Advanced book on Mathematics Olympiad

Geometry and Trigonometry 639

we recognize immediately the left-hand side to be(λ 1 −λ 2 )−→v 1 ·−→v 2. We obtain the desired
−→v 1 ·−→v 2 =0, which proves the orthogonality of the two surfaces. This completes the

solution.
(C. Ionescu-Bujor, O. Sacter,Exerci ̧tii ̧si probleme de geometrie analitica ̧si difer-
en ̧tiala(Exercises and problems in analytic and differential geometry), Editura Didactic ̆a
̧si Pedagogica, Bucharest, 1963) ̆

631.Using the algebraic identity

(u^3 +v^3 +w^3 − 3 uvw)=

1

2

(u+v+w)[ 3 (u^2 +v^2 +w^2 )−(u+v+w)^2 ],

we obtain

z− 3 =

3

2

xy−

1

2

x^3 ,

or

x^3 − 3 xy+ 2 z− 6 = 0.

This is the cubic surface from Figure 86.

Figure 86

(C. Co ̧sni ̧ta, I. Sager, I. Matei, I. Dragot ̆ ̆a,Culegere de probleme de Geometrie
Analitica ̆(Collection of Problems in Analytical Geometry), Editura Didactic ̆a ̧si Ped-
agogica, Bucharest, 1963) ̆

632.By the( 2 n+ 1 )-dimensional version of the Pythagorean theorem, the edgeLof the
cube is the square root of an integer. The volume of the cube is computed as a determinant
in coordinates of vertices; hence it is also an integer. We conclude thatL^2 andL^2 n+^1 are
both integers. It follows thatL^2 n+^1 /(L^2 )n=Lis a rational number. Because its square
is an integer,Lis actually an integer, as desired.