Prove analytically. The medians of a triangle intersect in a trisection point of each. The medians of a triangle intersect in a trisection point of each.
 A: Hint: Consider the analytic formulas relating the $\color{#00A000}{\text{vertices}}$, $\color{#C00000}{\text{midpoints}}$, and $\color{#0000FF}{\text{median}}$:
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A: Let the vertices of the triangle be at the points (vectors) $u$, $v$, and $w$. 
The midpoint $m$ of the side joining $u$ and $v$ is $\frac{u+v}{2}$.
The point on the line segment $mw$, one-third of the way up from $m$ towards $w$, is 
$$\frac{2m}{3}+\frac{w}{3}.\tag{1}$$  
Substitute $\frac{u+v}{2}$ for $m$ in (1). We get 
$$\frac{1}{3}(u+v+w).$$ 
This expression is symmetric in $u$, $v$, and $w$, so lies on all three medians.
Remark: To make the proof look more "analytic" we could uglify it. Let the coordinates of the vertices be $(u_x,u_y)$, $(v_x,v_y)$, $(w_x,w_y)$. Then as above we compute the coordinates of $m$, and then of trisection point. 
A: I can propose the following evidence. Let $ABC$ be an arbitrary triangle of a Cartesian plane. Let its vertices $A$, $B$, and $C$ have the coordinates $(x_1,y_1)$, $(x_2,y_2)$ and $(x_3, y_3)$ respectively. Then the midpoints $A_1$, $B_1$, and $C_1$ of the sides $BC$, $CA$, and $AB$ respectively have the coordinates $(\frac{x_2+x_3}2, \frac{y_1+y_3}2)$, $(\frac{x_1+x_3}2, \frac{y_1+y_3}2)$ and $(\frac{x_1+x_2}2, \frac{y_1+y_2}2)$ respectively. Shifting, if necessary, the zero $O$ of the Cartesian plane, without loss of generality, we may assume that $x_1+x_2+x_3=y_1+y_2+y_3=0$. Then all the determinants 
$\left|\begin{matrix}
x_1 & y_1 \\
\frac{x_2+x_3}2 & \frac{y_2+y_3}2\\
\end{matrix}
\right|,$
$\left|\begin{matrix}
x_2 & y_2 \\
\frac{x_1+x_3}2 & \frac{y_1+y_3}2\\
\end{matrix}
\right|$, and 
$\left|\begin{matrix}
x_3 & y_3 \\
\frac{x_1+x_2}2 & \frac{y_1+y_2}2\\
\end{matrix}
\right|$ 
are equal to zero. Therefore all the triangles $AOA_1$,  $BOB_1$, and $COC_1$ have zero area. Therefore $O$ is the intersection point of the medians of the triangle $ABC$. Moreover, ${\bf AO}=(-x_1,-y_1) =2(\frac{x_2+x_3}2, \frac{y_1+y_3}2)=2{\bf OA_1}$. Similarly, 
${\bf BO}=2{\bf OB_1}$ and ${\bf CO}=2{\bf OC_1}$. Thus $O$ is the trisection point of each of the segments $AA_1$, $BB_1$, and $CC_1$.
A: Here is the ugly solution:
First, note that $a,b,c$ are affinely independent, that is, $\binom{1}{a},\binom{1}{b},\binom{1}{c}$ are linearly independent. (Otherwise the three points line on a line.)
Pick the vertices $a,b$ first, and the corresponding midpoints on the opposite sides. We want to find the intersection of the medians, that is, find $\lambda_1, \lambda_2$ such that $$(1-\lambda_1)a + \frac{\lambda_1}{2}b + \frac{\lambda_1}{2}c = \frac{\lambda_1}{2}a + (1-\lambda_1)b + \frac{\lambda_1}{2}c$$
Rearranging gives
$$ a(1-\lambda_1-\frac{\lambda_2}{2}) + b(\frac{\lambda_1}{2}-(1-\lambda_2))+c(\frac{\lambda_1}{2}-\frac{\lambda_2}{2}) = 0$$
A quick computation shows that
$$1(1-\lambda_1-\frac{\lambda_2}{2}) + 1(\frac{\lambda_1}{2}-(1-\lambda_2))+1(\frac{\lambda_1}{2}-\frac{\lambda_2}{2}) = 0$$
from which we get the coefficients are all zero (because of affine independence). Solving gives $\lambda_1 = \lambda_2 = \frac{2}{3}$, and hence the intersection is $\frac{1}{3}(a+b+c)$. Repeating with the other two pairs results in the same answer.
A: The medians of an equilateral triangle must trisect by symmetry. Any other triangular shape can be found by performing a linear transformation on the vertices of the equilateral triangle, and such transformations preserve the medians along with their trisection point.
