• Join over 1.2 million students every month
  • Accelerate your learning by 29%
  • Unlimited access from just £6.99 per month
  1. 1
  2. 2
  3. 3
  4. 4
  5. 5
  6. 6
  7. 7
  8. 8
  9. 9
  10. 10
  11. 11
  12. 12
  13. 13
  14. 14
  15. 15
  16. 16
  17. 17
  18. 18
  19. 19
  20. 20

Koch Snowflake

Extracts from this document...









    The Koch Snowflake was created by the Swedish mathematician Niels Fabian Helge von Koch. In his 1904 paper entitled "Sur une courbe continue sans tangente, obtenue par une construction géométrique élémentaire" he used the Koch Snowflake to show that it is possible to have figures that are continuous everywhere but differentiable nowhere.

    In order to reach the Koch snowflake, we need to start with a Koch curve. A Koch curve is a straight line which is divided into three parts and an equilateral triangle is constructed keeping the base as the middle part of the divided line segment.


Finally, the base is removed which gives us the first iteration of the Koch curve (as shown in the figure above).

      From the Koch curve, comes the Koch snowflake. Now, instead of one line it starts with an equilateral triangle. The steps in Koch curve are applied now to each side of the equilateral triangle, which ultimately gives the shape of a “snowflake”.


       STAGE 0                       STAGE 1             STAGE 2              STAGE 3



Let Nn = Number of sides, Ln = the length of a single side, Pn = the length of the perimeter, An = the area of the snowflake; at the nth stage.

In this portfolio, the sign ‘*’ indicates multiplication whereas the sign ‘/’ indicates division.


...read more.


3 = (1/3)3 = 1/27 = 0.037037

Thus, the generalizations for Ln made above apply consistently to sets of values produced in the table and it also satisfies the graph above.

Perimeter of the Koch Snowflake

Iteration (n)

Perimeter (Pn)









To find the perimeter of any polygon, the easiest and the most convenient method is to multiply the number of the sides of the given polygon with the length of one side as the snowflake is a sort of a regular polygon. The same is the case of the koch snowflake. If we multiply the number of sides (Nn) with the length of one side (Ln), we would obtain the perimeter of the snowflake at every stage. With the help of this concept, we arrive at deriving a general formula to find the perimeter which is –


= 3*4n*(1/3)n

= 3*(4/3)n

Now here, we can again observe that the perimeter can also be found out using the method of geometric progression. The perimeter of the koch snowflake, like its number of sides (Nn) and the the length if one side (Ln), follows a geometric pattern which can be written in the following way –

Pn = P0*Rn

       = 3*(4/3)n

Since, P0 = 3 and R = 4/3 (as can be seen from the table above)

Verification –

For n = 0

P0 = 3*(4/3)0 = 3*1 = 3

For n = 1

P1 = 3*(4/3)1 = 4*1 = 4

For n = 2

P2 = 3*(4/3)2 = 16/3 = 5.333333

For n = 3

P3 = 3*(4/3)3 = 64/9 = 7.111111

Thus, the generalizations for Pn

...read more.


 = Ak + Uk+1


= Ak + Uk+1

From equation 1, we get

= Ak + (3(3/2)-2(n+1)*4(n+1)-2)

= Ak + (3(3/2)-2n-2*4n-1) = LHS

Thus, I have proved my general formula for the area of the Koch snowflake through induction.


Now that I have finished this extensively researched portfolio, I have actually got inspired to know more about this project of mine and I am sure that you will also, after reading my portfolio, get the surge to know more about the Koch snowflake. After finishing this portfolio I was so inspired that I searched on the internet for figures similar to the Koch Snowflake and have found several figures such as the Koch Antisnowflake, the Exterior Snowflake, the Pentaflake and the Sierpinski Curve. Finally, I hope that you like my portfolio and enjoy reading it.


Stages in Koch Antisnowflake


Some beautiful tillings can be made with iterations towards the Koch Snowflake.


Some beautiful modifications of the Koch Snowflake involving the constituent triangles with filled-in triangles inscribed, rotated at some angles.


An extraordinary Koch Snowflake created in LOGO.


