# Solution of Problem 2, Grade 9, RNO 2007

Let $ABC$ be an acute angled triangle and point $M$ chosen differently from $A,B,C$. Prove that $M$ is the orthocenter of triangle $ABC$ if and only if

$\dfrac{BC}{MA}\cdot\overrightarrow{MA}+\dfrac{CA}{MB}\cdot\overrightarrow{MB}+\dfrac{AB}{MC}\cdot\overrightarrow{MC}= \overrightarrow{0}.(*)$

Solution of my students.

If $M$ is the orthocenter of triangle $ABC$ then by a well-known theorem, we have $\tan A\cdot\overrightarrow{MA}+\tan B\cdot\overrightarrow{MB}+\tan C\cdot\overrightarrow{MC}=\overrightarrow{0} (1)$ By triangle is acute: $MA=2R\cos A, MB=2R\cos B, MC=2R\cos C$ therefore $\dfrac{BC}{MA}=\tan A, \dfrac{CA}{MB}=\tan B,\dfrac{AB}{MC}=\tan C$, so by $(1)$ we have $(*)$ is proved.

If $(*)$ is true then by $\dfrac{BC}{MA}>0, \dfrac{CA}{MB}>0,\dfrac{AB}{MC}>0$, $M$ in interior of the triangle $ABC$. By a well-known theorem, we have $S[MBC]\cdot\overrightarrow{MA}+S[MCA]\cdot\overrightarrow{MB}+S[MAB]\cdot \overrightarrow{MC}=\overrightarrow{0}(2)$

Use $(2)$ and $(*)$ we have $\dfrac{BC}{MA\cdot S[MBC]}=\dfrac{CA}{MB\cdot S[MCA]}=\dfrac{AB}{MC\cdot S[MAB]}$

But $2S[MBC]=MB\cdot BC\cdot\sin\angle MBC$ and

$2S[MCA]=MA\cdot CA\sin\angle MAC$, hence $\angle MBC=\angle MAC$. Similary, $\angle MBA=\angle MCA,\angle MAB=\angle MCB$. Now, easy see that $M$ is orthocenter of the triangle $ABC$.

Note: If $\sin x=\sin y$ and $0 then $x=y$.