# SA NC Doing Investigations/Chapter 6
## Examples of investigative activities in science and mathematics.

The investigations described so far are all projects and of the type where learners come up with an idea and then try to find out as much as they can about it. The purpose of investigations like these is to get learners into the processes of science (see Chapter 9 for a comprehensive list of the facets of scientific literacy). But the spirit of investigation can be found in much simpler activities and lives on in the saying of a great science educator of the past, Julius Sumner Miller: "more doing and less talking"! When introducing a new concept or simply when presenting learners with a remarkable phenomenon to stimulate their imaginative juices, investigations remain a very powerful tool in the hands of science and mathematics educators. Here are a few popular, investigative activities. Follow the instructions and work out what scientific concepts are being demonstrated. Investigative activities leading from each of these would be to find out where each phenomenon can be seen at work.
A useful acronym that summarises an approach to using these quick investigations is "OPE" or Observe, Predict and Explain. When doing them in class it is important to allow learners the time to observe carefully (possibly re-doing the demonstration in the process) to discuss their predictions and the theory behind their predictions and then their explanations of what happened, particularly if their predictions were incorrect.

### Quick investigations: important concepts in science
#### Lifting an ice cube without touching it.

You will need a glass of water, ice cubes, a piece of string and some salt. Step 1. Put an ice cube into a glass of water. It floats with about one tenth of the cube sticking out above the water.
Step 2. Put an end of the string on the exposed ice and sprinkle some salt on it. Let it stay there for a minute or two.
Step 3. Gently lift the string.
What do you expect? That the ice cube will remain in the water and the string will pull away. What actually happens? The string sticks to the ice cube strongly enough to lift it from the water. How does it work? Salt lowers the freezing point of water to below 0&deg C. When the ice melts where the end of the string lies, the melted ice dilutes the salt and the ice freezes again. The warmth of the string will first cause a small hole to form in the ice and the water-salt mixture then freezes around it again when the string is inside the ice cube.

#### Heavy paper.

You will need a double sheet of newspaper and a ruler. Step 1. Place the ruler on a table with a few centimetres sticking over the edge.
Step 2. Cover the ruler on the table with the sheet of newspaper.
Step 3. Hit the piece of ruler that extends over the edge of the table hard, with your hand.
What do you expect? You expect the paper to leap into the air as you hit the ruler. What actually happens? The paper remains stationary on the table. It does not even tear. How does it work? The weight of air on the palm of your hand, turned upwards, is about 45 kg. What is the weight of air on a double sheet of newspaper? Certainly more than a ruler can lift!

#### How strong are your lungs?

You will need a balloon and an empty cold drink bottle.
Step 1. Place the balloon in the neck of the bottle.
Step 2. Wrap the opening of the balloon around the opening of the bottle.
Step 3. Blow into the balloon which is inside the bottle.
What do you expect? The balloon should inflate and fill the bottle. What actually happens? The balloon inflates very slightly but then it is impossible to inflate it any more. How does it work? As the balloon inflates, the air in the bottle compresses into the remaining space. This means that the pressure back on the blower's lungs also increases. It is very difficult to inflate the balloon, even a little!

#### It's not so easy to move coins.

You will need a heavy water glass, a strip of newspaper, two large coins and a ruler.
Step 1. Put one end of a strip of paper over the rim of the glass.
Step 2. Balance the two coins on the edge of the glass.
Step 3. Make sure the coins are just touching the paper but are actually balancing on the rim and not on the paper.
Step 4. Lift the free end of the paper and hold it horizontal.
Step 5. Strike the paper with a ruler about 5 cm from the edge of the glass.
What do you expect? You expect the coins to drop into the glass. What actually happens? The paper is whipped out from under the coins and they remain on the lip of the glass. 'How does it work? All bodies resist being put into motion (i.e. being made to move). The more mass in a body, the more difficult it is to make it move. So the light paper moves but the heavier coins do not. (Find out what this property of matter is called.)

#### Paper scorch and char.

You will need a spirit burner, matches, a rod of wood and metal (make one with a dowel rod and a piece of aluminium foil wrapped around one half of it a few times) and a piece of writing paper.
Step 1. Wrap a single layer of paper around the wood-metal rod.
Step 2. Light the spirit burner.
Step 3. Play the flame lightly on the paper over the boundary between the metal and the wood on the rod.
What do you expect? The paper will burn. What actually happens? The paper chars on the half that covers the wood but not the aluminium foil. How does it work? The metal conducts the heat away from the paper while the wood does not. So the temperature of the paper over the wood soon rises to the point at which paper scorches and chars. This does not happen over the metal.

