| T.R | Title | User | Personal
Name | Date | Lines |
|---|
| 524.1 | ah, now I understand the problem | CLT::GILBERT | Juggler of Noterdom | Thu Jun 26 1986 13:50 | 8 |
| After coming up with the following...
Draw a line through a and b. Draw a parallel line through c and
a perpendicular line through d. This gives three sides to a square;
supply the fourth.
I realized that the problem intended that each point be on one (extended)
side of the square, and each (extended) side of the square had one point.
|
| 524.2 | clarification | CACHE::MARSHALL | | Mon Jun 30 1986 10:20 | 6 |
| re .1:
Exactly, when the constuction is complete, no two points share the
same (extended) side of the square.
sm (b.t.f.d)
|
| 524.3 | | CLT::GILBERT | Juggler of Noterdom | Tue Jul 01 1986 21:17 | 16 |
| Looking at the diagram in .0, I can imaging the parallel lines
through a and b rotating, so the distance between them varies.
Similarly, the parallel lines through c and d can rotate, as they
stay perpendicular to the lines through a and b.
Note that the distance between the a and b lines is 0 when they coincide,
and (if cd is not perpendicular to ab) the distance between the c and d
lines in this case is non-zero. So, the c-d distance >= the a-b distance.
Similarly, it will happen that the a-b distance >= the c-d distance.
Note that as the lines rotate, the distances are changing smoothly.
Thus, there must be some place where the a-b distance = the c-d distance,
and here we have a square.
How about a constructive proof?
|
| 524.4 | cautions | CACHE::MARSHALL | beware the fractal dragon | Wed Jul 02 1986 09:51 | 8 |
| re .3:
a valid analysis, but remember, You are given only the four
points. How do you decide which two are A & B ? Or, can you show
that a square can be constructed regardless of how you assign the
parallel pairs?
sm
|
| 524.5 | | CLT::GILBERT | Juggler of Noterdom | Wed Jul 02 1986 11:16 | 3 |
| The analysis is independent of which points are A & B, and which are C & D.
As the analysis is valid, it doesn't matter which are chosen -- a square
can be constructed in any case.
|
| 524.6 | | CLT::GILBERT | Juggler of Noterdom | Wed Jul 02 1986 22:35 | 48 |
| Let Dab be the distance between points a and b, and Dcd the distance
between c and d. Let the (directed) line ab be at an angle Tab from
the x axis, and let cd be at angle Tcd from the x axis.
Consider two parallel lines through a and b, that make an angle of Uab
with the (directed) line ab. The distance between these lines is given
by: Dab |sin(Uab)|. Similarly, for c and d, we have Dcd |sin(Ucd)|.
Now we want these lines to be perpendicular, so Tab+Uab = Tcd+Ucd � pi/2.
And the distances must be equal, so Dab |sin(Uab)| = Dcd |sin(Ucd)|.
Thus, we require that Dab |sin(Uab)| = Dcd |sin(Tab-Tcd+pi/2+Uab)|, so
we must find a solution Uab to this equation (all other parts of the
equation are constants defined by the problem). It should be easy for
someone familiar with geometric problems to construct a solution x to:
k1 sin(x) = k2 sin(x+k3),
but I'm not that ... someone?
Omigosh! I see it! The construction follows.
Translate the line segment CD to C'D' so that C' and A coincide -- in
other words draw a line through A parallel to CD, and mark off point D'
at distance CD from A. Note that, if we find a solution to this problem
(i.e., given the points A,B,C' and D', with C'=A, find the square), we
can easily construct the solution to the original problem by translating
the parallel lines that pass through C' and D' back to C and D.
Construct a line through A perpendicular to AD', and mark a point D"
at distance CD from A. Draw the line BD", and another through A and
parallel to BD" (these are two of the lines in our solution!). Drop
a perpendicular from C' to this last line, meeting it at E.
[ The distance between the last two parallel lines is Dab |sin(Uab)|
from the above analysis, and it equals Dcd |sin(Tab-Tcd+pi/2+Uab)|.
This follows from noticing that angle(BAD") = Tab-Tcd+pi/2, and letting
angle(EAB) = Uab. ]
Now the length of AE is equal to the distance from D" to the line AE
(by the two similar triangles with sides AD' and AD"), and this is
also equal to the distance between the last two parallel lines (that
form part of the solution).
Altogether, the four lines in the solution are: BD", the line parallel
to BD", and passing through A, and the two perpendiculars to these lines
that pass through points C and D.
|
| 524.7 | RIGHT! but... | CACHE::MARSHALL | beware the fractal dragon | Thu Jul 03 1986 11:25 | 15 |
| I think you got it! but I don't understand the following:
>the length of AE is equal to the distance from D" to the line AE<
(by the two similar triangles with sides AD' and AD"), and this is
also equal to the distance between the last two parallel lines (that
form part of the solution).
When I performed the construction, this just was not so, but I did
get a square (by eye anyway)
looking at it, maybe you meant D' instead of D"?
