ofA 1 ,B 1 ,C 1 are in the same region.
Suppose now that one ofA 1 ,B 1 ,C 1 (sayA 1 ) lies in one of the regions (sayRA).
Because min{A 1 B, A 1 C}≥BC, we have that∠BA 1 C≤ π 3. We know thatB 1 does
not lie inRA. Also, because the polygon is convex,Bdoes not lie in the interior of the
triangleAA 1 B 1 , andCdoes not lie in the interior of triangleAA 1 B 1. It follows thatB 1
lies in the closed region bounded by the rays|A 1 Band|A 1 C. So doesC 1. Therefore,
π
3 =∠B^1 A^1 C^1 ≤∠BA^1 C≤
π
3 , with equalities ifB^1 andC^1 lie on rays|A^1 Band|A^1 C.
Because the given polygon is convex, this is possible only ifB 1 andC 1 equalBandC
in some order, in which caseBC=B 1 C 1. This would imply that trianglesABCand
A 1 B 1 C 1 are congruent.
The remaining situation occurs when none ofA 1 ,B 1 ,C 1 are inRA∪RB∪RC,in
which case they coincide withA, B, Cin some order. Again we conclude that the two
triangles are congruent.
We have proved that the distance between any two vertices of the given polygon is
the same. Therefore, given a vertex, all other vertices are on a circle centered at that
vertex. Two such circles have at most two points in common, showing that the polygon
has at most four vertices. If it had four vertices, it would be a rhombus, whose longer
diagonal would be longer than the side, a contradiction. Hence the polygon can only be
the equilateral triangle, the desired conclusion.
(Romanian Mathematical Olympiad, 2000)
67.Because
a^2 +b^2 =(
a+b
√
2) 2
+
(
a−b
√
2