Locate $\frac{1}{3}\vec{u} + \frac{1}{3}\vec{v} + \frac{1}{3}\vec{w}$ and $\frac{1}{2}\vec{u} + \frac{1}{2}\vec{w}$ on the figure. I was studying Gilbert Strang's linear algebra book (An Introduction to Linear Algebra (4e)) and came across a problem that I couldn't understand. I've noticed someone's asked the same question here but the only answer didn't address the question.
The "figure" that I'm referring to in the title is this:


Locate $\frac{1}{3}\vec{u} + \frac{1}{3}\vec{v} + \frac{1}{3}\vec{w}$ and $\frac{1}{2}\vec{u} + \frac{1}{2}\vec{w}$ on the figure. Challenge problem: Under what restrictions on $c$, $d$, $e$ will the combinations $c\vec{u} + d\vec{v} + e\vec{w}$ fill in the dashed triangle?

It's not hard finding answers to problems, but even with the answer I cannot grasp the solution. The answer is that "$\frac{1}{3}\vec{u} + \frac{1}{3}\vec{v} + \frac{1}{3}\vec{w}$ is the center point of the dashed triangle," "$\frac{1}{2}\vec{u} + \frac{1}{2}\vec{w}$ is the point between $\vec{u}$ and $\vec{w}$."
I can somewhat understand how the second answer came to be, but I'm really having difficulty understanding how the sum of vector thirds leads to the center point of the triangle.
Would anybody kind enough to explain?
Thank you.
 A: It depends on what you mean by "the centre" of a triangle. There are thousands of points which people call a "centre" of a triangle. In this case, the "centre" being referred to is the Centroid.
The most common characterisation of the centroid is the common points of intersection between the lines from each vertex to the midpoint of its opposite side. The midpoint of $u$ and $v$ is $\frac{u}{2} + \frac{v}{2}$, so the line through $w$ and $\frac{u}{2} + \frac{v}{2}$ is given by
$$x(t) = w + t\left(\frac{u}{2} + \frac{v}{2} - w\right).$$
Note that
$$x\left(\frac{2}{3}\right) = \frac{u}{3} + \frac{v}{3} + \frac{w}{3}.$$
Also note that this expression is completely symmetric, so doing the same with, say, $v$ and $\frac{u}{2} + \frac{w}{2}$ will produce a line with the same point on it. Therefore, all three possible lines will all pass through this one point.
A: I work on that book, too. If the challenge part is solved, the case of $\frac{1}{3} \vec{u} + \frac{1}{3}\vec{v} + \frac{1}{3}\vec{w}$ will resolve itself.
Assume a Point $P$ on the said triangle, and we let,
$$
\vec{OP} = c\vec{u}+ d\vec{v} + e\vec{w}
$$
and assume a Point $Q$ on one of the edge, say the edge $\vec{u} - \vec{w}$. I think you must understand the Problem 15-19, so,
\begin{align}
&\vec{OQ} = a\vec{u} + b\vec{w}, \text{ and},\\
&a+b = 1
\end{align}
and,
$$
\vec{v} - \vec{OP} = \lambda(\vec{v} - \vec{OQ})
$$
plug in $\vec{OQ}$, we get,
\begin{align}
&\vec{OP} = \lambda a\vec{u} + (1-\lambda)\vec{v} + \lambda b\vec{w}, \text{ and},\\
&c + d + e = \lambda a + (1-\lambda) + \lambda b = 1
\end{align}
Hence, Point $P$ is on the triangle.
Further, you can find that in this case, $\lambda = \frac{2}{3}, a = b = \frac{1}{2}$, so Point $P$ is the CENTROID of the triangle.
