Mathematics SL Parellels and Parallelograms. This task will consider the number of parallelograms formed by intersecting m horizontal parallel lines with n parallel transversals; we are to deduce a formula that will satisfy the above.

                                          Table 1.0

At this point, I realized that the ‘First difference between terms’ was somewhat similar to what happens with the terms in a triangle sequence:

Also since the number of parallelograms created as the number of transversals increased each had a Second Order difference of 1, it was immediately known that the general formula must be a quadratic equation. So to find the formula I followed the following steps:

Then solve them as simultaneous equations

Solve the two new equations as simultaneous equations

Answer:  

Replace the values in the following formula

But when I tried out the formula I found out that the answers didn’t match:

U2 = 3 (3 +1)  = 6

          2

U3 = 4 (4 +1)  = 10

          2

The results we had have moved one term so instead of adding 1 to “n”, we need to subtract 1 to “n” so the values can match. Now our final formula is:

Un = n (n -1)

          2

Once again I tested the above:

U2 = 2 (2 -1) = 1

          2

U3 = 3 (3 -1) = 3

          2

8. In order to find the general formula for the parallelograms formed by m horizontal parallel lines intersected by n parallel transversals; I decided to further the investigation by considering the number of parallelograms formed by three horizontal parallel lines intersected by a pair of parallel transversals (Figure 8). Three parallelograms were formed: A1, A2, and A1∪A2

9. When a third transversal was added to the above figure, nine parallelograms were formed (Figure 9). A1, A2, A3, A4, A1∪A2, A3∪A4, A1∪A3, A2∪A4, A1∪A4

10. One more transversal was added to Figure 9, to form 18 parallelograms (figure 10). A1, A2, A3, A4, A5, A6, A1∪A2, A3∪A4, A5∪A6, A1∪A3, A3∪A5, A2∪A4, A4∪A6, A1∪A5, A2∪A6, A1∪A4, A3∪A6, A1∪A6

11. When I added the fifth transversal to Figure 10, it became incredibly difficult to count the parallelograms; luckily by then a pattern had emerged and I was able to predict the next few terms (recorded in table 1.1):

  Table 1.1

Now the ‘Second Order Difference’ is 3 – triple the first set of parallelograms (pair of parallels intersecting with parallel transversals). Due to the second order being three, I deduced and found true that the number of parallelograms was increasing in multiples of three. The difference between the multiples is 2,3,4,5, and 6; which again follows the triangle number sequence. This meant that subsequently the formula for m horizontal parallel lines intersected by n parallel transversals would include the formula for the triangle sequence.

12. To make my statements broader I expanded my table to the following (table 1.2):

Table 1.2

See graph 1 to get a more graphical view of how the number of parallelograms formed by m horizontal parallel lines intersected by n parallel transversals are very sequential, by looking at the graph, the sequence is much more evident.

13. When I compared all the formulas I got, I found the relationship which led me to find the general formula.

2 horizontal lines: 1n (n -1)

                          2

3 horizontal lines: 3(n (n -1))

                          2

4 horizontal lines: 6(n (n -1))

                          2

5 horizontal lines: 10(n (n -1))

                                                  2

The number and sequence repeat the formula (n (n -1)) and multiplies by the first term.                                                                  2

14. From this realization I was able to find that the final formula for calculating the number of parallelograms formed when m horizontal parallel lines are intersected by n parallel transversals:

m(m -1)   x   n(n -1)

    2                    2

15. To test the validity of the formula I tested it against previously counted parallelograms (Figure 10), the intersection of 4 transversals with 3 horizontal parallel lines should form 18 parallelograms:

        Using the formula:

m(m -1)   x   n(n -1)

    2                    2

3(3 -1)   x    4(4 -1)

    2                  2

3(2)  x  4(3)

  2          2

6  x  12

 2      2

3  x  6

= 18 parallelograms

Limitations

It is very difficult to test the validity of the formula when there are lots of parallel transversals intersecting lots of horizontal parallels, because even though we will get a number, to prove there are so many parallelograms is confusing and difficult; therefore we can only assume that the answer might be right.