A lot of time, there are questions from triangles become too complicated to visualize and the textbook way of solving them is through the concept of similarity between triangles. In this article, I will introduce the concept of solving these questions through mass point geometry.

What exactly is mass point geometry? In simple words, it is an application of how masses behave around a lever associated with the concept of center of mass of a system. The concept uses a formula similar to that of center of mass:

If there are masses m1, m2, m3… mn at a distance of x1, x2, x3… xn from a fixed point, the center of mass will be at a distance x from that fixed point where

Sounds easy till now? Let’s take a few questions and explore the concept of mass point geometry further.

Let’s say you have a system of weights at points A and B such that there is a 3 kg weight at A and a 5 kg weight at B and the distance between A and B is 80 units. Let’s say that the center of mass of this system is at a point C which is at a distance of x from point A. According to the formula,

x = (3*0 + 5*80)/(3 + 5)

x = 50 units

Diagrammatically, it would be represented as below:

In other words, the ratio of the distances from the center of mass would be inverse of that of the weights at the end points. Also, the weight at C will be the sum of weights at points A and B. In other words, this entire system can be denoted by a single point C that has a mass equal to 8 kg.

Let’s try a few questions that have a direct application of the concept of mass point geometry.

### Simple ratio based questions

If in a triangle ABC, there are points D and E on AB and AC respectively such that AD/BD = 4/3 and AE/CE = 5/2, find the ratio of DF/CF and BF/EF.

If we put a 4x kg weight at B and a 3x kg weight at A, the system would balance itself with the center of mass at D equivalent to 7 kg. Similarly, we can balance the system by putting weights 2y at A and 5y at C and getting the resultant at E. But as the weight at A should be unique, we put say 6 kg at A (using the LCM of 2 and 3) and adjust the rest accordingly. The new figure would look like this:

Now, it is easy to see that DF/CF will be 15/14 and BF/EF will be 21/8 to balance the weights.

Simple, isn’t it? You can try the same question using the conventional ways and see how much time you can save.

### Application in finding the missing areas in triangles

Find the area of triangle ABC if the areas of the three smaller triangles are 5 units, 8 units and 10 units as shown:

If you have no clue about mass point geometry (and the ladder theorem), the probable way of solving this would be as follows:

Consider triangles BDF and BFC. As they share a common vertex, their areas would be in the same ratio as their bases.

So,

DF/CF = 5/10

Similarly, BF/EF = 10/8

Now, if I join AF, and say that areas of triangles AFD and AFE are x and y, we get:

(5+x)/y = 10/8 and

(8+y)/x = 10/5

So, we get

40+8x = 10y

40+5y = 10x

On solving, we get

10y – 8x = 10x – 5y -> 15y = 18x -> y = 1.2x

So, x = 10, y = 12

And total area is 10 + 12 + 5 + 8 + 10 = 45.

Now, to solve it using mass point geometry, we will first write down the ratio of the bases as we have done in the first method:

Now consider the system D – F – C. If there were say a 10 kg weight at D and a 5 kg weight at C, the system would balance itself at E.

Now, the weight at F should be the same for both the cases. To keep the weight at E constant, we check the two ratios

DF/CF = 1/2 and BF/EF = 5/4. So, we multiply and divide the first fraction by the sum of the numerator and denominator of the second fraction and multiply and divide the second fraction by the sum of the numerator and denominator of the first fraction. So,

DF/CF = 1/2 = 9/18

BE/EF = 5/4 = 15/12

So, our new figure will be:

Now, we can see that the weight at F will be 27 in both the cases.

Applying the concept of mass point geometry we can see that if there were an 18 kg weight at D and a 9 kg weight at C, they would balance themselves out. Similarly, if there were a 12 kg weight at B and a 15 kg weight at E, the system would balance itself. Applying this, the new figure would be:

Now, if we look at the system, if D is 18, A + B has to be 18 and so, A has to be 6. Because we have already balanced F, we can see that A = 6 matches the A – E – C system as well.

Going by the reverse logic, we can see that AE/CE will be 9/6 and will be same as the ratio of the areas of triangles ABE and BCE. So, we get A(ABE)/A(BCE) = 9/6

A(ABE) = 9*18/6 = 27 and so, the area of the triangle will be 27 + 18 i.e. 45 units.

This might sound a bit tedious at the start but once you get the hang of it, the steps are extremely mechanical and you would get the result in no time. I will outline the steps for you once again:

- Get the ratio of the bases of the triangles
- Balance the point of intersection by multiplying and dividing the ratios by the sum of the numerator and denominator of the other ratio in its purest form
- Use the concept of mass point geometry to balance weights at the ends of the segments
- Get the weight at the third vertex
- Find the ratio of the distances and equate it to the ratio of the areas

You can try out a couple of questions to understand this concept in a better manner.

I will be discussing the ladder theorem in the next article which is probably a shorter way to do these questions.

You can solve a few questions from the topic to practice the concept.

Q.1 In triangle ABC, points D and E lie on BC and AC respectively. If AD and BE intersect at T, so that AT/DT = 3 and BT/ET = 4, what is the value of CD/BD?

Q.2 In triangle ABC, medians AD and CE intersect at P. If PE = 1.5, PD = 2 and DE = 2.5, what is the area of quadrilateral AEDC? (Mind you this can very well be done without using mass point geometry and the solution is a beautiful piece of art in itself)

Q.3 Triangle ABC has AB = 21, AC = 22 and BC = 20. Points D and E are located on AB and AC respectively such that DE is parallel to BC. The incenter I of the triangle ABC lies on the segment DE. If the length of DE is in the form of m/n where m and n are integers and m/n is the ratio in the simplest form, find the value of m + n.

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