User:Phalkunlim

Phalkun’s theorem www.mathtoday.wordpress.com Theorem1 Let $$X$$ be an arbitrary point in the plane of triangle $$ABC$$, then $$\overrightarrow{XH}=(\cot B\cot C)\,\overrightarrow{XA}+(\cot A\cot C)\overrightarrow{XB}+(\cot A\cot B)\,\overrightarrow{XC}\,\,\,\,\left( 1 \right)$$ where $$H$$ is the orthocenter of triangle. Proof Let $${{p}_{A}}=\overset{\xrightarrow – }{\mathop{AB}}\,.\overset{\xrightarrow – }{\mathop{AC}}\,\,\,,\,\,{{p}_{B}}=\overset{\xrightarrow – }{\mathop{BA}}\,.\overset{\xrightarrow – }{\mathop{BC}}\,\,\,,$$$${{p}_{c}}=\overset{\xrightarrow – }{\mathop{CA}}\,.\overset{\xrightarrow – }{\mathop{CB}}\,$$ and$$BC=a\,\,,\,\,AC=b\,\,,\,AB=c$$. We have $$\overset{\xrightarrow – }{\mathop{AK}}\,\bot \overset{\xrightarrow – }{\mathop{BC}}\,\Leftrightarrow \overset{\xrightarrow – }{\mathop{AK}}\,.\overset{\xrightarrow – }{\mathop{BC}}\,=0$$  by $$\overset{\xrightarrow – }{\mathop{AK}}\,=\overset{\xrightarrow – }{\mathop{AB}}\,+\overset{\xrightarrow – }{\mathop{BK}}\,=\overset{\xrightarrow – }{\mathop{AB}}\,+\lambda \,\,\overset{\xrightarrow – }{\mathop{BC}}\,\,\,,\,\lambda \in \mathbb{R}$$ (because $$\overset{\xrightarrow – }{\mathop{BK}}\,//\overset{\xrightarrow – }{\mathop{BC}}\,$$ ). Then $$\left( \overset{\xrightarrow – }{\mathop{AB}}\,\,\,+\lambda \,\overset{\xrightarrow – }{\mathop{BC}}\, \right).\overset{\xrightarrow – }{\mathop{BC}}\,=0\Leftrightarrow \lambda =\frac{\overset{\xrightarrow – }{\mathop{BA}}\,.\overset{\xrightarrow – }{\mathop{BC}}\,}{B{{C}^{2}}}\,\,\,\,=\frac$$ or  $$\overset{\xrightarrow – }{\mathop{AK}}\,=\overset{\xrightarrow – }{\mathop{AB}}\,\,\,\,+\frac\,\,\overset{\xrightarrow – }{\mathop{BC}}\,\,\,\,\,\left( 2 \right)$$ We have $$\overset{\xrightarrow – }{\mathop{AH}}\,//\overset{\xrightarrow – }{\mathop{AK}}\,$$ then $$\overset{\xrightarrow – }{\mathop{AH}}\,=\mu \,\,\overset{\xrightarrow – }{\mathop{AK}}\,$$,  by $$\overset{\xrightarrow – }{\mathop{AH}}\,=\overset{\xrightarrow – }{\mathop{AB}}\,\,\,+\overset{\xrightarrow – }{\mathop{BH}}\,$$ then $$\,\overset{\xrightarrow – }{\mathop{BH}}\,=\mu \overset{\xrightarrow – }{\mathop{AK}}\,-\overset{\xrightarrow – }{\mathop{AB}}\,\,$$ $$\overset{\xrightarrow – }{\mathop{BH}}\,\bot \overset{\xrightarrow – }{\mathop{AC}}\,\Leftrightarrow \overset{\xrightarrow – }{\mathop{BH}}\,.\overset{\xrightarrow – }{\mathop{AC}}\,=0$$ or $$\left( \mu \,\overset{\xrightarrow – }{\mathop{AK}}\,-\overset{\xrightarrow – }{\mathop{AB}}\, \right).\overset{\xrightarrow – }{\mathop{AC}}\,=0\Leftrightarrow \mu =\frac{\overset{\xrightarrow – }{\mathop{AB}}\,.\overset{\xrightarrow – }{\mathop{AC}}\,}{\overset{\xrightarrow – }{\mathop{AK}}\,.\overset{\xrightarrow – }{\mathop{AC}}\,}\,\,\,\,=\frac{\overset{\xrightarrow – }{\mathop{AK}}\,.\overset{\xrightarrow – }{\mathop{AC}}\,}$$ $$\mu =\frac{\left( \overset{\xrightarrow – }{\mathop{AB}}\,\,\,+\frac\,\,\overset{\xrightarrow – }{\mathop{BC}}\, \right)\,.\overset{\xrightarrow – }{\mathop{AC}}\,}=\frac{\overset{\xrightarrow – }{\mathop{AB}}\,.\overset{\xrightarrow – }{\mathop{AC}}\,\,\,+\frac\,\,.\,\overset{\xrightarrow – }{\mathop{BC}}\,.