| T.R | Title | User | Personal
Name | Date | Lines |
|---|
| 809.1 | Equal area, or same shape, too? | ZFC::DERAMO | Frustrated personal name composer | Thu Dec 24 1987 14:36 | 13 |
| What about dividing it up this way, where all three angles
are supposed to be 120 degrees:
|
|
|
center --> |
/ \
/ \
/ \
/ \
Dan
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| 809.2 | Straightedge/compass (no proof) | SMURF::JMARTIN | Joseph A. Martin, ULTRIX kernel | Thu Dec 24 1987 15:40 | 12 |
| 1) Inscribe a couple non-parallel chords.
2) Construct their perpendicular bisectors. Where the bisectors intersect
is the center.
3) Set compass from center to circumference.
4) Mark a point on the circumference.
5) Pivot the compass on this point and mark the point intersected by
other leg.
6) Repeat the above pivoting on the new point, proceeding around the
circle until you reach the point marked in 4).
The circle is now divided into six equal parts and thus into three as well.
--Joe
|
| 809.3 | Let's have two categories for these ... | KEEPER::KOSTAS | He is great who confers the most benefits. | Mon Dec 28 1987 00:10 | 16 |
| re. .1, .2
We could have two categories: 1) One in which we have both equal
areas and identical shapes, and 2) one in which we have equal areas
but not identical shapes.
Both reply .1 and .2 are of the 1st categorie. I know a few more
for the 1st categorie, namely the Chinese Kung Fu symbol (Yin Yang).
This symbol divides the circle in two equal parts of equal shapes.
And one for category 2 which similar to the Yin Yang but divides
the circle in three equal parts of diferent shapes.
/Kostas
|
| 809.4 | Easier Solution | LABC::FRIEDMAN | | Tue Jan 05 1988 17:44 | 3 |
|
1) Purchase a Mercedes-Benz.
2) Look at its insignia.
|
| 809.5 | | CLT::GILBERT | Builder | Wed Jan 06 1988 14:25 | 5 |
| The circle of radius r can be divided into three parts having equal area
with two parallel lines. The lines are at distance x from the center of
the circle, where x is a solution to:
arccos(x/r) - (x/r) * sqrt(1 - (x/r)^2) = pi / 3
|