2 3 4 5 6
1 1 1 1
Table 2: Two Horizontal Parallel Lines
General statement
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.
If there are n transversals and two horizontal lines, then p = sum of all integers from 1 to (n - 1).
P=12nn-1
or
P=n2-n÷2
TEST OF VALIDITY for n=10
Figure 7: ten transversals
A1 , A2 , A3, A4, A5 , A6, A7, A8, A9, A1 ᴗ A2, A1 ᴗ A3, A1 ᴗ A4 , A1 ᴗ A5, A1 ᴗ A6 , A1 ᴗ A7, A1 ᴗ A8, A1 ᴗ A9 , A2 ᴗ A3, A2 ᴗ A4, A2 ᴗ A5, A2 ᴗ A6, A2 ᴗ A7, A2 ᴗ A8, A2 ᴗ A9, A3 ᴗ A4, A3 ᴗ A5, A3 ᴗ A6, A3 ᴗ A7, A3 ᴗ A8, A3 ᴗ A9, A4 ᴗ A5, A4 ᴗ A6, A4 ᴗ A7, A4 ᴗ A8, A4 ᴗ A9, A5 ᴗ A6, A5 ᴗ A7, A5 ᴗ A8, A5 ᴗ A9, A6 ᴗ A7, A6 ᴗ A8, A6 ᴗ A9, A7 ᴗ A8, A7 ᴗ A9, A8 ᴗ A9.
Adding a tenth transversals gives us a total of forty-five parallelograms.
p=n2-n÷2
p=102-10÷2
=90÷2
=45
Adding a third horizontal parallel line with two transversals
Figure 8: two transversals
A1 , A2 , A1 ᴗ A2 .
Adding a third horizontal parallel line with two transversals will give us a total of three parallelograms.
Adding a third horizontal parallel line with three transversals
Figure 9: three transversals
A1 , A2, A3, A4, A1 ᴗ A2, A1 ᴗ A3, A3 ᴗ A4, A2 ᴗ A4, A1 ᴗ A4 .
Adding a third horizontal parallel line with three transversals will give us a total of nine parallelograms.
Adding a third horizontal parallel line with four transversals
Figure 10: four transversals
A1 , A2, A3, A4, A5, A6, A1 ᴗ A2, A1 ᴗ A3, A1 ᴗ A4, A2 ᴗ A3, A2 ᴗ A5, A3 ᴗ A6, A4 ᴗ A5, A4 ᴗ A6, A5 ᴗ A6, A1 ᴗ A5, A2 ᴗ A6, A1 ᴗ A6 .
Adding a third horizontal parallel line with four transversals gives us a total of eighteen parallelograms.
Adding a third horizontal parallel line with five transversals
Figure 11: five transversals
A1 , A2, A3, A4, A5, A6, A7, A8, A1 ᴗ A2, A1 ᴗ A3, A1 ᴗ A4, A1 ᴗ A5, A2 ᴗ A3, A2 ᴗ A4, A2 ᴗ A6, A3 ᴗ A4, A3 ᴗ A7, A4 ᴗ A8, A5 ᴗ A6, A5 ᴗ A7, A5 ᴗ A8, A6 ᴗ A7, A6 ᴗ A8, A7 ᴗ A8, A1 ᴗ A6, A1 ᴗ A7, A1 ᴗ A8, A2 ᴗ A7, A2 ᴗ A8, A3 ᴗ A8.
Adding a third horizontal parallel line with five transversal lines gives us a total of thirty parallelograms.
