Advanced book on Mathematics Olympiad

626 Geometry and Trigonometry

=n−

∑n

k= 1

cos

(

α+

2 (k− 1 )π

n

)

=n−cosα

∑n

k= 1

cos

(

2 (k− 1 )π

n

)

+sinα

∑n

k= 1

sin

(

2 (k− 1 )π

n

)

=n.

By the Cauchy–Schwarz inequality,
(
1
1 −cosα 1

+

1

1 −cosα 2

+···+

1

1 −cosαn

)

≥

n^2

( 1 −cosα 1 )+( 1 −cosα 2 )+···+( 1 −cosαn)

=

n^2

n

=n.

The equality case would imply that allαk’s are equal, which is impossible. Hence the
inequality is strict, as desired.
(Romanian Mathematical Olympiad, 2004, proposed by C. Popescu)

613.We solve part (e). Choose a coordinate system such thatB=(− 1 , 0 ),C=( 1 , 0 ),
S=( 0 ,

√

3 ),S′=( 0 ,−

√

3 ). Assume that the ellipse has vertices( 0 ,±k)withk>

√

3,

so its equation is

x^2

k^2 − 3

+

y^2

k^2

= 1.

If we setr=

√

k^2 −3, then the ellipse is parametrized byA=(rcosθ,ksinθ). Parts
(a) through (d) are covered by the degenerate situationk=

√

3, when the ellipse becomes
the line segmentSS′.
LetA=(rcosθ,ksinθ)withθnot a multiple ofπ. Consider the pointsD,E,F,
respectively, onBC,AC,AB, given by

D=((r+k)cosθ, 0 ),

E=

(

( 2 k^2 +rk− 3 )cosθ+k−r

r+ 2 k+3 cosθ

,

k( 2 r+k)sinθ

r+ 2 k+3 cosθ

)

,

F=

(

( 2 k^2 +rk− 3 )cosθ−k+r

r+ 2 k−3 cosθ

,

k( 2 r+k)sinθ

r+ 2 k−3 cosθ

)

.

The denominators are never zero sincer≥0 andk≥

√

3. The linesAD,BE, andCF
   intersect at the point

P=

(

r+ 2 k

3

cosθ,

2 r+k

3

sinθ

)

,

as one can verify, usingr^2 =k^2 −3, that