Because o is the outer center, AO = BO = CO

Then m, n and PNP are vertically divided into a, b and c respectively;

According to Pythagorean theorem, there are

m=√(r^2-(a/2)^2)； n=√(r^2-(b/2)^2)； p=√(r^2-(c/2)^2)；

R is the radius of the circumscribed circle of the triangle; Find by sine theorem:

r = a/(2 Sina)= b/(2 sinb)= c/(2 sinc)；

Sina is solved by cosine theorem:

There is COSA = (B2+C2-A2)/(2bc);

By who? cos^2？ Answer? +? sin^2？ Answer? = 1? Looking for Sina; Then calculate r from this; So you can find out? Man: n:p? =cosA:cosB:cosC

2. Because X= 1 holds, the root of the original equation is 1. According to Vieta's theorem, the sum of two roots is C-B.

Let a/b = x.

2(x^2)(b^2)+ 1234567890xb+3=0

2x(b^2)+ 1234567890b+3/x=0

It is known that 2x=3.

x=3/2

4、

See the figure below: