How many isosceles, right-angled, INSCRIBED triangles exist in an ellipse

euclidean-geometrygeometrytriangles

Are there more than four isosceles, right-angled, INSCRIBED triangles in an ellipse?
By inscribed, I mean all three vertexes should lie ON the ellipse.

I am attaching a simplified picture that shows my count of four such triangle.
Two of them form the only inscribed square of an ellipse in the middle.
The other two would exist at each end.

Are there any proof or studies showing that these are the only four, or are there more?

*Edit: As Deepak pointed, when I said four, I was not careful. If you draw the diagonal line in the inscribed square the other way, you can also have two others.

To give you a better idea, I'm specifically wondering whether there are isosceles right angled inscribed triangles at different locations entirely, as is shown in my supplementary diagram that has a question mark.

Note that in that diagram, I'm using a FAKE isosceles triangle (the two sides are not actually of same lengths), as I cannot find a true isosceles triangle. But I'm wondering whether a real one exists that looks similar to what is drawn.enter image description here

enter image description here

Best Answer

Consider the following diagram.

Here $D$ is an arbitrary but fixed point on the ellipse, without loss of generality in the first quadrant. $E$ and $G$ are variable points counterclockwise of $D$, and $E'$ and $G'$ are their $90^\circ$ rotations. $DEE'$ and $DGG'$ are then isosceles right-angled triangles. We find the following:

  • When the second point is close to $D$, as with $E$, then its rotation ($E'$) lies inside the ellipse.
  • When the second point is far from $D$, as with $G$, then its rotation ($G'$) lies outside the ellipse.

So, since motions are continuous, there must be a position for the second point where the third point lies exactly on the ellipse, making an inscribed triangle. Hence there are infinitely many right-angled isosceles triangles inscribed in the ellipse.

As Misha Lavrov says, the exact position for $E$ can be found by rotating the ellipse $90^\circ$ around $D$. $E$ is another ellipse intersection (there may be up to three of them):

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