This is a UK proposal by Sam Bealing (me). Solution 1: Synthetic Let the vertices of the squares be $AC_1C_2B$, $BA_1A_2C$, $CB_1B_2A$ and label $\triangle ABC$ anticlockwise. We will make repeated use of the following lemma: Lemma: $BB_2=CC_1$ and $BB_2 \bot CC_1$. Proof: Notice that by rotating $\triangle AC_1C$ by $90^{\circ}$ we get $\triangle ABB_2$ proving the lemma. Claim: $AO_a \bot O_b O_c$ Proof: Let $M$ be the midpoint of $AB$. By our lemma applied at vertex $C$ we get $AA_2=BB_1$ and they are perpendicular. By homothety of factor $2$ at $A$ and then $B$ we get: $$MO_b=\frac{1}{2} BB_1=\frac{1}{2} AA_2=MO_a \quad \text{and} \quad MO_b \parallel BB_1 \; , \; MO_a \parallel AA_2$$Hence $MO_a,MO_b$ are also perpendicular so in fact $\triangle O_bMO_a$ is an isosceles right triangle. This is also trivially the case for $\triangle AMO_c$. Now applying our lemma to $\triangle AMO_b$ at vertex $M$ we get $O_bO_c$ and $AO_a$ are perpendicular which is exactly what we wanted. Similarly we get $BO_b \bot O_aO_c$ and $CO_c \bot O_aO_b$ so lines $AO_a,BO_b,CO_c$ concur at $H$, the orthocentre of $\triangle O_a O_b O_c$. As $A$ lies on $\omega$ and on $O_aH$ it follows $A$ is the reflection of $H$ in line $O_bO_c$. Claim: $H=B$ or $H=C$. Proof: Assume not. By the previous observations we get: $$O_cH=O_cA=O_cB$$Hence as $O_{c}O_a \bot BH$ and $B \neq H$ this means $B$ is the reflection of $H$ in $O_cO_a$ so $B$ lies on $\omega$. Similarly, $C$ lies on $\omega$. But then we get: $$\angle ACB=180^{\circ}-\angle BO_cA=90^{\circ} \quad \text{and} \quad \angle CBA=180^{\circ}-\angle AO_bC=90^{\circ}$$so $\angle ACB+\angle CBA=180^{\circ}$ which is absurd so in fact one of $B,C$ is equal to $H$. WLOG $B=H$. As $A,H,O_a$ and $C,H,O_c$ are collinear this means in fact $B$ lies on these lines. Hence: $$\angle AO_cC=\angle AO_cB=90^{\circ}$$Also $\angle AO_bC=90^{\circ}$ hence $C$ also lies on $\omega$ and $\omega$ in fact has diameter $AC$ and so its circumcentre is the midpoint of $AC$ which lies on the perimeter of $\triangle ABC$. Solution 2: Cartesians We use Cartesians with $A=(0,0)$, $B=(2,0)$, $C=(2\alpha,2 \beta)$. Then we calculate: $$O_{A}=\left(1+\alpha+\beta,1-\alpha+\beta\right) \quad O_{B}=\left(\alpha-\beta,\alpha+\beta\right) \quad O_{C}=\left(1,-1\right)$$Now if we label the points $A,O_C,O_B,O_A$ to have coordinates $(x_i,y_i)$ for $1 \leq i \leq 4$ then the condition that they are concyclic is equivalent to (by expanding out the circle equation $(x-x_0)^2+(y-y_0)^2=r^2$ and converting to a linear problem for $x_0,y_0,r^2,1$ for which we seek non-trivial solutions): $$ 0=\det{ \begin{pmatrix} 1 & 2 x_1 & 2y_1 & -x_1^2-y_1^2 \\ 1 & 2 x_2 & 2y_2 & -x_2^2-y_2^2 \\ 1 & 2 x_3 & 2y_3 & -x_3^2-y_3^2 \\ 1 & 2 x_4 & 2y_4 & -x_4^2-y_4^2 \end{pmatrix}} $$Using $x_1=x_2=0$ and expanding about the top row this reduces to: $$ 