Number 3 is very easy.
Dividng the given equations give x^{2}y^{2}z^{2}=(x^{2}-xy+y^{2})(y^{2}-yz+z^{2})(z^{2}-zx+x^{2}).
Equality is confirmed by A.M-G.M.
question 3:
http://www.mathlinks.ro/viewtopic.php?p=1745211#1745211
(solved by prophet sir)
question 6:
http://www.mathlinks.ro/viewtopic.php?t=325459
(solved by prophet sir)
question 1:
http://www.mathlinks.ro/viewtopic.php?t=325446
Number 3 is very easy.
Dividng the given equations give x^{2}y^{2}z^{2}=(x^{2}-xy+y^{2})(y^{2}-yz+z^{2})(z^{2}-zx+x^{2}).
Equality is confirmed by A.M-G.M.
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 (a^{3}+b^{3}) and (a^{3}-b^{3})
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.
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
I got the answer of second question as all primes, twice the primes, 8 and 9.
Did someone else get the same?
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.