\overset{\xrightarrow – }{\mathop{AC}}\,}=\frac{{{p}_{a}}+\frac\,{{p}_{c}}}=\frac{{{a}^{2}}{{p}_{a}}+{{p}_{b}}{{p}_{c}}}$$ by  $$\,\overset{\xrightarrow – }{\mathop{BH}}\,=\mu \overset{\xrightarrow – }{\mathop{AK}}\,-\overset{\xrightarrow – }{\mathop{AB}}\,\,$$  then  $$\overset{\xrightarrow – }{\mathop{XH}}\,-\overset{\xrightarrow – }{\mathop{XB}}\,=\mu \,\,\left( \overset{\xrightarrow – }{\mathop{XK}}\,-\overset{\xrightarrow – }{\mathop{XA}}\, \right)-\left( \overset{\xrightarrow – }{\mathop{XB}}\,-\overset{\xrightarrow – }{\mathop{XA}}\, \right)$$ where $$X$$ be an arbitrary point in the plane of triangle. so $$\overset{\xrightarrow – }{\mathop{XH}}\,=\left( 1-\mu  \right)\overset{\xrightarrow – }{\mathop{XA}}\,+\mu \,\overset{\xrightarrow – }{\mathop{XK}}\,\,\,\,\,\left( 3 \right)$$ from $$\left( 2 \right)\,\,:$$ $$\overset{\xrightarrow – }{\mathop{XK}}\,-\overset{\xrightarrow – }{\mathop{XA}}\,=\overset{\xrightarrow – }{\mathop{XB}}\,-\overset{\xrightarrow – }{\mathop{XA}}\,\,\,+\frac\left( \overset{\xrightarrow – }{\mathop{XC}}\,-\overset{\xrightarrow – }{\mathop{XB}}\, \right)$$ or $$\overset{\xrightarrow – }{\mathop{XK}}\,=\frac\overset{\xrightarrow – }{\mathop{XC}}\,+\left( 1-\frac \right)\,\overset{\xrightarrow – }{\mathop{XB}}\,\,\,\,\left( 4 \right)$$ from$$\left( 3 \right)\,\And $$$$\left( 4 \right)$$ គេបាន $$\overset{\xrightarrow – }{\mathop{XH}}\,=\left( 1-\mu \right)\overset{\xrightarrow – }{\mathop{XA}}\,+\mu \,\,\left[ \frac\,\overset{\xrightarrow – }{\mathop{XC}}\,+\left( 1-\frac \right)\,\overset{\xrightarrow – }{\mathop{XB}}\, \right]$$ or $$\overset{\xrightarrow – }{\mathop{XH}}\,=\left( 1-\mu  \right)\overset{\xrightarrow – }{\mathop{XA}}\,+\mu \left( 1-\frac \right)\,\overset{\xrightarrow – }{\mathop{XB}}\,+\mu \frac\,\overset{\xrightarrow – }{\mathop{XC}}\,$$  by $$\mu =\frac{{{a}^{2}}{{p}_{A}}+{{p}_{B}}\,{{p}_{C}}}$$ $$\overset{\xrightarrow – }{\mathop{XH}}\,=\frac{{{a}^{2}}{{p}_{A}}+{{p}_{B}}{{p}_{C}}}\overset{\xrightarrow – }{\mathop{XA}}\,+\frac{{{p}_{A}}({{a}^{2}}-{{p}_{B}})}{{{a}^{2}}{{p}_{A}}+{{p}_{B}}{{p}_{C}}}\overset{\xrightarrow – }{\mathop{XB}}\,+\frac{{{a}^{2}}{{p}_{A}}+{{p}_{B}}{{p}_{C}}}\overset{\xrightarrow – }{\mathop{XC}}\,$$ We have $$\left\{ \begin{align} & {{p}_{a}}=\overset{\xrightarrow – }{\mathop{AB}}\,.\overset{\xrightarrow – }{\mathop{AC}}\,=bc\cos A \\ & {{p}_{b}}=\overset{\xrightarrow – }{\mathop{BA}}\,.\overset{\xrightarrow – }{\mathop{BC}}\,=ac\cos B \\ & {{p}_{c}}=\overset{\xrightarrow – }{\mathop{CA}}\,.\overset{\xrightarrow – }{\mathop{CB}}\,=ab\cos C \\ \end{align} \right.$$ $$\overset{\xrightarrow – }{\mathop{XH}}\,=\frac{{{a}^{2}}bc\cos B\cos C}{{{a}^{2}}bc\cos A+{{a}^{2}}bc\cos B\cos C}\overset{\xrightarrow – }{\mathop{XA}}\,+\frac{{{a}^{2}}bc\cos A-ab{{c}^{2}}\cos A\cos B}{{{a}^{2}}bc\cos A+{{a}^{2}}bc\cos B\cos C}\overset{\xrightarrow – }{\mathop{XB}}\,+\frac{ab{{c}^{2}}\cos A\cos B}{{{a}^{2}}bc\cos A+{{a}^{2}}bc\cos B\cos C}\overset{\xrightarrow – }{\mathop{XC}}\,$$ $$\overset{\xrightarrow – }{\mathop{XH}}\,=\frac{\cos B\cos C}{\cos A+\cos B\cos C}\overset{\xrightarrow – }{\mathop{XA}}\,+\frac{\cos A-\frac{c}{a}\cos A\cos B}{\cos A+\cos B\cos C}\overset{\xrightarrow – }{\mathop{XB}}\,+\frac{\frac{c}{a}\cos A\cos B}{\cos A+\cos B\cos C}\overset{\xrightarrow – }{\mathop{XC}}\,$$ $$\overset{\xrightarrow – }{\mathop{XH}}\,=\frac{\cos B\cos C}{\sin B\sin C}\overset{\xrightarrow – }{\mathop{XA}}\,+\frac{\cos A-\frac{\cos A\cos B\sin C}{\sin A}}{\sin B\sin C}\overset{\xrightarrow – }{\mathop{XB}}\,+\frac{\frac{\cos A\cos B\sin C}{\sin A}}{\sin B\sin C}\overset{\xrightarrow – }{\mathop{XC}}\,$$ So  $$\overset{\xrightarrow – }{\mathop{XH}}\,=(\cot B\cot C)\,\overset{\xrightarrow – }{\mathop{XA}}\,+(\cot A\cot C)\overset{\xrightarrow – }{\mathop{XB}}\,+(\cot A\cot B)\,\overset{\xrightarrow – }{\mathop{XC}}\,\,\,\,$$ completed proof. Theorem2 If $$X\equiv H\,\,$$then $$(\cot B\cot C)\,\overrightarrow{HA}+(\cot A\cot C)\overrightarrow{HB}+(\cot A\cot B)\,\overrightarrow{HC}\,\,\,\,\,=\overset{\to }{\mathop{O}}\,\,\,\,\left( 5 \right)$$ Theorem3 $$\overrightarrow{OA}\,\,\sin 2A+\overrightarrow{OB}\,\,\sin 2B+\overrightarrow{OC}\,\,\sin 2C=\overrightarrow{O\,}\,\,\,\left( 6 \right)$$ Proof: If $$X\equiv O$$ then $$\overrightarrow{OH}=(\cot B\cot C)\,\overrightarrow{OA}+(\cot A\cot C)\overrightarrow{OB}+(\cot A\cot B)\,\overrightarrow{OC}\,\,\,\,\,=\overset{\to }{\mathop{O}}\,\,\,\,\left( 7 \right)$$ Where $$O$$ be the circumcentre of triangle $$ABC$$. By $$\overrightarrow{OH}=\overrightarrow{OA}+\overrightarrow{OB}+\overrightarrow{OC}\,\,\,\left( 8 \right)$$ then $$\overrightarrow{OA}+\overrightarrow{OB}+\overrightarrow{OC}=(\cot B\cot C)\,\overrightarrow{OA}+(\cot A\cot C)\overrightarrow{OB}+(\cot A\cot B)\,\overrightarrow{OC}\,\,\,\,\,$$ Or $$(\cot B\cot C-1)\,\overrightarrow{OA}+(\cot A\cot C-1)\overrightarrow{OB}+(\cot A\cot B-1)\,\overrightarrow{OC}\,\,=\overset{\to }{\mathop{O}}\,\,\,\,$$ By $$\cot B\cot C-1=\cot A\cot B+\cot A\cot C=\cot A\left( \cot B+\cot C \right)=-\frac{\sin 2A}{2\sin A\sin B\sin C}$$ $$\cot A\cot C-1=-\frac{\sin 2B}{2\sin A\sin B\sin C}\,\,\,,\,\,\cot A\cot B-1=-\frac{\sin 2C}{2\sin A\sin B\sin C}$$ Then $$-\frac{\sin 2A}{2\sin A\sin B\sin C}\,\overrightarrow{OA}-\frac{\sin 2B}{2\sin A\sin B\sin C}\overrightarrow{OB}-\frac{\sin 2C}{2\sin A\sin B\sin C}\,\overrightarrow{OC}\,\,=\overset{\to }{\mathop{O}}\,\,\,\,$$ $$\overrightarrow{OA}\,\,\sin 2A+\overrightarrow{OB}\,\,\sin 2B+\overrightarrow{OC}\,\,\sin 2C=\overrightarrow{O\,}\,\,\,$$ completed proof. Theorem4 Let $$X$$ be an arbitrary point in the plane of triangle $$ABC$$, then $$\left( \cot B\cot C \right)X{{A}^{2}}+\left( \cot A\cot C \right)X{{B}^{2}}+\left( \cot A\cot B \right)X{{C}^{2}}=X{{H}^{2}}+8{{R}^{2}}\cos A\cos B\cos C\,\,\,\left( 8 \right)\,\,\,\,$$, where $$H$$ is the orthocenter of triangle and $$R$$ is circumradius. Proof: If $$X\equiv H$$ then $$\left( \cot B\cot C \right)\,\overset{\xrightarrow – }{\mathop{HA}}\,+\left( \cot A\cot C \right)\,\overset{\xrightarrow – }{\mathop{HB}}\,+\left( \cot A\cot B \right)\,\overset{\xrightarrow – }{\mathop{HC}}\,=\overset{\to }{\mathop{O}}\,\,\,\,\,\left( 9 \right)$$ Let $${{\alpha }_{a}}=\cot B\cot C\,\,,\,\,{{\alpha }_{b}}=\cot A\cot C\,\,,\,\,{{\alpha }_{c}}=\cot A\cot B$$ $${{\alpha }_{a}}\overset{\xrightarrow – }{\mathop{HA}}\,+{{\alpha }_{b}}\,\overset{\xrightarrow – }{\mathop{HB}}\,+{{\alpha }_{c}}\,\overset{\xrightarrow – }{\mathop{HC}}\,=\overset{\to }{\mathop{O}}\,\,\,\,\left( 10 \right)$$ Let $$X$$ be an arbitrary point in the plane of triangle $$ABC$$ We have $$\left\{ \begin{align} & {{\alpha }_{a}}X{{A}^{2}}={{\alpha }_{a}}{{\left( \overset{\xrightarrow – }{\mathop{XA}}\,+\overset{\xrightarrow – }{\mathop{HA}}\, \right)}^{2}}={{\alpha }_{a}}X{{H}^{2}}+{{\alpha }_{a}}H{{A}^{2}}+2{{\alpha }_{a}}\overset{\xrightarrow – }{\mathop{XH}}\,.