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Trisect each side of an equilateral triangle of side length 2 and construct an equilateral triangle on the center third of each side. Omit the side of each new triangle that was once a part of the original triangle. Repeat the trisection, construction, and omission steps on each segment formed above, creating the figure below. When this process is repeated indefinitely, a new figure is approached. The square of the area of this new figure can be written as
where p and q are relatively prime positive integers. Find p + q. ps. the figure is a koch flake. pretty sure the answer is 223, but i some conformation. thank you :)
 

 Jun 27, 2024
 #1
avatar+1075 
+1

I think it's just a glitch but the problem is missing some values. 

We need to know that the area of the square can be written as for p and q to solve this problem. 

I think the latex didn't go through. 

 

Can you please just fix the part for the equation? 

I'll try to solve it then!

 

Thanks! ::)

 Jun 27, 2024
 #2
avatar+85 
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Trisect each side of an equilateral triangle of side length 2 and construct an equilateral triangle on the center third of each side. Omit the side of each new triangle that was once a part of the original triangle. Repeat the trisection, construction, and omission steps on each segment formed above, creating the figure below. When this process is repeated indefinitely, a new figure is approached. The square of the area of this new figure can be written as p/q
where p and q are relatively prime positive integers. Find p + q. ps. the figure is a koch flake. pretty sure the answer is 223, but i some conformation. thank you :)

this is the edited version. thank u :)
 

 Jul 2, 2024
 #3
avatar+1075 
+1

This is a tough questipn! It takes some thought to solve! Good one!

Now, let's note something really important first. 

The area of the koch flake can be written in the form

where a_0 was the original area and n was the iteration number. 

The limit of the area in terms of sidelength s is

 

Thus, plugging in 2, we have

\(\frac{8\sqrt3}{5}\)

 

Now, we square it and see what we get. We have

\(\frac{64(3)}{25} = \frac{192}{25}\)

 

So \(p=192; q=25\)

 

Thus, \(p+q = 25+192 = 217\)

Hmmm, I'm not sure if I'm correct. Can you explain your thought process. I think I misunderstood my own logic...

I am pretty sure I'm wrong...ummm, is there a way you can check?

 

Thanks! :)

 

~NTS

 Jul 2, 2024
edited by NotThatSmart  Jul 2, 2024
edited by NotThatSmart  Jul 2, 2024

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