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        riddles >> hard >> sweeping a square
        
(Message started by: JocK on Dec 5^th, 2004, 5:41am)

Title: sweeping a square
Post by JocK on Dec 5^th, 2004, 5:41am

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?

Title: Re: sweeping a square
Post by towr on Dec 5^th, 2004, 8:30am

an equilatetal triangle with a side equal to the squares side should be a good first guess I think, one can do better though..

Title: Re: sweeping a square
Post by rmsgrey on Dec 5^th, 2004, 11:01am

::[hide]
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.
[/hide]::

Title: Re: sweeping a square
Post by JocK on Dec 5^th, 2004, 11:16am

on 12/05/04 at 11:01:58, rmsgrey wrote:

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.. ;)

Title: Re: sweeping a square
Post by rmsgrey on Dec 5^th, 2004, 11:38am

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.

Title: Re: sweeping a square
Post by Barukh on Dec 6^th, 2004, 1:28am

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. ;D

Title: Re: sweeping a square
Post by JocK on Dec 6^th, 2004, 1:36pm

on 12/06/04 at 01:28:02, Barukh wrote:

First of all, I assume that the shape can do other movements inside the square, not just rotate around a single point.

Correct. Thanks. I have updated the text of the riddle slightly so as to stress this point.

on 12/06/04 at 01:28:02, Barukh wrote:

But - having in mind Jock's keenness - I don't believe it's optimal. ;D

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!

Title: Re: sweeping a square
Post by SWF on Dec 6^th, 2004, 9:43pm

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.

Title: Re: sweeping a square
Post by Barukh on Dec 7^th, 2004, 1:30am

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? ;)

Title: Re: sweeping a square
Post by JocK on Dec 7^th, 2004, 10:40am

on 12/07/04 at 01:30:43, Barukh wrote:

What would you say, Jock? ;)

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? ;)