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On Thu, 6 Apr 2006 14:27:28 +0000 (UTC), magidin@xxxxxxxxxxxxxxxxx
(Arturo Magidin) wrote:
>In article <1ch832t9ml04qhkh6votm666mrb29p4657@xxxxxxx>,
>quasi <quasi@xxxxxxxx> wrote:
>
>I did make one mistake, though it was inconsecuential.
>
>>On Wed, 5 Apr 2006 22:22:17 +0000 (UTC), magidin@xxxxxxxxxxxxxxxxx
>>(Arturo Magidin) wrote:
>>
>>>In article <4ue832p724s7nv77ofo6p6mlia6ucrd38n@xxxxxxx>,
>>>quasi <quasi@xxxxxxxx> wrote:
>>>>What is the minimum volume of a parallelepiped such that the distance
>>>>between any pair of vertices is at least 1?
>>>
>>>There is no minimum, and the infimum is zero.
>>>
>>>Consider the family of parallelograms with one vertex on (0,0), one
>>>vertex on (2,0), one vertex at (cos(t),sin(t)) with 0<t<=pi/2, and the
>>>upper right vertex at (cos(t)+2,sin(t)).
>>>
>>>The distance between any two vertices is always at least 1: the only
>>>distance that decreases as t ranges from pi/2 to 0 is the distance
>>>between (cos(t),sin(t)) and (2,0), and that distance is
>>>sqrt(2)*sqrt(cos(t)+1)>= sqrt(2)>1.
>
>This is a mess. The distance is sqrt(5-4cos(t)). Still larger than 1
>for any t>0.
I didn't catch that error since the geometric view of your
construction makes it clear without calculation that all distances are
at least 1.
For example, it's immediate that the distance from (2,0) to the line
x=1 is 1, so the distance from (2,0) to any point in the xy-plane to
the left of the line x=1 must exceed 1, which takes care of the
distance from (2,0) to (cos(t),sin(t)) in your construction.
quasi
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