Question

Prove that the coordinates of the centre of the circle inscribed in the triangle whose angular points are $(1,2)$, $(2,3)$, and $(3,1)$ are $\frac{8+\sqrt{10}}{6}$ and $\frac{16-\sqrt{10}}{6}$.
Find also the coordinates of the centres of the described circles.

Answer 1

$BC=a=\sqrt{{(2-3)}^2+{(3-1)}^2}=\sqrt{5}CA=b=\sqrt{{(3-1)}^2+{(1-2)}^2}=\sqrt{5}AB=c=\sqrt{{(1-2)}^2+{(2-3)}^2}=\sqrt{2}$

Coordinates of incentre are

$\left(\frac{aA+bB+cC}{a+b+c}\right)(\frac{\sqrt{5}\times 1+\sqrt{5}\times 2+\sqrt{2}\times 3}{\sqrt{5}+\sqrt{5}+\sqrt{2}},\frac{\sqrt{5}\times 2+\sqrt{5}\times 3+\sqrt{2}\times 1}{\sqrt{5}+\sqrt{5}+\sqrt{2}})(\frac{3\sqrt{5}+3\sqrt{2}}{2\sqrt{5}+\sqrt{2}},\frac{5\sqrt{5}+\sqrt{2}}{2\sqrt{5}+\sqrt{2}})(\frac{3\sqrt{5}+3\sqrt{2}}{2\sqrt{5}+\sqrt{2}}\times \frac{2\sqrt{5}-\sqrt{2}}{2\sqrt{5}-\sqrt{2}},\frac{5\sqrt{5}+\sqrt{2}}{2\sqrt{5}+\sqrt{2}}\times \frac{2\sqrt{5}-\sqrt{2}}{2\sqrt{5}-\sqrt{2}})(\frac{24+3\sqrt{10}}{18},\frac{48-3\sqrt{10}}{18})(\frac{8+\sqrt{10}}{6},\frac{16-\sqrt{10}}{6})$

Coodinates of $O_1$ are

$\left(\frac{-aA+bB+cC}{-a+b+c}\right)(\frac{-\sqrt{5}\times 1+\sqrt{5}\times 2+\sqrt{2}\times 3}{-\sqrt{5}+\sqrt{5}+\sqrt{2}},\frac{-\sqrt{5}\times 2+\sqrt{5}\times 3+\sqrt{2}\times 1}{-\sqrt{5}+\sqrt{5}+\sqrt{2}})(\frac{\sqrt{5}+3\sqrt{2}}{\sqrt{2}},\frac{\sqrt{5}+\sqrt{2}}{\sqrt{2}})(\frac{6+\sqrt{10}}{2},\frac{2+\sqrt{10}}{2})$

Coordiantes of $O_2$ are

$\left(\frac{aA-bB+cC}{a-b+c}\right)(\frac{\sqrt{5}\times 1-\sqrt{5}\times 2+\sqrt{2}\times 3}{\sqrt{5}-\sqrt{5}+\sqrt{2}},\frac{\sqrt{5}\times 2-\sqrt{5}\times 3+\sqrt{2}\times 1}{\sqrt{5}-\sqrt{5}+\sqrt{2}})(\frac{-\sqrt{5}+3\sqrt{2}}{\sqrt{2}},\frac{-\sqrt{5}+\sqrt{2}}{\sqrt{2}})(\frac{6-\sqrt{10}}{2},\frac{2-\sqrt{10}}{2})$

Coordinates of $O_3$ are

$\left(\frac{aA+bB-cC}{a+b-c}\right)(\frac{\sqrt{5}\times 1+\sqrt{5}\times 2-\sqrt{2}\times 3}{\sqrt{5}+\sqrt{5}-\sqrt{2}},\frac{\sqrt{5}\times 2+\sqrt{5}\times 3-\sqrt{2}\times 1}{\sqrt{5}+\sqrt{5}-\sqrt{2}})(\frac{3\sqrt{5}-3\sqrt{2}}{2\sqrt{5}-\sqrt{2}},\frac{5\sqrt{5}-\sqrt{2}}{2\sqrt{5}-\sqrt{2}})(\frac{8-\sqrt{10}}{6},\frac{16+\sqrt{10}}{6})$