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I'd like to "resurrect" this not long time ago deleted question. It looks interesting and not immediately obvious (unless I'm missing something trivial).

enter image description here

Given an acute triangle $ABC$ and its circumscribed circle centered at $O$. A variable point $X$ is placed on the minor arc $AB$ of the circle; segments $CX$ and $AB$ meet at $D$. The circumcenters of $\triangle ADX$ and $\triangle BDX$ are $Y$ and $Z$, respectively. How can we find the location of the point $X$ for which the area of $\triangle OYZ$ is minimized?

Numerical tests suggest that $\triangle OYZ$ is always similar to the reference $\triangle ABC$, $\angle ZOY=\angle BCA$, and

\begin{align} \min_{X\in AB}S_{OYZ}(X) &=\tfrac14\,S_{ABC} \end{align}

when $CX\perp AB$.

Complex numbers/coordinate geometry approach with unit circle centered at the origin using known function for line/line intersection and the location of the circumcenter based on the coordinates of the three vertices lead to too unreasonably overcomplicated expressions.

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  • $\begingroup$ I don't have the time to fully calculate this, but you can WLOG define triangle ABC so that the points are on the unit circle, line AB is parallel to the x axis, and line AB has a negative y coordinate. We can then define points D, Y, and Z as a function on the x coordinate of A, the x coordinate of C, and the x coordinate of X. Then, using Gauss's shoelace formula, we can find the area of triangle OYZ as a function of these three variables. Then, optimize the area by taking the partial derivative with respect to the x coordinate of X and solving to find minima and maxima $\endgroup$
    – Moko19
    Jun 22 '20 at 13:14
  • $\begingroup$ @Moko: that's exactly what I tried to do, but as I mentioned, expressions become too complex. $\endgroup$
    – g.kov
    Jun 22 '20 at 13:22
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  1. How do we find $Y$ and $Z$? $Y$ lies on perpendicular bisectors of $AX$ and $DX$, $Z$ lies on perpendicular bisectors of $XB$ and $DX$. Note, that $O$ lies on perpendicular bisectors of $AX$ and $XB$.

figure for perpendicular bisectors

  1. Perform homothety centred at $X$ with coefficient $2$. Midpoints of $AX$, $DX$ and $XB$ go to points $A$, $D$, $B$ respectively. Perpendicular bisectors go to perpendicular lines, passing through these points, and $O$ goes to diametrically opposite to $X$, say $X'$.

figure for homothety and the solution

  1. $\angle X'CX=90^\circ$ $\Rightarrow$ $CX'\perp CD$ and $Y'Z'\perp CD$ $\Rightarrow$ $CX'||Y'Z'$
  2. Though there are some cases of points positioning, but I'm feeling lazy to deal with it. Sometimes an angle is simply that angle but sometimes $180^\circ-$angle.
    $\angle A=180^\circ-\angle CX'B=\angle X'Z'Y'$
    $\angle B=\angle CX'A=\angle X'Y'Z'$
    $\Rightarrow$ $\triangle X'Z'Y'\sim \triangle CAB$
  3. $CX'||Y'Z'$ $\Rightarrow$ length of $CD$ is the length of height from $X'$ in the $\triangle Y'X'Z'$.
    In the $\triangle ACB$ length of the corresponding height is not more than $CD$ (and equal when $CD$ is the height), i.e. the similarity coefficient $\displaystyle\frac{CD}{h}\ge 1$ and $S_{X'Y'Z'}=4S_{OYZ}$.
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    $\begingroup$ Downvoters please care to explain why. Thanks. $\endgroup$ Jun 22 '20 at 11:52
  • $\begingroup$ @Alexey Burdin: agree. My wild guess - perhaps, some oversized images? Just in case, I crop them a little. $\endgroup$
    – g.kov
    Jun 22 '20 at 13:15
  • $\begingroup$ I have access to the exact geogebra sources -- here and here -- in case you have an idea how can the images be made better. Now they appears oversized)) I'd leave height="250" alone))) Thanks. Is solution clear? $\endgroup$ Jun 22 '20 at 13:21
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enter image description here

Using standard notation for the angles of $\triangle ABC$. inscribed into the unit circle, $\angle XOA=\theta\in(0,2\gamma)$, coordinates of the points are

\begin{align} O&=(0,0) ,\quad A=(-\sin\gamma,\,-\cos\gamma) ,\quad B=(\phantom{-}\sin\gamma,\,-\cos\gamma) \tag{1}\label{1} ,\\ C&=(\sin(2\alpha+\gamma),\,-\cos(2\alpha+\gamma)) ,\quad X=(\sin(\theta-\gamma),\, -\cos(\theta-\gamma)) \tag{2}\label{2} ,\\ D&=\left( \frac{ 2\sin(\alpha+\gamma)\sin\tfrac12\theta -\sin\gamma\sin(\alpha+\tfrac12\theta) }{\sin(\alpha+\tfrac12\theta)} ,\, -\cos\gamma \right) \tag{3}\label{3} . \end{align}

Corresponding coordinates of the points $Y$ and $Z$ can be found as

\begin{align} Y&=\left( -\sin\gamma +\frac{\sin(\alpha+\gamma)\sin\tfrac12\theta}{\sin(\alpha+\tfrac12\theta)} ,\, \phantom{-}\frac{\sin\alpha\sin(\theta-2\gamma)} {2\,\sin(\alpha+\tfrac12\theta)\sin(\gamma-\tfrac12\theta)} \right) \tag{4}\label{4} ,\\ Z&=\left( \phantom{-}\sin\gamma +\frac{\sin\alpha\sin(\tfrac12\,\theta-\gamma)}{\sin(\alpha+\tfrac12\,\theta)} ,\, -\frac{\sin(\alpha+\gamma)\sin\theta}{2\,\sin(\alpha+\tfrac12\,\theta)\sin\tfrac12\,\theta} \right) \tag{5}\label{5} . \end{align}

Given the coordinates of $O,Y,Z$, the squared area of $\triangle OYZ$ in terms of $\theta$ can be found as

\begin{align} S_{OYZ}^2(\theta)= &\left( \sin\tfrac12\theta\sin(\gamma-\tfrac12\theta) \, ( \sin^2\alpha\, \sin(2\, \gamma-\theta)+\sin^2\beta\, \sin\theta ) \right. \\ &\left. -\sin\gamma\, \sin(\alpha+\tfrac12\theta) \, ( \sin\alpha\, \sin(\tfrac12\theta)\, \sin(2\, \gamma-\theta) +\sin\beta\, \sin\theta\, \sin(\gamma-\tfrac12\theta) ) \right)^2 \\ &\left/(4\, \sin\tfrac12\theta\, \sin(\gamma-\tfrac12\theta)\, \sin^2(\alpha+\tfrac12\theta))^{2} \right. \tag{6}\label{6} . \end{align}

Expression for $\frac d{d\theta}(S_{OYZ}^2(\theta))$ is rather complicated, but it indeed reaches zero at $\theta=\pi-2\alpha$, when $CX\perp AB$.

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