Advanced book on Mathematics Olympiad

Geometry and Trigonometry 617

m−a

b−a

=

n−c

b−c

=

p−c

d−c

=

q−a

d−a

=,

where=cosπ 3 +isinπ 3. Therefore,

m=a+(b−a), n=c+(b−c),

p=c+(d−c), q=a+(d−a).

It is now easy to see that^12 (m+p)=^12 (n+q), meaning thatMPandNQhave the
same midpoint. So either the four points are collinear, or they form a parallelogram.
(short list of the 23rd International Mathematical Olympiad, 1982)

601.We refer everything to Figure 77. The triangleBAQis obtained by rotating the
trianglePACaroundAby the angleα. Hence the angle between the linesPCandBQ

is equal toα. It follows that in the circumcircle ofBRC, the measure of the arc


BRCis
equal to 2α, and this is also the measure of∠BOC. We deduce thatOis obtained from
Bthrough the counterclockwise rotation aboutCby the complement ofαfollowed by
contraction by a factor of 2 sinα.

A

P

O

B

R

C

Q

Figure 77

Now we introduce complex coordinates with the origin atA, with the coordinates
ofBandCbeingbandc. Setω =eiα, so that the counterclockwise rotation byα
is multiplication byω, and hence rotation by the complement ofαis multiplication by
i/ω=iω ̄. Then the coordinatezofOsatisfies

z−c

b−c

=

1

2 sinα

·

i

ω

,

from which we compute

z=

b−c

2 sinα

·

i

ω

+c=

b−c

−i(ω− ̄ω)

·

i

ω

+c=

b−c

1 −ω^2

.