#### A balancing trick at the dining room table - if your mother lets you do it!

You will need a cork, a pin, two forks and a heavy glass.
Step 1. Place the pin in one end of the cork.
Step 2. Stick the forks into the side of the cork so that they hang downwards &shy their handles being below the bottom of the cork that has the pin in it.
Step 3. Balance the pin on the bottom of an upturned glass.
What do you expect? You expect the whole arrangement to fall over because it looks so unstable. What actually happens? The arrangement balances on the pin and although it may wobble at first it soon settles down. How does it work? Every body has a point called the Centre of Gravity (CoG) associated with it. It is the point at which the entire mass of the object appears to be concentrated. If the CoG is below the balance point the object will be very stable.

#### Superboy and Supergirl!!

You will need two broomsticks and a length of rope.
Step 1. Let Superboy/Supergirl hold two broomsticks vertically, one in each hand, and pull inwards while two other people try to pull the broomsticks (and the holder's arms) outwards.
Step 2. Now, while the two "pullers" each hold a broomstick vertically, tie one end of a piece of rope onto one broomstick (about one third of the way up) and then wind the rope about 5 times around both, working in an upwards direction.
Step 3. Let the two pullers try to pull the broomsticks outwards again while Superboy/Supergirl pulls on the loose end of the rope.
What do you expect? You expect the two people to pull the broomsticks apart with ease. What actually happens? The rope pulls the two people inwards quite easily. How does it work? This acts like a series of pulleys (or block and tackle). Every turn of rope almost doubles the pulling power of the person on the end of the rope.

#### Tight fit.

You will need a single sheet of A4 paper and a pair of scissors.
Step 1. Give each learner a pair of scissors and a piece of A4 paper.
Step 2. Set them the problem: "Can you cut the paper so that it fits over your body?"
Step 3. Fold the paper in half short end to short end.
Step 4. Cut out a narrow rectangle from the folded end as shown (X).
Step 5. Make an odd number of cuts according to this pattern:
What do you expect? Can't do it! What actually happens? It works!! How does it work? Look carefully at the arrangement and see why it works.
Step 6. You can experiment using another piece of paper. Try disobeying the instructions, for instance by making an even number of cuts or not cutting right to the edge of the paper!

## Investigating Number Patterns

This series of activities was prepared by Professor Otis Thompson of the University of Montana Western in his capacity as Principal Investigator and Mathematics Specialist on the Tlhatloga Project. The activities were presented at a series of workshops held in 2003. Professor Thompson has had a distinguished career as a mathematics educator and regards his three years working with South African educators as the highlight of his professional life.

### Educator's Guide

Mathematics has been defined as the study of patterns. Patterns are everywhere &shy on wallpaper, floors and wall tiles in artistic works in plant and animal growth and behavior in traffic and even in television schedules. Scientists look for patterns in order to isolate variables so that they can reach valid conclusions in their researches. Learners can gain experience with patterns by investigating numerical and geometric patterns and expressing these patterns mathematically in words or symbols. To analyze patterns we first determine the structure of the pattern and how the pattern changes. We then organize the information systematically and finally develop generalizations about the mathematical relationships in the patterns. In these activities learners can explore number patterns and learn some methods for analyzing them. They can then use their analyses to express the patterns mathematically. Learners can also investigate how pure number patterns apply to the real world.
Activity 1 In this activity, the learner will explore some familiar number patterns and will name the set of numbers described.
Activity 2 In this activity, the learner will explore arithmetic sequences, geometric sequences, and sequences of figurate numbers. An arithmetic sequence is one in which each successive term is obtained from the previous term by the addition or subtraction of a fixed number called the difference. A geometric sequence is one in which each successive term is obtained from the previous term by multiplying by a fixed number called the ratio. Figurate numbers are numbers that can be represented by dots arranged in the shape of certain geometric figures.
Activity 3 The Fibonacci sequence will be introduced using the original problem that Leonardo Fibonacci presented in 1202. Then the learner will explore some other problems that lead to the Fibonacci sequence. The learner will discover some of the interesting relationships within this sequence and express these relationships mathematically. The learner will explore how plant growth patterns are aligned with the Fibonacci numbers.
Activity 4 1+ \frac{1}{\frac{1}{1} + \frac{1}{1 + L} } Learners will explore the continued fraction and see that stopping at a given point always leads to a ratio of Fibonacci numbers. When converted to a decimal these ratios approach 1.61803..., the golden ratio. The learners will then experiment to find a "most pleasing rectangular shape" and learn that the ratio of length to width is approximately this golden ratio. The learners will then locate rectangles in their surroundings that are "golden rectangles."
Materials needed
* Paper and pencil
* Calculator (optional)
* A whole pineapple (optional)
* Compass
* Straight edge
* Scissors or cutting knife

