Mathematical Problem

1. with the help of a ruler, the following method can be used,
using a point A on the circle as center, use the compass to draw circle A (any radius) passing through the original circle at point B,
Use point B as center with the same radius, draw another circle passing through circle A at point C and D, these two points C and D have the same distance between point A and B (CA=CB=DA=DB).
Joining points C and D, a diameter of the original circle is obtained.
Then, repeat the above steps to obtain another diameter of the original circle, the interception point of the two diameters is the center of the original circle.

2.
Let D be a point on the circumscribed circle (not on arc BAC),
O be the center of the circumscribed circle,
AB=AC,angle ABC=30*, then angle ACB = 30*, angle BAC=120*
angle BDC=180-120=60*
angle BOC = 2 x 60 = 120*
So, triangle BOC congruent to triangle BAC
Thus, radius of the circumscribed circle = AB = (3sqrt(3)/2) / cos30* = 3
Therefore, length of arc BAC = 2pi(3) x (120/360) =2pi

3.
2x^2+4x+2
=2(x^2+2x+1)
=2(x+1)^2