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        riddles >> medium >> Parallelogram
        
(Message started by: william wu on Jan 5^th, 2003, 1:43pm)

Title: Parallelogram
Post by william wu on Jan 5^th, 2003, 1:43pm

From e-mail. Haven't done it myself.

Construct a ~~parallelogram~~ trapezoid with the following lengths for its top, left, bottom, and right line segments respectively: 5, 4, 9, and 3.

Title: Re: Parallelogram
Post by Icarus on Jan 5^th, 2003, 2:44pm

What definition of "parallelogram" are you using? The only one I know of comes with a theorem that the opposite sides have equal lengths, so it can't be done with the numbers given!

Title: Re: Parallelogram
Post by william wu on Jan 5^th, 2003, 3:27pm

I think the sender meant trapezoid. "(top and bottom parallel)"

Supposedly this is an "old chestnut".

Title: Re: Parallelogram
Post by lukes new shoes on Jan 5^th, 2003, 9:07pm

can the left and right sides be parallel instead of the top and bottom sides?

Title: Re: Parallelogram
Post by william wu on Jan 5^th, 2003, 10:44pm

I'll ask for clarification ...

Title: Re: Parallelogram
Post by towr on Jan 6^th, 2003, 2:50am

what's supposed to be difficult about this?
I just cut out some pieces of said length, and laid them out in trapezoid shape.

Title: Re: Parallelogram
Post by towr on Jan 6^th, 2003, 2:56am

on 01/05/03 at 21:07:00, lukes new shoes wrote:

can the left and right sides be parallel instead of the top and bottom sides?

No it's impossible given the length and places of the sides..

Title: Re: Parallelogram
Post by James Fingas on Jan 6^th, 2003, 12:27pm

I would agree that this riddle seems pretty easy. I was able to find the coordinates of the four vertices of the trapezoid pretty quickly (didn't even need the quadratic formula!)

Here's a question though: if the lengths are a,b,c, and d, can you always make a trapezoid? Can you always make a trapezoid with a and c parallel? If not, what conditions must be satisfied so that you can?

Title: Re: Parallelogram
Post by SWF on Jan 6^th, 2003, 5:10pm

I think the question is asking how to construct the trapezoid with straightedge and compass, not just to figure out coordinates of all the points. Even though you can figure out the height and construct it from that, I think using the fewest number of steps is best.

Here is one way which makes use of some of the special characteristics of this particular trapezoid:

Draw two perpendicular lines meeting at point A. Find point B on the horizontal line a distance of 3 from point A. Find point C on the vertical line such that it's distance from point B is 8. Extend line BC to point D (with C between B and D) such that D is distance 9 from point B. Find the other point (B is the first such point) on line AB which is distance 9 from point D, call this E. Construct a line parallel to BD that passes through A and instersects line DE at F. The trapezoid is ABDF. Note how point F will work out to be the right distance from D and A.

Title: Re: Parallelogram
Post by towr on Jan 7^th, 2003, 12:26am

on 01/06/03 at 12:27:11, James Fingas wrote:

Here's a question though: if the lengths are a,b,c, and d, can you always make a trapezoid?

no, for example: 4, 1, 1, 1

Quote:

Can you always make a trapezoid with a and c parallel? If not, what conditions must be satisfied so that you can?

no, a trapezoid is a triangle and a parallellogram put together. So to get a and c parallel, |a-c|, b and d have to form a triangle. This isn't always possible (every two sides together have to be longer than the remaining one).

   ______
 /\     \
/__\_____\

Title: Re: Parallelogram
Post by Cyrus on Jan 7^th, 2003, 8:42am

on 01/07/03 at 00:26:41, towr wrote:

No, for example: 4, 1, 1, 1

No, a trapezoid is a triangle and a parallellogram put together. So to get a and c parallel, |a-c|, b and d have to form a triangle. This isn't always possible (every two sides together have to be longer than the remaining one).

So, in order to make a trapezoid (a+b+c)>d ?
But explain again the constraints in making a trapezoid with 2 parallel sides. I understand that to make a triangle every 2 sides together must be longer than the remaining one, but how does this help? Maybe I just need a little more time to think about it.

Title: Re: Parallelogram
Post by towr on Jan 7^th, 2003, 9:22am

Well, a+b+c > d is one constraint, but also a+b+d > c, and a+c+d > b and b+c+d > a has to hold true to be able to construct them into a trapezoid or any other quadrilateral for that matter..

