circles and centers of triangles Let circle $O$ have radius $10$ and let $H$ be a point such that $OH=6$. Call an arbitrary point $X$ beautiful that lies on or inside all possible triangles $ABC$ with circumcenter $O$ and orthocenter $H$. Find the area of the region that consists of all beautiful points.

One possibility is shown above. $X$ can be on or inside this triangle. I have tried to work with this but cannot seem to find any relation between the given items and the area of this triangle. Also, I am not sure how to bound the area of the total region.
 A: All points in $\odot O$ are beautiful. I used complex numbers. We have
$|A|=|B|=|C|=10$. Wlog $H=6$. Now by the following method we can find all $\triangle ABC$s.
First we prove that $H=A+B+C$. The orthocenter of $\triangle ABC$ is unique and the point $H'=A+B+C$ meets the definition of an orthocenter because
$$AH'\perp BC\Leftrightarrow\frac{B+C}{B-C}\in i\Bbb R.$$
Taking a conjugate,
$$
\overline{\left(\dfrac{B+C}{B-C}\right)}
=\dfrac{\overline{B} + \overline{C}}{\overline{B} - \overline{C}}
=\dfrac{\frac{100}B+\frac{100}C}{\frac{100}B-\frac{100}C}
=\dfrac{B+C}{C-B}=-\dfrac{B+C}{B-C},
$$
where we used the the identity $\overline{X} = \frac{|X|^2}{X}, \forall X \in \mathbb C\setminus\{0\}$.
Done.
Pick any $C\in O(0,0)$ and put $D = H-C$. Then $A = H - (B+C) = D-B$ and likewise $B = H - (A+C) = (H-C)-A = D-A$. Therefore both $A$ and $B$ lie on the cicle centered at $D$ with radius 10, and so they can be found by intersecting this circle with $O(0,0)$. This shows that $A$ and $B$ with desired properties can be constructed for any choice of $C$ on $O(0,0)$.
Consider the middle of the triangle that is between $C$ and $C'$, the middle of the segment $AB$. We have $C' = (A+B)/2 = (0,3)-C/2$. Then the point $C'':= (C+2C')/3 = (0,2)$ is also on the midline. We conclude that the segment between the point $E=(0,2)$ and $C$ is beautiful. As $C$ moves around, the segments $EC$ cover the whole circle $O(0,0)$.

So the answer should be $$S=10^2\cdot\pi=\boxed{100\pi}.$$
A: I thought of another, more interesting, way to interpret this question: find the locus of the beautiful points lying inside ALL triangles. And it turns out that this locus is an ellipse.
To see why, remember that the foot of each altitude of triangle $ABC$ lies on the nine-point circle, which is a circle centred at $N$ (the midpoint of $OH$) and with radius ${1\over2}OA=5$ (see figure below). Hence points $B$ and $C$ are fixed once $A$ has been chosen: the intersection $A'$ between line $AH$ and the nine-point circle gives the foot of the altitude, and $BC$ is perpendicular to that line at $A'$.
But the sides of triangle $ABC$ are tangent to the ellipse with foci $O$, $H$ and major axis $OA=10$ (orange in figure below). This follows from a well-known construction: given a point external to the ellipse (point $C$, for instance), the intersections between the auxiliary circle of the ellipse (which in this case is the nine-point circle) and the circle of diameter $HC$ lie on the lines through $C$ tangent to the ellipse. But those intersections are just the feet of the altitudes $A'$ and $B'$ (by Thales's theorem), hence those tangents are the same as sides $BC$ and $AC$.
The ellipse defined above is then the intersection of all possible triangles and is the locus to be found. Its area is $20\pi$, given its semi-axes have lengths $5$ and $4$.

