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what is an easy way to prove that use mass point geometry to solve a problem in the link i provide that is involving cevians in a triangle is same as using the other way in euclidean geometry or coordinate geometry or analytic geometry or trignometry? i google it, but doesn't have any proof...Any help would be appreciated! My main concern is how do we derive the mass point geometry to solve those problem relate to triangle and cevian? reference: http://en.wikipedia.org/wiki/Mass_point_geometry

i suspect if the mass point geometry method is right: How do we prove that all the step to solve this problem by using mass point geometry is valid? i don't understand why the mass point is the inverse of the length between two pont, also why the mass point is define as the inverse of the length is the inverse ratio of the mass point. why it is useful to define that way?

here is an example from wikipedia with solution: '''Problem.''' In triangle $ABC$, $E$ is on $AC$ so that $CE = 3AE$ and $F$ is on $AB$ so that $BF = 3AF$. If $BE$ and $CF$ intersect at $O$ and line $AO$ intersects $BC$ at $D$, compute $\tfrac{OB}{OE}$ and $\tfrac{OD}{OA}$.

'''Solution.''' We may arbitrarily assign the mass of point $A$ to be $3$. By ratios of lengths, the masses at $B$ and $C$ must both be $1$. By summing masses, the masses at $E$ and $F$ are both 4. Furthermore, the mass at $O$ is $4 + 1 = 5$, making the mass at $D$ have to be $5 - 3 = 2$ Therefore $\tfrac{OB}{OE}$ $= 4$ and $\tfrac{OD}{OA} = \tfrac{3}{2}$. See diagram: http://upload.wikimedia.org/wikipedia/en/1/13/Mass_points4.png

i think the solution is not true because my teacher told me, so i want you guys to verify it that it is or not. The vector and basic geometry approach is not neat, so i rather not post it here.*My problem is: why the mass of a point is defined as the inverse ratio of the line segment of the transversal*?

All i could find is example in books and internet, but no clear explanation or proof toward the mass point geometry from those source, so your help could really help the Global people to understand this subject!i expect a mathmatical approach.Thanks in advance!

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Do you guys think start a bounty for this question is a good idea? –  Victor Oct 25 '11 at 20:44
4  
It would be a good idea to improve the wording of the question. For example, what is "the other way"? –  Phira Oct 25 '11 at 20:54
3  
Well, honestly, I do not understand the question. First you can start off by trying to clarify what you mean. If the question is still not answered after you explained it better, then adding a bounty might help. –  Srivatsan Oct 25 '11 at 20:55
    
@SrivatsanNarayanan - i mean do you guys know how to derive the mass point geometry method by other math subject? is it elementary? –  Victor Oct 25 '11 at 21:03
    
I don't know anything about mass-point geometry, but since "all problems that can be solved using mass point geometry can also be solved using either similar triangles, vectors, or area ratios" I'd be shocked if it weren't elementary. –  user7530 Oct 25 '11 at 23:20

1 Answer 1

up vote 3 down vote accepted
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The best way to understand this kind of calculation is to first define a barycentre or center of gravity.

Definition: The barycentre of $n$ points $x_1,\dots,x_n$ with (strictly positive, for now) masses $m_1,\dots,m_n$.is $\dfrac{m_1x_1+\dots+m_nx_n}{m_1+\dots+m_n}$.

The motivation is that this definition has exactly the properties that one wants from a physical center of gravity, that is, in many applications, one can replace the $n$ points by a single point at the center of gravity with mass $m_1+\dots+m_n$.

In particular, the property that we want to use is:

Lemma 1: The barycentre of $n$ points $x_1,\dots,x_n$ with masses $m_1,\dots,m_n$ is the barycentre of the first $n-1$ points with mass $m_1+\dots+m_{n-1}$ and the point $x_n$ with mass $m_n$.

Proof:

The second barycentre in the lemma is given by:

$$(m_1+\dots+m_{n-1})\cdot \frac{m_1x_1+\dots+m_{n-1}x_{n-1}}{m_1+\dots+m_{n-1}} + m_nx_n= m_1x_1+\dots+m_{n-1}x_{n-1}+m_nx_n,$$

which is clearly the expression for the barycentre of all points.


Corollary: In the calculation of a barycentre, we can replace an arbitrary subset of points by their barycentre with the sum of their masses.

Proof: Just apply the lemma repeatedly.


Lemma 2: The barycentre of two points lies on the line containing the two points.

