26 Apr
2006
26 Apr
'06
4 p.m.
On Wednesday 26 April 2006 16:22, Eugene Salamin wrote:
Adjoin two congruent triangles along their c edge to obtain a parallogram of area 2A = a b sinC. (2A)^2 = a^2 b^2 (1 - (cosC)^2). Using the law of cosines, A^2 can be expressed as a polynomial P(a,b,c) of degree 4. P vanishes when c=a+b, etc. So
P(a,b,c) = (-a+b+c)(a-b+c)(a+b-c)Q(a,b,c).
Q is of degree 1, and since P is symmetric, so is Q. Then Q is a multiple of a+b+c. The fixed constant can be found by consdering e.g. an equilateral triangle.
Entertaining! -- g