The smallest of nine consecutive integers is 2012. These nine integers are placed in the circles to the right. The sum of the three integers along each of the four lines is the same. If this sum is as small as possible, what is the value of u?
One version:
1--Label the circles a, b, c, d, u, f, g, h and j.
2--By inspection, a+b = d+u, c+d = f+g, u+f = f+j
3--u must be the median value which must be 16.
4--Working outward from u = 16, let d = 17 and f = 18.
5--u+d = 33 and u+f = 34.
6--Therefore, c = 15 and g = 14 resulting in sums of 48.
7--a+b must therefore equal 33 and h+j must therefore be 32.
8--Thus, a = 20, b = 13, h = 17 and j = 15.
9--The sum of each of the 4 legs is 48.
Just another viewpoint.
The thought entered my mind that the number 2012 might be the lowest possible value for "u".
We'll see.
