User:Phalkunlim/sandbox

Phalkun's theorem

Theorem1

Let $$X$$ be an arbitrary point in the plane of triangle $$ABC$$, then $$\overrightarrow{XH}=(\cot B\cot C)\,\overrightarrow{XA}+(\cot A\cot C)\overrightarrow{XB}+(\cot A\cot B)\,\overrightarrow{XC}\,\,\,\,$$, where $$H$$ is the orthocenter of triangle. If $$X\equiv H\,\,$$then $$(\cot B\cot C)\,\overrightarrow{HA}+(\cot A\cot C)\overrightarrow{HB}+(\cot A\cot B)\,\overrightarrow{HC}\,\,\,\,\,=\overset{\to }{\mathop{O}}\,$$ ,

Theorem2

Let $$X$$ be an arbitrary point in the plane of triangle $$ABC$$, then $$\left( \cot B\cot C \right)X{{A}^{2}}+\left( \cot A\cot C \right)X{{B}^{2}}+\left( \cot A\cot B \right)X{{C}^{2}}=X{{H}^{2}}+8{{R}^{2}}\cos A\cos B\cos C\,\,\,\,$$, where $$H$$ is the orthocenter of triangle and $$R$$ is circumradius.

Remarque: If I is incenter of triangle $$ABC$$ then $$I{{H}^{2}}=2{{r}^{2}}-4{{R}^{2}}\cos A\cos B\cos C$$

So $$\left( \cot B\cot C \right)X{{A}^{2}}+\left( \cot A\cot C \right)X{{B}^{2}}+\left( \cot A\cot B \right)X{{C}^{2}}=X{{H}^{2}}+2\,\,I{{H}^{2}}+4{{r}^{2}}\,\,\,\,$$.

$$r\,$$denote the inradius. www.mathtoday.wordpress.com