| T.R | Title | User | Personal
Name | Date | Lines |
|---|
| 1142.1 | needle in a haystack | AKQJ10::YARBROUGH | I prefer Pi | Thu Nov 02 1989 16:41 | 20 |
| >For example to make it more concrete, take a fixed perimeter of 24 inches;
>find the shape and size of a piece of paper that will cover all possible
>figures that can be constructed with that fixed perimeter.
You only need consider convex shapes, since concave shapes have as convex
hull a larger figure with a shorter perimeter.
Seems like a semiellipse of minor axis 12 and major axis of 8*sqrt(3) ~ 13.86
( = twice the altitude of an equilateral triangle of perimeter 24), halved
along the minor axis, will do. That will certainly cover all the triangles
of perimeter 24, and a circle of diameter 24/pi ~ 7.64 will certainly fit;
and those dimensions are needed to enclose the degenerate triangle and the
equilateral triangle. Anything else should be somewhere between the circle
(infinitely many sides) and the degenerate triangle (two sides).
There are probably symmetry arguments to the effect that one quadrant of
the ellipse can be reduced somewhat. I'm too tired to dig that far right
now.
Lynn Yarbrough
|
| 1142.2 | please clarify dimensions | TROU10::RITCHE | From the desk of Allen Ritche... | Sun Nov 05 1989 22:08 | 15 |
| >Seems like a semiellipse of minor axis 12 and major axis of 8*sqrt(3) ~ 13.86
>( = twice the altitude of an equilateral triangle of perimeter 24), halved
>along the minor axis, will do. That will certainly cover all the triangles
>of perimeter 24, and a circle of diameter 24/pi ~ 7.64 will certainly fit;
Lynn,
Could you please clarify this shape for me. Is this a halved ellipse or halved
semiellipse? What are the lengths (b and a) of the semi-minor and semi-major
axes of original ellipse?
With my interpretation of the above, I'm having trouble squeezing in a 7.64
circle.
Allen
|
| 1142.3 | What's half a haystack? It's a haystack... | AKQJ10::YARBROUGH | I prefer Pi | Mon Nov 06 1989 13:52 | 14 |
| > -< needle in a haystack >-
> ... a semiellipse of minor axis 12 and major axis of 8*sqrt(3) ~ 13.86
>( = twice the altitude of an equilateral triangle of perimeter 24), ...
> halved along the minor axis,
The figure I suggest is haystack-shaped, half of an ellipse, the ellipse
halved along the minor axis. Yes, I erred in figuring the height of the
equilateral triangle, so it appears the height of the haystack should be
24/pi = 7.64 to accomodate the circle, i.e. the major axis of the ellipse
is 15.28. Sorry about that!
Lynn
|