If the odd points: 1, 3 and 5 are moved a bit towards the centre, the longest distance gets shorter quickly and the shortest distance doesn't shrink so fast, so it looks like the ratio would decrease until you get to:
.......1
...2.....3
4.....5.....6
?
Puzzle doesn't give or require an estimate of actual distances. It only cares about the the relative value or ratio of y to x. Let's asume the unit of measure is a speck of dust. Seems to follow that y would equal two specks and x one speck. x/y = 2/1 = 2.0. Believe the solution to the puzzle is that simple, and that attempting to complicate the soulution with considerations not already given only succeeds in compromising the logic and introducing error to the solution.
In which 2346 is a square (x=1), the maximum distance between 3 and 1 (or 4 and 1, 3 and 5, 2 and 5), is y = sqrt ( 0,5^2 + (1 + sqrt(1-0,5^2))^2 ) = 1,932
I am a little confused by this. When I analyze your configuration of points, I get that angle DFB must be around 24.27 degrees, give or take. From that I get that the ratio of one side of square ABFE to AC (=DC=CE=DB=DF) is roughly 1.823, as you state. But we are instead interested in y/x, where here y should be the square's diagonal. So, I think y/x for this configuration would have to be ~2.58, or about sqrt(2)*1.823. I may be missing something...
Originally posted by iamatiger ...1...6...
2.........5
...3...4...
If the odd points: 1, 3 and 5 are moved a bit towards the centre, the longest distance gets shorter quickly and the shortest distance doesn't shrink so fast, so it looks like the ratio would decrease until you get to:
.......1
...2.....3
4.....5.....6
In that case, the shortest distance is 4 to 5, which is the same as 5 to 6; and the longest distance is 4 to 6, which is twice as long. Hence, this ratio is still exactly 2.
Originally posted by Thomaster ...1
2....6
........5
3....4
In which 2346 is a square (x=1), the maximum distance between 3 and 1 (or 4 and 1, 3 and 5, 2 and 5), is y = sqrt ( 0,5^2 + (1 + sqrt(1-0,5^2))^2 ) = 1,932
That's still worse than 4*sin(54)*cos(54) = 1.90211... .
Suppose you have five different points. Suppose that the smallest distance between any two of the five points is x. Suppose that the largest distance between any two of the five points is y. Under this scenario, what is the minimum possible value of y/x?
Originally posted by LemonJello I am a little confused by this. When I analyze your configuration of points, I get that angle DFB must be around 24.27 degrees, give or take. From that I get that the ratio of one side of square ABFE to AC (=DC=CE=DB=DF) is roughly 1.823, as you state. But we are instead interested in y/x, where here y should be the square's diagonal. So, I think y/x ...[text shortened]... is configuration would have to be ~2.58, or about sqrt(2)*1.823. I may be missing something...
This seems correct. I don't know if this is how you got your values, but if s is the side length of the outer square, then the distance from C to one of the horizontal sides is s/2.
A_______B
________
__C__D__
________
E_Z_____F
Using the point where it intersects the horizontal side (Z), we can draw a right triangle CZE, with the remaining side of length (s-x)/2. Then: x^2 = (s/2)^2+((s-x)/2)^2 which has only one positive solution s = (1+sqrt(7))/2*x. This is approximately the 1.823*x, which means the diagonal of the square AF is sqrt(s)*1.823*x.
Originally posted by Shallow Blue Here's a nice variation:
Suppose you have five different points. Suppose that the smallest distance between any two of the five points is x. Suppose that the largest distance between any two of the five points is y. Under this scenario, what is the minimum possible value of y/x?
Richard
Interesting! My initial intuition tells me 1/2(sqrt(5)+1) = 1.618...
Originally posted by iamatiger If the odd points: 1, 3 and 5 are moved a bit towards the centre, the longest distance gets shorter quickly and the shortest distance doesn't shrink so fast, so it looks like the ratio would decrease until you get to:
.......1
...2.....3
4.....5.....6
?
Nice to see some food for thought...
I think the problem with such configurations is that you have 3 points on a line, so that can never work as y will be at least 2x...
Seems that if the number of points n is even, the best is to take a regular polygon with n-1 sides and the centroid. If n is odd then the best is to take the regular polygon with n sides. Is this correct?
For 4 points:
equilateral triangle with centre point:
longest = side x.
shortest = "inradius" = x*sqrt(3)/6
so longest/shortest = 6/sqrt(3) = 3.46410162
if the 4 points are in a square:
shortest = side x.
longest = x*sqrt(2)
so longest/shortest = sqrt(2) = 1.41421356
For N = 4 the rule that even N is best with a centre point does not work
Originally posted by iamatiger For 4 points:
equilateral triangle with centre point:
longest = side x.
shortest = "inradius" = x*sqrt(3)/6
so longest/shortest = 6/sqrt(3) = 3.46410162
if the 4 points are in a square:
shortest = side x.
longest = x*sqrt(2)
so longest/shortest = sqrt(2) = 1.41421356
For N = 4 the rule that even N is best with a centre point does not work
...[text shortened]... lateral triangle) 1/1 = 1
4 points (square) sqrt(2)/1 = 1.41421356
5 points ?
6 points ?
Yeah, you're correct that the square is better, although something's wrong with your calculation, it should be sqrt(3) the ratio (for the triangle with the centroid).
Six Points
02 Sep '11 18:20
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