Thanks,
sm (Blast! is EVE a pain!)
|
| 524.8 | | CLT::GILBERT | Juggler of Noterdom | Thu Jul 03 1986 18:12 | 22 |
| Huh? Before translating the result C'D' (aka AD') back to CD,
edge AE is one of the sides of the square! The distances had
better be equal!
Here's a diagram (well, the points, anyway):
.D'
.D" .B
.F .A .E
Recall that we started with A and B, AD' is equivalent to CD (same size
and orientation), and AD' and AD" are perpendicular with equal lengths.
Line FAE is parallel to BD", and AED and AFD" are right angles.
I assert that the lengths of AE and D"F are equal, because AD'E and AD"F
are similar triangles.
The sides of the square are on the lines FAE, D"B, D'E, and the fourth
side is on a line parallel to D'E, and passing through A.
Does this help?
|
| 524.9 | My brain hurts | CACHE::MARSHALL | beware the fractal dragon | Mon Jul 07 1986 09:22 | 24 |
| I hate to be so contrary, but I'm getting more confused than ever.
When I perform the construction, AE is not one of the sides of the
square. It is parallel to a side so is equal in length, but it is
not one of the sides. Let me describe how I laid out my points.
B = 0,0
C = -7,2
A = -4,12
D = 8,2
first step of the construction puts D' at 11,12
D" at -4,-3
E ends up in the vicinity of ~3.3,2.5
The line passing through C |- BD" just misses D"
The line passing through D |- BD" is pretty far from E.
I am sure that your construction is correct. I'm just confused by
the proof.
sm
|
| 524.10 | a proof | CACHE::MARSHALL | beware the fractal dragon | Mon Jul 07 1986 10:30 | 57 |
| First, forget point B.
Show that any and all parallel lines drawn through a & d" will yield a square,
when perpendiculars are constructed through c & d.
\ \/
\ /\ /
\ a / \ /
\ * \ /
\ /| \ /
\ /z| X
\ / | / \
\ / | / \
\ / | / y\ d
c *-/-----+--@-------*
X | / \
/ \ |/ \
/ \ * \
/ \ / d" \
/ \/ \
/ /\ \
m n p r
excuse the distortion. imagine the X's are right angles
z and y are the interior angles
m,n,p,r are the lines passing through a,d",c,d respectively
GIVEN:
> p || r
> m || n
> p |- m , p |- n
> r |- m , r |- n
> ad" |- cd
__ __
> ad" = cd (lengths equal by construction)
PROVE: <z = <y (angle :== "<" )
The intersection marked "@" is the key to proving z = y
<d",@,+ = <d,@,X
<@,+,d" = <@,X,d = |_ (right angles)
.: <+,d",@ = <X,d,@
<+,d",@ = <y (substitution)
<+,d",@ = <z (interior angles of a line through parallel lines)
.: <z = <y
__ __
Now, with <z = <y and ad" = cd,
_ _ _ _
the distance between m and n must be equal to the distance between p and r
.: the sides of the constructed figure are of equal length and perpendicular
and therefore must be a square. QED
|
| 524.11 | | CLT::GILBERT | Juggler of Noterdom | Mon Jul 07 1986 13:24 | 30 |
| Argggh! There's a typo in 524.6. The paragraph should read:
Construct a line through A perpendicular to AD', and mark a point D"
at distance CD from A. Draw the line BD", and another through A and
parallel to BD" (these are two of the lines in our solution!). Drop
a perpendicular from D' to this last line, meeting it at E.
(that is, the last line erroneously had C' instead of the correct D').
Thus, E is at 5.6,19.2, and the point you constructed (at 3.2,2.4 exactly)
happens to be another corner of the square (along with A and E).
Although I think it can be simplified, I like your proof *much* better.
Roughly, AD" and CD are perpendicular and of equal length. Parallels
through A and D" that are perpendicular to parallels through C amd D
form a square, since the sides are of equal length, as can be seen be
rotating AD" (and the lines through A and D") by 90�, noting that this
diagram coincides with the one for CD, so that the distance between
each set of parallels is the same.
Or another way to show this is to notice that when you take a square,
and draw parallel lines through opposite corners, with the parallel
lines for one pair of opposite corners perpendicular to the other
corners' parallel lines, another square is formed. Now apply this,
after translating CD to bisect AD" (so that CADD" is a square).
The reason I prefer the approach in 524.10 (perhaps as restated above)
is that a solution to the (original) problem follows so easily from it.
|
| 524.12 | Thanks | CACHE::MARSHALL | beware the fractal dragon | Tue Jul 08 1986 15:42 | 13 |
| Well...
Thank you for the praise, I too think that my proof could probably
be simplified, I think I was a little redundant in a few places.
But mostly, thank you for providing the construction itself, and
for solving a problem that's been bothering me for years (although
the real problem is why I couldn't rederive the construction after
having already done it in high school)
Guess I'm getting old and senile ( and only 28! ;-) )
sm
|