# Centroid

The centroid of a plane figure is, roughly speaking, its center of mass. If the plane figure is cut out from uniform cardboard, say, and you connected a string to its centroid and held the other end of the string, the figure would be perfectly balanced. (The centroid does not have to be in the figure, however. A condition under which the centroid must be inside the figure is when the figure is convex.)

Of particular interest to students of olympiad geometry is the centroid of a triangle. This is the point of intersection of the medians of the triangle and is conventionally denoted $G$ (mnemonic: gravity). The centroid has the special property that, for each median, the distance from a vertex to the centroid is twice that of the distance from the centroid to the midpoint of the side opposite that vertex. Also, the three medians of a triangle divide the triangle into six regions of equal area.

The coordinates of the centroid of a coordinatized triangle are $(a,b)$ where $a$ is the arithmetic mean of the $x$-coordinates of the vertices of the triangle and $b$ is the arithmetic mean of the $y$-coordinates of the triangle. $[asy] pair C=dir(0); pair A=dir(90); pair B=dir(180); pair M=(B+C)/2; pair N=(A+C)/2; pair O=(A+B)/2; pair G=centroid(C,A,B); D(MP("A",A,D(A))--MP("B",B,D(B))--MP("C",C,D(C))--cycle); D(MP("M",M,dir(270)));D(MP("N",N,D(N)));D(MP("O",O,D(O)));D(M); D(A--M--C--N--B--O--C);MP("G",G,D(G)); [/asy]$

## Proof of concurrency of the medians of a triangle

Note: The existence of the centroid is a trivial consequence of Ceva's Theorem. However, there are many interesting and elegant ways to prove its existence, such as those shown below.

### Proof 1

Readers unfamiliar with homothety should consult the second proof.

Let $D,E,F$ be the respective midpoints of sides $BC, CA, AB$ of triangle $ABC$. We observe that $DE, EF, FD$ are parallel to (and of half the length of) $AB, BC, CA$, respectively. Hence the triangles $ABC, DEF$ are homothetic with respect to some point $G$ with dilation factor $-\frac{1}{2}$; hence $AD, BE, CF$ all pass through $G$, and $AG = 2 GD; BG = 2 GE; CG = 2 GF$.

### Proof 2

Let $ABC$ be a triangle, and let $D,E,F$ be the respective midpoints of the segments $BC, CA, AB$. Let $G$ be the intersection of $BE$ and $CF$. Let $E',F'$ be the respective midpoints of $BG, CG$. We observe that both $EF$ and $E'F'$ are parallel to $CB$ and of half the length of $CB$. Hence $EFE'F'$ is a parallelogram. Since the diagonals of a parallelogram bisect each other, we have $GE = E'G = BE'$, or $BG = 2GE$. Hence each median passes through an appropriate trisection point of each other median and the medians concur.

We note that both of these proofs give the result that the distance of a vertex of a point of a triangle to the centroid of the triangle is twice the distance from the centroid of the triangle to the midpoint of the opposite side.