### Learner activities
#### Number Patterns Activity 1

Thembisa had to learn a large list of spelling words for her final spelling test for the term. The first day she learned two words, and each day thereafter she learned to spell two more words. If she had the list learned by the 15th day, how many words were on her list?
To solve this problem, complete the following table:
| Day | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 | 11 | 12 | 13 | 14 | 15 |
| Accumulated no. of words | 2 | | | | | | | | | | | | | | |
So we see that Thembisa's list contained _____________ words.
The list of numbers you created (called a sequence) is part of a very special set of whole numbers. This set, written as {0, 2, 4, 6, 8, 10, 12, 14, 16, ...}, where the ... (called ellipsis) means the pattern continues forever. What is the name of this set of whole numbers?
There are other sequences of whole numbers for which you probably have a name. Let's explore and see which ones you know. In each sequence below, fill in the next three terms and give a name to this sequence.
1. 1, 3, 5, 7, 9, ______ , ______ , ______ , ...
Name: ________________________________
2. 0, 1, 4, 9, 16, 25, ______ , ______ , ______ , ...
Name: ________________________________
3. 0, 3, 6, 9, 12, 15, ______ , ______ , ______ , ...
Name: _______________________________
4. 2, 3, 5, 7, 11, 13, 17, _____ , ______ , ______ , ...
Name: ______________________________

#### Number Patterns Activity 2

A large theatre is set up in such a way that there are 20 seats in the first row and 4 additional seats in each consecutive row. The last row has 144 seats. How many rows are there in the theatre?
If we try to solve this problem by creating a list as we did in Activity One, it might take us a while to finish the list. Let's explore some of the properties of a sequence such as this one to try to simplify our method of solution.
The sequence we need to create to solve this problem is called an arithmetic sequence. An arithmetic sequence is one in which each successive term is obtained from the previous term by the addition or subtraction of a fixed number called the difference. Write the first five terms of the sequence that correspond to the first five rows: What would be the tenth term in this sequence? What would be the twentieth term in this sequence? Write a rule for finding the nth term of this sequence?
Now use this rule to find the number of rows in the theatre. To determine if a sequence is an arithmetic sequence, one simply looks to see if the difference between consecutive terms is constant. Which of the following sequences are arithmetic sequences? If the sequence is an arithmetic sequence, write the 100th term of this sequence.
1. 5, 8, 11, 14, 17, ...
Arithmetic ? Yes/No
100th term = ________________
2. 4, 6, 9, 13, 18, ....
Arithmetic ? Yes/No
100th term = ________________
3. 1040, 1032, 1024, 1016, 1008, ....
Arithmetic? Yes/No
100th term = _____________
4. The set of even whole numbers
Arithmetic? Yes/No
100th term = _____________
Now the question arises regarding the introductory problem in Activity 2: How many seats are there in the theatre?
To answer this question, first let's consider a simpler problem. What is the sum of these numbers: 1 + 2 + 3 + 4 + ... + 100? (Note this is an arithmetic sequence with a difference of 1.)
Write out the sum: 1 + 2 + 3 + ... + 98 + 99 + 100
Below it write the Sequence "backwards.": 100 + 99 + 98 + ... + 3 + 2 + 1
Now if we add the pairs above and below each other, the sum is 101, and since there are 100 such pairs, the total sum of the two rows is 100 x 101. But since this contains every number in the sequence twice, we will divide this product by 2 giving us 1 + 2 + 3 + ... + 99 + 100 = 5050.
(Note: To find the sum of a finite number of terms of an arithmetic sequence, you need to know the first term, the last term, and the number of terms. Write a rule using this information for finding such a sum and then use this rule to find the number of seats in the theatre.)
Find the sum the following sums: 1 + 2 + 3 + 4 + ... + 1000 = _______________
2 + 4 + 6 + 8 + ... + 72 = _______________
8 + 11 + 14 + 17 + ... + 305 = _______________
Not all sequences are arithmetic. Other types of sequences exist. A geometric sequence is a sequence in which each successive term is obtained from its predecessor by multiplying by a fixed number called the ratio. Find the next three terms in these geometric sequences:
1. 2, 4, 8, 16, ______ , ______ , ______ , ...
2. 729 486 162, _____ , _____ , _____ , ...
3. 1, 0.4, 0.16, 0.064, _____ , _____ , _____, ...
Figurate numbers (numbers that can be represented by dots arranged in the shape of certain geometric figures) provide examples of sequences that are neither arithmetic nor geometric. Below is the first four terms of the sequence of triangular numbers.
Explain why the next triangular number is 15.
The sequence 1, 3, 6, 10, 15, ... is not an arithmetic sequence (since no common difference) nor a geometric sequence (since no constant multiplier.) But if we look at the difference between consecutive terms and then difference between consecutive differences, we see a pattern that should help us find the next three terms of this sequence.
Here are the first three pentagonal numbers. Find the next three pentagonal numbers using the method you used with the triangular numbers.
When asked to find a pattern for a given sequence, first look for some easily recognizable pattern and determine if the sequence is arithmetic or geometric. If a pattern is unclear, taking successive differences (as done with the figurate numbers) may help. It is possible that none of these methods will reveal a pattern.

