Isn't that interesting? What a neat little problem.
The middle number between a and b = (a + b)/2 The middle number between b and c = (b + c) / 2 The middle number between c and d = (c + d)/2 The middle number between d and a = (d + a)/2
The sum of the numbers in the corners of diagram 1 = a + b + c + d Do you agree.
Now look at diagram two. Start by putting a, b, c and d in the corners. Now remove the brackets from what I found above. diagram 2 = a/2 + b/2 + b/2 + c/2 + c/2 + d/2 + d/2 + a/2 Now collect all the like terms. diagram 2 = a/2 + a/2 + b/2 + b/2 + c/2 + c/2 + d/2 + d/2 a/2 + a/2 = a does it not? b/2 + b/2 = b c/2 + c/2 = c d/2 + d/2 = d
The sum of the middle numbers in diagram 2 = a + b + c + d But that's the same sum as diagram 1, which was what you were asked to prove. You cannot come up with a counter example that will give a different result, at least in the positive integers.
The question provides you with room for a written answer. You are going to have to reproduce in some form what I've put in italics.
