No problem but please show what I wanted in my first post of the thread
There is no need for a rigorous proof I know it

I think you are

Thank You for your efforts

Yes, Please tell me
Also explain why the perimeter is fixed

To produce the demo do this:

1) In the input bar enter (2,2). Point A will be created.

2) In the input bar enter (10,2). Point B will be created. These will be the foci of the ellipse.

3) In the input bar enter (6,2+sqrt(48)). Point C will be created. This is to scale the ellipse correctly.

4) Use the ellipse tool and click A, B and C in that order and a nice ellipse will be created.

5) Hide C by right clicking it and unchecking show object.

6) Use the point on object tool to create D on the ellipse.

7) Move D around first with the mouse and then with your + and - keys. Notice it stays riveted to the ellipse.

8) Use the polygon tool and click A,B,D and back to A. poly1 will be created in the algebra pane and a number will be associated with it. This value is the area of the triangle.

9) Go into options-> rounding and click 5 decimal places.

10) In the algebra pane look under Segment and right click a. Then go into object properties and check show label. In the input box change it from name to value. Immediatey when you close the property box you will see the BD has a length for its label.

11) Repeat step 10 for b and d.

12) Use the move tool to position the length labels nicely around the triangle.

13) Enter in the input bar:
perimeter = a+b+d The variable "perimeter" will show in the algebra pane. It will have a value of 24.

14) Move D around and notice that though AD and BD change, the perimeter stays constant. This is a consequence of the definition of an ellipse.

15) Now comes the hard part. Move D around with your mouse watching poly1 closely. Try to get the largest value you can in poly1.

16) I got

poly1 = 27.7128
a = 7.99633
b = 8.00367
d = 8

This is highly suggestive that the largest area occurs at sides of length 8. Can we do a little better with the accuracy? Yes!

17) Right click D and go into properties -> Algebra and set increment to .001. Set in options rounding to 10 decimal places.

18) Hold down the shift key and press + or - until poly1 is the largest you can get it. On mine I get

poly1 = 27.7128129163
a = 7.9999258439
b = 8.0000741561
d = 8.

It does look like the equilateral triangle with sides of 8 is the best one. You should have something like figure 1. That ends the demonstration.

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Hi;

You will need a little background.

The foci are the two points that are used to create the ellipse.

Def: An ellipse is a locus (collection) of points such that: For any point P(x,y) on the ellipse, the sum of the distances to two given points F1 and F2 (the foci) is a constant.

http://www.valleyview.k12.oh.us/vvhs/de … guide.html

http://www.mathsisfun.com/geometry/ellipse.html

Like to see a video?

Also, please go through my geogebra tut post #11. It is especially for you and will answer and raise many questions. Geogebra is such a powerful tool for doing geometry. If you work with it you will have good results.

Hi Agnishom;

Anything else I can do? Did you get around to the maximum area tut? Need some help with it?

Hi Agnishom;

Let me know how you do.

Yes, it does.

Hi Agnishom;

Do you have an example you would like shown?

Hi;

I am working on something simple that can be done on geogebra. Please hold on.

Hi;

I suppose the simplest is to show the polar grid and the creation of a single point A.

In polar coordinates geogebra specifies a point a little different than in cartesian.

1) Open geogebra and set the grid for the x and y axis to polar.

2) In the input bar type A = (3;1). The point A will be created in (r,θ) format. See the drawing below.

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