- Level: International Baccalaureate
- Subject: Maths
- Word count: 2222
How many pieces? In this study, the maximum number of parts obtained by n cuts of a four dimensional object will be analyzed by looking at the patterns of the maximum number of segments made by n cuts on a one dimensional line, the maximum number of regio
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Introduction
IB Mathematics HL Internal Assessment
How many pieces?
Candidate Number:
Emily Wang
Mr. Dramble
30 January 2012
Much of mathematics involves finding rules by looking at patterns. In this study, the maximum number of parts obtained by n cuts of a four dimensional object will be analyzed by looking at the patterns of the maximum number of segments made by n cuts on a one dimensional line, the maximum number of regions made by n cuts on a two dimensional circle, and the maximum number of parts made by n cuts on a three dimensional cuboid. The sketches, cuts and results, conjectures and proofs will be shown, and the conjecture of the formula for the maximum number of cuts for the four dimensional objects will be made according to the previous studies.
- One Dimensional Object
For a line segment, to obtain the rule which relates the maximum number of segments (S) obtained from n cuts, we could look at examples and draw our results on a table.
n = 1, S = 2
n = 2, S = 3
n = 3, S = 4
n = 4, S = 5
n = 5, S = 6
n | 1 | 2 | 4 | 4 | 5 |
S | 2 | 3 | 4 | 5 | 6 |
It can be seen that this shows an arithmetic sequence, and by using the arithmetic sequence formula, an = a1 + (n – 1)d, (an = S) it can be seenthat the common difference, d, is 1 since the values of S increase by 1 as n increases by 1.
Middle
11
16
Find pattern for Rn.
R1= 2 → 1 + 1 → R0+ S0
R2 = 4 → 2 + 2 → R1+ S1
R3= 7 → 4 + 3 → R2+ S2
R4= 11 → 7 + 4 → R3+ S3
R5= 16 → 11 + 5 → R4+ S4
Rn= Rn - 1 + Sn – 1
R= X + S→ X = R – S
Plug R and S: X = (½)n 2 + (½)n + 1 – (n + 1)→ = (½)n 2 – ½n
R = (½)n 2 – ½n + (n + 1 ) = (½)n 2 + (½)n + 1
- Three Dimensional Cuboid
For a finite, three-dimensional cuboid, the maximum number of parts (P) that are obtained with n cuts are
Code on Wolfram Alpha Mathematica: Graphics3D[{Green,Opacity[.3],Cuboid[{-10,-10,-10},{10,10,10}],Blue,Opacity[1],Polygon[{{-15,0,-15},{-15,0,15},{15,0,15},{15,0,-15}}]}]
Code on Wolfram Alpha Mathematica:
Graphics3D[{Green,Opacity[.3],Cuboid[{-10,-10,-10},{10,10,10}],Blue,Opacity[1],Polygon[{{-15,0,-15},{-15,0,15},{15,0,15},{15,0,-15}}], Orange, Opacity [1], Polygon [{{0, 15, -15}, {0, 15, 15}, {0, -15, 15}, {0, -15, -15}}]}]
Graphics3D[{Green,Opacity[.3],Cuboid[{-10,-10,-10},{10,10,10}],Blue,Opacity[1],Polygon[{{-15,0,-15},{-15,0,15},{15,0,15},{15,0,-15}}], Orange, Opacity [1], Polygon [{{0, 15, -15}, {0, 15, 15}, {0, -15, 15}, {0, -15, -15}}], Purple, Opacity[1], Polygon [{{-15, -15, 0}, {-15,15, 0}, {15, 15, 0}, {15,-15, 0}}]}]
Code on Wolfram Alpha Mathematica: Graphics3D[{Green,Opacity[.3],Cuboid[{-10,-10,-10},{10,10,10}],Blue,Opacity[1],Polygon[{{-15,3,-15},{-15,4,15},{15,4,15},{15,4,-15}}], Orange, Opacity [1], Polygon [{{-1, 15, -15}, {-1, 15, 15}, {-1, -15, 15}, {-1, -15, -15}}], Purple, Opacity[1], Polygon [{{-15, -15, 4}, {-15,15, 4}, {15, 15, -9}, {15,-15, -9}}], Yellow, Polygon [{{-6, -15, -15}, {-6, 15, -15}, {15, 15, 15}, {15, -15, 15}}]}]
Code on Wolfram Alpha Mathematica: Graphics3D[{Green,Opacity[.3],Cuboid[{-10,-10,-10},{10,10,10}],Blue,Opacity[1],Polygon[{{-15,3,-15},{-15,4,15},{15,4,15},{15,4,-15}}], Orange, Opacity [1], Polygon [{{-1, 15, -15}, {-1, 15, 15}, {-1, -15, 15}, {-1, -15, -15}}], Purple, Opacity[1], Polygon [{{-15, -15, 4}, {-15,15, 4}, {15, 15, -9}, {15,-15, -9}}], Yellow, Polygon [{{-6, -15, -15}, {-6, 15, -15}, {15, 15, 15}, {15, -15, 15}}], Red, Polygon [{{-15, -15, -15}, {-15, 15, -15}, {15, 15, 1}, {15, -15, 1}}]}]
n | 1 | 2 | 3 | 4 | 5 |
P | 2 | 4 | 8 | 15 | 26 |
n | 1 | 2 | 3 | 4 | 5 |
R | 2 | 4 | 7 | 11 | 16 |
Conclusion
Q = Z + (1/6)n 3 + (5/6)n + 1
Plug in (1/24)n 4 – (1/12)n3 + (11/24)n2 + (7/12)n + 1 for Q
(1/24)n 4 – (1/12)n3 + (11/24)n2 + (7/12)n + 1 = Z + (1/6)n 3 + (5/6)n + 1
Solve for Z: Z = (1/24)n 4 – (1/12)n3 + (11/24)n2 + (7/12)n + 1 – ((1/6)n 3 + (5/6)n + 1)
Z = (1/24)n 4– (1/4)n 3 + (11/24)n2 –(1/4)n
Q = (1/24)n 4– (1/4)n 3 + (11/24)n2 –(1/4)n + (1/6)n 3 + (5/6)n + 1
Q = (1/24)n 4 – (1/12)n 3 + (11/24)n2 + (7/12)n + 1
By using the first dimension, second dimension and third dimension, I was able to find a rule using the recursive formulae from the previous dimensions to find a pattern for the fourth dimension. The equation Qn = Qn-1+Pn-1 can be used to find the n amount of cuts for the fourth dimension, and assuming this pattern is accurate, the n amount of cuts can be found for the fifth, sixth, seventh…infinite dimensions. For example, using G as the maximum amount of cuts for the fifth dimension, the pattern for the segments resulted from n amount of cuts can be found by doing Gn = Gn-1+Qn-1, assuming that G1 is 2 because of the previous patterns. The resulting numbers can be found by using a regression test, for example, the equation for the fourth dimension, Qn = (1/24)n 4 – (1/12)n3 + (11/24)n2 + (7/12)n + 1, was found by using a quartic regression. This use of finding a pattern and using technology to find the equation for the pattern and verifying the pattern can help find the maximum number of parts made by n number of cuts for an object with infinite dimensions.
This student written piece of work is one of many that can be found in our International Baccalaureate Maths section.
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