Advanced book on Mathematics Olympiad

Geometry and Trigonometry 619

Remark.Let us show how this geometric identity can be used to derive a trigono-
metric identity. Forn= 2 m+1,man integer,A 0 A 1 ·A 0 A 2 ···A 0 Am =A 0 A 2 m·
A 0 A 2 m− 1 ···A 0 Am+ 1 ; henceA 0 A 1 ·A 0 A 2 ···A 0 Am=

√

2 m+1. On the other hand, for
i= 1 , 2 ,...,m, in triangleA 0 OAi,AAi=2 sin 2 m^2 π+ 1. We conclude that

sin

2 π

2 m+ 1

sin

4 π

2 m+ 1

···sin

2 mπ

2 m+ 1

=

1

2 m

√

2 m+ 1.

(J. Dürschák,Matemaikai Versenytételek, Harmadik kiadás Tankönyviadó, Budapest,
1965)

604.First solution: We assume that the radius of the circle is equal to 1. Set the origin
atBwithBAthe positivex-semiaxis andtthey-axis (see Figure 78). If∠BOM=θ,
thenBP=PM=tanθ 2. In trianglePQM,PQ=tanθ 2 /sinθ. So the coordinates of
Qare

⎛

⎜

⎝

tan

θ

2

sinθ

,tan

θ

2

⎞

⎟

⎠=

(

1

1 +cosθ

,

sinθ

1 +cosθ

)

.

Thexandycoordinates are related as follows:

(

sinθ

1 +cosθ

) 2

=

1 −cos^2 θ

( 1 +cosθ)^2

=

1 −cosθ

1 +cosθ

= 2

1

1 +cosθ

− 1.

Hence the locus ofQis the parabolay^2 = 2 x−1.

O

M

P

A B

Q

Figure 78

Second solution: With∠BOM=θwe have∠POM=∠POB=θ 2. SincePQis
parallel toOB, it follows that∠OPQ=θ 2. So the triangleOPQis isosceles, and