Advanced book on Mathematics Olympiad

Geometry and Trigonometry 605

=f^2 (−→v ×−→v′)×(−→v ×−→v′′).

By thecab-bacformula this is further equal to

f^2 (−→v′′·(−→v ×−→v′)−→v −−→v ·(−→v ×−→v′)−→v)=f^2 ((−→v ×−→v′)·−→v′′)−→v.

The equation reduces therefore to

f^2 ((−→v ×−→v′)·−→v′′)−→v =−→v.

By hypothesis−→v is never equal to

−→

0 , so the above equality implies

f=

1

√

(−→v ×−→v′)·−→v′′

.

So the equation can be solved only if the frame(−→v,−→v′,−→v′′)consists of linearly inde-
pendent vectors and is positively oriented and in that case the solution is

−→u =√^1

Vo l(−→v,−→v′,−→v′′)

−→v ×−→v′,

where Vol(−→v,−→v′,−→v′′)denotes the volume of the parallelepiped determined by the three
vectors.
(Revista Matematica din Timi ̧soara ̆ (Timi ̧soara Mathematics Gazette), proposed by
M. Ghermanescu) ̆

579.(a) Yes: simply rotate the plane 90◦about some axis perpendicular to it. For example,
in thexy-plane we could map each point(x, y)to the point(y,−x).
(b) Suppose such a bijection existed. In vector notation, the given condition states that

(−→a −

−→

b)·(f (−→a)−f(

−→

b))= 0

for any three-dimensional vectors−→a and

−→

b.
Assume without loss of generality thatfmaps the origin to itself; otherwise,g(−→p)=
f(−→p)−f(

−→

0 )is still a bijection and still satisfies the above equation. Plugging

−→

b =
( 0 , 0 , 0 )into the above equation, we obtain that−→a ·f(−→a)=0 for all−→a. The equation
reduces to

−→a ·f(−→b)−−→b ·f(−→a)= 0.

Givenany vectors−→a,

−→

b,−→c and any real numbersm, n, we then have

m(−→a ·f(

−→

b)+

−→

b ·f(−→a))= 0 ,