INMO 2010

Pls don't provide the links bfore the qsns are solved here.

Number 3 is very easy.

Dividng the given equations give x²y²z²=(x²-xy+y²)(y²-yz+z²)(z²-zx+x²).

Equality is confirmed by A.M-G.M.

provided you know something about the signs of x,y,z

Very similar to the inequality u posted here a few days back.
Basically only possiblity is that only 2 reals can be negative.
So best way is again to proceed by modulus.

Can anyone confirm my ans to the 2nd one?

Primes and nos. of the form 2p, where p is a prime.

can anyone tell me the ans of 4? i got 16 tuples.........

n i got the same eq in 3 as soumik replied........
bt i used the identity (a³+b³) and (a³-b³)

http://www.mathlinks.ro/resources.php?c=78&cid=46&year=2010

this is hwere u can get all the questions and the answers..

hit on the question number to get the solution.

where the hell?? i can't find the solution of 4........

in 4, i got the condition that a1=a3=a5 and a2=a4=a6........

then tuples were of form p,q,p,q,p,q where p,q are from {1,2,3,4} ........

its been answered a little later:

see this q4:

http://www.mathlinks.ro/viewtopic.php?p=1746110&sid=0a2e71e9caa941aacecd0ccbae696798#1746110

5)

Easy to see that PQ bisects OH

Its wellknown that the midpoint of OH is the ninepoint center N of triangle ABC

Let the midpoints of the segments AB,AC,BH,CH be P,T,S,R respectively.

Then PTRS is a rectangle. Since RS is the perpendicular OK , the circumcenter P of triangle OKH lies on the line RS.

Since N is the nine point center of ABC, a half turn about N maps S to T and R to P. So N is the center of the rectangle PTRS. Since P lies in RS, and NP= NQ, we obviously have that Q lies on PT and we are done.