Advanced book on Mathematics Olympiad

650 Geometry and Trigonometry

With the substitutionc+^1 c=x, the inequality becomes

x^3 − 3 x+ 1 ≤ 3 (x− 1 )^3 , forx≥ 2.

But this reduces to

(x− 2 )^2 ( 2 x− 1 )≥ 0 ,

which is clearly true. Equality holds if and only if∠A= 60 ◦andc=1(AB=AC),
i.e., when the triangleABCis equilateral.
(proposed by T. Andreescu for the USA Mathematical Olympiad, 2006)

652.Denote bya, b, c, d, e, f, g, hthe lengths of the sides of the octagon. Its angles are
all equal to 135◦(see Figure 92). If we project the octagon onto a line perpendicular to
sided, we obtain two overlapping segments. Writing the equality of their lengths, we
obtain

a

√

2

2

+b+c

√

2

2

=e

√

2

2

+f+g

√

2

2

.

Becausea, b, c, e, f, gare rational, equality can hold only ifb =f. Repeating the
argument for all sides, we see that the opposite sides of the octagon have equal length.
The opposite sides are also parallel. This means that any two consecutive main diago-
nals intersect at their midpoints, so all main diagonals intersect at their midpoints. The
common intersection is the center of symmetry.

b

d

f

e

g

a c

h

Figure 92

653.Let us assume that the three diagonals do not intersect. Denote byMthe intersection
ofADwithCF,byNthe intersection ofBEwithCF, and byPthe intersection ofAD
withBE. There are two possibilities: eitherMis betweenAandP,orPis betweenA
andM. We discuss only the first situation, shown in Figure 93, and leave the second,
which is analogous, to the reader.
LetA(x)denote the area of the polygonx. FromA(BCDE)=A(ABEF )it fol-
lows that