已知三角形的內心到三頂點的距離分別為6，4，3，
求此三角形面積為何？

設\(\overline{IA}=x\)、\(\overline{IB}=y\)、\(\overline{IC}=z\)、內切圓半徑\(r\)
\( \alpha+\beta+\gamma=90^{\circ} \)
\( sin \gamma=sin(90^{\circ}-\alpha-\beta)=cos(\alpha+\beta)=cos\alpha cos \beta-sin \alpha sin \beta \)
\( \displaystyle \frac{r}{z}=\sqrt{1-(\frac{r}{x})^2}\cdot \sqrt{1-(\frac{r}{y})^2}-\frac{r}{x}\cdot \frac{r}{y} \)
\( \displaystyle \frac{r}{z}+\frac{r}{x}\cdot \frac{r}{y}=\sqrt{1-(\frac{r}{x})^2}\cdot \sqrt{1-(\frac{r}{y})^2} \)
\( \displaystyle (\frac{r}{z}+\frac{r}{x}\cdot \frac{r}{y})^2=\left[1-(\frac{r}{x})^2 \right] \cdot \left[1-(\frac{r}{y})^2 \right] \)
\( \displaystyle \frac{r^4}{x^2y^2}+2\frac{r^3}{xyz}+\frac{r^2}{z^2}=1-\frac{r^2}{x^2}-\frac{r^2}{y^2}+\frac{r^4}{x^2y^2} \)
\( \displaystyle 2\frac{r^3}{xyz}+\frac{r^2}{z^2}+\frac{r^2}{x^2}+\frac{r^2}{y^2}-1=0 \)
\( 2xyzr^3+(x^2y^2+y^2z^2+z^2x^2)r^2-x^2y^2z^2=0 \)

設\(\overline{IA}=x=6\)、\(\overline{IB}=y=4\)、\(\overline{IC}=z=3\)
\( 144r^3+1044r^2-5184=0 \)
\( 4r^3+29r^2-144=0 \)
但實根\(r\)只能用公式解，預估面積也是很醜的式子

感謝你!