#### Number patterns Activity 3

In 1202 Leonardo Fibonacci presented the following problem in his book Liber abaci (Book of the abacus): A certain man put a pair of baby rabbits in a place surrounded on all sides by a wall. How many pairs of rabbits can be produced from that pair in a year, if it is supposed that every month each pair begets a new pair which, from the second month on, becomes productive?
To solve this problem, let's make a table and keep track of how many pairs we have each month. Remember, we will start with one baby pair. At the start of the second month, they are now an adult pair of rabbits, so they breed. The gestation period for rabbits is one month so at the start of the third month, we now have an adult pair and a baby pair.
| Month | Adult pair | Baby pair | Total pairs |
| 1 | 0 | 1 | 1 |
| 2 | 1 | 0 | 1 |
| 3 | 1 | 1 | 2 |
| 4 | | | |
| 5 | | | |
| 6 | | | |
| 7 | | | |
| 8 | | | |
| 9 | | | |
| 10 | | | |
| 11 | | | |
| 12 | | | |
Now complete the table to arrive at the answer as to how many pairs we have at the end of the year. Carefully analyze the sequence of numbers you obtained for the total number of pairs. Describe how one can determine the next term in this sequence.
(Note: this sequence is neither an arithmetic nor a geometric sequence. The method of successive differences doesn't work like it did for the figurate numbers, but it may give a clue to how to build successive terms in this sequence.)
Now let's explore this sequence in greater detail and make some conjectures about it. 1. Complete the following table for the first 25 terms of the Fibonacci sequence: There is nothing anywhere here to explain what the Fibonacci numbers are or how they are obtained?
* Symbol
* F1
* F2
* F3
* F4
* F5
* F6
* F7
* F8
* F9
* F10
* F11
* F12
* F13
* F14
* F15
* F16
* F17
* F18
* F19
* F20
* F21
* F22
* F23
* F24
* F25
2. If your table was completed correctly, the twenty- fifth Fibonacci term, F25, should be 75025.
3. Find the following sums:
a. F$_1$ + F$_2$
b. F$_1$ + F$_2$ + F$_3$
c. F$_1$ + F$_2$ + F$_3$ + F$_4$
d. F$_1$ + F$_2$ + F$_3$ + F$_4$ + F$_5$
e. F$_1$ + F$_2$ + F$_3$ + F$_4$ + F$_5$ + F$_6$
f. F$_1$ + F$_2$ + F$_3$ + F$_4$ + F$_5$ + F$_6$ + F$_7$
g. Write a rule that would let you give the answer to the sum of the first n terms in terms of a Fibonacci number.
h. Use your rule to find the sum of the first twenty terms of this sequence.
4. Find the following sums (note: we are just using the odd numbered terms):
a. F$_1$ + F$_3$
b. F$_1$ + F$_3$ + F$_5$
c. F$_1$ + F$_3$ + F$_5$ + F$_7$
d. F$_1$ + F$_3$ + F$_5$ + F$_7$ + F$_9$
e. F$_1$ + F$_3$ + F$_5$ + F$_7$ + F$_9$ + F$_{11}$
f. Write a rule that would let you give the answer to the sum of the first n odd numbered terms in terms of a Fibonacci number.
g. Use your rule to find the sum of the first ten odd numbered terms.
5. Find the following sums (note: here we are just using the even numbered terms):
a. F$_2$ + F$_4$
b. F$_2$ + F$_4$ + F$_6$
c. F$_2$ + F$_4$ + F$_6$ + F$_8$
d. F$_2$ + F$_4$ + F$_6$ + F$_8$ + F$_{10}$
e. F$_2$ + F$_4$ + F$_6$ + F$_8$ + F$_{10}$ + F$_{12}$
f.Write a rule that would let you give the answer to the sum of the first n even numbered terms in terms of a Fibonacci number.
g. Use your rule to find the sum of the first ten even numbered terms.
6. Find the sum of the square of two consecutive Fibonacci numbers -This is written symbolically as: (F_n)^2 + (F_{n+1})^2. The result should be another Fibonacci number. Which one is it?
7. a. Take any three consecutive Fibonacci numbers from your list. Multiply the first number by the third number and subtract the square of the middle number. -This is written symbolically as: F_{n-1} X F_{n+1} - {F_n}^2 Write a rule that would explain what the answer will be.)
b. Without actually doing the computations, what is the value of F$_{24}$ x F$_{26}$ &shy F$_{25}$^2
8. a. Now compute F_{n-2} x F_{n+2} - {F_n}^2 for various values of n.
b. Write a rule that will explain what the answer should be each time.
9. (For the "experts.") Write a rule that will give the answer to F_{n-k} x F_{n+k} - {F_n}^2.
10. All the Fibonacci numbers are divisible by F_1 and F_2 (since F_1 = F_2 = 1 and every whole number is divisible by 1.)
a. Which Fibonacci numbers are divisible by F_3 ?
b. Which Fibonacci numbers are divisible by F_4 ?
c. Which Fibonacci numbers are divisible by F_5 ?
d. Describe the relationship between the Fibonacci numbers that are divisible by other Fibonacci numbers.
11. Take any 10 consecutive Fibonacci numbers and add them together. Divide this sum by 11. What was the result? Does this work every time?
12. Find the sum of the squares of the first n Fibonacci numbers. While the sums do not seem to be related to Fibonacci numbers directly, is it possible that they might have Fibonacci numbers as factors? If so, write the factors. Now write a rule for finding this sum of squares.
13. Every natural number is either a Fibonacci number or can be expressed uniquely as a sum of Fibonacci numbers whereby no two are Fibonacci numbers. Here is one way to do it:
