Maths Olympiad Prep

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Algebra Difficulty 5.0 AIME, harder Find the answer

Let ff be a function that takes in a triple of integers and outputs a real number. Suppose that ff satisfies the equations f(a,b,c)=f(a+1,b,c)+f(a−1,b,c)2f(a, b, c) =\frac{f(a+1, b, c)+f(a-1, b, c)}{2}, f(a,b,c)=f(a,b+1,c)+f(a,b−1,c)2f(a, b, c) =\frac{f(a, b+1, c)+f(a, b-1, c)}{2}, f(a,b,c)=f(a,b,c+1)+f(a,b,c−1)2f(a, b, c) =\frac{f(a, b, c+1)+f(a, b, c-1)}{2} for all integers a,b,ca, b, c. What is the minimum number of triples at which we need to evaluate ff in order to know its value everywhere?

A number or a short expression. Spacing and $ signs are ignored.

Solution

Note that if we have the value of ff at the 8 points: (0,0,0),(1,0,0),(0,1,0),(0,0,1),(0,1,1),(1,0,1),(1,1,0),(1,1,1)(0,0,0),(1,0,0),(0,1,0),(0,0,1),(0,1,1),(1,0,1),(1,1,0),(1,1,1), we can calculate the value for any triple of points because we have that f(a+1,b,c)−(a,b,c)f(a+1, b, c)-(a, b, c) constant for any aa, if bb and cc are fixed (and similarly for the other coordinates). To see why we cannot do this with less points, notice that we need to determine what the value of these 8 points anyways, and there is no "more efficient" way to determine them all in fewer evaluations.

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