\overset{\xrightarrow – }{\mathop{HA}}\, \\ & {{\alpha }_{b}}X{{B}^{2}}={{\alpha }_{b}}{{\left( \overset{\xrightarrow – }{\mathop{XH}}\,+\overset{\xrightarrow – }{\mathop{HB}}\, \right)}^{2}}={{\alpha }_{b}}X{{H}^{2}}+{{\alpha }_{b}}H{{B}^{2}}+2{{\alpha }_{b}}\overset{\xrightarrow – }{\mathop{XH}}\,.\overset{\xrightarrow – }{\mathop{HB}}\, \\ & {{\alpha }_{c}}X{{C}^{2}}={{\alpha }_{c}}{{\left( \overset{\xrightarrow – }{\mathop{XH}}\,+\overset{\xrightarrow – }{\mathop{HC}}\, \right)}^{2}}={{\alpha }_{c}}X{{H}^{2}}+{{\alpha }_{c}}H{{C}^{2}}+2{{\alpha }_{c}}\overset{\xrightarrow – }{\mathop{XH}}\,.\overset{\xrightarrow – }{\mathop{HC}}\, \\ \end{align} \right.$$ $${{\alpha }_{a}}X{{A}^{2}}+{{\alpha }_{b}}X{{B}^{2}}+{{\alpha }_{c}}X{{C}^{2}}=\left( {{\alpha }_{a}}+{{\alpha }_{b}}+{{\alpha }_{c}} \right)X{{H}^{2}}+{{\alpha }_{a}}H{{A}^{2}}+{{\alpha }_{b}}H{{B}^{2}}+{{\alpha }_{c}}H{{C}^{2}}+2\overset{\xrightarrow – }{\mathop{XH}}\,.\left( {{\alpha }_{a}}\overset{\xrightarrow – }{\mathop{HA}}\,+{{\alpha }_{b}}\overset{\xrightarrow – }{\mathop{HB}}\,+{{\alpha }_{c}}\overset{\xrightarrow – }{\mathop{HC}}\, \right)$$ by $${{\alpha }_{a}}+{{\alpha }_{b}}+{{\alpha }_{c}}=\cot A\cot B+\cot B\cot C+\cot C\cot A=1$$ and $${{\alpha }_{a}}\overset{\xrightarrow – }{\mathop{HA}}\,+{{\alpha }_{b}}\,\overset{\xrightarrow – }{\mathop{HB}}\,+{{\alpha }_{c}}\,\overset{\xrightarrow – }{\mathop{HC}}\,=\overset{\to }{\mathop{O}}\,$$ then $${{\alpha }_{a}}X{{A}^{2}}+{{\alpha }_{b}}X{{B}^{2}}+{{\alpha }_{c}}X{{C}^{2}}=X{{H}^{2}}+{{\alpha }_{a}}H{{A}^{2}}+{{\alpha }_{b}}H{{B}^{2}}+{{\alpha }_{c}}H{{C}^{2}}\,\,\,\,\,\left( 11 \right)$$  we find $${{\alpha }_{A}}H{{A}^{2}}+{{\alpha }_{B}}H{{B}^{2}}+{{\alpha }_{C}}H{{C}^{2}}$$ We have $${{\alpha }_{a}}X{{A}^{2}}+{{\alpha }_{b}}X{{B}^{2}}+{{\alpha }_{c}}X{{C}^{2}}=X{{H}^{2}}+{{\alpha }_{a}}H{{A}^{2}}+{{\alpha }_{b}}H{{B}^{2}}+{{\alpha }_{c}}H{{C}^{2}}$$ If $$X\equiv \{\,\,A,B,C\,\}$$ then $$\left\{ \begin{align} & {{\alpha }_{b}}A{{B}^{2}}+{{\alpha }_{c}}A{{C}^{2}}=(1+{{\alpha }_{a}})H{{A}^{2}}+{{\alpha }_{b}}H{{B}^{2}}+{{\alpha }_{c}}H{{C}^{2}} \\ & {{\alpha }_{a}}A{{B}^{2}}+{{\alpha }_{c}}B{{C}^{2}}=(1+{{\alpha }_{b}})H{{B}^{2}}+{{\alpha }_{a}}H{{A}^{2}}+{{\alpha }_{c}}H{{C}^{2}} \\ & {{\alpha }_{a}}A{{C}^{2}}+{{\alpha }_{b}}B{{C}^{2}}=(1+{{\alpha }_{c}})H{{C}^{2}}+{{\alpha }_{a}}H{{A}^{2}}+{{\alpha }_{b}}H{{B}^{2}} \\ \end{align} \right.