a. Write down a natural number.
b. Find the largest Fibonacci number that does not exceed your number. That Fibonacci number is the first term of your sum.
c. Now subtract that Fibonacci number from your number and look at this new number.
d. Next find the largest Fibonacci number that does not exceed this new number. That Fibonacci number is the second number in your sum.
e. Continue this process until the remainder becomes a Fibonacci number (the last term of your sum.)
14. The game, Fibonacci nim, is played with two people: Person 1 and Person 2. All that is needed is a pile of sticks. Person 1 moves first by taking any number of sticks (at least one but not all) away from the pile. After Person 1 moves, it is Person 2's move, and the moves continue to alternate between them. Each person (after the first move) may take away as many sticks as he or she wishes with the restrictions that he or she must take at least one stick but no more than two times the number of sticks the previous person took. The player who takes the last stick wins the game. Player 1 can always win using the following strategy. Make sure that the initial number of sticks is not a Fibonacci number. Write the number of sticks in the pile as a sum of nonconsecutive Fibonacci numbers (see problem 13.) Figure out the smallest Fibonacci number occurring in the sum and as Player 1, remove that many sticks from the pile. Now your opponent (Player 2) plays. No matter what Player 2 does, you repeat the preceding procedure. That is, once Player 2 is done, Player 1 expresses the number of remaining sticks as a sum of nonconsecutive Fibonacci numbers, and then removes the number of sticks that equal the smallest Fibonacci number in the sum. It is a fact that, no matter what Player 2 does, Player 1 will always be able to remove that number of sticks without breaking the rules and eventually Player 1 will be able to remove the last stick and win. Now let's explore some other problems.
1. A child is trying to climb a staircase. The maximum number of steps he climbs at a time is two that is, he can climb either one step or two steps at a time. If there are n steps in total, in how many different ways can he climb the staircase?
If there is only one step (n = 1), clearly there is only one way to climb it. If there are two steps, the child can either climb the two steps at once or take them one at a time thus there are two ways. What if there were 3, 4, 5, ... steps?
2. Suppose we have two glass plates made of slightly different types of glass mounted face to face. If we shine light through the plates, the light rays can (in principle) reflect internally at four reflective surfaces before emerging. More specifically, they can pass through without reflecting at all, or they can undergo one internal reflection, two internal reflections, three internal reflections, and so on before emerging (see the diagrams below.) Count the number of beams that emerge when there are n internal reflections.
3. In a honeybee colony a drone (male) bee gets his start from an unfertilized egg and thus has just one parent (the queen.) The queen bee gets her start from a fertilized egg and thus has two parents (a queen and a drone.) If we went back 10 generations in a drone bee's "family tree," how many great-great-great-great-great-great-great-great grandparents would this drone bee have?
4. Here is an algorithm for creating a sequence known as the Golden Sequence. Start with the number 1, and then replace 1 by 10. From then on, replace each 1 by 10 and each 0 by 1. How many 1's are there in the sequence of 0's and 1's after you have done this iteration process 12 times?
The above problems all lead to the Fibonacci sequence. The Fibonacci sequence is everywhere. It even occurs in plants. Petal counts and petal arrangements of some flowers harbor Fibonacci numbers. Most field daisies have thirteen, twenty-one, or thirty-four petals (all Fibonacci numbers.) The leaves along a twig of a plant or the stems along a branch tend to grow in positions that would optimize their exposure to the sun, rain, and air. The passage from one leaf to the next (or from one stem to the next along a branch) is characterized by a screw-type displacement around the stem. For example, on the pear tree it takes three complete turns to pass through eight stems, a pair of alternate members of the Fibonacci sequence. The hexagonal scale on the surface of a pineapple is a part of three different spirals. Most pineapples have five, eight, thirteen, or twenty-one spirals of increasing steepness on their surface. All of these are Fibonacci numbers. When you look at the head of a sunflower, you will notice both clockwise and counterclockwise spiral pattern. In the pictures below, there are 21 clockwise spirals and 34 counterclockwise spirals, again Fibonacci numbers.
File:DoingInvestigations chp6 sunflower.jpg
A pinecone's exterior is composed of two sets of interlocking spirals, usually 5 spirals in one direction and 8 in the other. What examples of Fibonacci numbers can you find in the plants in this location?

