Advanced book on Mathematics Olympiad

606 Geometry and Trigonometry

n(−→a ·f(−→c)+−→c ·f(−→a))= 0 ,

a·f(m

−→

b +n−→c)+(m

−→

b +n−→c)·f(−→a)= 0.

Adding the first two equations and subtracting the third gives

−→a ·(mf (−→b)+nf (−→c)−f(m−→b +n−→c))= 0.

Because this is true for any vector−→a, we must have

f(m

−→

b +n−→c)=mf (

−→

b)+nf (−→c).

Therefore,fislinear, and it is determined by the images of the unit vectors

−→

i =( 1 , 0 , 0 ),

−→

j =( 0 , 1 , 0 ), and

−→

k =( 0 , 0 , 1 ).If

f(

−→

i)=(a 1 ,a 2 ,a 3 ), f (

−→

j)=(b 1 ,b 2 ,b 3 ), and f(

−→

k)=(c 1 ,c 2 ,c 3 ),

then for a vector−→x we have

f(−→x)=

⎡

⎣

a 1 b 1 c 1

a 2 b 2 c 2

a 3 b 3 c 3

⎤

⎦−→x.

Substituting inf(−→a)·−→a =0 successively−→a =

−→

i,

−→

j,

−→

k, we obtaina 1 =b 2 =c 3 =

0. Then substituting in−→a·f(

−→

b)+

−→

b·f(−→a),(−→a,

−→

b)=(

−→

i,

−→

j),(

−→

j,

−→

k),(

−→

k,

−→

i),

we obtainb 1 =−a 2 ,c 2 =−b 3 ,c 1 =−a 3.

Settingk 1 =c 2 ,k 2 =−c 1 , andk 3 =b 1 yields

f(k 1

−→

i +k 2

−→

j +k 3

−→

k)=k 1 f(

−→

i)+k 2 f(

−→

j)+k 3 f(

−→

k)=

−→

0.

Becausefis injective andf(

−→

0 )=

−→

0 , this implies thatk 1 =k 2 =k 3 =0. Then

f(−→x)=0 for all−→x, contradicting the assumption thatfwas a surjection. Therefore,

our original assumption was false, and no such bijection exists.

(Team Selection Test for the International Mathematical Olympiad, Belarus, 1999)

580.The important observation is that

A∗B=AB−

1

2

tr(AB),

which can be checked by hand. The identity is therefore equivalent to

CBA−BCA+ABC−ACB=−

1

2

tr(AC)B+

1

2

tr(AB)C.