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2. Teachers Notes This is a brief but very interesting look, at the Von Koch Snowflake Curve. After introducing the curve and discussing its generation, the students are simply asked to derive the perimeter formula for nth iteration. We then move on to discuss the curve’s finite area and reveal, (by reference to the formula) its infinite perimeter. Students are encouraged to generate a spreadsheet from the formula for the first 50 terms in the sequence to convince themselves of the infinity of the perimeter. (Spreadsheet is at slide 16). There is a printable worksheet if needed at slide 18 (some students may wish to make jottings/notes on it). If you want to extend still further for the very able, then you might wish to see if they can work out the area A n for the nth iteration.

3. The Von Koch Snowflake

8. 1 3 L 1 3 L 1 3 L The curve is generated from an equilateral triangle by trisecting the sides and constructing this smaller equilateral triangle on each of the sides. This is then repeated ad infinitum. P 0 = L The Von Koch Snowflake

9. Thinking about the increased length of this side, what will the first new perimeter, P 1 be? 1 3 L 1 3 L 1 3 L P 0 = L P 1 = 4 3 L The Von Koch Snowflake

10. 1 3 L 1 3 L 1 3 L Derive a general formula for the perimeter of the n th curve in this sequence, P n. P 1 = 4 3 L P 0 = L P 2 =( ) 2 4 3 L The Von Koch Snowflake

11. 1 3 1 3 1 3 Derive a general formula for the perimeter of the n th curve in this sequence, P n. P 1 = 4 3 L P 0 = L P 2 =( ) 2 4 3 L P 3 =( ) 3 4 3 L P n =( ) n 4 3 L The Von Koch Snowflake

12. The area A n of the nth curve is finite. This can be seen by constructing the circumscribed circle about the original triangle as shown. P 1 = 4 3 L P 0 = L P 2 =( ) 2 4 3 L P 3 =( ) 3 4 3 L A0A0 A1A1 A2A2 A3A3

13. It is a surprising fact therefore that the perimeter of the curve is infinite. A0A0 A1A1 A2A2 A3A3 P 1 = 4 3 L P 0 = L P 2 =( ) 2 4 3 L P 3 =( ) 3 4 3 L P n =( ) n 4 3 L

14. Whatever fixed value you care to make the perimeter of any curve in the sequence it can always be exceeded by choosing a large enough value for n. A0A0 A1A1 A2A2 A3A3 P 1 = 4 3 L P 0 = L P 2 =( ) 2 4 3 L P 3 =( ) 3 4 3 L P n =( ) n 4 3 L

15. Use a spreadsheet to compute the first 50 values for the perimeter. Set P 0 = 1. A0A0 A1A1 A2A2 A3A3 P 1 = 4 3 L P 0 = L P 2 =( ) 2 4 3 L P 3 =( ) 3 4 3 L P n =( ) n 4 3 L

16. P01 P11.333P261771.769 P21.778P272362.359 P32.370P283149.812 P43.160P294199.749 P54.214P305599.666 P65.619P317466.221 P77.492P329954.961 P89.989P3313273.282 P913.318P3417697.709 P1017.758P3523596.945 P1123.677P3631462.593 P1231.569P3741950.125 P1342.092P3855933.499 P1456.123P3974577.999 P1574.831P4099437.332 P1699.775P41132583.110 P17133.033P42176777.480 P18177.377P43235703.306 P19236.503P44314271.075 P20315.337P45419028.100 P21420.449P46558704.133 P22560.599P47744938.844 P23747.465P48993251.792 P24996.620P491324335.722 P251328.827P501765780.963

17. The Von Koch Snowflake The perimeter of the Von Koch Snowflake Curve is infinite. Just as the coast line of the UK is infinite. The smaller the ruler that you use to measure the coast line, the longer it becomes. Coast line  11 000 miles