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Let $\Delta ABC$ and any point $M$ in the triangle. $A_1,B_1$ and $C_1$ be symmetric points of $M$ through the midpoints of $BC, CA, AB$. Prove that $AA_1, BB_1,CC_1$ are concurrent at $O$ and $M,O,G$ are collinear.


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I proved that $BCB_1C_1$ is parallelogram then it has done but i am studying about vector then i want to solve it by vector. Otherwise i thought we will assume $AA_1\cap BB_1=O$ then $C_1,O,C$ are collinear. Indeed i tried to prove $\overrightarrow{CO}=k\overrightarrow{CC_1}$ but failed. Help me

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2 Answers 2

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The solution by vectors.

Let $\overrightarrow{MA}=\vec{a}$, $\overrightarrow{MB}=\vec{b}$, $\overrightarrow{MC}=\vec{c}$, $A_2$, $B_2$ and $C_2$ be midpoints of $BC$, $AC$ and $AB$ respectively

and let $M_1$ be a point such that $\overrightarrow{MA}=\overrightarrow{A_1M_1}.$

Thus, $$\overrightarrow{BM}=\overrightarrow{BC_2}+\overrightarrow{C_2M}=\overrightarrow{C_2A}+\overrightarrow{C_1C_2}=\overrightarrow{C_1A}.$$ By the same way we obtain: $$\overrightarrow{MB}=\overrightarrow{AC_1}=\overrightarrow{B_1M_1}=\overrightarrow{CA_1}=\vec{b},$$ $$\overrightarrow{MA}=\overrightarrow{BC_1}=\overrightarrow{A_1M_1}=\overrightarrow{CB_1}=\vec{a}$$ and $$\overrightarrow{MC}=\overrightarrow{AB_1}=\overrightarrow{C_1M_1}=\overrightarrow{BA_1}=\vec{c}.$$ Now, let $AA_1\cap BB_1=\{O\}$.

Thus, $$\overrightarrow{CO}=\frac{1}{2}(\overrightarrow{CB}+\overrightarrow{CB_1})=\frac{1}{2}(-\vec{c}+\vec{b}+\vec{a})=\frac{1}{2}(\overrightarrow{B_1C_1}+\overrightarrow{BC_1})=\overrightarrow{OC_1},$$ which says that $O$ is a common point of $AA_1$, $BB_1$ and $CC_1$.

Now, $$\overrightarrow{MG}=\overrightarrow{MB}+\overrightarrow{BG}=\vec{b}+\frac{2}{3}\cdot\frac{1}{2}(\overrightarrow{BC}+\overrightarrow{BA})=$$ $$=\vec{b}+\frac{1}{3}(-\vec{b}+\vec{c}-\vec{b}+\vec{a})=\frac{1}{3}(\vec{a}+\vec{b}+\vec{c}).$$ Also, $$\overrightarrow{MO}=\frac{1}{2}\left(\overrightarrow{MB}+\overrightarrow{MB_1}\right)=\frac{1}{2}(\vec{b}+\vec{c}+\vec{a})$$ and we got that $M$, $O$ and $G$ are placed on the same line.

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Since $BMCA_1$, $BMAC_1$ and $MCB_1A$ are parallelograms, we obtain that $A_1C_1AC$, $A_1B_1AB$ and $CB_1C_1B$ are parallelograms, which gives that $AA_1$, $BB_1$ and $CC_1$ are concurrent because any two of them they are diagonals of one parallelogram from three last parallelograms.

Now, we can see the picture as a projection of a parallelepiped $A_1BMCM_1C_1AB_1,$ where $O$ is a midpoint of $MM_1.$

$AA_1$, $BB_1$, $CC_1$ and $MM_1$ they are diagonals of the parallelepiped.

Now, easy to get also the second part of the problem because $MM_1\cap(ABC)=\{G\}$

and we got also that $MG=\frac{1}{3}MM_1$ and $MO=\frac{1}{2}MM_1$.

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  • $\begingroup$ Thank you Michael but i want a solution by vector. $\endgroup$
    – DVdivi
    Aug 5, 2019 at 14:53

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