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Topic: sweeping a square (Read 719 times) |
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JocK
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sweeping a square
« on: Dec 5th, 2004, 5:41am » |
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What is the largest area that can be rotated freely inside a unit square, and by rotating and moving inside the square can sweep it completely?
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| « Last Edit: Dec 6th, 2004, 1:41pm by JocK » |
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solving abstract problems is like sex: it may occasionally have some practical use, but that is not why we do it.
xy - y = x5 - y4 - y3 = 20; x>0, y>0.
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towr
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Re: sweeping a square
« Reply #1 on: Dec 5th, 2004, 8:30am » |
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an equilatetal triangle with a side equal to the squares side should be a good first guess I think, one can do better though..
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| « Last Edit: Dec 5th, 2004, 8:33am by towr » |
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Wikipedia, Google, Mathworld, Integer sequence DB
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rmsgrey
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Re: sweeping a square
« Reply #2 on: Dec 5th, 2004, 11:01am » |
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The "triangle" of constant diameter formed by drawing an arc centered at each corner, connecting the other two corners will work - just ask anyone who's ever drilled a square hole... Not sure whether it's maximal though.
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JocK
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Re: sweeping a square
« Reply #3 on: Dec 5th, 2004, 11:16am » |
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on Dec 5th, 2004, 11:01am, rmsgrey wrote:::
The "triangle" of constant diameter formed by drawing an arc centered at each corner, connecting the other two corners will work - just ask anyone who's ever drilled a square hole... Not sure whether it's maximal though.
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Nope, a so-called Reuleaux triangle doesn't do the job: it sweeps only a fraction [pi]/6 + 2[sqrt]3 - 3 [approx] 98.77% of the square.
Maybe engineers can drill a square hole, mathematicians can't.. 
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solving abstract problems is like sex: it may occasionally have some practical use, but that is not why we do it.
xy - y = x5 - y4 - y3 = 20; x>0, y>0.
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rmsgrey
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Re: sweeping a square
« Reply #4 on: Dec 5th, 2004, 11:38am » |
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Of course it doesn't - the corners are too wide - 120 degrees.
You should be able to do something similar with circular arcs on the sides of an equilateral triangle so that the corners are only 90 degrees - the radii being at 75 degrees to the side at each corner. Since that's smaller than the Reuleaux triangle, it will still be able to turn within the square, and will be able to get right to the corners.
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Barukh
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First of all, I assume that the shape can do other movements inside the square, not just rotate around a single point.
The equilateral triangle proposed by towr has an area [sqrt]3/4 = 0.433… This is less than the area of a square with a unit diagonal (0.5).
I beleive the attached shape satisfies the conditions of the problem, its area is [pi]/3 – [sqrt]3/4 = 0.614…
But - having in mind Jock's keenness - I don't believe it's optimal. 
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JocK
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Re: sweeping a square
« Reply #6 on: Dec 6th, 2004, 1:36pm » |
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on Dec 6th, 2004, 1:28am, Barukh wrote:| First of all, I assume that the shape can do other movements inside the square, not just rotate around a single point. |
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Correct. Thanks. I have updated the text of the riddle slightly so as to stress this point.
on Dec 6th, 2004, 1:28am, Barukh wrote:But - having in mind Jock's keenness - I don't believe it's optimal.
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Your intuition is correct... ... one certainly can do better than removing a meniscus-shaped portion of the Reuleaux triangle leaving an area of 0.614... Good guess though!
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| « Last Edit: Dec 6th, 2004, 3:27pm by JocK » |
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solving abstract problems is like sex: it may occasionally have some practical use, but that is not why we do it.
xy - y = x5 - y4 - y3 = 20; x>0, y>0.
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SWF
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Re: sweeping a square
« Reply #7 on: Dec 6th, 2004, 9:43pm » |
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Just a semi-circle of diameter 1 with a (sqrt(2)/2, sqrt(2)/2, 1) right triangle in place of the other half of the circle has area 0.6427, and looks like there is still extra area that can be added to the curved portion of this shape.
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Barukh
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Of course, SWF, you are right! I considered your example but somehow miscalculated its area.
Applying your idea, here's another shape (attached). All the indicated arcs have unit radii. The area calculations are somewhat messy, I get 0.6866... (will elaborate upon request).
What would you say, Jock?
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JocK
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Re: sweeping a square
« Reply #9 on: Dec 7th, 2004, 10:40am » |
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on Dec 7th, 2004, 1:30am, Barukh wrote:| What would you say, Jock? |
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I am impressed..!
The question now is: is this the maximum area?
(I think it is... but maybe some genius here can find an even larger shape?)
And Barukh.... I guess you know what the follow-up question will be?
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solving abstract problems is like sex: it may occasionally have some practical use, but that is not why we do it.
xy - y = x5 - y4 - y3 = 20; x>0, y>0.
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