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Subject: "geometry"     Previous Topic | Next Topic
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Conferences The CTK Exchange High school Topic #359
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mar
guest
Feb-18-07, 01:59 PM (EST)
 
"geometry"
 
   IS THERE ANY FOUR-SIDE SHARE WITH EQUAL THE OPPOSSITE ANGLES AND SIDES BUT IT'S NOT PARALLELOGRAMME;IF YES,COULD YOU DESCRIBE IT TO ME9PROBABLY YES).I HAVE MADE MANY THOUGHTS AND THIS QOESTINO HAS PROBABLY REFERENCE TO THEBAULT'S LEMME(NOT THEOREM)


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  Subject     Author     Message Date     ID  
geometry mar Feb-18-07 TOP
  RE: geometry alexb Feb-18-07 1
  RE: geometry Kenneth Ramsey Feb-18-07 2
  RE: geometry mar Feb-18-07 3
     RE: geometry sfwc Feb-19-07 4
         RE: geometry mar Feb-24-07 8
             RE: geometry sfwc Feb-25-07 9
     RE: geometry alexb Feb-19-07 5
         RE: geometry Kenneth Ramsey Feb-20-07 6
             RE: geometry alexb Feb-20-07 7
  RE: geometry mpdlc Feb-25-07 10

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alexb
Charter Member
1966 posts
Feb-18-07, 02:08 PM (EST)
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1. "RE: geometry"
In response to message #0
 
   >IS THERE ANY FOUR-SIDE SHARE WITH EQUAL THE OPPOSSITE ANGLES

Fact 1: a quadrilateral with two pairs of equal opposite angles is a parallelogram. (Because then the opposite sides are parallel.)

>AND SIDES

Fact 2: a quadrilateral with two pairs of equal opposite sides is a parallelogram. (Because of SSS when you draw one of the diagonals.)

There is a quadrilateral with two pairs of equal sides: a kite. It does have a pair of opposite equal angles.

BUT IT'S NOT PARALLELOGRAMME;

Not every kite is a parallelogram.


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Kenneth Ramsey
guest
Feb-18-07, 07:24 PM (EST)
 
2. "RE: geometry"
In response to message #0
 
   It is not a square since a square is a parallelogram

Maybe you are talking of something three-dimensional?


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mar
guest
Feb-18-07, 07:24 PM (EST)
 
3. "RE: geometry"
In response to message #0
 
   >IS THERE ANY FOUR-SIDE SHApE WITH EQUAL only one pair of THE OPPOSSITE ANGLES
>AND only one pair of the oppossite SIDES BUT IT'S NOT PARALLELOGRAMME;(i mean that i want an anti-example to the suggestion:if a four-side shape has only one pair of equal the oppossite sit's and only one pair of equal the oppossite angles,then it's parallelogramme)IF YES,COULD YOU
>DESCRIBE IT TO ME PROBABLY YES).I HAVE MADE MANY THOUGHTS
>AND THIS QuESTIon HAS PROBABLY REFERENCE TO THEBAULT'S
>LEMME(NOT THEOREM)


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sfwc
Member since Jun-19-03
Feb-19-07, 08:59 AM (EST)
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4. "RE: geometry"
In response to message #3
 
   Let ABC be isosceles with AB = AC. Pick D on BC. Let C' be the reflection of C in the perpendicular bisector of AD. ABDC' has two opposite sides the same length and two opposite angles equal but is not a parallelogram if D isn't the midpoint of AB. This construction gives all such quadrilaterals.

Thankyou

sfwc
<><


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mar
guest
Feb-24-07, 11:50 PM (EST)
 
8. "RE: geometry"
In response to message #4
 
   I think that's the solution demanded.However,ADC'C is rectangular,isn't it-AD and CC' are parallel and in case they "meet" in point K,after comparing the triangles ADK and KC'C they are equal (AAA) so C'C and AD are equal-.How that's possible?


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sfwc
Member since Jun-19-03
Feb-25-07, 08:17 AM (EST)
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9. "RE: geometry"
In response to message #8
 
   >I think that's the solution demanded.However,ADC'C is
>rectangular,isn't it?
No. See this page for a diagram

>AD and CC' are parallel and in case
>they "meet" in point K,after comparing the triangles ADK and
>KC'C they are equal (AAA) so C'C and AD are equal-.How
>that's possible?
ADK and KC'C are similar (by AAA), but not congruent. So we cannot deduce that C'C = AD.

