ThorganHazard

\(\textrm{Let}\quad\overline{BC}=a,\quad\overline{CA}=b,\quad\overline{AB}=c\\ \frac{a}{sinA}=\frac{b}{sinB}=\frac{c}{sinC}=2R\quad (\textrm{R=radius of the circumscribed circle of triangle ABC})\\ \therefore sinA=\frac{a}{2R},\quad sinB=\frac{b}{2R},\quad sinC=\frac{c}{2R}\\ cosA=\frac{b^2+c^2-a^2}{2bc},\quad cosB=\frac{c^2+a^2-b^2}{2ca},\quad cosC=\frac{a^2+b^2-c^2}{2ab}\\ \therefore \frac{c}{2R}=\frac{\frac{a}{2R}+\frac{b}{2R}}{\frac{b^2+c^2-a^2}{2bc}+\frac{c^2+a^2-b^2}{2ca}}\\ \quad c=\frac{2abc(a+b)}{a(b^2+c^2-a^2)+b(c^2+a^2-b^2)}=\frac{2abc(a+b)}{ab^2+c^2a-a^3+bc^2+a^2b-b^3}=\frac{2abc(a+b)}{c^2(a+b)-(a-b)^2(a+b)}=\frac{2abc}{c^2-(a-b)^2}\\ \therefore c^2-(a-b)^2=2ab\\ \quad c^2=2ab+(a-b)^2=a^2+b^2\\ \therefore \textrm{Triangle ABC is a right triangle that has angle C as the right angle.}\)

Let the crossing point between AQ and PM B.

PA+AB+PB=PA+AQ+QB+PB

              =(PA+AQ)+(PB+BQ)

              =(PA+AN)+(PB+BM)

              =PN+PM

              =22cm

\(f(x)=((x+3)^2)^{50}-4x+3=(x+3)^{100}-4x+3\)

\((r_i+3)^{100}-4r_i+3=0 (i=1, 2, ...,100)\)

\((r_i+3)^{100}=4r_i-3\)

Therefore, \((r_1+3)^{100}+(r_2+3)^{100}+...+(r_{100}+3)^{100}=(4r_1-3)+(4r_2-3)+...+(4r_{100}-3)\)

\( =4(r_1+r_2+...+r_{100})-300\)

\( =4*(-300)-300=-1500\)

(By Vieta's formulas, \(-(r_1+r_2+...+r_{100})\) equals the coefficient of \(x^{99}\) of \(f(x)\), which is \(nCr(100, 1)*3^1=300\) by the binomial theorem.

Therefore, \(r_1+r_2+...+r_{100}=-300\).)