Medians of a Triangle (for Carolina)

February 24, 2025

Let $u, v$ and $w$ be the position vectors of the vertices of a triangle in $n$-space. Find an expression for the intersection of the medians of the triangle. A median is a line segment from a vertex to the midpoint of the opposite side. Don't use results from planar geometry. Use convex combinations of the form $(1-t)p + tq$ to set up expressions for two medians and find their common position vector.

hi carolina i cant seem to fit this in a tweet. we need to find the intersection point of the medians of a triangle in $n$-dimensional space (the centroid), but we want to use convex combinations and we cant just use geometry :((

so we parameterize the two medians — one from $u$ to the midpoint of $vw$, one from $v$ to the midpoint of $uw$:

$$ (1 - t)\,u + t\!\left(\frac{v + w}{2}\right) \qquad\qquad (1 - s)\,v + s\!\left(\frac{u + w}{2}\right) $$

then set them equal and solve

$$\begin{align} (1 - t)\,u + \frac{t}{2}(v + w) &= (1 - s)\,v + \frac{s}{2}(u + w) \\[6pt] 1 - t &= \tfrac{s}{2}, \qquad \tfrac{t}{2} = 1 - s, \qquad \tfrac{t}{2} = \tfrac{s}{2} \\[6pt] t &= s = \tfrac{2}{3} \end{align}$$

then plug that back in

$$\begin{align} \frac{1}{3}u + \frac{1}{3}v + \frac{1}{3}w &= \frac{u + v + w}{3} \end{align}$$

giving us

$$\boxed{\dfrac{u + v + w}{3}}$$

i think this makes sense, because like a convex combination is like a line from $p$ to $q$ as $t$ varies from 0 to 1 — when ur at $t=0$ ur at $p$ and $t=1$ is at $q$ so the values in the middle represent points along the line. also it just makes sense that its all 3 divided by 3 because that seems like what the center would be i think.

$(1-t)p + tq$ is like smushing (scaling) a value from $p$ to $q$ to $[0, 1]$, the same way you can represent rgb colors from 0–255 or 0f to 1f. like if i took rgb (200, 200, 200) (it's offwhite) and ran that on its interpolation from 0 to 1 ($t$) we get the og value.

the reason u do this in this problem: the simplest way to solve it would just be to recognize that the intersection of medians of a triangle is their mean (via geometry) — BUT! the problem says ur not allowed to do that. so they force u to find the distance from each point to the median of the other two, by parameterizing. the reason u parameterize is because parameterization lets you describe the paths of the medians explicitly using convex combinations.

— vish