A geometry problem by Yasin Habib Abir

Geometry Level 3

if a b c are the length of a triangle and A B C are the angles of the triangle then find the value of

a^3 cos(B-C)+b^3 cos(C-A)+c^3 cos(A-B)

abc 3abc 0 a^2 b^2 c^2

1 solution

Nguyễn Phát
Sep 28, 2014

We have to find the value of P= symmetric sigma(s.s)of (a^3 cos(B-C)).
I will use these formulas in trigonometry :
- a/sin(A)=2R or 2Rsin(A)=a with R is radius of the circumscribed circle of triangle ABC.
- sin(2B)+sin(2C)=2sin(B+C)cos(B-C) and sin(A+B)=sin(C).
- cos(A)=(b^2+c^2-a^2)/(2bc)
We have :
- P=(s.s) of a^3cos(B-C) =(s.s) of a^3 2cos(B-C)sin(B+C)/(2sin(A)) =(s.s) of R a^2 (sin(2B)+sin(2C)) =(s.s) of a^2 ( 2R sin(B)cos(B)+2R sin(C)cos(C)) =(s.s) of a^2 ( b cos(B)+c cos(C)) =(s.s) of cos(A) a(b^2+c^2) =(s.s) of a^2 (b^2+c^2)(b^2+c^2-a^2)/(2abc) = Q/(2abc)
With :
- Q=(s.s) of a^2 (b^2+c^2)(b^2+c^2-a^2) =(s.s) of x(y+z)(y+z-x) (i suppose x= a^2;y=...) =(s.s) of x(y+z)^2- x^2 (y+z) =6xyz or 6a^2b^2c^2
This leads to my result: +P=Q/(2abc)= 3abc.