$$ $$\left( {{\alpha }_{a}}+{{\alpha }_{b}} \right)A{{B}^{2}}+\left( {{\alpha }_{a}}+{{\alpha }_{c}} \right)A{{C}^{2}}+\left( {{\alpha }_{b}}+{{\alpha }_{c}} \right)B{{C}^{2}}=\left( 1+3{{\alpha }_{a}} \right)H{{A}^{2}}+\left( 1+3{{\alpha }_{b}} \right)H{{B}^{2}}+\left( 1+3{{\alpha }_{c}} \right)H{{C}^{2}}$$ $${{\alpha }_{a}}H{{A}^{2}}+{{\alpha }_{b}}H{{B}^{2}}+{{\alpha }_{c}}H{{C}^{2}}=\frac{\left( {{\alpha }_{a}}+{{\alpha }_{b}} \right)A{{B}^{2}}+\left( {{\alpha }_{a}}+{{\alpha }_{c}} \right)A{{C}^{2}}+\left( {{\alpha }_{b}}+{{\alpha }_{c}} \right)B{{C}^{2}}}{3}-\frac{H{{A}^{2}}+H{{B}^{2}}+H{{C}^{2}}}{3}\,\,\left( 12 \right)$$ by $${{\alpha }_{a}}+{{\alpha }_{b}}=\cot B\cot C+\cot A\cot C=\left( \cot A+\cot B \right)\cot C=\frac{\sin (A+B)\cos C}{\sin A\sin B\sin C}$$ and $$\sin (A+B)=\sin C$$, We have$${{\alpha }_{a}}+{{\alpha }_{b}}=\frac{\cos C}{\sin A\sin B}$$ and similar $${{\alpha }_{a}}+{{\alpha }_{c}}=\frac{\cos B}{\sin A\sin C}\,\,;\,\,{{\alpha }_{b}}+{{\alpha }_{c}}=\frac{\cos A}{\sin B\sin C}$$. We have $$H{{A}^{2}}+H{{B}^{2}}+H{{C}^{2}}=12{{R}^{2}}-({{a}^{2}}+{{b}^{2}}+{{c}^{2}})$$. $${{\alpha }_{a}}H{{A}^{2}}+{{\alpha }_{b}}H{{B}^{2}}+{{\alpha }_{c}}H{{C}^{2}}=\frac{1}{3}\left( \frac{{{c}^{2}}\cos C}{\sin A\sin B}+\frac{{{b}^{2}}\cos B}{\sin A\sin C}+\frac{{{a}^{2}}\cos A}{\sin B\sin C} \right)-4{{R}^{2}}+\frac{{{a}^{2}}+{{b}^{2}}+{{c}^{2}}}{3}$$ By $$\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}=2R$$, we have $${{\alpha }_{a}}H{{A}^{2}}+{{\alpha }_{b}}H{{B}^{2}}+{{\alpha }_{c}}H{{C}^{2}}=\frac{4}{3}{{R}^{2}}\,\,\frac{{{\sin }^{3}}A\cos A+{{\sin }^{3}}B\cos B+{{\sin }^{3}}C\cos C}{\sin A\sin B\sin C}-4{{R}^{2}}+\frac{{{a}^{2}}+{{b}^{2}}+{{c}^{2}}}{3}$$ Let $$T={{\sin }^{3}}A\cos A+{{\sin }^{3}}B\cos B+{{\sin }^{3}}C\cos C$$  ,from $${{\sin }^{3}}x\cos x=\frac{2\sin 2x-\sin 4x}{8}$$ We have $$T=\frac{\sin 2A+\sin 2B+\sin 2C}{4}-\frac{\sin 4A+\sin 4B+\sin 4C}{8}$$ by $$\sin 2A+\sin 2B+\sin 2C=4\sin A\sin B\sin C$$ and $$\sin 4A+\sin 4B+\sin 4C=-4\sin 2A\sin 2B\sin 2C=-32\sin A\sin B\sin C\cos A\cos B\cos C$$ so $$T=\sin A\sin B\sin C\left( 1+4\cos A\cos B\cos C \right)$$. we have $${{\alpha }_{a}}H{{A}^{2}}+{{\alpha }_{b}}H{{B}^{2}}+{{\alpha }_{c}}H{{C}^{2}}=\frac{4}{3}{{R}^{2}}\,\,\left( 1+4\cos A\cos B\cos C \right)-4{{R}^{2}}+\frac{{{a}^{2}}+{{b}^{2}}+{{c}^{2}}}{3}$$ or     $${{\alpha }_{a}}H{{A}^{2}}+{{\alpha }_{b}}H{{B}^{2}}+{{\alpha }_{c}}H{{C}^{2}}=\frac{{{a}^{2}}+{{b}^{2}}+{{c}^{2}}}{3}-\frac{8}{3}{{R}^{2}}+\frac{16}{3}{{R}^{2}}\cos A\cos B\cos C$$ by $$\cos A\cos B\cos C=\frac{{{a}^{2}}+{{b}^{2}}+{{c}^{2}}}{8{{R}^{2}}}-1$$ , so $${{\alpha }_{a}}H{{A}^{2}}+{{\alpha }_{b}}H{{B}^{2}}+{{\alpha }_{c}}H{{C}^{2}}=\frac{{{a}^{2}}+{{b}^{2}}+{{c}^{2}}}{3}-\frac{8}{3}{{R}^{2}}+\frac{16}{3}{{R}^{2}}\left( \frac{{{a}^{2}}+{{b}^{2}}+{{c}^{2}}}{8{{R}^{2}}}-1 \right)={{a}^{2}}+{{b}^{2}}+{{c}^{2}}-8{{R}^{2}}$$ $${{\alpha }_{a}}H{{A}^{2}}+{{\alpha }_{b}}H{{B}^{2}}+{{\alpha }_{c}}H{{C}^{2}}={{a}^{2}}+{{b}^{2}}+{{c}^{2}}-8{{R}^{2}}=8{{R}^{2}}\cos A\cos B\cos C\,\,\,\,\left( 13 \right)$$ So $${{\alpha }_{a}}X{{A}^{2}}+{{\alpha }_{b}}X{{B}^{2}}+{{\alpha }_{c}}X{{C}^{2}}=X{{H}^{2}}+8{{R}^{2}}\cos A\cos B\cos C\,\,\,\,\,$$ Or $$\left( \cot B\cot C \right)X{{A}^{2}}+\left( \cot A\cot C \right)X{{B}^{2}}+\left( \cot A\cot B \right)X{{C}^{2}}=X{{H}^{2}}+8{{R}^{2}}\cos A\cos B\cos C\,$$. 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 $$\left( \cot B\cot C \right)X{{A}^{2}}+\left( \cot A\cot C \right)X{{B}^{2}}+\left( \cot A\cot B \right)X{{C}^{2}}=X{{H}^{2}}+2\,\,I{{H}^{2}}+4{{r}^{2}}\,\,\,\,$$. where R and r denote the circumradius and inradius respectively.

