MHB How can you solve the eight-digit challenge?

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Place the digits 1 through 8 in the boxes
so that no two consecutive digits are adjacent
(not vertically, horizontally or diagonally).
. . \begin{array}{cccccccccc}&& * & - & * & - & * \\ && | && | && | \\ * &-& * &-& * &-& * &-& * \\ | && | && | && | && | \\ * &-& * &-& * &-& * &-& * \\ && | && | && | \\ && * & - & * & - & * \end{array}
 
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The two central boxes are adjacent to the most other boxes, so it makes sense to put 1 and 8 in them to maximize our options (since 0 and 9 are not valid digits). By horizontal symmetry, it doesn't really matter what order we put them in. So we start with:​

Code:
  ? ?
? 1 8 ?
  ? ?

Here we automatically deduce that 7 must go in the leftmost box and 2 in the rightmost box, otherwise there would be two adjacent consecutive numbers. Hence:

Code:
  ? ?
7 1 8 2
  ? ?

Hence we have 3, 4, 5, and 6 left to place. We see that 6 must be in one of two boxes on the right, and 3 must be in one of the boxes on the left. Furthermore, 4 and 5 cannot be put adjacent to one another. This forces the following configuration (up to symmetry):

Code:
  3 5
7 1 8 2
  4 6

And the puzzle is solved.
 
Seemingly by some mathematical coincidence, a hexagon of sides 2,2,7,7, 11, and 11 can be inscribed in a circle of radius 7. The other day I saw a math problem on line, which they said came from a Polish Olympiad, where you compute the length x of the 3rd side which is the same as the radius, so that the sides of length 2,x, and 11 are inscribed on the arc of a semi-circle. The law of cosines applied twice gives the answer for x of exactly 7, but the arithmetic is so complex that the...
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