geometry
posted by Anonymous...please help! on .
7 points are placed in a regular hexagon with side length 20. Let m denote the distance between the two closest points. What is the maximum possible value of m?

Is there a figure that you were given? This question does not make much sense to me.

no.....there was no figure

What exactly is the question asking?
It would make sense if they were asking the greatest distance between 2 points but asking what shortest distance between 2 points does not make sense.
Perhaps I'm not understanding the question properly. If you could expand on what the question is asking perhaps I could show you how to do the math involved. 
i have another problem can u try it???

Sure; post it and I'll give it a go.

ok

How many numbers from 1 to 100 are multiples of 3 but not 5?

OK.
So there are 33 multiples of 3 that are < or = 100.
3,6,9,12,15,18,21,24,27,30,33, 36,39,42,45,48,51,54,57,60,63,66,69,72,75,78,81,84,87,90,93,96,99 ect.
There are 20 multiples of 5 that are < or = 100.
5,10,15,20,25,30,35,40,45,50,55,60,65,70,75,80,85,90,95,100
The ones that match are the one's you would cancel out: 15,30,45,60,75,90
336 = 27
NOW THE EASIER WAY TO DO THIS:
I think the easier way to do this would be to figure out the # of multiples of 3 that are < = 100 which is 33 and then subtract all of the multiples of 15 (3x5) up to 100 and there are 6 (15,30,45,60,75,90)
336 = 27 
Not sure how to do the hexagon problem, but it's asking how to figure the greatest minimum separation of points in the hexagon.
If you try placing them at the vertices and the center, then the smallest distance is the side length. Surely they can be distributed in such a way that the two closest points are farther apart than that.
Gotta think on it some. Maybe work my way up from fewer points or a different polygon. 
Actually, it appears that the center and vertices are the optimal positions, since the radius of the circumcircle is the same as the side length.
If you move any point on the vertices, it will be closer to the center. If you move the point on the center, it will be closer to one of the vertices.
I'm sure this happy arrangement only works for the hexagon, where the radius is equal to the side length. 
steveur right.. we can place six points on each of it's veritices...nd if we put the other point on it's center,the maximum distance between the two closest points will be same as the length of it's side length...so the answer is 20.