You are usign the highest sum you can find in the grid. In the 2x2 case, that is 6. If you arrange the numbers any other way, you'll get a higher maximum sum. For example:
For the second, 4x4 grid, the highest sum inside it is 19. To prove that this is the minimal highest sum, we would need to prove that for all other arrangements of numbers from 1 to 16 into the grid, there will be two adjacent numbers which sum to 19 or higher.
You look at all possible fillings of the grid and for each filling you see what the maximum sum of adjacent ellements is. Then you find the minimum of those numbers.
For example, for n=1, you are filling a 2x2 grid with numbers 1,2,3,4. The lowest adjacent sum, which is 6, can be obtained with the filling:
For n=2, a filling with S=19 *can* be found, but I am not sure if there is a lower one (probably not):
These threads are a fantastic idea! Nice work.
An intersting fact about the Nine Point Circle (I also know it under the name of Euler's Circle) is that when you apply inversion around the Nine Point Circle, the circumcircle gets mapped into the incircle and vice versa!
You can prove that there is at least one number c in (a,b) such that f(c)=C by the Intermediate Value Theorem. To prove that there is only one such number, we will take any two numbers, c1 and c2, such that f(c1)=f(c2)=C. If c1 were less than c2, than f(c1) would be less than f(c2) which is not possible. Also, if c1 were greater than c2, then f(c1) would be greater than f(c2), which is impossible as well. Thus, c1 must be equal to c2, which proves that there is exactly one such number.