ভরবিন্দু কী?

Discovering Mass Points

Let's start out with a problem:

Let there be a $\triangle ABC$ in the plane with point $F$ on $\overline{AB}$ and point $E$ on $\overline{AC}$ such that $\dfrac{AF}{FB} = 2$ and $\dfrac{AE}{EC} = 1/5$. Let the intersection of $BE$ and $CF$ be $P$. Find $\dfrac{BP}{PE}$. We look at the problem, and draw a diagram:

At first, this problem may seem hard. However, there's a simple, neat problem solving technique to solve this, known as mass points.

Definition

A mass point is a point that is assigned a positive integer mass. Some examples are: $2A, 4C, 4E$. A system of mass points is a set of mass points. An example of a system of mass points would be $4A+5K+1L$.

Every system of mass points has a unique center of mass. Basically, if you had those points arranged on an indefinitely thin table (the plane), and you had rocks of the corresponding masses lying on those points, then the point at which you would put a rod under the table to perfectly balance it is at the center of mass.

We let similar systems of mass points $S_1$ and $S_2$ be two systems of mass points that have the same center of mass. We denote this by $S_1 = S_2$.

Finding centers of masses

Now, we've seen some definitions of this, but how do we actually calculate the center of masses of systems? Well, first, we define the center of mass of a system of one mass point to be the point itself. Then, we give the "see-saw" rule: $$mA + nB = (m+n)P$$ Where $P$ is the point on $AB$ such that $\dfrac{AP}{PB} = \dfrac nm$.

Now, here is the claim that gets everything in mass points started. If we replace two points in a system of mass points with the center of mass of those two points, then the resulting system is similar to the original! This proof is non-trivial and won't be gone over in this light discussion, but it's quite a near exercise with vectors.

So, for example, to find the center of mass of the system $4A + 2B + 1C$, we would take the first two and add them with the see saw rule to get $6K + C$ for some point $K$ such that $\dfrac{AK}{KB} = \dfrac 12$. Then, we apply it again to get that the center is $7P$ for some point $P$ such that $\dfrac{KP}{PC} = \dfrac 16$.

Calculating the Mass in Multiple Ways

In our example of calculating the mass, nothing said that the mass couldn't be calculated by first summing $B$ and $C$ then adding $A$ to it! If we do it this way, we get:

The system $4A + 2B + 1C$ is equivalent to the system $4A + 3L$ for $L$ such that $\dfrac{BL}{LC} = \dfrac 12$. However, the center of mass of the system is unique, so point $P$ is also a point such that $\dfrac{AP}{PL} = \dfrac 34$.

Does this situation and diagram sound familiar? Think about it for a while. What else in this explanation does this look like?

Oh boy, yes it does! It's essentially the same as the problem from the beginning!

Basic Problem Solving with Mass Points

Let's go back to the problem at the beginning of this explanation.

Let there be a $\triangle ABC$ in the plane with point $F$ on $\overline{AB}$ and point $E$ on $\overline{AC}$ such that $\dfrac{AF}{FB} = 2$ and $\dfrac{AE}{EC} = 1/5$. Let the intersection of $BE$ and $CF$ be $P$. Find $\dfrac{BP}{PE}$.

Let's consider the system of mass points with $A$, $B$, and $C$ as the "points." We want to assign masses to these such that $mA+nB = (m+n)F$ and $mA+pC = (m+p)E$. Thus, we consider the system of mass points $5A + 10B + 1C$. Let the center of mass of this system be $P'$. Combining masses $5A$ and $10B$ this system is equivalent to $15F+1C$. However, we can also combine $5A+1C$ to get that it is equivalent to $10B+6F$. Since the center of mass $P'$ lies on both $BE$ and $CF$, it is the intersection of them, so thus $P=P'$. Therefore, we can use the see-saw rule to get that $\frac{BP}{PE} = \frac 6{10} = \boxed{\frac 35}$.

That's all there is to the basics of mass points. Thanks for reading!

The median is the line connecting a vertex of a triangle to the midpoint of the opposite side.

Here, D,E,F are the midpoints of the sides of ABC.

Notice medians $AD,BE,CF$ are concurrent.

We assume $AD,BE$ intersect at G and $CX$ goes through G for some point X on AB. Then we will try to show $X=F$. We can use mass points technique, to eventually find $BX=AX$. Thus, X is the midpoint of AB and the three medians are concurrent.

(Ceva's Theorem also gives us a trivial proof of this)

Another interesting property of medians and midpoints is that $DEF$ forms the medial triangle . It has sides parallel to those of ABC, and each side is half the length. For example, $DF = 0.5 AC, DE = 0.5 AB$. This is useful in many problems.

Consider the diagram, where $D,E,F$ are midpoints of the sides of triangle ABC. The medians concur at the centroid G.

We use a technique called mass points to prove this. Mass points consists of assigning "weights" to certain vertices on a triangle, and making all the weights balance at a point in a triangle.

It makes sense that if a is the mass on point A, b is the mass point B, and C is in the middle, then for everything to balance we must have $a \cdot BC = c \cdot AB$.

Thus, we assign masses of $1,1,1$ to points $A,B,C$.

It's 2 because if there are weights 1 and 1 on B and C, then these weights combine and $1+1=2$.

Thus, $D,E,F$ all have weights of 2. Finally G has a weight of $2+1=3$. This means that the ratios $BG:GE,CG:GF,AG:GD$ are all 2 to 1.

We also have a different proof of this fact:

Yes... so we know $[ACF]=[BCF], [AGF]=[BGF]$. We subtract to get $[AGC]=[BGC]$

Since $[AGC]=2[AEG],[BGC]=2[BGD]$, we have $[AEG]=[BGD]$.

By this logic, all six triangles have equal area.

Now, ratio $[ABC]:[AGB]$ is three to one. Since the bases are equal, this implies the heights have ratio three to one! We can draw the altitude form C to AB. It's not hard to find that if the ratio of the altitudes is 3 to 1, then the ratio $CF$ to $GF$ is three to one, so we are done.