0=\begin{vmatrix} 2 & -2 & -2 \\ 2(\alpha-\beta) & 2(\alpha+\beta) & -2(\alpha^2+\beta^2) \\ 2(\alpha+\beta+1) & 2(1-\alpha+\beta) & 2\left(\alpha^2+(\beta+1)^2\right) \end{vmatrix} =8 \begin{vmatrix} 1 & -1 & -1 \\ \alpha-\beta & \alpha+\beta & -\alpha^2-\beta^2 \\ 2 \beta & 2(1-\alpha) & -2 \beta \end{vmatrix} $$$$ \cdots=16 \begin{vmatrix} 1 & 0 & 0 \\ \alpha-\beta & 2 \alpha & -\alpha^2-\beta^2-\alpha+\beta \\ \beta & 1+\beta-\alpha & 0 \end{vmatrix} =16(\alpha^2+\beta^2+\alpha-\beta)(1+\beta-\alpha) $$So $A$ lies on $\omega$ iff: \begin{align*} 0&=1+\beta-\alpha \tag{1} \\ 0&=\alpha^2+\beta^2+\alpha-\beta \tag{2} \end{align*}Now we consider the circumcentre $O$ of $\odot AO_{C}O_{B}$. By the fact that it lies on the perpendicular bisector of $AO_{C}$ it has form: $$O=\left(\frac{1}{2}+\lambda,-\frac{1}{2}+\lambda \right)$$Now the condition that $O$ is on the perp bisector of $AO_{B}$ becomes: \begin{align*} 0&=\left(\alpha-\beta,\beta+\alpha \right) \cdot \left(1+2 \lambda-\alpha+\beta,-1+2 \lambda-\alpha-\beta \right) \\ &=\left(-2\beta-(\alpha-\beta)^2-(\alpha+\beta)^2\right)+4 \alpha \lambda \\ &=-2\left(\beta+\alpha^2+\beta^2\right)+4 \alpha \lambda \end{align*}so: $$\lambda=\frac{\alpha^2+\beta^2+\beta}{2\alpha} \Rightarrow O=\left(\frac{\alpha^2+\beta^2+\alpha+\beta}{2\alpha},\frac{\alpha^2+\beta^2+\beta-\alpha}{2\alpha}\right)$$Now we see $(2)$ corresponds to the $y$-coordinate of $O$ being $0$ i.e. $O$ is on $BC$. To see what $(1)$ corresponds to consider: \begin{align*} AO \parallel AC \Leftrightarrow 0&=\left(\frac{\alpha^2+\beta^2+\alpha+\beta}{2\alpha},\frac{\alpha^2+\beta^2+\beta-\alpha}{2\alpha}\right) \cdot \left(-\beta,\alpha\right) \\ &=\frac{(\alpha-\beta)\left(\alpha^2+\beta^2\right)-\left(\alpha^2+\beta^2\right)}{2\alpha}\\ &=-\frac{(1+\beta-\alpha)\left(\alpha^2+\beta^2\right)}{2\alpha} \\ \Leftrightarrow & 0=1+\beta-\alpha \end{align*}so we see $(1)$, $(2)$ correspond exactly to $O$ lying on $AB,AC$ respectively. Solution 3: Cartesians If we instead set $A=(0,0)$, $B=(2,2b)$, $C=(-2,2c)$ we get: $$O_{A}=\left(c-b,2+b+c\right) \quad O_{B}=(-1-c,-1+c) \quad O_{C}=(1+b,b-1)$$Expanding out a similar determinant condition for $AO_AO_BO_C$ cyclic gives: $$0=16\left(2+b+c+b^2-bc\right)\left(2+b+c+c^2-bc\right)$$The circumcentre $O$ of $\odot AO_{B}O_{C}$ has coordinates: $$O=\frac{1}{2(bc-1)} \left((b-c)(bc-b-c-1),2+b+b+b^2+c^2+b^2c+bc^2\right)$$and we have: \begin{align*} AO \parallel AC &\Leftrightarrow 0=(2+b+c-bc+b^2)(1+c^2) \Leftrightarrow 2+b+c-bc+b^2=0 \\ AO \parallel AB & \Leftrightarrow 0=(2+b+c+c^2-bc)(1+b^2) \Leftrightarrow 2+b+c+c^2-bc=0 \end{align*}which is exactly equivalent to $A$ lying on $\omega$. Note: We should worry about if $bc=1$ as then we're dividing through by $0$ to find $O$. This case corresponds to $\angle A=90^{\circ}$ in which case $A,O_B,O_C$ are collinear so $A$ cannot lie on $\omega$.