87. An interesting vectorial identity.

by Virgil Nicula, Aug 17, 2010, 11:32 PM

http://www.artofproblemsolving.com/Forum/viewtopic.php?f=46&t=362204
Quote:
Denote the projections $D$ , $E$ , $F$ of an interior point $M$ on the sides of equilateral $\triangle ABC$ .
Prove that that $\overrightarrow{ME}+\overrightarrow{MD}+\overrightarrow{MF}=\frac{3}{2}\overrightarrow{MO}$ , where $O$ is the center of triangle.
Quote:
Extension. Consider a triangle $ABC$ and a point $M$ for which denote its projections $D$ , $E$ , $F$ on the sidelines $BC$ , $CA$ , $AB$
respectively. Then exists the identity $\boxed {\ \frac {\cos A}{\sin B\sin C}\cdot \overrightarrow{MD}+\frac {\cos B}{\sin C\sin A}\cdot \overrightarrow{ME}+\frac {\cos C}{\sin A\sin B}\cdot \overrightarrow{MF}=\overrightarrow{MH}\ }\ (*)$ .
Proof. If $H$ is orthocenter of $\triangle ABC$ and $X\in AH\cap BC$ , then $\overrightarrow {AH}=\cot A(\cot B+\cot C)\cdot\overrightarrow {AX}$ because $(\cot B\cot C,\cot C\cot A,\cot A\cot B)$ are the normalized barycentrical coordinates of $H$ w.r.t. $\triangle ABC$ . If $(x,y,z)$ are the normalized barycentrical coordinates of $M$ w.r.t. $\triangle ABC$ , then $\overrightarrow{MD}=x\cdot \overrightarrow{AX}$ , i.e. $\cot A(\cot B+\cot C)\cdot\overrightarrow{MD}=x\cdot \overrightarrow{AH}$ . From the well-known identity $\sum x\cdot\overrightarrow{AH}=\overrightarrow {MH}$ obtain $\sum\cot A(\cot B+\cot C)\cdot\overrightarrow{MD}=\overrightarrow {MH}$ . Using the simple relations $\cot A(\cot B+\cot C)=\frac {\cos A}{\sin B\sin C}$ a.s.o. obtain the proposed identity.
Remark. The identity $(*)$ is equivalently with $\boxed {\ \sin 2A\cdot\overrightarrow{MD}+\sin 2B\cdot\overrightarrow{ME}+\sin 2C\cdot\overrightarrow{MF}=\frac {S}{R^2}\cdot \overrightarrow{MH}\ }$ , where $S=[ABC]$ .
Quote:
Generalization. Let $P$ be a point with the normalized barycentrical coordinates $(\alpha , \beta , \gamma )$ w.r.t. $\triangle ABC$ .
For a point $M$ denote $\left\|\begin{array}{cc} D\in BC\ , & MD\parallel AP\\\ E\in CA\ , & ME\parallel BP\\\ F\in AB\ , & MF\parallel CP\end{array}\right\|$ . Then exists the identity $\boxed {\ \sum (\beta +\gamma )\cdot\overrightarrow{MD}=\overrightarrow{MP}\ }$ .

Particular cases.
$1\blacktriangleright$ If $P:=G\left(\frac 13,\frac 13,\frac 13\right)$ , then $\boxed {\ \overrightarrow {MD}+\overrightarrow {ME}+\overrightarrow {MF}=\frac 32\cdot \overrightarrow {MG}\ }$ (an easy extension to any triangle of the initial proposed problem).
$2\blacktriangleright$ If $P:=N\left(\frac {p-a}{p}, \frac {p-b}{p},\frac {p-c}{p}\right)$ (Nagel's point), then $\boxed {\ a\cdot \overrightarrow {AD}+b\cdot\overrightarrow {BE}+c\cdot\overrightarrow {CF}=2p\cdot\overrightarrow {IS}\ }$ , where $S$ is midpoint of $[MN]$ .
This post has been edited 11 times. Last edited by Virgil Nicula, Nov 23, 2015, 2:14 PM

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2 Comments

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Pls, correct your posts' numbering sequence, the last three lost a 50!
Otherwise, as usual, a huge useful work, with a lot of golden ideas!

With kind regards,
S. Fulger

by sunken rock, Aug 18, 2010, 4:45 AM

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Multumesc, Sunken Rock ! Iti multumesc pentru aprecierile acestui my (math)blog. Si cand acestea vin si de la tine inseamna ca efortul meu va da si roade, chiar daca se identifica cu un hobby, adica este si placerea mea.
This post has been edited 1 time. Last edited by Virgil Nicula, Aug 19, 2010, 1:40 AM

by Virgil Nicula, Aug 19, 2010, 12:40 AM

Own problems or extensions/generalizations of some problems which was posted here.

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  • Blog created: Apr 20, 2010
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