• Join over 1.2 million students every month
  • Accelerate your learning by 29%
  • Unlimited access from just £6.99 per month

In this portfolio, I am required to investigate the number of regions obtained by making cuts in one, two and three dimensional objects

Extracts from this document...





A line segment is a finite one-dimensional object. Find a rule which relates the maximum number of segments (S) obtained from n cuts. Comment on your results.

A circle is a finite two-dimensional object. Investigate the maximum number of regions (R) obtained when n chords are drawn.

A cuboid is a finite three-dimensional object. Investigate the maximum number of parts (P) that are obtained with n cuts.

If a finite four-dimensional object exists and the procedure is repeated what would you expect to find?

In this portfolio, I am required to investigate the number of regions obtained by making cuts in one, two and three dimensional objects. By finding a rule in all the three cases, I need to develop a rule for a four-dimensional object by searching for a definite pattern.

The data is as follows-

One – dimensional object

No of cuts made (n)

No of segments formed (S)












In the data table, we can see that in all the 5 cases, the number of segments formed is 1 more than the number of cuts made.

...read more.


After drawing all the sketches and finding the number of regions obtained, I proceeded to find the pattern of the sequence.

First I found the difference between the terms of the sequence of number of regions.

i.e –   2, 4, 7, 11, 16

The differences are - 2, 3, 4, and 5

And the difference is equal to the number of chords drawn.


Number of regions   =   Number of regions obtained   +   number of chords drawn

 obtained                          in the previous sketch

We can also consider the number of regions obtained as a function of the number of chords drawn.

Therefore, the recursive rule can be stated as follows-



However, to find the generalization, I need to eliminate the function and prepare an equation only with two variables – n and R. Therefore, I plotted the figures in the data table in the XY data set of geometry software Autograph to find the equation of the line. The results was –


In the result box, we can see that the equation of the curve is –

Y = 0.5image02.png

2 +0.5image02.png

 + 1

Therefore, substituting R and n back, I got the generalized rule as follows –


Three – dimensional object-

For three-dimensional graphing, I again used the Autograph software.

...read more.



Substituting x and y as n and (P) respectively, I got –



After getting the equation, I also found the value for 5 cuts-

(P) = 2(15) – 8 + 4

(P) = 26

Therefore, the final results obtained from all the 3 dimensions are –

No of cuts (n)

No of segments in 1d (S)

No of regions in 2d (R)

No of parts in 3d (P)





















The sequence of the values for 1d line is - 2, 3, 4, 5, and 6.

And, the sequence of the values for 2d line is – 2, 4, 7, 11, and 16.

The sequence of the difference in values for 2d circle is – 2, 3, 4, and 5 which is the same as the sequence of values for 1d line.

Therefore, the values of a 2d figure can be simply obtained by just adding the previous term and the respective term for a 1d figure.

i.e.  11 (answer needed) = 7 (previous answer) + 4 (answer in 1d for the previous number of cuts)

 The same thing is applicable between 2d and 3d figure.

i.e.  26 (answer needed) = 15 (previous answer) +11 (answer in 2d for the previous number of cuts)

Therefore, the same should be applicable for a 4d figure as well.

Hence, the recursive formula for a 4d figure can be written as –



where the image10.png

 term is from a 3d sequence.

...read more.

This student written piece of work is one of many that can be found in our International Baccalaureate Maths section.

Found what you're looking for?

  • Start learning 29% faster today
  • 150,000+ documents available
  • Just £6.99 a month

Not the one? Search for your essay title...
  • Join over 1.2 million students every month
  • Accelerate your learning by 29%
  • Unlimited access from just £6.99 per month

See related essaysSee related essays

Related International Baccalaureate Maths essays

  1. Math Portfolio: trigonometry investigation (circle trig)

    240 -0.866025404 -0.5 1.732050808 260 -0.984807753 -0.173648178 5.67128182 270 -1 -1.83772E-16 5.44152E+15 280 -0.984807753 0.173648178 -5.67128182 300 -0.866025404 0.5 -1.732050808 320 -0.64278761 0.766044443 -0.839099631 340 -0.342020143 0.939692621 -0.363970234 360 -2.4503E-16 1 -2.4503E-16 The number with E is error made by MS Excel calculations.

  2. How many pieces? In this study, the maximum number of parts obtained by n ...

    rule to the maximum amount of regions, we have to view the R cases. R1 = 2 R2 = 4 --> R1 + 2 R3 = 7 --> R2 + 3 R4 = 11 --> R3 + 4 R5 = 16 --> R4 + 5 So the recursive formula for generating the maximum amount of regions is Rn=Rn-1+ n.

  1. In this essay, I am going to investigate the maximum number of pieces obtained ...

    values for n gives: n=12=a+b+c n=24=4a+2b+c n=37=9a+3b+c Solve these three equations for a, b, c gives a=1/2, b=1/2, c=1 Thus Rn =1/2n�+1/2n+1 When n=5, R5=1/2*5^2+1/2*5+1=16, which is also corresponds to the tabulated value for n=5 above. Rn =1/2n�+1/2n+1 is the same with Rn=(n^2+n+2)/2 So far we have formed a conjecture

  2. Statistics project. Comparing and analyzing the correlation of the number of novels read per ...

    The 30 girls surveyed: Modal mark is 61-80. Most read a higher amount of books of 2 per week than boys, but is still quite low. Their median value for the marks is 40-60 and their average mark is 66.7.

  1. Math HL portfolio

    which is shown in the graph below: 3. Investigate your conjecture for any real value of a and any placement of the vertex. Refine your conjecture as necessary and prove it. Maintain the labelling convention used in parts 1 and 2 by having the intersections of the first line to be and , and the intersections with the second line to be and .

  2. The segment of a polygon

    The side of the smaller (inner) triangle is 3.93cm. It has the ratio of 1.527:1 or 1:0.655 to the side length of the outer triangle (outer:inner) . The outer equilateral triangle's area obtained from the "Geometer's Sketchpad" is 15.59 cm� while the inner equilateral triangle's area is 6.688cm�. The sketchpad ratio of the two area is 1.53:1 or 7:3 (outer : inner).

  1. The segments of a polygon

    I use program to get the function of a curve, which goes through this points. And the program gives me this function: Now I have to replace x with n and f(x) with ratio of areas and I get: = (d)

  2. The aim of the assignment is to examine the gold medalist heights for the ...

    and at (1948, 198). A cubic function is effective for this set of data specifically because of the concave up and down shape, although if it were for Gold Medal High Jump results in the future, a logarithmic model as it illustrates humans infinitely increasing ability.

  • Over 160,000 pieces
    of student written work
  • Annotated by
    experienced teachers
  • Ideas and feedback to
    improve your own work