Advanced book on Mathematics Olympiad

364 Algebra

=

(

x

y

+

y

x

) 2

+

(

y

z

+

z

y

) 2

+

(

z

x

+

x

z

) 2

− 4.

Hence

m^2 +n^2 +p^2 =mnp+ 4.

Adding 2(mn+np+pm)to both sides yields

(m+n+p)^2 =mnp+ 2 (mn+np+pm)+ 4.

Adding now 4(m+n+p)+4 to both sides gives

(m+n+p+ 2 )^2 =(m+ 2 )(n+ 2 )(p+ 2 ).

It follows that

(m+ 2 )(n+ 2 )(p+ 2 )= 20042.

But 2004= 22 × 3 ×167, and a simple case analysis shows that the only possibilities are

(m+ 2 ,n+ 2 ,p+ 2 )=( 4 , 1002 , 1002 ),( 1002 , 4 , 1002 ),( 1002 , 1002 , 4 ). The desired

triples are( 2 , 1000 , 1000 ),( 1000 , 2 , 1000 ),( 1000 , 1000 , 2 ).

(proposed by T. Andreescu for the 43rd International Mathematical Olympiad, 2002)

94.LetM(a,b)=max(a^2 +b, b^2 +a). ThenM(a,b)≥a^2 +bandM(a,b)≥b^2 +a,

so 2M(a,b)≥a^2 +b+b^2 +a. It follows that

2 M(a,b)+

1

2

≥

(

a+

1

2

) 2

+

(

b+

1

2

) 2

≥ 0 ,

henceM(a,b)≥−^14. We deduce that

min

a,b∈R

M(a,b)=−

1

4

,

which, in fact, is attained whena=b=−^12.

(T. Andreescu)

95.Leta= 2 xandb= 3 x. We need to show that

a+b−a^2 +ab−b^2 ≤ 1.

But this is equivalent to

0 ≤

1

2

[

(a−b)^2 +(a− 1 )^2 +(b− 1 )^2

]

.