#### Number Patterns Activity 4

Now let us look at a very different type of never-ending expression, this time involving fractions: 1 + \frac{1}{1 + \frac{1}{1 + \frac{1}{1 + \frac {1}{1 + \frac {1}{...}}}}}
This is a special case of mathematical entities known as continued fractions. How would we compute the value of this fraction? One way is to "truncate" the continued fraction as various points, simplify the fraction, and look for a pattern in our results.
1. Express as an improper fraction. 1 + \frac{1}{1}
2. Express as an improper fraction. 1 + \frac{1}{1 + \frac{1}{1}}
3. Express as an improper fraction. 1 + \frac{1}{1 + \frac{1}{1 + \frac{1}{1}}}
4. By this time you may have discovered a "short cut" for doing this. So let's do one more. Express as an improper fraction. 1 + \frac{1}{1 + \frac{1}{1 + \frac{1}{1 + \frac {1}{1}}}}
5. Have you seen the numbers that appear in the numerator and denominator of these improper fractions before? What are they? (F$_{n+1}$)/F$_n$
6. Use a calculator to get a decimal approximation to the fraction for larger and larger values of n where Fn is the nth Fibonacci number. This number is known as the Golden Ratio and is usually identified by , the Greek letter phi. It can be shown mathematically that this number is exactly .
Now let's try a little experiment. You will be given a piece of paper with several rectangles drawn on it. You are to select the one rectangle that is the most "pleasing, attractive, and elegant" of the set. The results for the group will be tabulated. Let us now construct a rectangle that has some special appeal to many people.
1. Draw square ABCD.
2. Locate the midpoint of segment DC.
3. Place the compass point at M and using radius MB, draw arc BE.
4. Point E is the point where DC extended intersects the arc.
5. Locate point F so that AFED is a rectangle. File:DoingInvestigations chp6 7.png
It can be shown mathematically that in rectangle AFED, the ratios of sides frac{AF}/{DE} and {BC}/{BF} are both equal to , the Golden Ratio. Hence rectangle AFED is known as the Golden Rectangle. Much has been written about how various artists and architects have allegedly used the Golden Rectangle in designing their various works. Near the center of a sheet of paper, construct a Golden Rectangle. Carefully cut the rectangle from the sheet leaving the outside of the sheet intact. Now use this "template" to find objects around you that are close to being "Golden Rectangles."

### Investigating Combinatorial Mathematics

This activity is based on an idea derived from "The Man Who Loved Only Numbers: the story of Paul Erdös and the search for mathematical truth," by Paul Hoffman (1998) and "The Combinatorial Mathematics of Scheduling," Scientific American, March 1978. The activity was written by Dr Peter Glover and was first used in June 2001 during the Tlhatloga Project's Summer Institute in Dillon, Montana, USA.

