<s>[INST] Area of the triangle formed by circumcenter, incenter and orthocenter Lets say we have $\triangle$$ABC$ having $O,I,H$ as its circumcenter, incenter and orthocenter. How can I go on finding the area of the $\triangle$$HOI$.
I thought of doing the question using the distance (length) between $HO$,$HI$ and $OI$ and then using the Heron's formula, but that has made the calculation very much complicated. Is there any simple way to crack the problem?
 [/INST] A área do triângulo HOI do triângulo ABC de lados a=|BC|, b=|AC| e c=|AB| é dada por:
$AreaHOI=|\frac{1}{4}.\frac{\left|\begin{array}{} a & a^3 & 1 \\ b & b^3 & 1 \\ c & c^3 & 1 \\ \end{array}  \right| }{\left|\begin{array}{} 0 & a^2 & b^2 &1\\ a^2 & 0 & c^2&1 \\ b^2 & c^2 & 0&1 \\1&1&1&0 \end{array}  \right| }|$
Dedução: Seja o triângulo ABC, com o vértice A na origem do Sistema Cartesiano e C no semi eixo positivo x. Assim:
$A=(0,0)$
$B=(\frac{-a^2+b^2+c^2}{2b},\frac{2S}{b})$
$C=(b,0)$
$S=\sqrt{p(p-a)(p-b)(p-c)}$
$p=\frac{a+b+c}{2}$
$AreaHOI=3.AreaGIO$
$AreaGIO=\frac{1}{2}\left|\begin{array}{} xG & yG & 1 \\ xI & yI & 1 \\ xO & yO & 1 \\ \end{array}  \right| $
"Obtendo as coordenadas de G, I e O através  de suas respectivas fórmulas e fazendo substituições obtemos:"
$AreaHOI=3.\frac{1}{2}\left|\begin{array}{} \frac{-a^2+3b^2+c^2}{6b} & \frac{2S}{3b} & 1 \\ \frac{-a+b+c}{2} & \frac{2S}{a+b+c} & 1 \\ \frac{b}{2} & \frac{b(a^2-b^2+c^2)}{8S} & 1 \\ \end{array}  \right| $
"Desenvolvendo, substituindo S, p e fatorando obtemos:"
$AreaHOI=\left|\frac{(a-b)(a-c)(b-c)(a+b+c)}{16S}  \right| $
"Substituindo as expressões por determinantes (S no formato de área de Cayley-Menger), obtemos:"
$AreaHOI=|\frac{1}{4}.\frac{\left|\begin{array}{} a & a^3 & 1 \\ b & b^3 & 1 \\ c & c^3 & 1 \\ \end{array}  \right| }{\left|\begin{array}{} 0 & a^2 & b^2 &1\\ a^2 & 0 & c^2&1 \\ b^2 & c^2 & 0&1 \\1&1&1&0 \end{array}  \right| }|$
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