Koch Snowflake – Silbury Hill, England

A Beauty Indeed!


http://www.nonoscience.info/2006/08/28/the-koch-curve-and-visual-resolution - 12/7/2007

http://www.mathworld.wolfram.com/KochSnowflake.html - 12/7/2007

http://ecademy.agnesscott.edu/~lriddle/ifs/ksnow/ksnow.htm - 20/7/2007

http://www.math.ubc.ca/~cass/courses/m308-05b/projects/fung/page.html - 20/7/2007

http://math.arizona.edu/~caine/chaos.html - 25/7/2007

http://mathworld.wolfram.com/KochAntisnowflake.html - 30/7/2007

http://jeffreymunro.com/amazingpics/index_old.php?showimage=318 - 1/8/2007

http://www.seasip.demon.co.uk/Unix/Joyce/screens.html - 1/8/2007

...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. Extended Essay- Math

    m-��2��� ��&T'$�IMJæd2g���i��:�0�/:|����xeo�\ib�l...G�X��(c)og�jq� �}�wkh1/2���<Þ¢q���Õk����;;�"o߬1/4E�"��k��"g�'��<X}6�~����1�'�Æ��{(r)��p��+��o^1/2�y->=2#�!i��1/4��"O= [_��\t_r_����Mh���{�j���kY?�?Ϭ֣×on"o�m�m��|�%�E�j�Z�V�N�3/43/4c��@%Å�@��������� �<3/41/21/2^1/21/21/2Y��7�����+v���(tm)�Ét'���7���� ��s�c�b pHYs �� IDATx� "\�y�"��3/4�ޭV"��...6�AH�%�����-1`�D`�>(tm)�I� 1c"�(tm)�8 ��1����x��6F� YBR*���1/2"��U�"�x*����Z��#ݺ���{�է{��-�\.g���h4V�� �M'�f �4;2 �H t()J�'w"� ����rd$P:"�c�;'�v PRhwn92(J�ұ�[email protected]"()�;� "�%E�X�N$ ]"Ú[��JG�'�t,y'�.J ��-GF�#@IQ:-1/4 h-%...v�-##�� �(KÞ�K�'B"s�''@�PR"�%�D�%@I�ݹ��H t()J�'w"� ����rd$P:"�c�;'�v PRhwn92(��F�~���{������Wn�t�F�;' "[email protected]�$��7-m� B.�1/4F!?�-�(��xw ����c`�{s�%k2笶�`�8=��~��\ï¿½ï¿½Û "'@...$��>�Å)�~1 yaa�fM����' "qp1/45 �@�THR�\�j�d H f2�� d����!Z]��aï¿½ï¿½ï¿½Û "�@...$�w�iʤLb� f�w�d6�9�)g��=u��K$P"'���?��%2�29��"�LØX�{�j_|�?-h81/4 �@YTNR �]���֡~S2f�e Èl"�Â3[��(���$P:��(tm){?k�fpL�3�s&!g2�Æ�_��O��D$Pb�-�3/4���"����!8�?�J<2ÞH t� )�~í¦TBÈ&1�1CO�� V��.� x' �����Ì�!#B-iÌ�=�Ù�X��"��xp1/4 �@�TGR<~��`�A�yUE� �t�Y�A�[email protected]%&PI�i�j!...H�b3�SS�hJÛ¸)���v$P*�'�1/2�(tm)�i!��yE~0&!�t� �T���@i

  2. Math IA - Logan's Logo

    This can, in turn, be attributed to the large uncertainty in obtaining the actual data. Because I was only able to read points off of the graph, smallest of 1 unit, my precision was limited to one decimal place, and the uncertainty was quite large (�0.5 units).

  1. The Koch Snowflake

    / ()) = 60� = 1/2 = tan 60 () x = Area of shaded small triangle = x 2 = Total area of Stage 2 = + = = Area of Stage 3 = Area of Stage 2 + Area of 48 small shaded triangles Height= cos -1 (()

  2. MAths HL portfolio Type 1 - Koch snowflake

    to be 1, the following table shows the values for the number of sides, length for each side, perimeter and area for the stages 0 to 3. Table 1 Nn Number of sides Ln Length of each side Pn Perimeter of the shape An Area of the shape n (stage)

  1. El copo de nieve de Koch

    La siguiente tabla muestra un resumen de los resultados de el n�mero de lados, la longitud del lado, el per�metro y �rea del copo de nieve para =0, 1 , 2 y 3. Tabla 1. Valores para el n�mero de lados, longitud de un lado, per�metro y �rea para n=0,

  2. Investigating the Koch Snowflake

    is equal to (V3)/2, which was found from the fact that the angels of the equilateral triangle are 60o, and using a segment which bisected the base and was normal to it. So using known famous triangles which the angles 30 o, 60 o, and 90o ** Each graph of

  1. Math Portfolio - The Koch snowflake investigation.

    There is a pattern present, and from that I observed that the number of sides increases by a factor 4 of the previous stage. n Nn 0 3 1 N0 ? 4 = 3 ? 4 = 12 2 N1 ?

  2. Maths Investigation: Pascals Triangles

    In this case, the height of the triangle is four rows. Explanation: Perhaps the triangular patterns are always formed on a base and height of (n-1) in a symmetrical pattern, like two sides of an equilateral triangle. Part 4: Pascal Petals 1.

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