Koch Snowflake Page
Apr 27th, 2011 by corricelli
Please use this page to post questions related to the Koch Snowflake.
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Mrs. Corricelli
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So Callum and I weren’t really sure about the snowflake question. Originally we found that the area of the first equilateral was √3/4 and then used this as G1 and used the infinite sum formula to get (3√3)/8. We also had the rate as being 1/3, but towards the end of class were leaning to it being 2/9. Any other theories?
Nick- My partner and I started out by finding the area too, and we got the same thing too, so thats good. But what you did sounds right now that i am thinking about it…
is this right
I don’t know my partner and I thought we should find the area of the second snowflake and the third, and then try and see if there was a ratio between them. But, we weren’t really sure, and we had some weird numbers.
so we are kind of stuck
I am not reading anything that sounds crazy… You are all on the right track. Everyone sounded good in class.
Your g1 is also correct at sqrt(3)/4.
Be careful about the 1/3 = r assumption; even though the side length is 1/3 the length of the prior triangle, this DOES not mean the area is 1/3 less. If you drove that forward the resulting area would be less than the original snowflake, which cannot be correct.
I do not want to post anything more – keep going with this, teams 2 and 7! The hardest part is getting r.
Enjoy the snow,
Mrs. Corricelli
Ok so I found g1 to also be sqrt3/4. I then found the area of g2 to be sqrt3/3. g2/g1= 4/3. I think r= 4/3. I’m about to find the area of g3 just to make sure that this is correct. Anyone else get this?
Wait actually…if r= 4/3, then 1-r would be negative for the infinite sum formula, and you can’t have the snowflake’s area converging at a negative, so this is probably wrong. Oh well.
Since the step between the first triangle and the set of three is the only part where three triangles are added, I think maybe we should set that part aside and focus on finding r starting with the three triangles and the following sets of two. From then on, only two triangles per pre-existing triangle are added. Just a thought.
yeah that is a good thought callum.
I found the first 3 areas g1=sqrt(3)/4 g2=1/sqrt(3) and g3=10/9sqrt(3)
Then i divided each one by the prior term to see if there was a constant rate and I found that from g1 to g2 it was 1 1/3 and from g2 to g3 it was 3 1/3. I’m trying to think of a way we could use n in the rate in order to account for this but I’m not sure that makes any sense at all…
Haley: for g2, I didn’t get sqrt3/3. I got the area of one new triangle to be sqrt3/72, the total area of the three small triangles to be sqrt3/24. The common ratio there is 1/6, unless I did something wrong.
Nick: Weird thing is, the ratio between g3 and g2 using the method I used was 2/9. I’m starting to see a pattern, but I need to see it first.
(Disregard my previous comments, I found mistakes in them)
Aside from g2 being 3 times as many triangles as g1, I got 12 triangles for g3 and 48 triangles for g4. You can see that 4 new line segments come up for every new triangle made after g2, making 4 times as much triangles as the last.
hey guys! I don’t know if its too late to be posting, but callum david and I thought we successfully found r in class by using the area of the second triangle minus that of the first. the first triangles area we found using half the base times the height and found it to be root 3 over 4. For the second are the are of the three little triangles can be found using absinC so (1/3 x 1/3 sin60degrees) which came out to be a long number. but if we take that multiply it by three since there are three small triangles connected to the original, so those three added to the original, minues the original will give us r.
That would be for an arithmetic sequence though.
I tried doing it and the rate of change between g1 and g2 v g2 and g3 is different. I think the sequence is either geometric or maybe exponential, but I’m not sure how to work that out.
Yes, Becca thats arithmetic and I think this problem is supposed to be geometric.
Meg, i agree with you. They aren’t the same if you do the ratios of the first to the second and then the second to the third. So, i don’t understand how to get r if they aren’t the same. I have been looking at this for a while and am not getting anywhere. :/ anybody else know how to get the r we need?? thanks!
I found some ratios!
We’ll the number of “layers” (I don’t know what else to call it) k.
G1=sqrt3/4. If you take the area of only ONE triangle on the next layer, you get sqrt3/36. Then the area of one triangle on the 3rd layer is sqrt3/324.
(sqrt3/36)/(sqrt3/4)=1/9
(sqrt3/324)/sqrt3/36)= 1/9
ratio of one triangle of every layer=1/9
ak=sqrt3/4(1/9)^(k-1)
Going from my previous comment, the number of triangles formed in each layer is 4 times more than the previous, except for 3 in the second layer. So let n= number of triangles,
gn=3(4)^n-1
Is there a way to combine them both so you have the area of all the triangles?
I am also unsuccessful finding r. I found the area of the first triangle to be sqrt(3)/4. For the second triangle, or g2, I got ((1 + 6sqrt(3))/ 24). For the area of the third triangle, or g3, I got a reallly huge number…like VERY HUGE! It made sense though, because each time you find the area of the next triangle, you are adding the previous one’s area and finding the smaller pieces (area of the smallest triangles) and adding them. I divided g3 by g2 to see if the common ratio would be the same as g2 divided by g1…but it isn’t….so I’m still confused…
For finding the Infinite geometric sum (since the process is indefinte), we need to know g1, and r. g1 is something we all landed upon….it seems reallly tricky to find r though!
This is as far as I got on the Koch snowflake last night.
I ended up trying to find the area a triangle being added to the figures, including the area of an additional triangle if there were a figure 4. The area of one the triangles being added in figures 2 and 3 are sqrt3/36 and sqrt3/324. I used 1/9 as r to assume that the area of a triangle added in a figure 4 would be sqrt3/2916. The amount of triangles added (which can be represented by gn=3(4)^n-1 as Mariah said) multiplied by the area of one triangle is the area of all triangles added to a figure. I was able to get that:
area of the exacting snowflake= sqrt3/4 + [3(4)^n-1][(rt3/4)(1/9)^n]; where n approaches infinity, is the proposed equation. I don’t know if this is right and I didn’t know where to go after this when trying to get the exact area.