#### PROBLEM 1

Divide a group of 5 weights (2, 2, 2, 3 and 3) into two piles so that each pile is as close as possible in total weight.
'SOLUTION 1-1' By inspection (i.e. just by looking at the weights) we can arrange them (3, 3) and (2, 2, 2). When dealing with a few weights the problem can be solved by "trial and error" i.e. by trying all the variations you can until you get the best answer. This is a problem in the branch of mathematics called "combinatorics" &shy the mathematics of making the best decision about what to put where and how to combine quantities in an optimal way to get a desired result. It is a relatively new branch of mathematics and it is finding applications everywhere from defense to cutting lengths of piping to packing of delivery trucks.
SOLUTION 1-2 Ronald Graham is one of the best combinatorial mathematicians today. Here is his algorithm* for packing weights most efficiently: "starting with the heaviest weight and working down to the lightest, put each weight into the pile that tends, at each step of the way, to keep the weights of the piles as equal as possible."
[* The word algorithm is taken from the ninth century Persian number theorist called Abu Ja'far Mohammed ibn Musa al-Khowarizmi, who also gave his name to "algebra." "An algorithm is ... a step-by-step procedure in which every step is explicitly stated so that a problem can be blindly solved by man or machine." (Hoffman. 1998).]
But using Graham's algorithm to solve PROBLEM 1 we get this solution:
File:DoingInvestigations chp6 8.png
We can see that this is not the best solution, which our trial-and-error solution was i.e. where each pile totals 6 weight units (i.e. 3, 3 and 2, 2, 2). But it is also not the worst combination, which it would have been if piles ranged in size from 2 to 10 (i.e. 3, 3, 2 and 2, 2 OR 3, 3, 2, 2 and 2).
Our computer programs are simply algorithms (many of them) strung together and written in a language that computers can decode, "understand" and act on. The best solution vs the algorithmic solution The best solution (which ours was and which was found by trial and error) is 3, 3 and 2, 2, 2 (i.e. 6 and 6). But Ron Graham's algorithm gives us 3, 2, 2 and 3, 2 (i.e. 7 and 5).
Therefore Graham's algorithm gives a solution that is out by 1/6 in each case, (i.e. 6 vs 7 and 6 vs 5) or &plusmn 16% off the best answer.
Graham was able to prove that for 2 piles and any distribution of any number of weights, his algorithm will never be off by more than 16%. In 1973, Jeffrey Ullman of Princeton University showed that the so-called "first-fit packing algorithm" &shy where you pack items in whatever order they arrive into whichever container you choose, as long as they fit &shy can be off by as much as 70%!

#### PROBLEM 2

You have the 33 weights shown below and bins with a capacity of 524 weight units. Using Graham's "first-fit, heaviest-to-lightest" algorithm, divide up the blocks provided into as few bins as possible.
1. Show any calculations you make.
2. Show how you packed your weights in the bins:

#### PROBLEM 3

You still have 33 weights, but now each one is half the weight of those used in PROBLEM 2. Where halving gives a 0.5, round up as shown in the examples. Your new bins have a capacity of 262. Before using the "first-fit, heaviest-to-lightest" algorithm, see if you can make any calculations to predict how many bins you will need to pack the weights. When you have made your prediction, pack your blocks into as few bins as possible.
Your prediction: _____ bins.
Show how you would pack your weights.

#### PROBLEM 4

Take the 33 weights as used in PROBLEM 3 except that wherever there was a "rounding up" for a 0.5, you should now round down any halves. The bins still have a capacity of 262. When you have predicted the number of bins required, use Graham's algorithm to pack your blocks into as few bins as possible.
Your prediction: _____ bins.
Show how you would pack your weights.

#### PROBLEM 5

Use the 33 weights you used in Problem 2 but discard the 46 weight. Predict the number of bins you will need now. (a) Use Graham's algorithm to pack the blocks into the fewest bins possible and (b) using a trial-and-error approach, try to improve on the solution your "first-fit, heaviest-to-lightest" strategy gave.
Your prediction: _____ bins.
Show how you would pack your weights.