Adding a third horizontal parallel line with six transversals
Figure 12: six transversals
A1 , A2, A3, A4, A5, A6, A7, A8, A9, A10, A1 ᴗ A2, A1 ᴗ A3, A1 ᴗ A4, A1 ᴗ A5, A1 ᴗ A6, A2 ᴗ A3, A2 ᴗ A4, A2 ᴗ A5, A2 ᴗ A7, A3 ᴗ A4, A3 ᴗ A5, A3 ᴗ A8, A4 ᴗ A5, A4 ᴗ A9, A5 ᴗ A10, A6 ᴗ A7, A6 ᴗ A8, A6 ᴗ A9, A6 ᴗ A10, A7 ᴗ A8, A7 ᴗ A9, A7 ᴗ A10, A8 ᴗ A9, A8 ᴗ A10, A9 ᴗ A10, A1 ᴗ A7, A1 ᴗ A8, A1 ᴗ A9, A2 ᴗ A8, A2 ᴗ A9 ,A2 ᴗ A10, A3 ᴗ A9, A3 ᴗ A10, A4 ᴗ A10, A1 ᴗ A10.
Adding a third horizontal parallel line with six transversal lines gives us a total of forty-five parallelograms.
Adding a third horizontal parallel line with seven transversals
Figure 13: seven transversals
A1 , A2, A3, A4, A5, A6, A7, A8, A9, A10, A11, A12, A1 ᴗ A2, A1 ᴗ A3, A1 ᴗ A4, A1 ᴗ A5, A1 ᴗ A6 , A1 ᴗ A7 A2 ᴗ A3, A2 ᴗ A4, A2 ᴗ A5, A2 ᴗ A6, A2 ᴗ A8, A3 ᴗ A4, A3 ᴗ A5, A3 ᴗ A6, A3 ᴗ A9, A4 ᴗ A5, A4 ᴗ A6, A4 ᴗ A10, A5 ᴗ A6, A5 ᴗ A11, A6 ᴗ A12, A7 ᴗ A8, A7 ᴗ A9, A7 ᴗ A10, A7 ᴗ A11, A7 ᴗ A12, A8 ᴗ A9, A8 ᴗ A10, A8 ᴗ A11, A8 ᴗ A12, A9 ᴗ A10, A9 ᴗ A11, A9 ᴗ A12, A10 ᴗ A11, A10 ᴗ A12, A11 ᴗ A12, A1 ᴗ A8, A1 ᴗ A9, A1 ᴗ A10, A1 ᴗ A11, A2 ᴗ A9 ,A2 ᴗ A10, A2 ᴗ A11, A2 ᴗ A12, A3 ᴗ A10, A3 ᴗ A11, A3 ᴗ A12, A4 ᴗ A11, A4 ᴗ A12, A5 ᴗ A12, A1 ᴗ A12.
Adding a third horizontal parallel line with seven transversal lines gives us a total of sixty-three parallelograms.
Table 3: Side by Side view of Corresponding Transversals to Parallelograms
Paralleograms:
3, 9, 18, 30, 45, 63, etc.
9-3 18-9 30-18 45-30 63-45
6 9 12 15 18
3 3 3 3
Table 4: Three Horizontal Parallel Lines
Now the second order difference is 3- triple the first set of parallelograms. Due to the second order being three, I deducted and found that the number of total parallelograms was increasing in multiples of three as shown by the fourth column of the table above.
General statement
Since the number of parallelograms created as the number of transversals increased each had a Second Order difference of 3, it was immediately known that the general formula must be a quadratic equation.
If there are n transversals and three horizontal lines, then p = sum of all integers from 1 to (n - 1).