Given a,b,c,d for which the above is true, to make a trapezoid with a and c parallel (rather than b and d parallel) you need to be able to make a triangle from |a-c| b and d:

   /\ 
 b  d
/____\
 |a-c|

(the drawing assumes c > a, else the triangle is upside down)
To complete the trapezium, add the parallelogram

    ___a___
  /\      \
 b  d      d
/____\______\
 |a-c|   a

(note that |a-c| +a = c in this case where c > a)

Title: Re: Parallelogram
Post by towr on Jan 7^th, 2003, 9:46am

heh, just edited in that quadrilateral bit the same minute you said it..
[edit]James sneakily deleted his post again :P[/edit]

Just to clarify, not all a,b,c,d that can form a quadrilateral can form a trapezium. For example 4,4,1,1 can't. Since |4-1|,4,1 can't form a triangle. (Keep in mind I don't count a flat line as triangle or trapezium, and in this case a flat one would work)

Title: Re: Parallelogram
Post by Cyrus on Jan 7^th, 2003, 10:44am

Ok, say again why 4,4,1,1 can't work ? To me that would form a perfect rectangle. With the top and bottom 4 units long and the 2 sides being 1 unit each. A rectangle is a form of a trapezium right?

Title: Re: Parallelogram
Post by James Fingas on Jan 7^th, 2003, 10:56am

towr,

Sorry about that! I reread your post and it seemed to answer the question. I was just trying to figure out whether or not every a,b,c,d could make a trapezoid or not (using your method of constructing a trapezoid). I think you've got the right answer, and I would be willing to accept a line 5 units long as a degenerate trapezoid from the quadrilateral family 4,4,1,1.

SWF,

towr's method of creating a trapezoid would seem to work well for the 5,4,9,3 trapezoid as well:

Construct triangle ABE so that AB=4, BE=4, EA=3. Draw a line EC, of length 5, extending BE. Point D is 3 away from C and 5 away from D, and the finished trapezoid is ABCD.
Cyrus,

a rectangle is a trapezoid, but we've been listing the sides in order. That is to say, the quadrilateral with sides 4,4,1,1 has the two length-4 sides side-by-side ( :P that was a little awkward! )

Title: Re: Parallelogram
Post by towr on Jan 7^th, 2003, 11:04am

euhm, yes.. But it is implied here that the numbers stand for: top, left, bottom, right side respectively. (Since that is how a,b,c,d were defined above)
So with 4 at the top, and 4 to the left, and 1 at the bottom and 1 to the right, you'll see that it won't be a rectangle, nor parallellogram, nor trapezium (but it is still a quadrilateral)..

And to clarify the hierarchy:
squares are a subset of rectangles,
rectangles a subset of paralellograms,
parallellograms a subset of trapeziums,
trapeziums a subset of quadrilaterals.

Title: Re: Parallelogram
Post by towr on Jan 7^th, 2003, 11:07am

Just in case anyone gets confused, trapezium and trapezoid is the same thing.. Trapezium is the British name (and in various other languages), trapezoid is the american name..

(It actually gets more confucing, since the american definition of trapezium is the opposite of the british definition)

http://mathworld.wolfram.com/Trapezoid.html
http://mathworld.wolfram.com/Trapezium.html

Title: Re: Parallelogram
Post by James Fingas on Jan 7^th, 2003, 11:24am

towr,

I like your heirarchy, but I also like being a little pedantic ;) There are many ways that you can drill down from quadrilaterals:

Your example:
quadrilaterals -> trapezoids -> parallelograms -> rectangles -> squares

An alternative:
quadrilaterals -> trapezoids -> parallelograms -> rhombs -> squares

If we define a few other shapes, then we could go another route (I'm just making up names for these, my apologies to the mathematical community at large):

mondiads: one pair of opposite corners have the same angle
kites: one pair of opposite corners have the same angle, and we have a mirror plane of symmetry (two isoceles triangles with their bases together)

now we can do this:
quadrilaterals -> mondiads -> parallelograms -> rectangles -> squares

or this:
quadrilaterals -> mondiads -> kites -> rhombs -> squares

Food for thought? Effectively, we remove one degree of freedom at each step. However, there's always a choice as to how we constrain the shape--effectively, which degrees of freedom we remove. And of course we don't have to end up with the square ...

Title: Re: Parallelogram
Post by towr on Jan 7^th, 2003, 11:44am

well a hierarchie is seldom linear.. You can fit all of them in without any problem..
The ordering by properties still remains.. a square has all attributes of a rectangle, and also of a rhombus. A rhombus has all properties of a parallelogram. And a rectangle has all properties of a paralellegram. Note however the convergence can be much further down the line (or up, depending on how you order your hierarchie)
In the end you go from the most specific shape to the most general. (Which of those is at the top of the hierarchie is a matter of taste)

This talk of rhombusses and other geometric shapes reminds me of Triangle and Robert (http://home.attbi.com/~pshaughn/tandr.html)

Title: Re: Parallelogram
Post by James Fingas on Jan 8^th, 2003, 11:27am

towr,

So every a,b,c,d can make a trapezoid, but here's the second part of my question: When can you make a trapezoid with a and c parallel? When can you make a trapezoid with b and d parallel? When do you have a choice about which you can make?

