I only need to know Fig 1 of all the proof. Then every parts of the shape can be some kind of convex. (I think Fig 1 has no problem but others), I think so, but what you are thinking
very sorry, URL allows here? Isoperimetric Theorem and Inequality
then, draw four points on the perimeter that quarter and average the perimeter, and if the points call 1, 2, 3, 4, connect 1, 3 to become a line a, and connect point 2 with a point in the line a that vertical with it, same apply to point 4, then you can see the lines cut the picture to 4 parts, but still with the same perimeter for each parts. then have the part (fan-shaped?) not smaller size than other 3 parts replace other 3 parts, this time you can see the picture is 4 parts symmetric...continue
The entire proof looks correct given in the link looks correct (with the addendum of the existence argument). I am not sure what you are asking. Why are you drawing four points on the perimeter and making symmetric parts? In your example, the symmetric parts will only line up correctly if the line connecting points 1 and 3 bisects the line connecting points 2 and 4 (and vice versa). You have not shown that to be true.
put the 4 parts symmetric shape into coordinate system, and mark a middle point on the perimeter of each quadrant, see <a href="http://tinypic.com?ref=r73jhv" target="_blank"><img src="http://i62.tinypic.com/r73jhv.jpg" border="0" alt="Image and video hosting by TinyPic"></a> and connect it with O point of coordinate system, the line cuts the shape into 8 parts(the red parts and grey parts). take away the grey parts and have the 2 red parts which locate in the 1st and 2nd quadrant stay in the position and have other red parts anticlockwise rotate to the edge of the red part. and also the grey parts, and put them together again to forum a new shape, you can see it only have the perimeter outside.
to Slip: Thank you for your reply and thank you for reading my English. I don't really understand what you say, and I draw a picture. the picture is not fully convex as fig 1, but more or less I think, any problem?
I guess I understand a little of what your mean, the line a is connected with 2 lines. The first line is point 2 connected with a point on line a, and the second line is point 2 connected with another point (if needed) on line a. And I don't think the entire proof is correct
for example, select 2 points on the perimeter, point_1 is pointing to right angle, point_2 is not, we adjust the perimeter and make point_2 pointed to the right angle ( the tip of right angle of right angle triange), while the point_1 is not right angle. So I can say any point on the perimeter should not be the tip of right angle
He randomly selects one shape-a from all the symmetric, convex shapes with same length perimeter (if u want to prove in his way should excluded circular) . And then, if anywhere on perimeter is not pointing the tip of right angle, generate a new shape, and there are so many generates, shape-1, shape-2, ...There is no proof the shape-a or its generations, anyone of them, can contain so many right angle triangles, so his reason is not good.
At first cut-knot want to select all kinds of shape to prove right triangle makes the shape larger in size than not-right triangle, cut-knot needs not-right triangle in the shape to prove right triangle's advantage in the shape, obviously there is no circular shape in his/her select list. And the world do exist the circular but circular does not obey the rule of selecting shapes.
my best level is that, next I will continue to talk about my method of isoperimetric which is no more than just some pictures.
saying the shape of red and grey, if you take to red part away and close the grey part together edge to edge, and make the red part edge to edge too, and again have the two parts close together (vamp? scratch?), the picture become like the last 2 ones in the picture.