## Tuesday, June 30, 2009

### Fox 70

This is a simple fractal. See original here: 1. 2. http://i36.tinypic.com/5bkle.gif

3. -8

4. I need someone's help. I thought I had the right answer but I'm wrong. If Y was the height of the triangle, and X was the width of the square at the bottom; wouldn't the perimeter be the same? 2X + 2Y should be the same as the total perimeter of the blue object right?

Well I got X to be 0.464 because the total height of the Triangle (Y) should equal X + the height of the smaller triangle at the top.

And I got Y to be 0.866 because 0.5 * tan(60 degrees) = sqrt(3)/2 or .866.

The final answer I got was 2.66.. Can someone show me where I might have screwed up?

5. I also got a different answer, but it's not on there. It was close though, did I just miss a trick?

http://i45.tinypic.com/24g6p9u.jpg

Interestingly, if you don't count the bottommost side (s1) as part of the shape, the answer will simplify to newzad's answer.

6. Yep, we're back!!
And I think we will need a correction for this one. We will post a corrected version soon.

In the last post: http://i45.tinypic.com/24g6p9u.jpg

2 (S_1 + h) is the answer (why?)

8foxes.com

7. Because the perimeter of the rectangle with sides S_1 and h, is the same as the perimeter of the blue object. Yes, the area is different; but the perimeter is the same.

8. Corrected! Thank you for the collective work!