Here's another problem--interesting but tricky: Given a circle of radius 4 and its center, construct the 5,4,9,3 trapezoid using only a straightedge (a ruler with no markings). Maybe this is closer to the intent of the original question...

Title: Re: Parallelogram
Post by towr on Jan 8^th, 2003, 12:16pm

on 01/08/03 at 11:27:28, James Fingas wrote:

So every a,b,c,d can make a trapezoid,

??
no, try 10000,1,1,1.
Not every a,b,c,d can form a quadrilateral, much less a trapezoid.
Quote:

but here's the second part of my question: When can you make a trapezoid with a and c parallel? When can you make a trapezoid with b and d parallel? When do you have a choice about which you can make?

a and c can be parallel when |a-c|,b and d can form a triangle, and by symetry b and d can be parallel when |b-d|,a,c can form a triangle.. You can choose when both conditions are satisfied.

I'm seem to keep repeating myself..

I'll have to think about that latest addition to the puzzle..

Title: Re: Parallelogram
Post by James Fingas on Jan 8^th, 2003, 1:43pm

towr,

So when are both conditions satisfied? I'm not trying to be a pain here, but there is a simpler statement of when you can make a trapezoid, that answers these questions.

Title: Re: Parallelogram
Post by towr on Jan 9^th, 2003, 11:04am

on 01/08/03 at 13:43:32, James Fingas wrote:

towr,

So when are both conditions satisfied? I'm not trying to be a pain here, but there is a simpler statement of when you can make a trapezoid, that answers these questions.

given:
|a-c| <= b + d
b <= |a-c| + d
d <= |a-c| + b
|b-d| <= a + c
a <= |b-d| + c
c <= |b-d| + a

assumptions:
a > c (otherwise flip)
b > d (otherwise flop)

thus:
a-c <= b + d
b <= a-c + d
d <= a-c + b
b-d <= a + c
a <= b-d + c
c <= b-d + a

thus:
1: a <= b+c+d
2: b+c <= a+d *
3: c+d <= a+b
4: b <= a+c+d
5: a+d <= b+c *
6: c+d <= a+b

thus:
{5 -> 1, 2 -> 4, assumptions -> 3 & 6}
a+d = b+c {from 2 and 5}

Which means there is either exactly one possible trapezoid (flat), or an infinite number of possible trapezoids (which are also parallellograms).

Title: Re: Parallelogram
Post by James Fingas on Jan 9^th, 2003, 11:23am

towr,

That's right, but I was thinking of these conditions:

1) You can make a quadrilateral if
|a-c| <= b+d and
|b-d| <= a+c

2) You can make a trapezoid with a ll c if
|a-c| > |b-d|

3) You can make a trapezoid with b ll d if
|b-d| > |a-c|

This shows pretty clearly why you can, in general, only make either a ll c or b ll d, and can only do both in the flat case, or the parallelogram case.

Title: Re: Parallelogram
Post by towr on Jan 9^th, 2003, 12:26pm

But then you always need the quadrilateral constraint as well.
Because with only:

Quote:

2) You can make a trapezoid with a ll c if
|a-c| > |b-d|

you fail with the 4,1,1,1 example again..

You'll need
|a-c| <= b + d and |b-d| <= |a-c|
to make a trapezoid with a ll c
In other words the expanded version my triangle-constraint,
which is equal to the first half of 1) + 2) from your constraints

Personally I think I prefer the geometric interpretation of triangle + parallellogram = trapezoid, since its easy to visualize..

Title: Re: Parallelogram
Post by James Fingas on Jan 9^th, 2003, 12:47pm

I guess that's true, but to check if you can make a triangle with |a-c|, b, and d, you need to know which is bigger of b and d. Whereas I'm assuming that you'll check to see if you can make a quadrilateral before you try to make a trapezoid. Either way, there are two checks you must make, and I find the comparison of |a-c| with |b-d| quite intuitive.

Title: Re: Parallelogram
Post by towr on Jan 9^th, 2003, 12:55pm

on 01/09/03 at 12:47:47, James Fingas wrote:

I guess that's true, but to check if you can make a triangle with |a-c|, b, and d, you need to know which is bigger of b and d.

Actually you don't..
b-d <= |b-d| and
d-b <= |b-d|
so if |b-d| <= |a-c| both are satisfied, and either d-b, or b-d is |b-d| so there's only two checks needed to see if you can form a triangle..

intuitively, you only need to know if b and d 'hinged' (changing the angle between b and d) can be smaller (or equal) and bigger (or equal) than |a-c|

Quote:

Whereas I'm assuming that you'll check to see if you can make a quadrilateral before you try to make a trapezoid.

Depends on what's asked, really.. but normally yes..