Question 389235
The sum of all the entries 1, 2, ..., 12 is equal to (12*13)/2 = 78. If we add each side length, as in (x_1 + ...x_4) + (x_4 +... + x_7) + (x_7 + ...+x_10) + (x_10 + ...+ x_1), this sum is equal to 22*4 = 88. However, this counts all x_i, but counts each corner square twice. Thus,


{{{sum(x_i, i=1, 4) + sum(x_i, i=4, 7) + sum(x_i, i=7, 10) + (x_10 + x_11 + x_12 + x_1) - sum(x_i, i=1, 12) = sum(x_3i-2, i=1, 4) = 88 - 78 = 10}}}


So the sum of the four corner entries is 10.