Activity 1: Mystic Rose In the picture there are 16 points on the circumference of the circle. The pattern for the number of lines compared to the number of point is related to the triangular numbers (1, 2, 6, 10, 15). The relation is such that if there are n points on the circumference of the circle there are  lines.
Since there are 16 points on the circumference there will be 8 x 15 or 120
lines in the pattern.
Holes will be present at the centre of the circles if the number of points on the circumference is an even number.
Change the numbers on one of the dice to 0, 0, 0, 6, 6, 6.
If the first two numbers are a and b, then the sequence of numbers are:
A simple way to multiply a 2-digit number by 11 is to add the two digits
together and place this number between the digits. If the sum is 10 or
greater, place the units digit of the sum between the two digits and add 1
to the tens digit of the number being multiplied by 11.
Activity 4: Seven Exponential 1999 
There were 18 girls and 18 boys when the bus set out. Heather and Hamish worked on their homework for 75 minutes.
1. If the jigsaw is square and has 2500 pieces, then it is 50 pieces across by 50 pieces down. It will have 4 pieces with two straight edges (corners), 192 pieces with 1 straight edges (sides) and 2304 pieces with no straight edges.
2. A square puzzle of n² pieces would need 4 corners, 4(n 2) side pieces and (n 2)² inside pieces.
 = (n2 4n + 4) + (4n 8) + 4 = n2
3. For every A-piece you also need a corresponding B-piece to interlock with it.
4. If you use only C-corner pieces, then there must be an even number of pieces along each edge.
1. If the jigsaw is square and has 625 pieces, then it is 25 pieces across by 25 pieces down. It will have 4 pieces with two straight edges (corners), 92 pieces with 1 straight edges (sides) and 529 pieces with no straight edges. 2. A square puzzle of n² pieces would need 4 corners, 4(n 2) side pieces and (n 2)² inside pieces (as in 2. of Jigsaw Puzzle One).
3. For every A-piece you need a corresponding B-piece to interlock with it. 4. As there is an even number of pieces (50 pieces) along each edge of a 2500- piece square jigsaw it would need 4 C-corner pieces. From Jigsaw One we know that it needs 192 type A or B pieces, and we know that it needs an equal number of each, so there must be 96 A-pieces and 96 B-pieces. It will also need 2304 centre pieces.
The shoelace is approximately 1175mm long.
The shoelace needs to have 55mm of lace to join the bottom pair of lace holes, 2 times 60mm for each subsequent pair of lace holes and 2 times 200 mm to allow for tying a bow. If n is the number of pairs of lace holes, then the length of lace needed in mm is:
55 + 120(n-1) + 400 = 55 + 120n 120 + 400 = 120n +335
The table could be like this:
 Students can graph the table. The formula would be something like that given in 2. above. The table would probably be the most useful as it is simple to read and requires no calculations.
The number of unshaded triangles at each stage (n) is 3n-1. At stage 4 there are 27, at stage 5 there are 81, and at stage 10 there are 39 or 19683 unshaded triangles.
If Kathy decides to purchase something less than $25 or more than $50 she should choose option 2, ie 10% off the price. If the price is between $25 and $50, Kathy should select option 1, and have $5 off the purchase price. 1. The table would look like this:
3. P = 15 x 4
4. The formula would not correctly describe the insect population after 50 years because eventually the number of insects would reach a threshold limited by availability of essentials such as food. This would occur well before 50 years.
A square is the most efficient shape in that it uses fewer tacks than a rectangle. A square 10 paper squares long by 10 paper squares wide would need 11x11 or 121 tacks.
 2. If the fish has 139 dots on its perimeter it is at stage 20. 7n 1 =139 x n = 20. 3. 7n 1
This investigation is mathematically very rich and can be explored at a variety of levels. What is interesting about the rectangles is that the length of the unknown line is independent of the length of the sides of the rectangle. It is related to the lengths of the other lines joining the corners of the rectangle to P by the relation PA² + PC² = PB² + PD² . Students can try to show, using Pythagoras, that their measurements are correct. A greater challenge is to show algebraically why this is true for all rectangles. Only very capable senior students would be able to do this. The table below gives the values for the lengths of the sides given in the investigation:
 
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It works because whatever the two numbers are, the sum of the lines will be
11 times the number in the 7th line.