Mass Point Geometry aka MPG
Mass point geometry is a problem-solving technique that is used in geometry which uses the concept of center of mass involving triangles and intersecting cevian. Cevian is a line with one endpoint on vertex and another end point on the opposite side. Medians, altitudes and angle bisectors are also cevians [special case of cevians]. The problems that can be solved using similar triangles, the ratio of areas are made simple if we use mass point geometry and such problems can be solved in half the time.
The catch in solving such a problem is: we are actually creating an equilibrium.
Here we assign mass to the first vertex (which can sometimes be arbitrary)
Point to be kept in mind while assigning values using ratios of the line is: force on each side cancel out
You will find many videos and explanation, which focuses on explaining the concept of mass point geometry, which to many students might seem complicated. So, here in this lesson, we will understand the use of mass point geometry and its concept with the help of examples. One and the only thing that needs to be kept in mind is: Mass point MPG is like a see-saw, the idea is to balance both the sides of sea-saw, by placing masses on both the sides.
Suppose you have a wire with 5 kg weights hanging on each side. So this structure is balanced because
- Weights on both the sides are equal
- Length AO = OB
Had the case been: 5kgs on side A and 10kgs on side B. Can we call it a balanced structure? NO!!! The wire will bend towards the 10kg mass.
Had the case been: 5 kg weights on both the sides and AO < OB, would we call the structure a balanced structure? The answer is NO. The structure will tilt on the side where the distance from the O is more
This balance depends on what? Mass point theorem says: if the weights on both the sides of balance are unequal, to balance the structure AO : OB = 10 : 5 = 2 : 1, then and only then the structure will be balanced.
In short, MPG says: To balance a structure
If the weights on both the sides are M1 and M2 and the distance of the weights from O is x and y respectively, then
And the total mass will be (M1+M2) at the center of the balance.
We will now learn how to use this in geometry
Q.1) ABC is a triangle and the O is the incenter E, F and G lies on BC, CA and AB respectively, Given AO: OE = 5: 4 and CO: OG = 3: 2. Find BO: OF.
Weight at O will be a multiple of (4+5) and (3+2) = 9*5 = 45
Now let’s solve this in parts, let us first consider line AE only
When we have only AE, the centre has a weight of 45 kg, since we have already established that it should be a multiple of 9 and 5
5k+4k=45, k = 5, so the weights at A and B should be 4k and 5k is 20 and 25
[if we take sum A+ B we will get 45]. Hence we can say this structure is balanced
Now let us consider only CG
Weight at O is 45. Since we have already established that it is a factor of 5 and 9. Sum 2m+3m = 45, m = 9
Therefore, at G the weight will be 3m = 27 and at C the weight will be 2m = 18
Now let us the see the whole figure
Weight at F = 18 + 20 = 38
Weight at B = 25- 18 = 7
The complete figure now looks like this
The line segments OB: OF will be in the opposite ratio of the weights = 38: 7
Let’s solve another question to check we have got the hang of the concept or not
Q.2) In the figure it is given that AG: GD = 3: 4 and BD: DC = 4: 7, AE = 24. Find the value of CE?
Weight at D from line AB is (7+4) = 11
And weight at D from Line AD is 3
So the weight at D is a multiple of 11 and 3
It is safe to say that weight at D = 33
For line BC 7k+4k = 33
Therefore, the weights at B and C are 7k and 3k which is 21 and 12
When 3m = 33
m = 11
Therefore, 4m = 44
Weight at A will be 44
For line AC
Weights are in the opposite ratio of the lengths of the lines separated by E
Split mass concept
Split mass concept is nothing but an extension of mass point geometry. This method is used when the triangle contains a transverse line in addition to Cevians. It may be treated as a normal mass point, the difference would be that it will have more than one mass used for each of the sides.
Let us try to understand this:
Weight at A = 15
Weight at B will be 6 therefore, weight at C = 10
Hence weight at F will be 15 + 6 = 21
Weight at E = 25
Now this weight 10 that is placed at C is to counter the weight on line CA. if we see line CB 10*2 is not equal to 6*3. This line is not balanced. The reason is 10 is already used to balance line CA so we will have to put more mass at C by treating it independently. 10 is only for the particular section AC. So in order to balance CB at point C, the weight will be 9.
This was all about Mass Point Geometry. I know, it is a complicated but revolutionary way of solving some very difficult problems. With more and more of practice, it will definitely assist you in solving very tricky problems and get a hold on this concept.