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: