Vector Calculus 7: Triangle Medians Are Concurrent, a Vector Algebra Proof

[chilly]

Very nice proof. I was thinking, what if we choose an origin that's not in the plane of the triangle? Then in a 3D space I think the three vectors A, B and C will necessarily be linearly independent (in fact, assuming that they are will force the origin out of the plane) and we can equate the coefficients to zero after all! Isn't that why that approach gives the correct values, rather than luck? Also the arbitrary origin is necessarily not at the midpoint of the triangle if it's out of the plane.

[mws]

I believe we have some problems with this video:

The matrix method is secretly equivalent to assuming each of the A, B and C coefficients are individually zero.
To see this, consider the sum of the three formulas (collected by the large curly bracket) before they are placed into matrix form. The sum of the left side will equal zero for any alpha or gamma, and the sum of the right side will give (beta)(a + b + c). So we have 0 = (beta)(a + b + c). For the matrix method to give a unique answer we saw that the determinant (-3/4(a+b+c)) could not work if a+b+c=0. We now see that this forces beta to equal zero, and hence each of our individual co-efficient formulas are equal to zero.

I believe the claim that a+b+c=0 would occur when the origin is chosen to be located at the centroid is also a minor error. It would be correct to say that A + B + C = 0 would give that situation (note the change in case), but this would occur when a=b=c, which cannot occur from a + b + c =0 (excluding the trivial case, which would tell us nothing).

I think we can recover a solution based of the current model without resorting to any of the other classic proofs or the tensor method (Side note - I was surprised that many of the the other online proofs start with the assumption that the ratio on the median is already known AND also assume that we can equate each coefficient).
I'm certainly no expert (one day I hope to save up for a proper qualification in maths) so be warned that the following may be completely wrong!

From our sum of the three curly-bracketed expressions - If we assume beta is not zero, then a+b+c is zero.
Let's consider the possibilities:
If a=b=c=0 then once again all three of the formulas are individually forced to be zero.
If only two of a, b, and c are zero, this also forces the third to be zero and the same result occurs. If only one of a, b or c are zero, then that tells us that one of them is just a multiple of the other, and so one of A, B or C is just a multiple of one of the others, and in all three scenarios that result, the large expression of A, B and C with it's coefficients involving alpha and gamma will reduce to a linear combination of only two vectors, each of which can be assumed to be linearly independent (otherwise all three corners would be colinear and we would have an improper triangle). Solving the two 'new' coefficients as equal to zero will give our desired value for alpha and gamma (unless the substituted corner is just 1x its buddy, but this is an improper triangle also).
Finally if a+b+c=0 but none of a, b or c are individually zero, then a = -b -c and from aA +bB +cC = 0 we get (-b-c)A +bB +cC =0 which gives b(B-A) = c(A-C) and suggests that two different sides of the triangle are parallel, which is not possible in a proper triangle.

Since the only valid solutions actually do force each of our expressions of alpha and gamma to individually equal zero (despite the clear flaw in assuming three vectors in a plane could be linearly independent) I feel as though this expresses something specific about the nature of the centroid - something practical, geometric and succinct that could be expressed in a high-school-maths-for-smart kids way that would mean my proof wasn't necessary. I'd be interested to know if that was the case.

[ericrawson]

12:18 determinant evaluates to -3/4(a+b+c)

[johnnybro]

This is awesome. I never really understood why but this proof is great!

[FernandoVinny]

Oh I don't know what is null space