#### Posers and Puzzles

1. 31 Aug '11 01:01
Suppose you have six different points in the plane. Suppose that the smallest distance between any two of the six points is x. Suppose that the largest distance between any two of the six points is y. Under this scenario, what is the minimum possible value of y/x?
2.  Palynka
Upward Spiral
31 Aug '11 09:08
The best I can think of gives me approximately 1.902...
3.  Grampy Bobby
31 Aug '11 19:26 / 1 edit
Originally posted by LemonJello

Suppose you have six different points in the plane. Suppose that the smallest distance between any two of the six points is x. Suppose that the largest distance between any two of the six points is y. Under this scenario, what is the minimum possible value of y/x?
y = 6 (on a scale of 1 to 6 largest distance between any two points in the plane).

x = 5 (one less, which is the minimum which could apply to all other five points).

y/x or 6/5 = 1.2 (largest distance/smallest distance between any two points).

.
4. 31 Aug '11 20:13
zero.......0/0
5. 01 Sep '11 02:04 / 1 edit
Originally posted by Grampy Bobby
y = 6 (on a scale of 1 to 6 largest distance between any two points in the plane).

x = 5 (one less, which is the minimum which could apply to all other five points).

[b]y/x or 6/5 = 1.2
(largest distance/smallest distance between any two points).

.[/b]
But are the 6 and 5 just chosen arbitrarily? If so, why not choose 11 and 10 for a ratio of 1.1. Or 101 and 100 for 1.01 and so on, ad infinitum.

I propose the following for consideration:
In order to obtain the smallest y/x ratio, y and x must be as close to equal as possible. Therefore, we want to have the 6 points as equidistant as possible.

Consider the following shape (not drawn to scale, but a helpful visual):

...1...6...
2.........5
...3...4...

Where the 6 points form a convex, regular hexagon where all 6 'edges' on the perimeter are the same distance apart. This is x. Also 1-6-4-3 forms a square. As do every set of 4 'vertices' whose 'edges' are parallel.

The largest distance here is a diagonal line along any of the squares (e.g. 1-4). This is y.

So y is the diagonal of a square and x is an edge. Therefore, the angle between them is pi/4 (45 degrees). We are only interested in the ratio y/x, so we can assign an arbitrary value to x without any loss of generality. Let x be 1. Thus y / x = y / 1 = y.

Using some trig, we now have:
cos (pi / 4) = x / y
sqrt(2) / 2 = 1 / y.

Therefore y / x = y = 2 / sqrt(2), which is approx. 1.4.
6. 01 Sep '11 03:10
Originally posted by Grampy Bobby
y = 6 (on a scale of 1 to 6 largest distance between any two points in the plane).

x = 5 (one less, which is the minimum which could apply to all other five points).

[b]y/x or 6/5 = 1.2
(largest distance/smallest distance between any two points).

.[/b]

But I can tell you that y/x = 1.2 is an impossibility here.
7. 01 Sep '11 03:14
Originally posted by TheSurgeon
But are the 6 and 5 just chosen arbitrarily? If so, why not choose 11 and 10 for a ratio of 1.1. Or 101 and 100 for 1.01 and so on, ad infinitum.

I propose the following for consideration:
In order to obtain the smallest y/x ratio, y and x must be as close to equal as possible. Therefore, we want to have the 6 points as equidistant as possible.

...[text shortened]... = x / y
sqrt(2) / 2 = 1 / y.

Therefore y / x = y = 2 / sqrt(2), which is approx. 1.4.
I think your analysis is flawed, since your 1-6-4-3 forms a rectangle, but not a square. For the regular hexagon you describe, I think y/x = 2.
8. 01 Sep '11 03:21
Originally posted by Palynka
The best I can think of gives me approximately 1.902...
I agree that works. But is it the minimum?
9.  Palynka
Upward Spiral
01 Sep '11 09:55
Originally posted by TheSurgeon
But are the 6 and 5 just chosen arbitrarily? If so, why not choose 11 and 10 for a ratio of 1.1. Or 101 and 100 for 1.01 and so on, ad infinitum.

I propose the following for consideration:
In order to obtain the smallest y/x ratio, y and x must be as close to equal as possible. Therefore, we want to have the 6 points as equidistant as possible.

...[text shortened]... = x / y
sqrt(2) / 2 = 1 / y.

