Let T(n) be a triangular portion of the triangular grid with n points on a side. It is an unsolved problem what the maximum number of points is that can be selected from T(n) so that no three selected points are the vertices of an equilateral triangle. For small values of n the maximum appears to be 2n-2. However, for T(12), shown below, I found more than 2*12-2=22 points, no three of which are the vertices of an equilateral triangle! How many can you find?
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Examples:
Here's an example of 5 points selected from T(4), no three of which are the vertices of an equilateral triangle.
It turns out that this selection of 5 cannot be increased to 6 without three of the selected points being the vertices of an equilateral triangle. If we add the first point in the second row, we get
Notice that three of the 6 s's are the vertices of an equilateral triangle.
There is a better selection of 6 points in T(4) no three of which are the vertices of an equilateral triangle.
1
2 3
4 5 6
7 8 9 10
11 12 13 14 15
16 17 18 19 20 21
22 23 24 25 26 27 28
29 30 31 32 33 34 35 36
37 38 39 40 41 42 43 44 45
46 47 48 49 50 51 52 53 54 55
56 57 58 59 60 61 62 63 64 65 66
67 68 69 70 71 72 73 74 75 76 77 78
1 / 4 / 11 / 22 / 37 / 56 (6 points).
3 / 8 / 17 / 30 / 47 / 68 (6 points).
9 / 18 / 31 / 48 / 69 (5 points).
10 / 19 / 32 / 49 / 70 (5 points).
26 / 60 (2 points).
27 / 42 (2 points).
Till now, 26 points (if there is no mistake).
There must be more, and certainly a formula or a better way to achieve them. I´m working with the diagonals, from right to left, downards.
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Posted by pcbouhid
on 2005-08-19 00:19:03 |
till now, 26
