Bariz - Problem

Source: Central American Olympiad 2000, problem 4

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Jutaro

09.08.2007 08:45

Write an integer on each of the 16 small triangles in such a way that every number having at least two neighbors is equal to the difference of two of its neighbors. Note: Two triangles are said to be neighbors if they have a common side. [asy][asy]size(100); pair P=(0,0); pair Q=(2, 2*sqrt(3)); pair R=(4,0); draw(P--Q--R--cycle); pair B=midpoint(P--Q); pair A=midpoint(P--B); pair C=midpoint(B--Q); pair E=midpoint(Q--R); pair D=midpoint(Q--E); pair F=midpoint(E--R); pair H=midpoint(R--P); pair G=midpoint(R--H); pair I=midpoint(H--P); draw(A--I); draw(B--H); draw(C--G); draw(I--D); draw(H--E); draw(G--F); draw(C--D); draw(B--E); draw(A--F);[/asy][/asy]

What about triangles with three neighbours? If the condition aplies to any pair of his neighbours, then if we call it's neighbours a, b and c and the number itself n, then we can conlude the following: Since n can't be 0 (because of riddler's and 10000th User's comments), a,b and c must be diferent and we can say that a is the smalest followed by b, then b-a=n and c-a=n, thus b=c which contradiscts itself and also means that, because n is also b-c, then n=0. So either 10000th User is wrong, and 0's are allowed or there must be some other condition. If 0's are allowed, then riddler's idea is one of the solutions, the other solutions are filling every triangle with three neighbours with a 0 and every other triangle with the same integer which may be 0.