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.
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.
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?
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.
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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.
ReplyDeleteThere 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.
ReplyDeletebleaug
At 22.5, quadrilateral degenerates into a 45-67.5-67.5 isoceles triangle the top of which is at left bottom vertex.
ReplyDeletedamn you're right... 22.5 is indeed a special solution, but there is no other one between 15 and 22.5 apparently.
ReplyDeletebleaug
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?
ReplyDeletepossible.
ReplyDelete22.5 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[.
ReplyDeleteIf 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
Thank you bleaug, I'll update the options accordingly.
ReplyDeleteDo not use all of these Private Money Lender here.They are located in Nigeria, Ghana Turkey, France and Israel.My name is Mrs.Ramirez Cecilia, I am from Philippines. Have you been looking for a loan?Do you need an urgent personal or business loan?contact Fast Legitimate Loan Approval he help me with a loan of $78.000 some days ago after been scammed of $19,000 from a woman claiming to be a loan lender from Nigeria but i thank God today that i got my loan worth $78.000.Feel free to contact the company for a genuine financial call/whats-App Contact Number +918929509036 Email:(fastloanoffer34@gmail.com)
ReplyDelete