Problem: Let $ABC$ be a fixed acute triangle
inscribed in a circle $\omega$ with center $O$.
A variable point $X$ is chosen on minor arc $AB$ of $\omega$,
and segments $CX$ and $AB$ meet at $D$.
Denote by $O_1$ and $O_2$ the circumcenters of
triangles $ADX$ and $BDX$, respectively.
Determine all points $X$ for which
the area of triangle $OO_1O_2$ is minimized.

Solution: We prove $[OO_1O_2] \ge \frac14 [ABC]$,
with equality if and only if $\ol{CX} \perp \ol{AB}$.

\paragraph{First approach (Bobby Shen).}
We use two simultaneous inequalities:
\begin{itemize}
  \ii Let $M$ and $N$ be the midpoints of $CX$ and $DX$.
  Then $MN$ equals the length of the $O$-altitude of $\triangle OO_1O_2$,
  since $\ol{O_1O_2}$ and $\ol{DX}$ meet at $N$ at a right angle.
  Moreover, we have \[ MN = \half CD \ge \half h_a \]
  where $h_a$ denotes the $A$-altitude.

  \ii The projection of $O_1 O_2$ onto line $AB$ has length exactly $AB/2$.
  Thus \[ O_1 O_2 \ge \half AB. \]
\end{itemize}
So, we find
\[ [OO_1O_2] = \half \cdot MN \cdot O_1 O_2
  \ge \frac 18 h_a \cdot AB = \frac14 [ABC]. \]
Note that equality occurs in both cases if and only if $\ol{CX} \perp \ol{AB}$.
So the area is minimized exactly when this occurs.


\paragraph{Second approach (Evan's solution).}
We need two claims.
\begin{claim*}
  We have $\triangle O O_1 O_2 \sim \triangle CBA$, with opposite orientation.
\end{claim*}
\begin{proof}
  Notice that $\ol{OO_1} \perp \ol{AX}$
  and $\ol{O_1 O_2} \perp \ol{CX}$,
  so $\dang OO_1O_2 = \dang AXC = \dang ABC$.
  Similarly $\dang OO_2O_1 = \dang BAC$.
\end{proof}

Therefore, the problem is equivalent to minimizing $O_1 O_2$.
\begin{center}
\begin{asy}
pair C = dir(130);
pair A = dir(200);
pair B = dir(340);
pair X = dir(280);
pair D = extension(C, X, A, B);
pair O_1 = circumcenter(A, D, X);
pair O_2 = circumcenter(B, D, X);
pair O = origin;
filldraw(O--O_1--O_2--cycle, opacity(0.1)+lightcyan, deepcyan);
filldraw(A--X--B--cycle, opacity(0.1)+lightred, red);
filldraw(A--B--C--cycle, opacity(0.1)+lightcyan, blue);
draw(X--C, red);
filldraw(O_1--X--O_2--cycle, opacity(0.1)+yellow, orange);
draw(unitcircle, gray);

dot("$C$", C, dir(C));
dot("$A$", A, dir(A));
dot("$B$", B, dir(B));
dot("$X$", X, dir(X));
dot("$D$", D, dir(245));
dot("$O_1$", O_1, dir(O_1));
dot("$O_2$", O_2, dir(20));
dot("$O$", O, dir(45));

/* TSQ Source:

C = dir 130
A = dir 200
B = dir 340
X = dir 280
D = extension C X A B R245
O_1 = circumcenter A D X
O_2 = circumcenter B D X R20
O = origin R45
O--O_1--O_2--cycle 0.1 lightcyan / deepcyan
A--X--B--cycle 0.1 lightred / red
A--B--C--cycle 0.1 lightcyan / blue
X--C red
O_1--X--O_2--cycle 0.1 yellow / orange
unitcircle gray

*/
\end{asy}
\end{center}

\begin{claim*}
  [Salmon theorem]
  We have $\triangle X O_1 O_2 \sim \triangle XAB$.
\end{claim*}
\begin{proof}
  It follows from the fact that $\triangle AO_1 X \sim \triangle BO_2 X$
  (since $\dang ADX = \dang XDB \implies \dang XO_1A = \dang XO_2B$)
  and that spiral similarities come in pairs.
\end{proof}

Let $\theta = \angle ADX$.
The ratio of similarity in the previous claim is equal to
$\frac{XO_1}{XA} = \frac{1}{2\sin \theta}$.
In other words,
\[ O_1 O_2 = \frac{AB}{2 \sin \theta}. \]
This is minimized when $\theta = 90\dg$,
in which case $O_1 O_2 = AB/2$
and $[OO_1O_2] = \frac14 [ABC]$.
This completes the solution.