666 Geometry and Trigonometry
xn=tan(
90 ◦−
30 ◦
2 n−^1)
=cot(
30 ◦
2 n−^1)
=cotθn, whereθn=30 ◦
2 n−^1.
A similar calculation shows thatyn=tan 2θn=2 tanθn
1 −tan^2 θn,
which implies that
xnyn=2
1 −tan^2 θn.
Because 0◦<θn< 45 ◦, we have 0<tan^2 θn<1 andxnyn>2. Forn>1, we have
θn< 30 ◦, and hence tan^2 θn<^13. It follows thatxnyn<3, and the problem is solved.
(Team Selection Test for the International Mathematical Olympiad, Belarus, 1999)
685.Leta=tanx,b=tany,c=tanz, wherex, y, z∈( 0 ,π 2 ). From the identity
tan(x+y+z)=tanx+tany+tanz−tanxtanytanz
1 −tanxtany−tanytanz−tanxtanzit follows thatabc=a+b+conly ifx+y+z=kπ, for some integerk. In this case
tan( 3 x+ 3 y+ 3 z)=tan 3kπ=0, and from the same identity it follows that
tan 3xtan 3ytan 3z=tan 3x+tan 3y+tan 3z.This is the same as
3 a−a^3
3 a^2 − 1·
3 b−b^3
3 b^2 − 1·
3 c−c^3
3 c^2 − 1=
3 a−a^3
3 a^2 − 1+
3 b−b^3
3 b^2 − 1+
3 c−c^3
3 c^2 − 1,
and we are done.
(Mathematical Olympiad Summer Program, 2000, proposed by T. Andreescu)
686.With the substitutionx=cosht, the integral becomes
∫
1
sinht+cosht
sinhtdt=
∫
et−e−t
2 et
dt=1
2
∫
( 1 −e−^2 t)dt=1
2
t+e−^2 t
4+C
=
1
2
ln(x+√
x^2 − 1 )+1
4
·
1
2 x^2 − 1 + 2 x√
x^2 − 1+C.
687.Suppose by contradiction that there exists an irrationalaand a positive integern
such that the expression from the statement is rational. Substitutea=cosht, wheretis
an appropriately chosen real number. Then