Equilateral triangle inscribed in a ellipse "Given any point on a ellipse, is it always possible to inscribe an equilateral triangle, with a vertex coincident with that point, in the ellipse?"
I thought I could use analytical geometry, but when I first come up with a line-long equation I gave it up. Do you think there are faster, and more elegant ways to solve this problem?
 A: @ajotatxe This is a neat approach but your method of picking the points $A_i,A_j$ won't always produce an equilateral triangle; consider for example an oblong ellipse resembling a line, and a point $P$ in the middle of this ellipse. Here is an attempt at improvement:
Fix a point $P$ on an ellipse $E$, and point $T≠P$ on the tangent line of $E$ at $P$. For each $0≤\theta≤120$ draw two lines $A_\theta P$ and $B_\theta P$ in a clockwise fashion such that $\angle A_\theta PT=\theta°$ and $\angle A_\theta PB_\theta =60°$. Consider the points $I_\theta ,J_\theta$ where $AP$ and $BP$, respectively, intersect $E$ ($I_\theta=P$ iff $\theta =0$, and similarly $J_\theta = 0$ iff $\theta = 120$), and define $f(\theta)=I_\theta P-J_\theta P$. 
Clearly $f$ is continuous on $[0,120]$ and $f(0)<0$, $f(120)>0$, so that $f(\phi)=0$ for some $0<\phi <120$. Then $I_\phi P=J_\phi P$ and $\angle I_\phi PJ_\phi =60°$so that $\triangle I_\phi J_\phi P$ is equilateral. Q.E.D.
It seems that this method works for any convex continuous shape, not just ellipses.
A: I have an idea, but a rigorous proof based on this idea may be a bit tough. Nevertheless, perhaps you find it useful.
For each $r>0$, consider the circle centered on the given point $P$ of the ellipse with radius $r$.
If the circle intersects the ellipse two points, they will be at a distance $d(r)$. We define $f(r)=d(r)/r$. If the circle intersects the ellipse more than two points $A_1,\ldots\,A_n$ take $A_i, A_j$ such that the arc $A_iPA_j$ doesn't cointain any other $A_k$. 
If the circle is tangent to the ellipse (being the ellipse within the circle), define $f(r)=0$.
If the circle and the ellipse don't share any points, $f$ remains undefined.
It seems that for small enough $r$, the triangle defined by the given point and the intersection has an obtuse angle, so $f(r)>1$. It also seems that $f$ is continuous where we have defined it. So there will be some $r$ such that $f(r)=1$.
A: You have your point P on the ellipse.  Imagine a point Q infinitesimally close to P, also on the ellipse, and a point R towards the center of the ellipse that forms an equilateral triangle with P and Q.  R is inside the ellipse.
Now slide Q around the ellipse to a point on the ellipse that is farthest from P, maintaining PQR as an equilateral triangle.  The distance between P and R will be the same as P and Q, so R must be either outside the ellipse or on the edge.  If R is on the edge then PQR is your inscribed equilateral triangle.  If R is outside the ellipse then at some point on Q's slide around the ellipse R moved from inside the ellipse to outside it, and then PQR was an equilateral triangle inscribed in the ellipse.
