Von Koch Curve

4759 /www/nrich/html/content/id/4759/ 0 2011-02-01T00:00:01 <mdoxml version="1.0"> <div>The Von Koch curve is a fractal. The rule for generating this curve is to start with an equilateral triangle and to replace each line segment by a zig-zag curve (a generator) made up of $4$ copies of the line segment it replaces, each reduced to one third of the original length.</div> <div>Click on the red button below to see the first six stages of the infinite process for generating the Von Koch curve.</div> <div><a href="/content/id/4759/vonKochcurve1.swf">Full Screen Version</a></div> <mdo:flash height="465" width="400"><param name="allowfullscreen" value="true" ></param><param name="movie" value="/content/id/4759/vonKochcurve1.swf" ></param><param name="flashplayerversion" value="7" ></param></mdo:flash> The original equilateral triangle has side $1$ unit. Work out the length of this curve in the first few stages and the length of the fractal curve formed when the process goes on indefinitely. Now suppose you made a poster for your classroom with coloured paper by drawing an equilateral triangle for Stage 0 and then cutting smaller equilateral triangles and sticking them on the edge. What is the total area of all the triangles you would stick on one edge if you could continue the process indefinitely to make the Von Koch curve? So what is the area inside the Von Koch curve? Find the dimension of the Von Koch curve using the formula $n=m^d$, where where $n$ is the number of self similar pieces in the generator and $m$ is the magnification factor. The diagrams below show Stages 0 to 5 in the evolution of the Von Koch curve. The Logo program for drawing this fractal is given in the Notes. See <a href="http://nrich.maths.org/8045">First Forward</a> for a ten part series giving an introduction to Logo programming for beginners. <mdo:image alt="stages 3 4 5" height="280" src="vonkoch%203%204%205.JPG" width="740"></mdo:image></mdoxml> <mdoxml version="1.0"> Thank you Shaun from Nottingham High School and Andrei from Tudor Vianu National College, Bucharest, Romania for your solutions. Both Shaun and Andrei used essentially the same method. To solve this problem I made a table with the first $3$ stages and then I determined analytical expressions for the quantities in the table as a function on $n$: <table border="1"> <tbody> <tr> <td style="text-align: center;">Stage</td> <td style="text-align: center;">Number of sides</td> <td style="text-align: center;">Length of curve</td> <td style="text-align: center;">Area contained inside the curve</td> </tr> <tr> <td style="text-align: center;">$0$</td> <td style="text-align: center;">$3$</td> <td style="text-align: center;">$3$</td> <td style="text-align: center;">${\sqrt 3\over 4}$</td> </tr> <tr> <td style="text-align: center;">$1$</td> <td style="text-align: center;">$3\times 4=12$</td> <td style="text-align: center;">$3\times{4\over 3}$</td> <td style="text-align: center;">${\sqrt 3\over 4}\left[1+3\left({1\over 9}\right)\right]$</td> </tr> <tr> <td style="text-align: center;">$2$</td> <td style="text-align: center;">$3\times 4^2=48$</td> <td style="text-align: center;">$3\times\left({4\over 3}\right)^2$</td> <td style="text-align: center;">${\sqrt 3\over 4}\left[1+3\left({1\over 9}\right)+12\left({1\over 9}\right)^2\right]$</td> </tr> </tbody> </table> Now I made the following observations: <ul> <li>The length of each edge at the $(n+1)$th stage will be one third of the length of the edgeat the $n$th stage.</li> <li>The number of edges at the ($n+1)$th stage is $4$ times greater then the number of edges at the $n$th stage.</li> <li>From the last two observations the length of the curve is increased by a factor ${4\over 3}$ at each stage.</li> <li>The area of each additional small triangle included inside the curve at the $(n+1)$th stage will be one ninth of the area of the additional triangles included at the $n$th stage.</li> <li>After stage $2$ the extra area included within the curve at the $(n+1)$th stage is ${4\over 9}$ths of the extra area included at the $n$th stage.