Recent content by bob012345

Could we say his ‘proof’ wasn’t even marginal?

The two solutions with square edges along the triangle sides cut the triangle into either two or three smaller 3-4-5 triangles. The case where only three points of the square touch the sides of the triangle cuts it into two triangles different from the 3-4-5 and one four sided shape.

I got the maximum and one relative maximum but missed the minimum. I had the diagonal of the square vertical which gives s=12/11 root(2).

It might be interesting to consider a square inscribed in a 3-4-5 triangle. If the square must touch all three sides, what is the largest, the smallest possible?

Apparently, this construction, the original square with semi-diagonals as @DaveC426913 drew in post #18, goes back thousands of years and has numerous properties of interest. There is a Wooden book called The Diagram all about it. This construction is also called the Helicon. Wooden books...

Yes, I rounded it to fit the grids of the graph paper from 2.828 to 2.75 since the grids were quarter inch. Made the drawing a lot easier!

It looks off but I computed the coordinates and checked the edge lengths which are ##2-√2##.

Finally, I think you meant this.

I thought you meant the squares inscribed in the irregular octagon.

For the irregular octagon, there are two squares inside it but they are not equal. The ratio of areas is 1.125.

Here is a simple drawing of the regular octagon within the square. If the square has corners at (±1,±1), the outside points are root(2) from the center. The angle of the lines to the sides is 22.5 degrees.

I was looking at how to make a regular octagon in the spirit of the OP but it’s difficult to construct the ratio you mentioned without invoking a compass or just measuring which I’m trying to avoid.

The edges are the same lengths but parallel edges are skewed wrt each other, they are parallelograms not rectangles.