Finding the centroid of a triangle

Problem:   Prove, using a vector method, that the medians of a triangle meet at a point which divides each median in the same ratio. Find this ratio.
This point is called the centroid of the triangle.
A median is a line which joins a vertex of a triangle to the midpoint of the opposite side.
a triangle, showing its medians

We have to prove that the medians AP, BQ and CR of triangle ABC do meet in a single point G and that AG/GP = BG/GQ = CG/GR = k and find the value of the constant, k.

Suppose that the position vectors of A, B and C with respect to some origin O are a, b and c.
O need not necessarily lie in the plane of the triangle.
DANGER   Students quite often think that a, b and c give the sides of the triangle. They don't! See the drawing below.

the triangle relative to the origin
The right-hand drawing gives the vectors for the sides of the triangle, found using vector addition.
For example, AB = -OA + OB.

We start by finding the position vector of the point where 2 of the medians cut each other.
We'll then have to show that the third median also passes through this point. I've chosen to work with AP and BQ.
Suppose that AP and BQ meet at G. We'll need to show that the position vector of G depends equally on a, b and c. Then, by similar working, G would also lie on CR proving that the 3 medians do all meet in a point.

finding where 2 medians intersect
The median AP lies in the extended line I've shown in my drawing. (See lines if necessary.)
Its vector equation can be written   r = a + sAP.
But  AP = AB + BP = -a + b + 1/2(-b + c) = -a + 1/2(b + c).
So the equation of line AP is   r = a + s (-a + 1/2(b + c)).
Show for yourself that the vector equation of line BQ is   r = b + t (-b + 1/2(c + a)).
The 2 lines meet where  r = a + s (-a + 1/2(b + c)) = r = b + t (-b + 1/2(c + a)).

To get the same vector r for each line, we must have the same quantity of each of a, b and c.
Matching up for c gives   1/2sc = 1/2tc so s = t.
Matching up for a gives   a - sa = 1/2ta so 1 - s = 1/2t  giving   t = 2/3 = s.
Therefore G lies 2/3 of the way down both AP and BQ so k = 2/3.
Also, putting s = 2/3 in the equation of line AP gives
r = g = a + 2/3(-a + 1/2(b + c ) = 1/3(a + b + c)).
We see that the position vector r of G depends equally on a, b and c so therefore, by similar working, G also lies 2/3 of the way down CR.

The point G is called the centroid of triangle ABC.
Its position vector g = 1/3 (a + b + c) where a, b and c are the position vectors of A, B and C.

Physically, if the triangle is made of a uniform thin heavy material, then G is where the whole weight of the triangle acts.
where the weight of the triangle acts

If the triangle is cut into tiny strips parallel to each side in turn, then the weight of each tiny strip acts at its midpoint. Therefore G, the point at which the weight of the whole triangle acts, must lie on each median. Therefore the medians must intersect.

Quickie version

If you are asked to show that the medians intersect at a point 2/3 of the way down each of them then there is a short cut method you can use.
finding G on median AP
Let OG = g and OP = p.
Then   g = p + 1/3 PA  but  p = b + 1/2BC = b + 1/2(-b + c) = 1/2(b + c).
Also  PA = -p + a = a - 1/2(b + c).
So  g = 1/2(b + c) + 1/3(a - 1/2(b + c)) = 1/3(a + b + c).
We see that g depends equally on a, b and c and is therefore independent of the choice of median. Therefore it lies on each median and gives their point of intersection.

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