Advanced book on Mathematics Olympiad

662 Geometry and Trigonometry

so

2 sin

7 a

4

sin

5 a

4

= 0 ,

meaning thata = 0 ,^47 π,^45 π. It follows that the solutions to the original equation are
x= 2 ,2 cos^47 π,−^12 ( 1 +

√

5 ).

677.The points(x 1 ,x 2 )and(y 1 ,y 2 )lie on the circle of radiusccentered at the ori-
gin. Parametrizing the circle, we can write(x 1 ,x 2 )=(ccosφ,csinφ)and(y 1 ,y 2 )=
(ccosψ, csinψ). Then

S= 2 −c(cosφ+sinφ+cosψ+sinψ)+c^2 (cosφcosψ+sinφsinψ)

= 2 +c

√

2

(

−sin

(

φ+

π

4

)

−sin

(

ψ+

π

4

))

+c^2 cos(φ−ψ).

We can simultaneously increase each of−sin(φ+π 4 ),−sin(ψ+π 4 ), and cos(φ−ψ)

to 1 by choosingφ=ψ=^54 π. Hence the maximum ofSis 2+ 2 c

√

2 +c^2 =(c+

√

2 )^2.

(proposed by C. Rousseau for the USA Mathematical Olympiad, 2002)

678.Leta=tanα,b=tanβ,c=tanγ,α, β, γ∈(−π 2 ,π 2 ). Thena^2 + 1 =sec^2 α,
b^2 + 1 =sec^2 β,c^2 + 1 =sec^2 γ, and the inequality takes the simpler form

|sin(α−β)|≤|sin(α−γ)|+|sin(β−γ)|.

This is proved as follows:

|sin(α−β)|=|sin(α−γ+γ−β)|

=|sin(α−γ)cos(γ−β)+sin(γ−β)cos(α−γ)|

≤|sin(α−γ)||cos(γ−β)|+|sin(γ−β)||cos(α−γ)|

≤|sin(α−γ)|+|sin(γ−β)|.

(N.M. Sedrakyan, A.M. Avoyan,Neravenstva, Metody Dokazatel’stva(Inequalities,
Methods of Proof), FIZMATLIT, Moscow, 2002)

679.Expressions of the formx^2 +1 suggest a substitution by the tangent. We leta=tanu,
b=tanv,c=tanw,u, v, w∈(−π 2 ,π 2 ).The product on the right-hand side becomes
sec^2 usec^2 vsec^2 w, and the inequality can be rewritten as

− 1 ≤(tanutanv+tanutanw+tanvtanw− 1 )cosucosvcosw≤ 1.

The expression in the middle is simplified as follows:

(tanutanv+tanutanw+tanvtanw− 1 )cosucosvcosw

=sinusinvcosw+sinucosvsinw+cosusinvsinw−cosucosvcosw