Advanced book on Mathematics Olympiad

664 Geometry and Trigonometry

And this is Jensen’s inequality applied to the tangent function, which is convex on( 0 ,π 2 ).

682.From the first equation, it follows that ifxis 0, then so isy, makingx^2 indeterminate;

hencex, and similarlyyandz, cannot be 0. Solving the equations, respectively, fory,z,

andx, we obtain the equivalent system

y=

3 x−x^3

1 − 3 x^2

,

z=

3 y−y^3

1 − 3 y^2

,

x=

3 z−z^3

1 − 3 z^2

,

wherex, y, zare real numbers different from 0.

There exists a unique numberuin the interval(−π 2 ,π 2 )such thatx=tanu. Then

y=

3 tanu−tan^3 u

1 −3 tan^2 u

=tan 3u,

z=

3 tan 3u−tan^33 u

1 −3 tan^23 u

=tan 9u,

x=

3 tan 9u−tan^39 u

1 −3 tan^29 u

=tan 27u.

The last equality yields tanu=tan 27u,souand 27udiffer by an integer multiple ofπ.

Therefore,u=kπ 26 for someksatisfying−π 2 <kπ 26 <π 2. Besides,kmust not be 0, since

x   =0. Hence the possible values ofkare± 1 ,± 2 ,...,± 12 ,each of them generating

the corresponding triple

x=tan

kπ

26

,y=tan

3 kπ

26

,z=tan

9 kπ

26

.

It is immediately checked that all of these triples are solutions of the initial system.

683.In the case of the sequence(an)n, the innermost square root suggests one of the

substitutionsan=2 sintnoran=2 costn, withtn∈[ 0 ,π 2 ],n≥0. It is the first choice

that allows a further application of a half-angle formula:

2 sintn+ 1 =an+ 1 =

√

2 −

√

4 −4 sin^2 tn=

√

2 −2 costn=2 sin

tn

2

.

It follows thattn+ 1 = t 2 n, which combined witht 0 =π 4 givestn = 2 nπ+ 2 forn≥0.

Therefore,an=2 sin 2 nπ+ 2 forn≥0.

For(bn)n, the innermost square root suggests a trigonometric substitution as well,

namelybn=2 tanun,n≥0. An easy induction shows that the sequence(bn)nis positive,

so we can chooseun∈[ 0 ,π 2 ). Substituting in the recursive formula, we obtain