1.5

Answer:

1.

First, when there are five points, the inscribed regular pentagon of a circle satisfies the condition.

Now let's prove that there must be no point distribution at 6 o'clock, and the distance between them is greater than the radius of the circle.

Because every two distances are greater than the radius, the center of the circle must not be one of these six points, because if the center of the circle is a point, the distances from other points to it are less than or equal to the radius.

Suppose these six points are A0~A5, starting from the center of the circle O and ending at A0~A5 respectively, and make six rays OA0, OA 1, ..., OA5,

Intersection B0, B 1, ..., B5 are respectively; Let B0~B5 be arranged clockwise on the circumference, thus forming six chords B0B 1, B 1B2, ..., B5B0, which connect A0A 1, A 1A2, ..., A5A0 and OA/kloc-0 in the triangle. So A0A 1 is the longest side of the triangle OAA 1, so the angle A0A 1 is greater than 60 degrees. Similarly, A 1OA2, ..., A5OA0 are all greater than 60 degrees, and the sum of these six angles is greater than 360 degrees, but the sum of the six angles is equal to 360 degrees, which is contradictory! The proof of more than 6 points is the same. So it is five points at most.

2.

In any quadrilateral ABCD, make △ABE make ∠BAE=∠CAD ∠ABE=∠ ACD, because △ABE∽△ACD, BE/CD=AB/AC, that is, be AC = AB CD (1

There is also a proportional formula AB/AC=AE/AD, ∠BAC=∠DAE.

So △ABC∽△AED is similar. BC/ED=AC/AD means ED AC = BC AD (2).

(1)+(2), get AC (be+ed) = ab CD+ad BC, and AC. BD = ab CD+ad BC, so E is on BD.

So ∠ABD=∠ ACD, so ABCD is a four-point circle (diagonally equal, four-point circle).

3.

Let the three sides of triangle A, B and C be A, B and C respectively, and the corresponding angular bisectors are AM, BN and CL respectively; Then the area of the triangle ABC =absin (angle BCA)/2=[CL.a.sin (angle BCA/2)+CL.b.sin (angle BCA/2)]/2, so CL=absin (angle BCA)/[(a+b)sin (angle BCA/2)]. And sin (Angel ·BCA)

So cl < 2ab/(a+b); Similarly, bn < 2ac/(a+c), am & lt2bc/(b+c); So am.bn.cl; =2 root number (ab), B+C > Root number (bc) of =2, a+c >; =2 roots (AC); (a+b)(b+c)(c+a > = 8 ABC, so am.bn.cl