P=12 3nn-1
or
P=3n2-n÷2
TEST OF VALIDITY for n=10
Figure 14: ten transversals
A1 , A2 , A3, A4, A5 , A6, A7, A8, A9, A10, A11, A12, A13, A14, A15, A16, A17, A18, A1 ᴗ A2, A1 ᴗ A3, A1 ᴗ A4 , A1 ᴗ A5, A1 ᴗ A6 , A1 ᴗ A7, A1 ᴗ A8, A1 ᴗ A9 , A1 ᴗ A10, A2 ᴗ A3, A2 ᴗ A4, A2 ᴗ A5, A2 ᴗ A6, A2 ᴗ A7, A2 ᴗ A8, A2 ᴗ A9, A2 ᴗ A11, A3 ᴗ A4, A3 ᴗ A5, A3 ᴗ A6, A3 ᴗ A7, A3 ᴗ A8, A3 ᴗ A9, A3 ᴗ A12, A4 ᴗ A5, A4 ᴗ A6, A4 ᴗ A7, A4 ᴗ A8, A4 ᴗ A9, A4 ᴗ A13, A5 ᴗ A6, A5 ᴗ A7, A5 ᴗ A8, A5 ᴗ A9, A5 ᴗ A14, A6 ᴗ A7, A6 ᴗ A8, A6 ᴗ A9, A6 ᴗ A15, A7 ᴗ A8, A7 ᴗ A9, A7 ᴗ A16, A8 ᴗ A9, A8 ᴗ A17, A9 ᴗ A18, A10 ᴗ A11, A10 ᴗ A12, A10 ᴗ A13, A10 ᴗ A14, A10 ᴗ A15, A10 ᴗ A16, A10 ᴗ A17, A10 ᴗ A18, A11 ᴗ A12, A11 ᴗ A13, A11 ᴗ A14, A11 ᴗ A15, A11 ᴗ A16, A11 ᴗ A17, A11 ᴗ A18, A12 ᴗ A13, A12 ᴗ A14, A12 ᴗ A15, A12 ᴗ A16, A12 ᴗ A17, A12 ᴗ A18, A13 ᴗ A14, A13 ᴗ A15, A13 ᴗ A16, A13 ᴗ A17, A13 ᴗ A18, A14 ᴗ A15, A14 ᴗ A16, A14 ᴗ A17, A14 ᴗ A18, A15 ᴗ A16, A15 ᴗ A17, A15 ᴗ A18, A16 ᴗ A17, A16 ᴗ A18, A17 ᴗ A18, A1 ᴗ A11, A1 ᴗ A12, A1 ᴗ A13, A1 ᴗ A14, A1 ᴗ A15, A1 ᴗ A16, A1 ᴗ A17, A1 ᴗ A18, A2 ᴗ A12, A2 ᴗ A13, A2 ᴗ A14, A2 ᴗ A15, A2 ᴗ A16, A2 ᴗ A17, A2 ᴗ A18, A3 ᴗ A13, A3 ᴗ A14, A3 ᴗ A15, A3 ᴗ A16, A3 ᴗ A17, A3 ᴗ A18, A4 ᴗ A14, A4 ᴗ A15, A4 ᴗ A16, A4 ᴗ A17, A4 ᴗ A18, A5 ᴗ A15, A5 ᴗ A16, A5 ᴗ A17, A5 ᴗ A18, A6 ᴗ A16, A6 ᴗ A17, A6 ᴗ A18, A7 ᴗ A17, A7 ᴗ A18, A8 ᴗ A18.
Adding a tenth transversals gives us a total of one hundred and thirty-five parallelograms.
p=3n2-n÷2
p=3102-10÷2
=3(90)÷2
=270 ÷2
=135
Adding a fourth horizontal parallel line with two transversals
Figure 15: two transversals
A1 , A2 , A3, A1 ᴗ A2, A2 ᴗ A3, A1 ᴗ A3.
Adding a fourth horizontal parallel line with two transversal lines gives us a total of six parallelograms.
Adding a fourth horizontal parallel line with three transversals
Figure 16: three transversals
A1 , A2 , A3, A4, A5, A6, A1 ᴗ A2, A1 ᴗ A3, A2 ᴗ A4, A2 ᴗ A6, A3 ᴗ A4, A3 ᴗ A5, A4 ᴗ A6, A5 ᴗ A6, A1 ᴗ A4, A1 ᴗ A5, A3 ᴗ A6, A1 ᴗ A6.
Adding a fourth horizontal parallel line with three transversal lines gives us a total of eighteen parallelograms.
