| T.R | Title | User | Personal
Name | Date | Lines |
|---|
| 2022.1 | | AUSSIE::GARSON | achtentachtig kacheltjes | Fri Dec 22 1995 01:34 | 5 |
| re .0
I guess the intuitively obvious answer is that two sides are pinched
together until the pentagon degenerates into an equilateral triangle
(of area sqrt(3)/4).
|
| 2022.2 | Christmas harbinger | HERON::BUCHANAN | | Sun Dec 24 1995 01:04 | 6 |
| re .1
The triangle was my first intuitively most obvious solution, but
then my intuition came up with a better one.
Andrew.
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| 2022.3 | | HANNAH::OSMAN | see HANNAH::IGLOO$:[OSMAN]ERIC.VT240 | Tue Dec 26 1995 11:32 | 12 |
|
I folded an index card into approximately 5 even sections, then viewed
it from edge to see pentagon segments.
Playing with it, my idea for minimal is similar to the triangle idea.
Except, instead of pinching two sides, fold them in concave first. Boy
I wish we had sketching ability in this dumb notes system.
Anyway, the thing looks a bit like a letter C.
/Eric
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| 2022.4 | | AUSSIE::GARSON | achtentachtig kacheltjes | Tue Jan 02 1996 21:43 | 11 |
| re .3
We do have sketching ability. Just go into DECwrite, draw up your
diagram and post it. Notes will understand. (Unfortunately as I'm on a
Mac I won't be able to see the diagram.)
Do you mean *like* a letter M with a line across the bottom or
alternatively two congruent isosceles triangles with the equal sides
all of the length 1 and the base of each length 1/2 ? If so, I make the
area sqrt(15)/8 which is slightly more than what I posted...or did I
misunderstand your configuration or screw up with the area calculation?
|
| 2022.5 | No calculus, but a handwave | WIBBIN::NOYCE | EV5 issues 4 instructions per meter | Wed Jan 03 1996 09:51 | 23 |
| > (b) If so, can it be shown without calculus?
_ _
| /\ /\ |
h1 / \ / \ h2
| / \ / \ |
| / x \/ 1-x \ |
- ------------------ -
Consider a configuration that has two triangles, each using part of a unit
base. One triangle has a base of x and a height of h1, and the other has
a base of 1-x and a height of h2. Assume WLOG that x>=1/2, so that h1<=h2.
x*h1 + (1-x)*h2
area = --------------- >= h1/2
2
It is "obvious" that h1 is minimized when x=1, and then the area is exactly
equal to h1/2 (since the other triangle has area h2*(1-x) = 1*0 = 0).
How do we count area for non-simple polyhedrons? If you take the configuration
with two isosceles triangles, and flip one over, do you get to negate its
area, so the total is zero?
|
| 2022.6 | naive definition of area | JOBURG::BUCHANAN | | Wed Jan 03 1996 20:45 | 9 |
| Re -.1
Define area to be that space enclosed by the polygon. So no
negative areas, and no areas with multiplicity > 1.
Cheers,
Andrew.
|
| 2022.7 | pentagram | TPOVC::BUCHANAN | A small bird called Offlogic | Tue Mar 19 1996 06:01 | 13 |
| The area of a regular pentagram is smaller than that of an equilateral
triangle of the same side length. But there's just a couple of per cent
in it.
I think that I can prove that the minimum area shape must be some
kind of pentagram, by showing that given any pentagon which is not a
pentagram, one can reduce the area by "flipping" two edges.
But I don't have any neat way of showing that the regular pentagram
then minimizes the area of what I've got left.
Regards,
Andrew.
|