Solve a problem using vectors

Question

The purpose of this problem is to use vectors to show that the medians of a triangle all meet at a point. First, I have to show that $P$ (see the picture below) must lie two-thirds of the way from $B$ to $M_1$ and two-thirds of the way from $C$ to $M_2$. Then I'll be able to show why all three medians must meet at $P$ (this is what my book says). I really can't figure it out...any help would be great.

Answer 1

Let $\vec{a}, \vec{b}, \vec{c}$ be the position vectors of the vertices of the triangle $\Delta A B C$ and $M_1$ is opposite $A$, $M_2$ opposite B, $M_3$ opposite $C$. Now, the sides $AC=\vec{c} - \vec{a}, BC=\vec{c}-\vec{b}, AB=\vec{b}-\vec{a}$. Therefore, the medians are $AM_1=\vec{b}-\vec{a}+ \frac{1}{2}(\vec{c}-\vec{b}) = \frac{1}{2}\vec{b}-\vec{a}+\frac{1}{2}\vec{c}, BM_2=\frac{1}{2}\vec{a}-\vec{b}+\frac{1}{2}\vec{c}, \text{and } CM_3=\frac{1}{2}\vec{a}-\vec{c}+\frac{1}{2}\vec{b}$

Now we will find the intersection point of two of the medians: let's say $AM_1$ and $BM_2$. The vector representation of the lines on which $AM_1$ and $BM_2$ are are: $$\vec{r_1}(p)=\vec{a}+p(\frac{1}{2}\vec{b}-\vec{a}+\frac{1}{2}\vec{c}) \\ \vec{r_2}(q)=\vec{b} + q(\frac{1}{2}\vec{a}-\vec{b}+\frac{1}{2}\vec{c})$$ We equate these two $\vec{r_1}=\vec{r_2}$. In order for them to be equal, the coefficients of $\vec{a}, \vec{b}, \vec{c}$ must be equal: $$\vec{b} + q(\frac{1}{2}\vec{a}-\vec{b}+\frac{1}{2}\vec{c}) = \vec{a}+p(\frac{1}{2}\vec{b}-\vec{a}+\frac{1}{2}\vec{c}) \\ \Rightarrow \frac{1}{2}q=\frac{1}{2}p \text{ and } 1-p=\frac{1}{2}q \Rightarrow p=q=2/3$$ Thus the intersection has position vector $\vec{r_1}(2/3)=\frac{1}{3}(\vec{a}+\vec{b}+\vec{c})$ Now note that the vector line for $CM_3$ is $$\vec{r_3}(s)=\vec{c}+s(\frac{1}{2}\vec{a}-\vec{c}+\frac{1}{2}\vec{b})$$ Now we want to see if we can find some value of $s=s_0$ such that $\vec{r_3}(s_0)=\frac{1}{3}(\vec{a}+\vec{b}+\vec{c})$. That would mean that the vector $CM_3$ also intersects with the other two at that same point. Again by equating the coefficients of the individual vectors, we get $$1-s=\frac{1}{3} \\ \frac{1}{2}s=\frac{1}{3} \\ \frac{1}{2}s=\frac{1}{3}$$ We see that $s=2/3=s_0$ satisfies this system. Thus, there is a common point at which all three medians meet.