Therefore y / x = y = 2 / sqrt(2), which is approx. 1.4.
But if that's a regular hexagon, the smallest distance x is the side length and the largest distance is 2x (if you draw the 6 equilateral triangles within the hexagon it becomes apparent).

I'm not sure what you mean by diagonal of a square, though. 1634 doesn't form a square, but a rectangle... 13 has size 2*height of an equilateral triangle of side x: that is sqrt(3)*x.

The value I reached above was by taking the five vertices of a regular pentagon, plus its centroid. The smallest distance in this case is the radius (x), the side length would be 2*x*sin(Pi/5) and the longest distances are the chords that connects each vertex to the two furthest away vertices. The length of these chords can be calculated using the golden ratio and the side length:

y = 1/2*(sqrt(5)+1)*2*x*sin(pi/5)

y/x = (sqrt(5)+1)*sin(pi/5) which is approximately 1.9021

But I have no clue if this is the smallest. I thought of keeping the circle of radius x and trying to put the five points on the circle in other ways but I always seem to increase the largest distance...
10. 01 Sep '11 13:23
Originally posted by LemonJello
I agree that works. But is it the minimum?
Yes, I think it is. At least, I can't find any better solution.

Richard
11.  Grampy Bobby
02 Sep '11 12:07 / 1 edit
Originally posted by LemonJello
Suppose you have six different points in the plane. Suppose that the smallest distance between any two of the six points is x. Suppose that the largest distance between any two of the six points is y. Under this scenario, what is the minimum possible value of y/x?
Lemonjello, let's let x, the smallest distance between any two points, equal one unit of whatever unit of measure is applicable to measuring the distance between the six points within the plane of the puzzle.

Let y equal two units, the minimum largest distance between two points within the finite distance calibration of the puzzle (recognizing that the y value could be 100 or 1,000 or a 1,000,000 units but the puzzle as stated must solve for the minimum value).

Then, y/x = 2/1 = 2.0 as the minimum possible value of the puzzle. Palynka's 1.902 was apparently the result of similar logic.

.
12. 02 Sep '11 15:51
Originally posted by TheSurgeon
But are the 6 and 5 just chosen arbitrarily? If so, why not choose 11 and 10 for a ratio of 1.1. Or 101 and 100 for 1.01 and so on, ad infinitum.

I propose the following for consideration:
In order to obtain the smallest y/x ratio, y and x must be as close to equal as possible. Therefore, we want to have the 6 points as equidistant as possible.

...[text shortened]... = x / y
sqrt(2) / 2 = 1 / y.

Therefore y / x = y = 2 / sqrt(2), which is approx. 1.4.
Looking at

...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

?
13. 02 Sep '11 17:37
Let's suppose we have five points in a circle and the sixth point is in the center of the circle with all five points touching each other and touching the center point. This would make the value of x as zero and the vulue of y as the diameter of the center point. So now we have an unknown value for y and zero for x. Since any number divided by zero becomes an undefined number that would approach infinity I don't see there is an answer.....?/0
14. 02 Sep '11 18:10
Originally posted by Grampy Bobby
Lemonjello, let's let [b]x, the smallest distance between any two points, equal one unit of whatever unit of measure is applicable to measuring the distance between the six points within the plane of the puzzle.

Let y equal two units, the minimum largest distance between two points within the finite distance calibration of the puzzle (rec ...[text shortened]... possible value of the puzzle. Palynka's 1.902 was apparently the result of similar logic.

.[/b]
Stipulating that x=1 (arbitrary unit) is fine; I have no problem with that. But why do you then stipulate that y=2? I guess I still do not understand your analysis.

But, at any rate, since Palynka has already demonstrated that y/x = 1.902... is possible; then y/x = 2 cannot be the answer we are looking for in this puzzle, since we are looking for the minimum possible.
15. 02 Sep '11 18:12
Originally posted by kiki46
Let's suppose we have five points in a circle and the sixth point is in the center of the circle with all five points touching each other and touching the center point. This would make the value of x as zero and the vulue of y as the diameter of the center point. So now we have an unknown value for y and zero for x. Since any number divided by zero becomes an undefined number that would approach infinity I don't see there is an answer.....?/0
Points in the plane that "touch each other" would have to be identical with each other, no? But, in the problem statement I tried to be careful to say that we are considering six "different" points. Hence, I am trying to rule out any cases where x=0.