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Kurre

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Registered: 2006-07-18

Posts: 280

Triangles center of mass at centroid

Does anyone have a good proof that the center of mass for a triangle is at the centroid? (the intersection of the medians)
I have managed to prove it in 3 slightly different ways, but I dont like them. What I wonder is if there is some elegant, simple proof using just some geometry/vectors that I have missed?

mathsyperson

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Registered: 2005-06-22

Posts: 4,900

Re: Triangles center of mass at centroid

How about a symmetry argument?
For any median, the centre of mass must lie somewhere on it, and so it must be at the intersection of them all.

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luca-deltodesco

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Registered: 2006-05-05

Posts: 1,470

Re: Triangles center of mass at centroid

which could also be seen as an argument for why the 3 medians MUST all intersect at a common point, because the triangle must have a single centre of mass

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Kurre

Member

Registered: 2006-07-18

Posts: 280

Re: Triangles center of mass at centroid

which could also be seen as an argument for why the 3 medians MUST all intersect at a common point, because the triangle must have a single centre of mass

Yea I also realized that. Its a really cool result

How about a symmetry argument?
For any median, the centre of mass must lie somewhere on it, and so it must be at the intersection of them all.

yes, if we can prove that it actually lie on one median. The masses of the two triangles that the median dividies the larger triangle obviously has the same mass, but that does not imply that the center of mass for the triangle is at the median. We need to prove that the distance from the median the the two center of masses for the smaller triangles are equal. That is relatively easy to do with the integral definition (its "obvious" in a sense that the mass after distance x m(x) are equal on both sides since the sides are both straight lines) but....I dont like it
Basically I made this thread because a friend told me that they proved this really easy by using vectors during a lecture, but today he showed the argument and it was in fact incorrect So maybe there is no simple method that I have missed...woho

Last edited by Kurre (2009-04-28 05:22:09)

whatismath

Member

Registered: 2007-04-10

Posts: 19

Re: Triangles center of mass at centroid

I think we could still somewhat apply an argument of symmetry,
and it might even be related to vectors. I write it just for the
sake of philosophical argument (but otherwise meaningless).
A simply proof is very unlikely but an intuitive, acceptable argument
is here depending on how we define the center of mass.

Aside, I think we are all sure vector algebra could show the
3 medians are concurrent: e.g.

lies on the median from A to the midpoint of D of BC.

If we define the CM as the point that balance a physical uniform triangular plate,
then the argument of symmetry is already a good one for me:
the median AD (D is the midpoint of BC) balances the plate, so does the median
BE (E is the midpoint of AC). Consider the (vector) moment rxF of the plate (at CM)
about the intersection M of AD and BC. Since AD and BC do not overlap, we can resolve
r into components along AD and BC, the same on rxF which become a sum of zero vectors.
So r must be the zero vector else there exist positions r not paralell to F (and r non-zero).

If we define the CM most fundamentally as a (2-D) integral (limit of Riemann sums),

I believe there is no simple proof. Using the linear equations of the sides of the
triangle already applied the Fundamental Theorem of Calculus.
Actually, the x- and y- directions in the definition of CM need not be perpendicular,
just any two non-overlapping lines work. Take AD and BE. Consider the definition
as Riemann sums, given any partition of

, we can produce
a finer partition by subdividing the two sub-partitions on both sides of AD (one side
imposed its mirror image parallel to BE on the other side of AD) by symmetry.
So the Riemann sum of the resulting partition is 0 along BE, and therefore so is

.
Similarly

Hence the CM lies on AD and BE, i.e. CM is the centroid.

Sorry for the long wind. Thanks for reading.

Kurre

Member

Registered: 2006-07-18

Posts: 280

Re: Triangles center of mass at centroid

If we define the CM as the point that balance a physical uniform triangular plate,
then the argument of symmetry is already a good one for me:
the median AD (D is the midpoint of BC) balances the plate, so does the median
BE (E is the midpoint of AC). Consider the (vector) moment rxF of the plate (at CM)
about the intersection M of AD and BC. Since AD and BC do not overlap, we can resolve
r into components along AD and BC, the same on rxF which become a sum of zero vectors.
So r must be the zero vector else there exist positions r not paralell to F (and r non-zero).

I dont get it. The bolded text is the difficult part of the proof, and you straight away assume its true? or do you prove it in the next part?

If we define the CM most fundamentally as a (2-D) integral (limit of Riemann sums),

I believe there is no simple proof. Using the linear equations of the sides of the
triangle already applied the Fundamental Theorem of Calculus.
Actually, the x- and y- directions in the definition of CM need not be perpendicular,
just any two non-overlapping lines work. Take AD and BE. Consider the definition
as Riemann sums, given any partition of

, we can produce
a finer partition by subdividing the two sub-partitions on both sides of AD (one side
imposed its mirror image parallel to BE on the other side of AD) by symmetry.
So the Riemann sum of the resulting partition is 0 along BE, and therefore so is

.
Similarly

Hence the CM lies on AD and BE, i.e. CM is the centroid.

Sorry for the long wind. Thanks for reading.

Im not sure I follow you, but if you use the same argument as my friends lecturer its not completely correct (ie we divide the triangle into a lot of trapezoids, and in the limit the center of mass of each trapezoid is in the middle, thus the cm of the triangle is on the median). this does not hold, consider for example a horseshoe. Apllying this argument there would show that the center of mass is in the horseshoe, which is obviously not correct.

I have a kind of cool proof now only using vectors, maybe I can post it tomorrow.

Kurre

Member

Registered: 2006-07-18

Posts: 280

Re: Triangles center of mass at centroid

Okey here is my proof:
Given triangle ABC, let D, M, N be midpoints of BC, AC, AB respectively (see picture). Insert a coordinate system along the median AD. We want to find the y component (perpendicular to AD) of the center of mass R for the triangle, which we call d. R will be the weighted sum (definition of center of mass) of the locations of the following three pieces: parallellogram ANDM, triangle BND, triangle DMC. The mass of ABC, BND, DMC and ANDM are m, m', m'' and m''' respectively.

But the center of mass of ANDM will lie on AD (can be seen by for example that ADN is congruent to DAM, so the y components will cancel), so we can skip that piece.

Furthermore, BND is congruent to DMC which both are similar to BAC, with a scaling factor of 1/2. Thus the distance to the center of masses R' and R'' for BND and DMC are a distance d/2 from their medians NE and MF (see picture). Also, since the areas for the triangles ADC and ABD are equal and they have the same base AD, the heights with base AD are equal, and equal h. Since the medians NE and EF are 1/2 of the height h away from AD, they have y components h/2 and -h/2. Also since BND and MDC are scaled by a factor 1/2, their areas (masses) are scaled by a factor 1/4, so m'=m''=m/4.

Now when im done explaining the picture we can come to the main part of the proof. Using the definition of center of mass one obtains (R''' being the location of the center of mass of ANDM):

We are only interested in the y components, and as noted earlier the parallellogram will not contribute, so we get:

Thus the center of mass lies on the median AD, and similar argument for the other medians shows that the center of mass is on all medians, and thus must be on the intersection.

cool eh?