Question Number 104364 by ajfour last updated on 21/Jul/20

Commented byajfour last updated on 21/Jul/20

Find the area of the △CGI ;  C is circumcentre of △PQR,  G its centroid and I its incentre.

Answered by mr W last updated on 21/Jul/20

c=(√(a^2 +b^2 ))  radius of incircle r  (((a+b+c)r)/2)=((ab)/2)  ⇒r=((ab)/(a+b+c))  G((a/3),(b/3))  I(r,r)  C((a/2),(b/2))  A_(ΔGCI) =(1/2)∣ determinant ((((a/2)−(a/3)),((b/2)−(b/3))),(((a/2)−r),((b/2)−r)))∣  =(1/2)∣[(a/6)((b/2)−r)−(b/6)((a/2)−r)]∣  =((ab∣a−b∣)/(12(a+b+(√(a^2 +b^2 )))))

Commented bymr W last updated on 21/Jul/20

yes, thanks!

Commented byajfour last updated on 21/Jul/20

sir  please check , i think    A_(△CGI)  = (((b−a)r)/(12))

Answered by ajfour last updated on 21/Jul/20

let Q be origin.  R(a,0) ;  P(0,b)    x_I  =((ax_1 +bx_2 +cx_3 )/(a+b+c))       = ((a×0+b×a+c×0)/(a+b+c)) =((ab)/(a+b+c))  y_I  = ((ay_1 +by_2 +cy_3 )/(a+b+c)) = ((ab)/(a+b+c))    G((a/3),(b/3))   ;  C((a/2),(b/2))  Area △CGI =(1/2) determinant (((a/2),(b/2),1),((a/3),(b/3),1),(((ab)/(a+b+c)),((ab)/(a+b+c)),1))     =(1/2){ (a/2)((b/3)−r)−(b/2)((a/3)−r)                             +r((a/3)−(b/3))}     = (r/2)((b/6)−(a/6)) = ((r(b−a))/(12))   △_(CGI)  = ((ab(b−a))/(12(a+b+(√(a^2 +b^2 ))))) ★