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why is the point where the medians of a triangle meet also the center of mass of the triangle.

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@BrianM.Scott: Thanks for catching my gross error. I have deleted my answer while I check my brain. –  copper.hat Jul 18 '12 at 23:54
    
It is not a simple average. It need to be weighted by mass/area. Let me see if I can find an explanation that does involve calculus. –  copper.hat Jul 19 '12 at 0:13
    
I have an explanation, but it will take me a minute to write it up. –  copper.hat Jul 19 '12 at 0:23
    
What is your definition of the center of mass? Does it involve calculus? :) –  Rahul Jul 19 '12 at 1:02
    
...And what is your definition of moment, please? –  Rahul Jul 19 '12 at 1:07

2 Answers 2

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I think I have it right this time; however the construction is not as simple as I had hoped. The idea is (was) to avoid calculus (explicitly, at least).

The basic idea is to use scaling and linearity to determine the center of mass (CM). If an object of mass $m$ with CM at $d$ is partitioned into smaller objects of mass $m_i$ (with $m = \sum m_i$, of course) each with CM at $d_i$, then we must have: $$d m = \sum d_i m_i.$$

Excuse the crudeness of my drawings below, I hope you get the idea. We want to find the CM of the big triangle and show that it is at the intersection of the medians.

We want to get the CM of the large triangle. Suppose, for simplicity, that one vertex is at the origin, and that the CM is at position $d \in \mathbb{R}^2$. Now take a similar triangle whose sides are $\frac{1}{3}$ of the big triangle. It should be clear that the CM of the little triangle is at position $\frac{d}{3}$ with respect to the corresponding vertex. Let $m$ be the 'mass' of the big triangle, then the mass of each of the smaller triangles is $\frac{m}{9}$.

Trisect the edges of the big triangle and connect the points to get $9$ similar small triangles. Let $a,b \in \mathbb{R}^2$ be the other two vertices of the big triangle.

By looking careful at the smaller similar triangles, it should be clear that the red dot is at the intersection of the medians, which is at position $\frac{a+b}{3}$.

Center of mass

Now consider the $6$ triangles that touch the red dot. By symmetry, the CM of the $6$ triangles is at the red dot, and the 'mass' of the $6$ triangles taken as a unit is $\frac{6}{9}m$.

By linearity, we can compute the CM of the combined $6$ units and the three blue triangles and this must equal $d$, the CM of the big triangle. Doing the computation gives: $$\frac{1}{m}\left[ \frac{6m}{9} \frac{a+b}{3} + \frac{m}{9}(0+\frac{d}{3}) + \frac{m}{9}(\frac{2a}{3}+\frac{d}{3}) + \frac{m}{9}(\frac{2b}{3}+\frac{d}{3}) \right] = d.$$ Simplifying gives: $$\frac{6(a+b)}{3}+\frac{2a+2b}{3}+d = 9d,$$ from which we get the formula $d = \frac{a+b}{3}$, which is the intersection of the medians.

Note: There is nothing special about splitting the triangle up into $9$ parts other than with this splitting, the CM and the median intersections fall nicely on the vertex of the smaller triangles.

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It’s not as simple as equal areas; think of a T-shaped region with equal areas in the vertical and horizontal strokes. What matters are the moments. –  Brian M. Scott Jul 18 '12 at 23:48
    
My God, I'm loosing it. I will delete my answer. –  copper.hat Jul 18 '12 at 23:50
    
I don't follow what you mean by "the[y] would fall to one side or the other". Also, any point on the triangle lies on some line through the vertex; I don't see how that tells you anything. What you are assuming later is that the line through the vertex and the center of mass is the same as the line through the midpoint and the center of mass, but you need to justify that. –  Rahul Jul 19 '12 at 0:58
    
I'm retreating again to think about this. Sorry for the noise. –  copper.hat Jul 19 '12 at 1:00
    
Finally I am having better moments :-). Thanks Brian & Rahul for catching my earlier mistakes. –  copper.hat Jul 19 '12 at 5:55

Can I give an alternate explanation ?

The key to this problem is the fact that the Center of Mass(CM) of any two areas lies on the line joining the CMs of the individual areas.

In this case let us divide the triangle $\bigtriangleup ABC$ ($A_1$) into a $\bigtriangleup ADE$ $(A_2)$ and a trapezoid $DECB$ $(A_3)$ by drawing a line $DE$ parallel to the base of $A_1$ say $BC$. Now all we know is that the CM of $A_1$ will lie on the line joining the CMs of $A_2$ and $A_3 $. Now if we keep making the trapezoid $A_3$ smaller and smaller by making the line $DE$ slide towards the base $BC$ then in the limit the CM of the trapezoid $A_3$ will coincide with the CM of the line $BC$ which is the mid-point of BC(Point $F$). Hence the line $l_2$ has to pass through the midpoint of BC.

Again if we start making the triangle $A_2$ smaller and smaller by making the line $DE$ smaller and smaller keeping it parallel to $BC$ all the time and making the $\bigtriangleup A_2$ approach the vertex $A$ of the triangle then in the limit the CM of the $\bigtriangleup $ $(A_2)$ coincides with that of the vertex $A$. So the line $l_2$ has to pass through the vertex A also. So the line $l_2$ has to pass through both $F$ and vertex $A$ essentially making $l_2$ coincide the median. So the CM of area $A_1$ must be on the median $AF$. This exercise shows that the CM of a $\bigtriangleup ABC$ should lie on each of its three medians and hence should be at the intersection of all the three medians. Triangle ABC

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