Adding a fourth horizontal parallel line with four transversals
Figure 17: four transversals
A1 , A2 , A3, A4, A5, A6, A7, A8, A9, A1 ᴗ A2, A1 ᴗ A3, A1 ᴗ A4, A1 ᴗ A7, A2 ᴗ A3, A2 ᴗ A5, A2 ᴗ A8, A3 ᴗ A6, A3 ᴗ A9, A4 ᴗ A5, A4 ᴗ A6, A4 ᴗ A7, A5 ᴗ A6, A5 ᴗ A8, A6 ᴗ A9, A7 ᴗ A8, A7 ᴗ A9, A8 ᴗ A9, A1 ᴗ A5, A1 ᴗ A6, A1 ᴗ A8, A1 ᴗ A9, A2 ᴗ A6, A2 ᴗ A9, A4 ᴗ A8, A4 ᴗ A9, A5 ᴗ A9.
Adding a fourth horizontal parallel line with four transversal lines gives us a total of thirty-six parallelograms.
Adding a fourth horizontal parallel line with five transversals
Figure 18: five transversals
A1 , A2 , A3, A4, A5, A6, A7, A8, A9, A10, A11, A12, A1 ᴗ A2, A1 ᴗ A3, A1 ᴗ A4, A1 ᴗ A5, A1 ᴗ A9, A2 ᴗ A3, A2 ᴗ A4, A2 ᴗ A6, A2 ᴗ A10, A3 ᴗ A4, A3 ᴗ A7, A3 ᴗ A11, A4 ᴗ A8, A4 ᴗ A12, A5 ᴗ A6, A5 ᴗ A7, A5 ᴗ A8, A5 ᴗ A9, A6 ᴗ A7, A6 ᴗ A8, A6 ᴗ A10, A7 ᴗ A8, A7 ᴗ A11, A8 ᴗ A12, A9 ᴗ A10, A9 ᴗ A11, A9 ᴗ A12, A10 ᴗ A11, A10 ᴗ A12, A11 ᴗ A12, A1 ᴗ A6, A1 ᴗ A7, A1 ᴗ A8, A1 ᴗ A10, A1 ᴗ A11 A1 ᴗ A12, A2 ᴗ A7, A2 ᴗ A8, A2 ᴗ A11, A2 ᴗ A12, A3 ᴗ A8, A3 ᴗ A12, A5 ᴗ A10, A5 ᴗ A11, A5 ᴗ A12, A6 ᴗ A11, A6 ᴗ A12, A7 ᴗ A12.
Adding a fourth horizontal parallel line with five transversal lines gives us a total of sixty parallelograms.
Table 5: Side by Side view of Corresponding Transversals to Parallelograms
Paralleograms:
6, 18, 36, 60, etc.
18-6 36-18 60-36
12 18 24
6 6
Table 6: Four Horizontal Parallel Lines
Now the second order difference is 6- double the second set of parallelograms. Due to the second order being six, I deducted and found that the number of total parallelograms was increasing in multiples of six as shown by the fourth column of the table above.
General statement
Since the number of parallelograms created as the number of transversals increased each had a Second Order difference of 6, it was immediately known that the general formula must be a quadratic equation.
If there are n transversals and four horizontal lines, then p = sum of all integers from 1 to (n - 1).
P=12 4nn-1
or
P=4n2-n÷2
Expanding…
Table 7: Number of Parallel Transversals
2 horizontal lines:1nn-1÷2
3 horizontal lines:3(nn-1)÷2
4 horizontal lines:6(nn-1)÷2
5 horizontal lines:10(nn-1)÷2
The number and sequence repeats the formula nn-1÷2 and multiplies by the first term.
The General Statement
From this realization I was able to find the final formula for calculating the number of parallelograms formed when m horizontal parallel lines are intersected by n parallel transversal:
m(m-1)2 × n(n-1)2
To test the validity of the formula, I tested it against a previously counted parallelograms (Figure 10), the intersection of four transversals with three horizontal parallel lines should form a total of eighteen parallelograms.
Figure 14: four transversals
Using the formula
m(m-1)2 × n(n-1)2
3(3-1)2 × 4(4-1)2
3(2)2 × 4(3)2
62 × 122
3 ×6
=18 Parallelograms
Scope/limitations
The formula will be valid for m, n≥2. if either value were to be 1 or 0, it would be impossible to create any parallelograms.
by
julianan (student)
