Advanced book on Mathematics Olympiad

614 Geometry and Trigonometry

A B

D C

H

G

E

F

Figure 75

LetEbe the origin of the rectangular system of coordinates, with lineEBas the
y-axis. Let alsoA(−a, 0 ),B( 0 ,b),C(c, 0 ), wherea, b, c >0. We have to prove that
b^2 =ac.
By standard computations, we obtain the following equations and coordinates:

lineGF: y=

c−a

2 b

x;

lineBC:

x

c

+

y

b

= 1 ;

pointF: xF=

2 b^2 c

2 b^2 +c^2 −ac

,yF=

cb(c−a)

2 b^2 +c^2 −ac

;

lineAB:−

x

a

+

y

b

= 1 ;

pointG: xG=

2 ab^2

− 2 b^2 +ac−a^2

,yG=

ab(c−a)

− 2 b^2 +ac−a^2

.

The conditionEG=EFis equivalent toxF=−xG, that is,

2 b^2 c

2 b^2 +c^2 −ac

=

2 ab^2

2 b^2 −ac+a^2

.

This easily givesb^2 =acora=c, and since the latter is ruled out by hypothesis, this
completes the solution.
(Romanian Mathematics Competition, 2004, proposed by M. Becheanu)

596.The inequality from the statement can be rewritten as

−

√

2 − 1

2

≤

√

1 −x^2 −(px+q)≤

√

2 − 1

2

,

or

√

1 −x^2 −

√

2 − 1

2

≤px+q≤

√

1 −x^2 +

√

2 − 1

2

.