A bit of homework here that I have no way of checking so I figured someone here can quickly refute or verify what I found.

 

Imagine an equalateral triangle 1 unit on a side. Inscribed in the triangle in one big circle. Then on each side of the cirle (moving towards each vertex) is another circle inscribed in the area enclosed by the two sides of the trianlgle plus the perimeter of the circle. Then moving towards each vertex again are three smaller cirlces. Then three more and so on for an infinite number of circles. The question is what is the area of these cirlces?

 

I got 13pi/36...I can show work better if this is incorrect but if it happens to be write I'd rather not write it all out now.

 

Thanks.

 

Ah...found a picture...http://nrich.maths.org/content/99/04/15plus3/circles.gif

you need to use an infinite series to find it.

That's a nice little problem; reminds me somewhat of the Koch Snowflake. I might have a go at it in a bit and see if I can confirm your answer.

A bit of homework here that I have no way of checking so I figured someone here can quickly refute or verify what I found.

 

Imagine an equalateral triangle 1 unit on a side. Inscribed in the triangle in one big circle. Then on each side of the cirle (moving towards each vertex) is another circle inscribed in the area enclosed by the two sides of the trianlgle plus the perimeter of the circle. Then moving towards each vertex again are three smaller cirlces. Then three more and so on for an infinite number of circles. The question is what is the area of these cirlces?

 

I got 13pi/36...I can show work better if this is incorrect but if it happens to be write I'd rather not write it all out now.

 

Thanks.

 

Ah...found a picture...http://nrich.maths.org/content/99/04/15plus3/circles.gif

I don't think thats right because your answer would be greater than the area of the triangle wouldn't it? I'm not 100% clear on the probelm but it invovles only inside the triangle right? Am I doing the math wrong or missing something obvious?

[math] \frac{13 \pi}{36} = 1.13446 [/math] and [math] A= \frac{1}{2}(1)(\frac{\sqrt{3}}{2}) = \frac{\sqrt{3}}{4}= 0.433013[/math]

And [math] 1.13446 > 0.433013 [/math]

I get a somewhat different answer of [math]\frac{11\pi}{96}[/math].

I get a somewhat different answer of [math]\frac{11\pi}{96}[/math'].

 

I got the same.

I can post working if it is needed, but you should get to a point where you have to calculate [math]\sum_{n=1}^{\infty} \left( \frac{1}{9} \right)^n[/math]