Theorem5 Let $$X$$ be an arbitrary point in the plane of triangle $$ABC$$, then $$O{{X}^{2}}+{{R}^{2}}=\frac{X{{A}^{2}}\,\sin 2A+X{{B}^{2}}\,\sin 2B+X{{C}^{2}}\,\sin 2C}{\sin 2A+\sin 2B+\sin 2C}$$, where $$R$$ is circumradius.

Proof:

We have $$\overrightarrow{OA}\,\,\sin 2A+\overrightarrow{OB}\,\,\sin 2B+\overrightarrow{OC}\,\,\sin 2C=\overrightarrow{O\,}\,\,\,$$ Let $$X$$ be an arbitrary point in the plane of triangle $$ABC$$ We have $$\left\{ \begin{align} & X{{A}^{2}}\sin 2A={{\left( \overset{\xrightarrow – }{\mathop{XO}}\,+\overset{\xrightarrow – }{\mathop{OA}}\, \right)}^{2}}\sin 2A=X{{O}^{2}}\sin 2A+O{{A}^{2}}\sin 2A+2\overset{\xrightarrow – }{\mathop{XO}}\,.\overset{\xrightarrow – }{\mathop{OA}}\,\sin 2A \\ & X{{B}^{2}}\sin 2B={{\left( \overset{\xrightarrow – }{\mathop{XO}}\,+\overset{\xrightarrow – }{\mathop{OB}}\, \right)}^{2}}\sin 2B=X{{O}^{2}}\sin 2B+O{{B}^{2}}\sin 2B+2\overset{\xrightarrow – }{\mathop{XO}}\,.\overset{\xrightarrow – }{\mathop{OB}}\,\sin 2B \\ & X{{C}^{2}}\sin 2C={{\left( \overset{\xrightarrow – }{\mathop{XO}}\,+\overset{\xrightarrow – }{\mathop{OC}}\, \right)}^{2}}\sin 2C=X{{O}^{2}}\sin 2C+O{{C}^{2}}\sin 2C+2\overset{\xrightarrow – }{\mathop{XO}}\,.\overset{\xrightarrow – }{\mathop{OC}}\,\sin 2C \\ \end{align} \right.$$

$$X{{A}^{2}}\sin 2A+X{{B}^{2}}\sin 2B+X{{C}^{2}}\sin 2C=\left( \sin 2A+\sin 2B+\sin 2C \right)X{{O}^{2}}+{{R}^{2}}\left( \sin 2A+\sin 2B+\sin 2C \right)$$because $$\overrightarrow{OA}\,\,\sin 2A+\overrightarrow{OB}\,\,\sin 2B+\overrightarrow{OC}\,\,\sin 2C=\overrightarrow{O\,}\,\,\,$$. So $$O{{X}^{2}}+{{R}^{2}}=\frac{X{{A}^{2}}\,\sin 2A+X{{B}^{2}}\,\sin 2B+X{{C}^{2}}\,\sin 2C}{\sin 2A+\sin 2B+\sin 2C}$$. Remarque: We have $$\sin 2A+\sin 2B+\sin 2C=4\sin A\sin B\sin C=\frac{abc}{2{{R}^{2}}}$$ $$\sin 2A=2\sin A\cos A=\frac{{{a}^{2}}\left( {{b}^{2}}+{{c}^{2}}-{{a}^{2}} \right)}{2R\,abc}\,\,\,,\,\,\sin 2B=\frac{{{b}^{2}}\left( {{c}^{2}}+{{a}^{2}}-{{b}^{2}} \right)}{2R\,abc}$$ And        $$\sin 2C=\frac{{{c}^{2}}\left( {{a}^{2}}+{{b}^{2}}-{{c}^{2}} \right)}{2R\,abc}$$. So $$O{{X}^{2}}+{{R}^{2}}=\frac{{{a}^{2}}\left( {{b}^{2}}+{{c}^{2}}-{{a}^{2}} \right)\,X{{A}^{2}}+{{b}^{2}}\left( {{c}^{2}}+{{a}^{2}}-{{b}^{2}} \right)\,X{{B}^{2}}+{{c}^{2}}\left( {{a}^{2}}+{{b}^{2}}-{{c}^{2}} \right)\,X{{C}^{2}}}{16{{S}^{2}}}$$ Where $$S=\frac{abc}{4R}$$ area of a triangle.  Distance between the incenter and circumcenter

