Advanced book on Mathematics Olympiad

Geometry and Trigonometry 649

O

NM

z

P

α/2

α/2

Figure 91

toBC, henceAB=BCandAD=DC. Exchanging the roles ofAandCwithBand
D, we find thatAB=ADandBC=CD. Therefore,AB=BC=CD=DEand
the quadrilateral is a rhombus.
The property is no longer true ifOis not the intersection of the diagonals. A coun-
terexample consists of a quadrilateral withAB=BC=3,BC=CD=4,BD=5,
andOonBDsuch thatOB=3 andOD=2.
(Romanian Team Selection Test for the International Mathematical Olympiad, 1978,
proposed by L. Panaitopol)

650.Assume by way of contradiction that the interiors of finitely many parabolas cover
the plane. The intersection of a line with the interior of a parabola is a half-line if that
line is parallel to the axis of the parabola, and it is void or a segment otherwise. There is
a line that is not parallel to the axis of any parabola. The interiors of the parabolas cover
the union of finitely many segments on this line, so they do not cover the line entirely.
Hence the conclusion.

651.Without loss of generality, we may assume thatAC=1, and let as usualAB=c.
We have

BC^2 =AB^2 +AC^2 − 2 AB·ACcos∠BAC≥AB^2 +AC^2 −AB=c^2 + 1 −c,

because∠BAC≥ 60 ◦. On the other hand,

CD^2 =AC^2 +AD^2 − 2 AC·ADcos∠CAD≥ 1 +c^6 +c^3 ,

because∠CAD≤ 120 ◦. We are left to prove the inequality

c^6 +c^3 + 1 ≤ 3 (c^2 −c+ 1 )^3 ,

which, after dividing both sides byc^3 >0, takes the form

c^3 + 1 +

1

c^3

≤ 3

(

c− 1 +

1

c

) 3

.