Find-the-cubic-equation-whose-roots-are-the-radius-of-three-escribed-circles-in-term-of-inradius-circumradius-and-semiperimeter-

Question Number 19055 by Tinkutara last updated on 03/Aug/17

Find the cubic equation whose roots

are the radius of three escribed circles

in term of inradius, circumradius and

semiperimeter .

Answered by behi.8.3.4.1.7@gmail.com last updated on 05/Aug/17

(x - r_{a}) (x - r_{b}) (x - r_{c}) = 0

\Rightarrow x^{3} - (Σ r_{a}) x^{2} + (Σ r_{a} r_{b}) x - Π r_{a} = 0

1) A + B = π - C \Rightarrow t g (\frac{A + B}{2}) = t g (\frac{π}{2} - \frac{C}{2})

\Rightarrow \frac{t g \frac{A}{2} + t g \frac{B}{2}}{1 - t g \frac{A}{2} t g \frac{B}{2}} = \frac{1}{t g \frac{C}{2}} \Rightarrow Σ t g \frac{A}{2} t g \frac{B}{2} = 1

2) Π t g \frac{A}{2} = \sqrt{\frac{(p - b) (p - c)}{p (p - a)} . \frac{(p - a) (p - c)}{p (p - b)} . \frac{(p - a) (p - b)}{p (p - c)}} =

= \sqrt{\frac{p (p - a) (p - b) (p - c)}{p^{4}}} = \frac{S}{p^{2}} = \frac{r}{p}

3) Σ t g \frac{A}{2} = Σ \frac{(p - b) (p - c)}{S} = \frac{p^{2} - p (b + c) + b c}{S} +

+ \frac{p^{2} - p (a + c) + a c}{S} + \frac{p^{2} - p (a + b) + a b}{S} =

= \frac{3 p^{2} - 2 p (a + b + c) + a b + b c + c a}{S} =

= \frac{a b + b c + c a - p^{2}}{S} = \frac{4 R r + r^{2}}{S} = \frac{4 R + r}{p}

4) r_{a} = p . t g \frac{A}{2} \Rightarrow {\begin{cases} Σ r_{a} = p Σ t g \frac{A}{2} = p . \frac{4 R + r}{p} = 4 R + r \\ Σ r_{a} r_{b} = p^{2} Σ t g \frac{A}{2} t g \frac{B}{2} = p^{2} \end{cases}

5) Π r_{a} = p^{3} . Π t g \frac{A}{2} = p^{3} . \frac{r}{p} = r . p^{2}

\Rightarrow \boldsymbol x^{3} - (4 \boldsymbol R + \boldsymbol r) \boldsymbol x^{2} + \boldsymbol p^{2} . \boldsymbol x - \boldsymbol r . \boldsymbol p^{2} = 0

Commented by Tinkutara last updated on 05/Aug/17

Thank you very much Sir!

Thank you very much Sir!

Commented by behi.8.3.4.1.7@gmail.com last updated on 05/Aug/17

c o r r e c t e d!

c o r r e c t e d!