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\centerline{\bf Math 254H  \hfill {\large Centroid Theorem} \hfill  Lect 1} 
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\noindent  % Prevent indentation at beginning of paragraph
A \emph{median} of a triangle is the line segment 
from a vertex to the midpoint of the opposite side. 

\vspace{1em}

\noindent  
\textsc{Proposition:} Given any triangle in the plane, the three medians intersect at a common point which is $\frac{2}{3}$ 
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of the way along each median.

\vspace{1em}

\noindent
\textsc{Proof:} We translate the proposition into the language of vector algebra. 
Let $\u, \v, \w$ be the vectors from the origin to the vertices of the triangle.. 

The vector from $\v$ to $\w$ is $\w-\v$,
and the midpoint of the corresponding side is:
$$
\v + \tfrac{1}{2}(\w-\v) \ = \ \tfrac{1}{2}\v + \tfrac{1}{2}\w.
$$
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The vector from $\u$ to this midpoint is $\tfrac12\v + \tfrac12\w - \u$,
and the median from $\u$ to the midpoint
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is the parametrized line segment:
$$
\l(t) = \u + t(\tfrac12\v + \tfrac12\w - \u) \ \ \ \text{for} \ \ \ 0 \leq t \leq 1.
$$
(That is, the points of the segment are the endpoints of the vectors $\l(t)$, 
written in standard form from the origin.)
The point $\frac23$ of the way along this median is:

$$\begin{array}{rcl}  
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\l(\tfrac23) & =  & \u + \tfrac23(\tfrac12\v + \tfrac12\w - \u)
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&=& \u + \tfrac13 \v + \tfrac13\w - \tfrac23\u
\\[1em]  
&=& (1-\tfrac23) \u + \tfrac13 \v + \tfrac13\w
\\[1em]  
&=& \tfrac13(\u + \v + \w),
\\[.5em]
\end{array}$$
where we expand and factor using 
the distributive property of scalar multiplication over vector addition.

The same computation for the other two medians
gives their $\frac23$ points:
$$
\v + \tfrac23(\tfrac12\u + \tfrac12\w - \v) 
\ = \  \tfrac13(\u + \v + \w),
$$
$$
\w + \tfrac23(\tfrac12\u + \tfrac12\v - \w) 
\ = \ \tfrac13(\u + \v + \w).
$$
Thus, all three medians contain a common point, the $\frac23$-point along each. 
\hfill $\square$
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\vspace{0em}

\noindent
\textsc{Note:} We call this common point the \emph{centroid} of the triangle.
The proof shows it is the vector average of $\u,\v,\w$.

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