There are n points in the unit circle, find the maximum possible value of the minimum value of the d

This sequence is a series of many cases: if n=1, it does not exist, between n=2 and 6, then divide the circumference into n parts, when n=2, the answer is 1, when n=6, the answer is, but when n=7, it is another series, and when n=7, the answer is still, (divide the circumference into 6 points of 6 parts, and add the center of the circle). I have referred to the answer of the third floor, and I found a little mistake, the third row should be divided into n*n small squares, not n, but this is only a small error, the most important mistake is that he should not use a square to divide, but a circle, and a square cannot replace a circle.

But his mind is very good.

It is almost equal to the diameter of the circle.

There are n points in the unit circle, find the maximum possible value of the minimum value of the distance between the two pairs of these n points unit circle, the area is , the area of the tangent square is 4, the square is divided into squares with a side length of sqrt[4 n], a total of n small squares are divided, if there are two points in a small square, the distance between these two points must be less than the diagonal sqrt[8 n] of the small square, if there are no more than two points in each small square, then there are points in each square.

The maximum distance between two adjacent squares is sqrt[20 n], from which the maximum distance is calculated to be sqrt[20 n].

Of course, at this time, the upper limit is not reached, and the estimated value of sqrt[20 n] is less than that of sqrt[4 n].

N points are the least distanced.

Summary. There is always a maximum, and there is always a maximum, a minimum. Calculate the normal equation for the ellipse at each point (which is simple by the polar equation at a point outside the ellipse), and when the normal of a point also passes a given point, the length of this normal segment is the maximum value of the distance; And there are only two such normals, which are exactly the two most valuable cases.

There is always the most valuable mold bucket, and there is always a maximum and a minimum value. Calculate the normal equation of the denier acre of the ellipse at each point (from the equation of the pole line at the point outside the ellipse, which is simple), when the normal of a point also passes a given point, the length of this normal segment is the maximum value of the distance; And there are only two such normals, which is the most valuable case of two stools.

If the perpendicular lines of AB are connected to a straight line through the center of the circle o and intersect at point P, point P is the point that meets the requirements.

<> point o should be divided into two kinds of vertical situations located inside the circle and outside As shown in Figure 1, when the point o is in the circle, the diameter of the imitation limb is 5+1=6cm, so the radius is 3cm;

Let the distance from p to the center of the circle be d

When p is outside the circle, 8=d+r, 6=d-r

Eliminate d to get r=1

When p is inside a circle, 8=d+r, 6=r-d

Eliminate the opening type d, and learn that hail guess r = 7

So the radius of the circle is 1 cm or 7 cm

Lou Mingyuan guessed that both upstairs and upstairs were positive relief types.

It's all cracked.

If the point is in a circle, then d=l1+l2=14cm

r=7cm, the point is outside the circle, then d=l1-l2=4cmr=2cm, and the combination is more than r=2 or 7cm

Solution: When p is outside the circle o, the farthest distance to the circle is 2r, so 2r 5 1 4, so Wang takes Sun so r 2

When p is in the circle O, the farthest distance to the circle + the closest distance 2r, so the trapped chain 2r 5+1 6, so the branch calls r 3

The maximum value of n is 5, and the minimum value is of course 2.

Since the length of the sides of the circle is equal to the radius of the circle, the distance between the two points is greater than the radius, only below the hexagon. So the maximum value is 5.

When the point p is outside the circle, the radius beam is dressed as (8-6) 2=1 When the point p is inside the circle, the radius of the slag base is (8+6) 2=7, and the straight line po has two intersection points A and B with the circle

PA and PB are the longest and the other shortest.

5 According to the nature of the regular hexagon inside the circle, the side length of the hexagon is equal to the radius of the circle, and there are 7 points (including the center of the circle). However, the problem requires that the distance between n points and pairs is greater than 1999, so it can only be a pentagon within a circle (not including the center of the circle).

Connect the p point with the center of the circle, and the two intersection points of the circle are the maximum and minimum distances, and there are two cases, carefully argue that when the p point is in the circle, the diameter is 7 + 3 = 10cm, and the radius is 5 When the p point is in the branch bench garden, the diameter of the Mengxiao Brigade is 7-3 = 4cm, and the radius is 2