Some blogs away I had posted several interesting, non-trivial, yet do-able-with-some-effort problems for preRMO and RMO.
A student of mine, RI has submitted the following beautiful solution to the chess problem. I am reproducing the question for convenience of the readers:
The sixty four squares of a chess board are filled with positive integers one on each in such a way that each integer is the average of the integers on the neighbouring squares. (Two squares are neighbours if they share a common edge on vertex. Thus, a square can have 8, 5 or 3 neighbours depending on its position.) Show that all the sixty four entries are in fact equal.
Answer (by RI):
Let us denote the set of all integers on the chess board by S (assume they are distinct). [Now, we can use the Well-ordering principle: every non-empty set of non-negative integers contains a least element. That is, every non-empty set S of non-negative integers contains an element a in S such that for all elements b of S}. So, also let “a” be the least element of set S here. As it is the average of the neighbouring elements, it can’t be less than each of them. But it can’t be greater than all of them also. So, all the elements of S are equal.
Three cheers for RI 🙂 🙂 🙂