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<x,y<\dfrac{\pi}{2} then x=y.

Link download http://tuan.nguyentrung.googlepages.com/Problem2RMO2007.pdf

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