Thankyou

sfwc
<><


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alexb
Charter Member
1966 posts
Feb-19-07, 10:04 AM (EST)
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5. "RE: geometry"
In response to message #3
 
   If the question is about quadrilaterals and not polygons, meaning that the side intersections are allowed, then, for an isosceles trapezoid ABCD with AB = CD, the quadrilateral ABDC has a pair of equal opposite sides and two pairs of equal opposite angles.


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Kenneth Ramsey
guest
Feb-20-07, 09:49 PM (EST)
 
6. "RE: geometry"
In response to message #5
 
   >If the question is about quadrilaterals and not polygons,
>meaning that the side intersections are allowed, then, for
>an isosceles trapezoid ABCD with AB = CD, the quadrilateral
>ABDC has a pair of equal opposite sides and two pairs of
>equal opposite angles.
My understanding is that an isosceles trapezoid by definition is a two dimensional figure and a polygon because it has more than three sides. Also, this might be the counter-example that Mar was looking for, since it has two opposite sides that are equal and two pairs of opposite angles that are equal.
However, I think that a three dimensional figure is a better counterexample since it can have two pairs of equal opposite sides and two pairs of equal opposite angles. I did not see the word polygon in the question.


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alexb
Charter Member
1966 posts
Feb-20-07, 10:05 PM (EST)
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7. "RE: geometry"
In response to message #6
 
   >>If the question is about quadrilaterals and not polygons,
>>meaning that the side intersections are allowed, then, for
>>an isosceles trapezoid ABCD with AB = CD, the quadrilateral
>>ABDC has a pair of equal opposite sides and two pairs of
>>equal opposite angles.
>My understanding is that an isosceles trapezoid by
>definition is a two dimensional figure and a polygon because
>it has more than three sides.

A triangle is a polygon too. But I should not have brought up the polygons, especially because I meant to say "a simple quadrilateral", which I an not sure is the definition of a 4-polygon.

>Also, this might be the
>counter-example that Mar was looking for, since it has two
>opposite sides that are equal and two pairs of opposite
>angles that are equal.

I believe, he looked for a single pair of each.

>However, I think that a three dimensional figure is a better
>counterexample since it can have two pairs of equal opposite
>sides and two pairs of equal opposite angles.

A matter of taste.

> I did not see the word polygon in the question.

Right. But somehow I had an image in my mind. I think we now have a complete solution: one for simple and the other for self-intersecting quadrilaterals.


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mpdlc
guest
Feb-25-07, 07:26 PM (EST)
 
10. "RE: geometry"
In response to message #0
 
   Being an old fashioned engineer the solution I describe below it is not mathematical demonstration but it will give you the procedure to draw our elusive quadrilateral in any case.

First I consider our potential quadrilateral like a four bar mechanism with two bars of equal length, I called them R1 and R2 .

The two other bars, without loosing generalization one I can draw horizontal I named it L, and the other remaining bar joining the two other edges of bars R1 and R2 , I called it B.

I called p the angle between R1 and L , and I called q the angle between R2 and L, both angles measured with a counterclockwise criteria , and I made q > p.

I called w the angle formed by B and L, it is obvious w equal to q - p


Now we take the horizontal and vertical projections of the polygonal represented by the four bars mechanism it will render the two following equations:

B cos (w) = R2 cos(q) - R1 cos (p) + L (1)
B sin (w) = R2 sin (q) - R1 sin (p) (2)

Since R1 equal R2 we can consider both as unit length then L and B will represent the ratio L/R and B/R

By division of the above equations and sub sequential ordering we get

cot (w) = (cos(q) - cos (p) + L) / (sin (q) - sin (p))

Remembering w = q-p now we can write

L = cos (p) - cos (q) + (sin (p) - sin (q)) cot (q-p)

So with this handy formula let us say we want to draw a quadrilateral two opposing side equal to 100 mm and the opposing angles forming p = 30 degrees and for example q = 45 degrees that will give us a value for L = 93.185 mm. and value for B amounts 80.02 mm



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