Proof: $$\dfrac{m_1x_1+m_2x_2}{m_1+m_2} = x_1+ \dfrac{m_2}{m_1+m_2}(x_2-x_1).$$

So the barycentre is the first point plus a multiple of the connecting vector $x_2-x_1$ which lies on the line.


Lemma 3: The barycentre $g$ of $x_1,x_2$ with masses $m_1,m_2$ has the property $\dfrac{\overline{x_1g}}{\overline{gx_2}}=\dfrac{m_2}{m_1}$.

Proof: $\begin{align} m_1\overrightarrow{x_1g}-m_2\overrightarrow{gx_2}&=m_1g-m_1x_1-m_2x_2+m_2g\\ &=g(m_1+m_2)-(m_1x_1+m_2x_2)\\ &=g(m_1+m_2)-(m_1+m_2)g=0 \end{align}$

Therefore, $m_1\overrightarrow{x_1g}=m_2\overrightarrow{gx_2}$. Dividing by the masses and taking the length of the vectors gives the desired result.


Example 1: In a convex quadrangle $ABCD$ the lines that connect the midpoints of opposite sides divide each other in parts of equal length and they meet in the centre of gravity of the quadrangle.

Why: To get the midpoint of $A$ and $B$ we have to put equal masses on $A$ and $B$.

So it is a natural approach to put masses of size 1 on every vertex and calculate their center of gravity $G$.

By the corollary above, we can calculate it in several ways:

Using lemma 1, we replace $A$ and $B$ with masses 1 by their midpoint with mass 2 and do the same for $C$ and $D$.

By lemma 2, the centre of gravity of these two points lies on the connecting line and since both masses are 2 it lies exactly in the middle.

Similarly, $G$ lies in the middle of the other connecting line of two midpoints.


Example 2: In the triangle $ABC$, $E$ is on $AC$ so that $CE=3AE$ and $F$ is on $AB$ so that $BF=3AF$. If $BE$ and $CF$ intersect at $O$ and line $AO$ intersects $BC$ at $D$, compute $\frac{\overline{OB}}{\overline{OE}}$ and $\frac{\overline{OD}}{\overline{OA}}$.

By lemma three, we want to place mass 3 on $A$ and a mass 1 on $C$ to have $E$ the center of gravity of the two points. Later, we want to connect $E$ to $B$, so we want to have this ratio 3 to 1 in the points $A$ and $C$. Similarly, we want the same ratio for $B$.

So, we want to investigate the three points $A$, $B$, $C$ with masses 3,1,1. (Note that if we wanted to regard the midpoint of $E$ and $F$ later, we would have chosen 6,1,1. The masses are freely chosen for the particular problem!)

Let the barycentre be $G$.

Now it is clear that $G$ is the barycentre of $B$ and $E$ with masses 3 and 2, respectively, so $G$ lies on $BE$. Similarly, $G$ lies on $CF$, so $G=O$ and $\frac{\overline{OB}}{\overline{OE}}=2/3$ by Lemma 3.

If we want to express $G=O$ as barycentre of $A$ and $D$, we just have to calculate first the barycentre $H$ of $BC$ which gets mass 2. Then we note that $H$ lies on $BC$, but by Lemma 2 also on $AO$, so actually, $H=D$.

And we conclude by Lemma 3 that $\frac{\overline{OD}}{\overline{OA}}=3/2$.


Remark:

If we are working with three points in the plane, every point of the plane is a barycentre of the three points if we place appropriate masses. If these masses are scaled to have some 1, they are the barycentric coordinates of the points of the plane. Since we allow negative "masses" in this case, one has to generalize the concept to the case where the sum of the "masses" is zero.

Barycentric coordinates or "mass point geometry" is the appropriate framework for affine geometry and is applicable whenever a geometry problem is only about ratios of lengths, that is, no absolute distances (including circles), no angles, no inner products.

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i think the proof in lemma 2 have quite a bit of problem... –  Victor Oct 30 '11 at 14:54
    
@Victor I corrected it. I did not delete all the terms after copying the fraction. –  Phira Oct 30 '11 at 15:39
    
May you explain why the points in lemma 2 is lie in the line because i am not familar with barycentric coordinate, however, i think i understand lemma 1 –  Victor Oct 30 '11 at 20:51
    
@Victor: The proof of lemma 2 does not use barycentric coordinates, it uses the parametric form of a line through P and Q which is exactly $P+t\cdot \overrightarrow{PQ}$ with $t$ a real number. And the vector $\overrightarrow{PQ}$ can be calculated by $Q-P$. –  Phira Oct 30 '11 at 23:31

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