# Minimizing the area of triangle based on three circumcenters

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).

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.

• 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 Jun 22 '20 at 13:14
• @Moko: that's exactly what I tried to do, but as I mentioned, expressions become too complex. Jun 22 '20 at 13:22

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 $$XB$$.

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 $$X'$$.

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 $$S_{X'Y'Z'}=4S_{OYZ}$$.
• Downvoters please care to explain why. Thanks. Jun 22 '20 at 11:52
• @Alexey Burdin: agree. My wild guess - perhaps, some oversized images? Just in case, I crop them a little. Jun 22 '20 at 13:15
• 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? Jun 22 '20 at 13:21

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 $$CX\perp AB$$.