1. ## in a triangle

in a triangle ABC and R is the radius of the circle into which the triangle is inscribed(in other words it is the circum radius of the triangle)

$\displaystyle \frac{a\cos A + b\cos B + c\cos C}{a\sin B + b\sin C + c\sin A} = \frac{a+b+c}{9R}$

then prove it is an equilateral triangle

2. When I simplified the left hand side, I got

$\displaystyle \frac{abc}{ab+bc+ca} = \frac{a+b+c}{9}$

By observation you can see the the equation is true only when a = b = c. But how to prove it?

3. Originally Posted by sa-ri-ga-ma
When I simplified the left hand side, I got

$\displaystyle \frac{abc}{ab+bc+ca} = \frac{a+b+c}{9}$

By observation you can see the the equation is true only when a = b = c. But how to prove it?
It took me quite a while to see how to simplify the left hand side to get $\displaystyle \frac{abc}{ab+bc+ca} = \frac{a+b+c}{9}$. (It's correct, but I'll leave it to sa-ri-ga-ma to provide the details. )

So we know that $\displaystyle (a+b+c)(ab+bc+ca) = 9abc$. Multiply out the left side to get $\displaystyle a(b^2+c^2) + b(c^2+a^2) + c(a^2+b^2) + 3abc$. Thus the equation becomes $\displaystyle a(b^2+c^2) + b(c^2+a^2) + c(a^2+b^2) = 6abc$, which you can write as $\displaystyle a(b-c)^2 + b(c-a)^2 + c(a-b)^2 = 0$. That obviously only holds if $\displaystyle a=b=c$.

4. ## i think i got it

i can simplify the denominator first
$\displaystyle a\sin B + b\sin C + c\sin A = a(b/2R) + b(c/2R) + c(a/2R)$
$\displaystyle = \frac{ab +bc+ca}{2R}$
$\displaystyle a\cos A + b\cos B + c\cos C = R(2\sin a\cos a+ 2\sin b\cos b + 2\sin c\cos c)$
$\displaystyle = R(\sin 2a+\sin 2b+\sin 2c)$
$\displaystyle = 4R(\sin a\sin b\sin c)$
$\displaystyle = \frac{abc}{2 R^2}$