## Friday, May 14, 2010

### Fox 287

We may have a single solution, but not sure if that's the whole story.

1. I looked over this for about 10 minutes and I believe that the range is from 0 degrees to 22.5 degrees (not including 22.5 degrees). We can see that if it was just 0 degrees, it would be the square itself. I'll show a proof of this later though. Like I said, I only looked at it for about 10 minutes.

2. There is a solution for any angle between 0 and 15°. At max angle, quadrilateral degenerates into a 30-75-75 isoceles triangle the top of which is at right bottom vertex.

bleaug

3. At 22.5, quadrilateral degenerates into a 45-67.5-67.5 isoceles triangle the top of which is at left bottom vertex.

4. damn you're right... 22.5 is indeed a special solution, but there is no other one between 15 and 22.5 apparently.

bleaug

5. I knew that 15 was a solution, so seems 0. Then, intuitively any angle between 0 and 15 should make the construction. I have some doubts about 22.5. Are we sure that 22.5 is possible in a "square"? Or does it demand a rectangle?

6. possible.

7. 22.5 is a solution if we accept a degenerate form to be a solution (it is inscribed in a square). However this solution cannot be approached by a sequence of solutions with angle in ]15, 22.5[.

If we allow the figure to become a rectangle, we can find an infinite number of WxL rectangles for any angle between 0 and 22.5. W=L only if angle is <=15 or = 22.5.

bleaug

8. Thank you bleaug, I'll update the options accordingly.

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