If $$X\equiv I$$ then $$O{{I}^{2}}+{{R}^{2}}=\frac{{{a}^{2}}\left( {{b}^{2}}+{{c}^{2}}-{{a}^{2}} \right)\,I{{A}^{2}}+{{b}^{2}}\left( {{c}^{2}}+{{a}^{2}}-{{b}^{2}} \right)\,I{{B}^{2}}+{{c}^{2}}\left( {{a}^{2}}+{{b}^{2}}-{{c}^{2}} \right)\,I{{C}^{2}}}{16{{S}^{2}}}$$ By $${{a}^{2}}\left( {{b}^{2}}+{{c}^{2}}-{{a}^{2}} \right)\,I{{A}^{2}}+{{b}^{2}}\left( {{c}^{2}}+{{a}^{2}}-{{b}^{2}} \right)\,I{{B}^{2}}+{{c}^{2}}\left( {{a}^{2}}+{{b}^{2}}-{{c}^{2}} \right)\,I{{C}^{2}}=32\left( {{R}^{2}}-rR \right){{S}^{2}}$$. So $$O{{I}^{2}}+{{R}^{2}}=2\left( {{R}^{2}}-rR \right)$$ or $$O{{I}^{2}}={{R}^{2}}-2rR$$ (Euler’s theorem).  Distance between the Orthocenter and Circumcenter $$\left( \cot B\cot C \right)X{{A}^{2}}+\left( \cot A\cot C \right)X{{B}^{2}}+\left( \cot A\cot B \right)X{{C}^{2}}=X{{H}^{2}}+8{{R}^{2}}\cos A\cos B\cos C\,\,\,\,$$ if $$X\equiv O$$ where $$O$$ be the circumcentre of triangle $$ABC$$. $$\left( \cot B\cot C \right)O{{A}^{2}}+\left( \cot A\cot C \right)O{{B}^{2}}+\left( \cot A\cot B \right)O{{C}^{2}}=O{{H}^{2}}+8{{R}^{2}}\cos A\cos B\cos C\,\,\,\,$$ We have $$\left\{ \begin{align} & OA=OB=OC=R \\ & \cot A\cot B+\cot B\cot C+\cot C\cot A=1 \\ & \cos A\cos B\cos C=\frac{{{a}^{2}}+{{b}^{2}}+{{c}^{2}}}{8{{R}^{2}}}-1\,\, \\ \end{align} \right.$$ so $${{R}^{2}}=O{{H}^{2}}+8{{R}^{2}}\left( \frac{{{a}^{2}}+{{b}^{2}}+{{c}^{2}}}{8{{R}^{2}}}-1 \right)$$ or $$O{{H}^{2}}=9{{R}^{2}}-\left( {{a}^{2}}+{{b}^{2}}+{{c}^{2}} \right)\,\,\,$$.  Distance between incentre and orthocentre We have $$\left( \cot B\cot C \right)X{{A}^{2}}+\left( \cot A\cot C \right)X{{B}^{2}}+\left( \cot A\cot B \right)X{{C}^{2}}=X{{H}^{2}}+8{{R}^{2}}\cos A\cos B\cos C\,\,\,\,$$ If $$X\equiv I$$ where $$I$$ be its incentre $$\left( \cot B\cot C \right)I{{A}^{2}}+\left( \cot A\cot C \right)I{{B}^{2}}+\left( \cot A\cot B \right)I{{C}^{2}}=I{{H}^{2}}+8{{R}^{2}}\cos A\cos B\cos C\,\,\,\,$$ by $$IA=\frac{r}{\sin \frac{A}{2}}\,\,\,,\,\,\,IB=\frac{r}{\sin \frac{B}{2}}\,\,\,,\,\,IC=\frac{r}{\sin \frac{C}{2}}$$ $${{r}^{2}}\left( \frac{\cot B\cot C}{{{\sin }^{2}}\frac{A}{2}}+\frac{\cot A\cot C}{{{\sin }^{2}}\frac{B}{2}}+\frac{\cot A\cot B}{{{\sin }^{2}}\frac{A}{2}} \right)=I{{H}^{2}}+8{{R}^{2}}\cos A\cos B\cos C$$ We have $$\frac{\cot B\cot C}{{{\sin }^{2}}\frac{A}{2}}+\frac{\cot A\cot C}{{{\sin }^{2}}\frac{B}{2}}+\frac{\cot A\cot B}{{{\sin }^{2}}\frac{A}{2}}=2+\frac{4{{R}^{2}}\cos A\cos B\cos C}$$ So $$I{{H}^{2}}=2{{r}^{2}}-4{{R}^{2}}\cos A\cos B\cos C\,\,\,\,$$  ។ Cartesian coordinates Let $$ABC$$ where $$A\left( {{x}_{a}},{{y}_{a}} \right)\,\,,\,\,B\left( {{x}_{b}},{{y}_{b}} \right)\,,\,C\left( {{x}_{c}},{{y}_{c}} \right)$$ and $$H$$ orthocenter of triangle. We have $$\overset{\xrightarrow – }{\mathop{XH}}\,=(\cot B\cot C)\,\overset{\xrightarrow – }{\mathop{XA}}\,+(\cot A\cot C)\overset{\xrightarrow – }{\mathop{XB}}\,+(\cot A\cot B)\,\overset{\xrightarrow – }{\mathop{XC}}\,\,\,\,\,\,$$ If $$X=O\left( 0,0 \right)$$ then  $$\overset{\xrightarrow – }{\mathop{OH}}\,=(\cot B\cot C)\,\overset{\xrightarrow – }{\mathop{OA}}\,+(\cot A\cot C)\overset{\xrightarrow – }{\mathop{OB}}\,+(\cot A\cot B)\,\overset{\xrightarrow – }{\mathop{OC}}\,\,\,\,\,\,$$ So $$\left\{ \begin{align} & {{x}_{h}}=\left( \cot B\cot C \right){{x}_{a}}+\left( \cot A\cot C \right){{x}_{b}}+\left( \cot A\cot B \right){{x}_{c}} \\ & {{y}_{h}}=\left( \cot B\cot C \right){{y}_{a}}+\left( \cot A\cot C \right){{y}_{b}}+\left( \cot A\cot B \right){{y}_{c}} \\ \end{align} \right.