</li> </ul> <table border="1"> <tbody> <tr> <td style="text-align: center;">Stage</td> <td style="text-align: center;">Number of edges</td> <td style="text-align: center;">Length</td> <td style="text-align: center;">Area contained inside the curve</td> </tr> <tr> <td style="text-align: center;">$n$</td> <td>$3\times 4^n$</td> <td>$3\times\left({4\over 3}\right)^n$</td> <td> <div>${\sqrt 3\over 4}\left [1+3\left({1\over 9}\right) \left(1 + \left({4\over 9}\right) + \left({4\over 9}\right)^2 + \ldots + \left({4\over 9}\right)^{n-1} \right)\right]$</div> </td> </tr> <tr> <td>$n \to \infty$</td> <td>$\infty$</td> <td>$\infty$</td> <td>${2\sqrt 3\over 5}$</td> </tr> </tbody> </table> To find the area contained inside the fractal curve we need the infinite sum of the geometric series: $$1 + \left({4\over 9}\right) + \left({4\over 9}\right)^2 + \left({4\over 9}\right)^3 + \ldots$$ which is $${1\over \left(1-{4\over 9}\right)} = {9\over 5}.$$ Hence the area inside the fractal curve is ${2\sqrt 3\over 5}$. In order to calculate the dimension $d$ in the formula $n=m^d$, where $n$ is the number of self similar pieces and $m$ is the magnification factor, I see that for this problem $n =4$ and $m = 3$. So, the dimension of the Von Koch Curve is $$d = {\log 4\over \log 3}= \log_3 4 \approx 1.262$$. </mdoxml> <mdoxml version="1.0"> This is a Logo program to draw the Von Koch curve. You can download a free copy of the MSW Logo software from <a href="http://www.softronix.com/logo.html">http://www.softronix.com/logo.html</a> to allvonkoch ;draws 6 stages on same screen cs pu fd 100 lt 90 fd 300 rt 90 pd vonkoch2 200 0 pu rt 90 fd 250 lt 90 pd vonkoch2 200 1 pu rt 90 fd 250 lt 90 pd vonkoch2 200 2 pu bk 300 lt 90 fd 500 rt 90 pd vonkoch2 200 3 pu rt 90 fd 250 lt 90 pd vonkoch2 200 4 pu rt 90 fd 250 lt 90 pd vonkoch2 200 5 end to side :x :y if :y=0 [fd :x stop] side :x/3 :y-1 lt 60 side :x/3 :y-1 rt 120 side :x/3 :y-1 lt 60 side :x/3 :y-1 end to vonkoch1 ;superimpose 6 stages cs pu bk 300 lt 90 fd 200 rt 90 pd vonkoch2 500 0 vonkoch2 500 1 vonkoch2 500 2 vonkoch2 500 3 vonkoch2 500 4 vonkoch2 500 5 end to vonkoch2 :x :y ;draws single curve size :x stage :y at current cursor position repeat 3 [side :x :y rt 120] end to vonkoch3 ;draws sixth stage cs pu bk 300 lt 90 fd 200 rt 90 pd vonkoch2 500 5 end </mdoxml> <mdoxml version="1.0">It helps to consider the area added to ONE edge first. You have to sum an infinite geometrical series. </mdoxml> <mdoxml version="1.0"> At each stage the length of the curve is increased by a factor ${4\over 3}$. So at Stage $n$ the length is $3 \times ({4\over 3})^n$. So the length of the curve keeps increasing without any bounds. The length of the curve is infinite. The area inside the curve is given by the area of the Stage $0$ triangle plus the sum of all the areas of the smaller triangles that we would add on at all the stages. Let us just consider the triangles added to ONE edge and, for simplicity, let's take the area of the triangles added at Stage $1$ to be $A$. (We'll compute $A$ later.) At each stage we add on triangles whose area is ${1\over 9}$th of the area at that stage, and the number of segments is multiplied by $4$ so we add on $1, 4, 16, 64, \ldots$ triangles (remember we are only considering ONE edge). So the sum of the areas added is: $$A + \left({4\over 9}\right)A+ \left({4\over 9}\right)^2A +\left({4\over 9}\right)^3A+ \cdots$$ The sum of this infinite geometric series is $$A{1\over 1-{4\over 9}}= {9A\over 5}$$ Now $A$ is ${1\over 9}$th of the area of the triangle at Stage $0$ so the total area added to all three edges is $$3\times {9A\over 5}$$ which is ${3\over 5}$th of the area of the original triangle. With side $1$ unit the area of the Stage $0$ triangle is ${\sqrt{3} \over 4}$ so the area inside the Von Koch curve is $${2\sqrt 3\over 5}.$$ The magnification factor is $3$ and the generator contains $4$ copies of the line segment it replaces so the dimension is given by $$3^d = 4$$ which gives $d={\log 4 \over \log 3} \approx 1.262$. </mdoxml> 2 5 0 0 0 0 1 Von Koch Curve Make a poster using equilateral triangles with sides 27, 9, 3 and 1 units assembled as stage 3 of the Von Koch fractal. Investigate areas & lengths when you repeat a process infinitely often. 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