| T.R | Title | User | Personal
Name | Date | Lines |
|---|
| 1145.1 | Clarification.. | SQM::SAXENA | | Sat Feb 22 1992 07:44 | 25 |
| 1145.2 | | BUSY::SLABOUNTY | HereComesTrouble&ItLooksLikeFun | Mon Feb 24 1992 17:06 | 13 |
| 1145.3 | | VMSMKT::KENAH | And became willing... | Tue Feb 25 1992 03:36 | 3 |
| 1145.4 | There is a solution. | TROOA::RITCHE | From the desk of Allen Ritche... | Fri May 29 1992 00:32 | 17 |
| 1145.5 | Background, hints, + spoiler... | TROOA::RITCHE | From the desk of Allen Ritche... | Fri May 29 1992 00:37 | 70 |
| 1145.6 | | BEING::EDP | Always mount a scratch monkey. | Fri May 29 1992 02:41 | 7 |
| 1145.7 | | SQM::SAXENA | Updated: 23-MAY-1992 10:23 | Fri May 29 1992 23:20 | 7 |
| 1145.8 | I don't think it matters if it's outside | TROOA::RITCHE | From the desk of Allen Ritche... | Tue Jun 30 1992 00:02 | 19 |
| 1145.9 | a direct trigonometric solution | NETCAD::ROLKE | The FDDI Genome Project | Sat Mar 15 1997 13:22 | 98 |
| Why isn't this solved yet? ;-)
Problem: Given a triangle with two equal angle bisectors then prove
the angles are equal.
The Diagram:
| \
| \'
| '\A
| ' \ <EBC = x
| ' \ <ABC = 2x
| ' \ BE = 1
| ' \ DC = 1
| ' \
| ' \
| ' \
| K ' \
H+---------+-----------------.\E
| ' . \
| ' . \
| D + . \
| '| + . \
| ' | . + \
| ' | . + \
| ' .| + \
| '. | + \
--+------+------------------------------+--
B| J C
|<-d-->|
Construction:
1. Label the origin (point 0,0) B.
2. From B draw line BE one unit long at angle x to point E.
3. From B draw line AB at angle 2x. Point A is on this line somewhere
(but point A is indefinite).
4. From point E draw a line parallel to the X axis back to the Y axis
at point H. Line HE intersects line AB at point K.
5. Pick a point D somewhere on line segment BK.
6. From point D draw a line one unit long to intersect the X axis at
point C (to the right of E).
7. From point D draw a line perpendicular to the X axis at point J.
Call the length of line segment BJ "d".
Observations:
1. Line BE has a slope of tan(x).
2. Line AB has a slope of tan(2x).
3. Point E is at coordinate (cos(x), sin(x)).
4. Point D is at coordinate (d, d tan(2x)).
5. The length of segment JC = sqrt(1 - (d tan(2x))^2).
- sin(x)
6. The slope of line CE (and AC) = ---------------
d + JC - cos(x)
Conclusions:
1. The slope of line CE may be derived from angle "x" and distance "d".
2. <DCB ("y") bisects <ACB ("2y") when d equals a value such that
tan(y) = (d tan(2x)) / sqrt(1 - (d tan(2x))^2)
AND
tan(2y) = sin(x) / (d + JC - cos(x))
The unique(1) solution for these equations is
d = sin(x)/tan(2x) or (d tan(2x)) = sin(x)
tan(y) = sin(x) / sqrt(1 - sin(x)^2) = sin(x)/cos(x) = tan(x)
tan(2y) = sin(x) / ( sin(x)/tan(2x) + cos(x) - cos(x) )
= sin(x) / ( sin(x)/tan(2x) )
= tan(2x)
3. When point D is at point K then CD and BE are equal angle bisectors.
Further, angle x = angle y and triangle ABC is isosceles.
4. When point D is not at point K then line CD does not bisect <ACB
and therefore it is not an angle bisector.
So, does this prove it or am I restating the obvious?
Chuck
(who presses the lever a lot just to get pellets)
1. Where point D lies along line segment AB is constrained by two
factors:
a. <DCB must be less than 45 degrees. Otherwise 2*<DCB is more
than 90 degrees and <DCB as an angle bisector does not make
sense. This limits distance (d tan(2x)) < sin(45).
b. BC must be greater than cos(x). Otherwise the slope of AC
goes positive and the triangle ABC loses its form. This is
only a problem when x > 30 degrees or so.
|