Inscribing Ellipse in a Parallelogram

jonbenedick shared this question 4 years ago
Answered

Given a parallelogram, what is the procedure in inscribing an ellipse inside?

Comments (18)

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hello


thus?


Ellipse in rectangular

http://tube.geogebra.org/ma...


Ellipse in diamond

http://tube.geogebra.org/ma...


mfg

günter

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How about if the free objects are the vertices of the parallelogram? In the link file, the free objects are the foci and a point along the ellipse.

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I think the problem is not entirely well defined since there are several inscribed ellipses possible.

My solution does not cover all cases.


I transformed a parallelogram back to the origin, used a little algebra to calculate the two parameters of the ellips and transformed back to the original position

https://ggbm.at/563531

.

There must be a more clever solution...


See also http://www.geogebratube.org...

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an ellipse is uniquely defined by five values​​:

x, y, width, height, rotation

or 5 points


a parallelogram is clearly defined with six values​​:

x, y, width, height, Angle1, Angle2

or 3 vectors of

or three points

A parallelogram is defined as opposed to the ellipse having a formula


constraint: 4 common points.


there are many combinations, which specified values ​​and which are sought.


can you please specify the task?

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A demonstration http://www.geogebratube.org... that there are many ellipses possible.

The algebraic way does not clearly reveils when there are solutiuon, when not and why not.


Anyone offering a geometric solution? :D

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Hi

here http://dmentrard.blogspot.f...

You can download the file and see the protocol :D

a79cadf8ca378a949a464716202e2084

Daniel

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I think the problem is not entirely well defined since there are several inscribed ellipses possible.

My solution does not cover all cases.


I transformed a parallelogram back to the origin, used a little algebra to calculate the two parameters of the ellips and transformed back to the original position[attachment=1]Inscribed_ellips_in_parallelogram.ggb[/attachment].

There must be a more clever solution...


See also http://www.geogebratube.org...

hello

your solution is very good. creating a rectangle with same base and height centered in origin then the ellipses and applying a matrix for tilting the rectangle you can get several solutions


saludosadd3cf6dd3fbd15c74ee9bbae09dfc1b

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I improved my solution in several ways. It is now stable.

See http://www.geogebratube.org/student/m51044

or

https://ggbm.at/563543

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I think a unique ellipse can be inscribed subject to the condition that it has the maximum area. Is that possible?

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I think a unique ellipse can be inscribed subject to the condition that it has the maximum area. Is that possible?

Did you try using the shear transformation [ e.g. (x,y) --> (x+my,y) ] ?

Note that shear mappings are area-preserving mappings.

https://ggbm.at/5635454227f89cffed6129895896d3ebe37238

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I think a unique ellipse can be inscribed subject to the condition that it has the maximum area. Is that possible?


It is trivial that this ellipse touches the midpoints.

Using the midpoints is the key in the geometric construction of Daniel Mentrard. However, his construction fails in case of a paralellogram.

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hello

begining with square and inscribed circle and (x,y)->(k*x,1/k*y) ->(k*x+m/k*y,1/k*y) we have the maximum area.


saludos

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Sure. The problem isn't a problem if a maximum area is required.

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I love these questions and those solutions.

the foci of the ellipse move on a hyperbole curve.

I can this not use for another solution unfortunately.

now I have the solution of PGvdVeen somewhat simplified and extended.

https://ggbm.at/563559

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hello

more specific solved problem

https://www.geogebra.org/fo...

saludos

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Here is another approach.


Any ellipse can be transformed in a circle by using orthogonal projection.

Circumscribed parallelogram is so transformed in a rhombus (pairs of parallel lines are transformed by orthogonal projection in another pair of parallel lines).

It is easy to see that pairs of segments joining adjacent tangency points are paralel. So if you know the position of one tangency point in the rhombus, the other three are uniquely determined.


The same applies for the original parallelogram (because of the parallelism property), so if you know the position of one of the tangency points you know the other three.


To find the position of a fifth point of the ellipse in order to plot it, I have used a very interesting property I read here:


http://www2.washjeff.edu/users/mwoltermann/Dorrie/43.pdf

(I have used midpoints of AG and IG)

https://ggbm.at/563571

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hello

my final work (generic quadrilateral)

i am remembering my classes of projective geometry and perspective


saludos

http://www.geogebratube.org...

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Hi everybody.

Here is another solution. The Brianchon's theorem is used a couple of times.

https://ggbm.at/5636890b0327af17f6533c86b80cfbfa0f75f5

Files: cc.png
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