1Yes , I meant " O " only , not zero .
Suppose we construct a plane such that it only contains points whose ordinates as well the abscissae are integers . Each point of this plane is labeled by a positive integer . Each of these numbers is -
1 > The arithmetic mean of its four neighbours ( up , down , right , left ) . Find a relation among all the labels if the origin is marked by " 0 " .
2 > Repeat the same task if the mean is a " geometric one " .
3 > Repeat the same task if the mean is a " harmonic one " .
Note : - Parts ( 2 ) and ( 3 ) are completely of my own device .
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UP 0 DOWN 0 0 12
12 Answers
21q seems unclear to me..
bcz the origin also has 4 neighbours , wch r all +ve.
so avg can't be zero..
62i guess we can change that to 3
or any other integer... and that he meant o not 0
1consider a hill.from peak u can only move down.
keeping this in mind when we concentrate on highest valued point and it's neighbour it becomes obvious that all numbers must be equal![1]
1" All the labels are same " - Does this hold true for all the three cases ?
21not for ur devices,[3] havn tried them (but i will)
the one out of ur device use Extremal principle (pick up the maximum/minimum element) and PHP 3.
1ricky bhaiya,
i understand that means are smaller than LARGEST entity involved...(not sure in ase of ur devices[3]..as mentioned by shubhodip bhaiya above)...so i think the analogy "from peak u can only come down" should hold regardless of what kind of mean we consider!
62A closed set bounded below has a minimal element.
So this whole plane will contain a local minima (atleast one)
We consider the neighbors of the element and we have to have have the remaining 4 elements to be equal. ...
Now we can proceed by induction and it is easy to prove that the whole plane will have equal elements.
1Shubhodip , Nishant Sir and Chinmaya - Great job done indeed .
To complete and construct a formal proof -
Consider a minimal label " m " , whose neighbours are - " a , b , c , d " .
1 . We have , m = a + b + c + d4
But we have , ( a , b , c , d ) ≥ m .
If any of these inequalities were to be strict , then we would have had -
4 m > 4 m . . . . . . - A contradiction !
Hence , all the indivdual numbers must be equal , and equal to " m " .
Since , we denoted the origin as " O " , so all the labels must be " O " only .
2 . By the same approach , we arrive at m 2 > m . . . . . . - A contradiction !
So the answer remains same .
3 . Here also , the answer is invariant .
Note that if the word " positive integer " is removed and " positive reals " is used instead , then we would have got into trouble .
Problem source : - Problem Solving Strategies : Arthur Engel
As I mentioned before , part " 2 and 3 " were easily deduced .
62chinmaya...
Initially when i used to look at prophet sir's post i would get clean bowled on how a person can think like this..
Then fortunately, i read this book...
(Trust me it is not at all needed for IIT JEE) but then if you have interest and love for maths and if your heart beat/ blood circulation increases after solving maths problems correctly, you could spend 10% time on this...
Read Arthur Angel (Problem Solving Strategies)