A non-trivial problem
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by udosuk » Tue May 02, 2006 6:14 pm
tarek wrote:I've got 12.5 units as the longest distance possible.
Starting point: Ground level, half-point between 2 corners of square
End Point: Opposite square .5 unit down the half point between 2 top corners.
Hi tarek, unfortunately you got it wrong on this one (Q4):
√146.25=12.093386622447824478549919845456
And here are the graphics for Q3:
Note √(256-64√3)=8√(4-√3)=12.04776943320180962759310746754
tarek wrote:If shorter means (shorter time), then covering less vertical surfaces & more horizontal surface would be better for the poor ant travelling against the forces of gravity
Truth is this whole set-up is inside a space shuttle out of this planet... So gravity isn't really an issue...
And it's every possibility that the brick is not touching the "ground" at all...
No evidence that this is the "first ever astronaut ant" though...
But still I'm thinking about Q4's solution... At the moment I think it's same as Q2...
- udosuk
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by SimulatedAnnealer » Tue May 02, 2006 6:54 pm
For Q4, I think the second point is 6/7 down from the mid-point of the top edge of the far square. (first point is mid-point of bottom edge of near square as stated).
And distance is 2*SQRT(1885)/7 = 12.404739
(One point must be the mid-point of the bottom edge of a square whilst the second must be somewhere on the vertical bisector of the other square. The two possible minimum distance routes are 1) directly via a side rectangular face and 2) via a side rectangular face and then the top face.)
And distance is 2*SQRT(1885)/7 = 12.404739
(One point must be the mid-point of the bottom edge of a square whilst the second must be somewhere on the vertical bisector of the other square. The two possible minimum distance routes are 1) directly via a side rectangular face and 2) via a side rectangular face and then the top face.)
- SimulatedAnnealer
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37 posts
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