$$ Where $$\cot A=\frac{{{b}^{2}}+{{c}^{2}}-{{a}^{2}}}{4S}\,\,,\,\,\cot B=\frac{{{c}^{2}}+{{a}^{2}}-{{b}^{2}}}{4S}\,,\,\,\,\cot C=\frac{{{a}^{2}}+{{b}^{2}}-{{c}^{2}}}{4S}$$ And $$\left\{ \begin{align} & a=BC=\sqrt{{{\left( {{x}_{c}}-{{x}_{b}} \right)}^{2}}+{{\left( {{y}_{c}}-{{y}_{b}} \right)}^{2}}} \\ & b=AC=\sqrt{{{\left( {{x}_{c}}-{{x}_{a}} \right)}^{2}}+{{\left( {{y}_{c}}-{{y}_{a}} \right)}^{2}}} \\ & c=AB=\sqrt{{{\left( {{x}_{b}}-{{x}_{a}} \right)}^{2}}+{{\left( {{y}_{b}}-{{y}_{a}} \right)}^{2}}} \\ \end{align} \right.$$, $$S=\sqrt{p\left( p-a \right)\left( p-b \right)\left( p-c \right)}$$  ( $$p=\frac{a+b+c}{2}$$ ). Pascal: find H(xh,yh): Var xa,ya,xb,yb,xc,yc,xh,yh,cotA,cotB,cotC,s,p,a,b,c:real; begin write('inpu xa= ');read(xa); write('inpu ya= ');read(ya); write('inpu xb= ');read(xb); write('inpu yb= ');read(yb); write('inpu xc= ');read(xc); write('inpu yc= ');read(yc); a:=sqrt(sqr(xc-xb)+sqr(yc-yb)); b:=sqrt(sqr(xc-xa)+sqr(yc-ya)); c:=sqrt(sqr(xb-xa)+sqr(yb-ya)); p:=(a+b+c)/2; s:=sqrt(p*(p-a)*(p-b)*(p-c)); cotA:=(b*b+c*c-a*a)/(4*s); cotB:=(a*a+c*c-b*b)/(4*s); cotC:=(a*a+b*b-c*c)/(4*s); xh:=(cotB)*(cotC)*xa+(cotA)*(cotC)*xb+(cotA)*(cotB)*xc; yh:=(cotB)*(cotC)*ya+(cotA)*(cotC)*yb+(cotA)*(cotB)*yc; writeln('xh= ',xh:8:8);readln; writeln('yh= ',yh:8:8);readln; End. For $$A\left( {{x}_{a}},{{y}_{a}},{{z}_{a}} \right)\,\,,\,\,B\left( {{x}_{b}},{{y}_{b}},{{z}_{a}} \right)\,,\,C\left( {{x}_{c}},{{y}_{c}},{{z}_{c}} \right)$$. Var xa,ya,za,xb,yb,zb,xc,yc,zc,xh,yh,zh,cotA,cotB,cotC,s,p,a,b,c:real; begin write('inpu xa= ');read(xa); write('inpu ya= ');read(ya); write('inpu za= ');read(za); write('inpu xb= ');read(xb); write('inpu yb= ');read(yb); write('inpu zb= ');read(zb); write('inpu xc= ');read(xc); write('inpu yc= ');read(yc); write('inpu zc= ');read(zc); a:=sqrt(sqr(xc-xb)+sqr(yc-yb)+sqr(zc-zb)); b:=sqrt(sqr(xc-xa)+sqr(yc-ya)+sqr(zc-za)); c:=sqrt(sqr(xb-xa)+sqr(yb-ya)+sqr(zb-za)); p:=(a+b+c)/2; s:=sqrt(p*(p-a)*(p-b)*(p-c)); cotA:=(b*b+c*c-a*a)/(4*s); cotB:=(a*a+c*c-b*b)/(4*s); cotC:=(a*a+b*b-c*c)/(4*s); xh:=(cotB)*(cotC)*xa+(cotA)*(cotC)*xb+(cotA)*(cotB)*xc; yh:=(cotB)*(cotC)*ya+(cotA)*(cotC)*yb+(cotA)*(cotB)*yc; zh:=(cotB)*(cotC)*za+(cotA)*(cotC)*zb+(cotA)*(cotB)*zc; writeln('xh= ',xh:8:8);readln; writeln('yh= ',yh:8:8); writeln('zh= ',zh:8:8); readln; End.

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.

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