#### A Pipe Cutting Problem: another combinatorial investigation

Fiberglass insulation comes in 1-meter precut sections. A plumber must install insulation in a basement on piping that is often interrupted by joints. The distances (measured in cm) between the joints on the stretches of pipe that must be insulated are 30, 40, 45, 30, 20, 25, 40, 50, 30, 35, 50, 55, 60, 70, 95, 10, 70, 30, 80, 85, 40, 45, 30, 20, 15, 90, 60, and 50. How many precut sections would the plumber have to use to provide insulation. A stretch of pipe must be covered by one solid piece of insulation. No joining of two shorter pieces to make a longer piece is allowed.
This problem and the one presented earlier are known as bin-packing problems. Bin packing refers to finding the minimum number of bins of weight capacity W into which weights w1, w2, w3, ..., wn (each less than or equal to W) can be packed. Graham's algorithm, "first fit, heaviest to lightest" is also known as the First Fit Decreasing (FFD) algorithm. This algorithm works quite nicely if you know all the weights prior to assigning to bins so they can be arranged in order from largest to smallest. In some bin packing problems, we do not have this luxury. In these cases other algorithms must be used. Here are the definitions of the more commonly used algorithms for bin packing.
First Fit (FF)
The next weight to be packed is placed in the lowest-numbered bin already opened into which it will fit. If it fits in no open bin, a new bin is opened.
First Fit Decreasing (FFD)
The first fit algorithm is applied to the list of weights sorted so that they appear in decreasing order.
Next Fit (NF)
A new bin is opened if the weight to be packed next will not fit in the bin that is currently being filled. No previously opened bin can be used.
Next Fit Decreasing (NFD)
The next fit algorithm is applied to the list of weights sorted so that they appear in decreasing order.
Worst Fit (WF)
The next weight to be packed is placed into the open bin with the largest amount of room remaining. If the weight fits in no open bin, a new bin is opened.
Worst Fit Decreasing (WFD)
The worst fit algorithm is applied to the list of weights sorted so that they appear in decreasing order.
Best Fit (BF)
The next weight to be packed is placed into the open bin that will leave the least room left over after it is placed in that bin. If the weight fits in no open bin, a new bin is opened.
Best Fit Decreasing (BFD)
The best fit algorithm is applied to the list of weights sorted so that they appear in decreasing order.
Try to solve the pipe-cutting problems at the top of this page using the different bin packing algorithms. Study the different answers you get from the various algorithms and see if you can understand why the answers differ. Which of the algorithms appears to be most useful? Why? The chart of pipe lengths will be useful.

#### Pipes and Bubbles: An investigation

Regard this as an educator's guide for teaching the lesson. Use the pages provided but do not distribute descriptions to the learners before time. The activity only works if an element of surprise is retained.
* Four farmers' houses are exactly at the corners of a large, square field that they share.
* They decide to put in a windmill at the very centre of the field to pump water for the farmhouses.
* Naturally they want to use the shortest length of piping possible to reduce their costs.
* The farmers discuss the matter and come to the obvious conclusion:
four pipes leading directly from the pump to each farmhouse will give the shortest total length of piping &shy and therefore the lowest cost.
Do the measurements. Measure to the centres of the circles and the square and write down the total length of piping in ARRANGEMENT A is ............. cm.
Now ...
* ... using wire and pliers make a cube from wire (a cubic frame). The wires at the corners can be twisted together using the pliers. Make a handle so that the frame can be dipped into soap solution.
* A cube is a "block" shape with 6, square faces. In the case of the cube you are making, it is really a framework that is not filled in &shy like a small version of a climbing frame at a playground.
* When you and/or the learners have made a cubic frame, dip it in soap solution. (A good recipe for soap solution: 1 part dishwashing detergent, 2 parts water, ? part glycerine. The glycerine makes the soap bubbles stronger. Try these proportions and make adjustments to ensure that good bubbles can be blown. If you have no glycerin, add more detergent.)
* Each face has a thin film of soap on it. You will also notice that some soapy film forms inside the cube. Shake your cube gently until the soap film settles. Do this a few times until you get the soapy film settling into the same shape of film each time.
* Now look carefully at how the soap film is arranged inside the cube.
* If you wind all the edges of the wire frame with knitting wool you will find that the soap films last longer. Make sure that the wool is well soaked with soap solution.
* The shape of the soap films in the cube gives a clue as to how to arrange the water pipes on the farm? Let learners draw an arrangement of pipes that is different from ARRANGEMENT A and which might solve the farmers' problem.
* Measure the total length of the pipes in ARRANGEMENT B is ............. cm.
Could the soap bubbles have helped the farmers to save piping?
ARRANGEMENT A Measure the total length of the four pipes, each going directly from the windmill(the sqaure in the image) to each farmhouse (the sircles in the image).
ARRANGEMENT B Measure the total length of the four pipes going from each farmhouse to the centre and the piece of pipe laid crossways, through the position of the windwill at the centre of the field.