BMO Short List

Let the vertices of the squares be $A{C}_1{C}_{2}B$, $B{A}_1{A}_{2}C$, $C{B}_1{B}_{2}A$.

Lemma: $B{B}_2=C{C}_1$ and $B{B}_2\perp C{C}_1$.

Proof: Notice that by rotating $△A{C}_{1}C$ by $90^{\circ }$ we get $△AB{B}_2$ proving the lemma.

Claim: $A{O}_a\perp O_b{O}_c$

Proof: Let $M$ be the midpoint of $AB$. By our lemma applied at vertex $C$ we get $A{A}_2=B{B}_1$ and they are perpendicular. By homothety of factor $2$ at $A$ and then $B$ we get: $$M{O}_b=\frac{1}{2}B{B}_1=\frac{1}{2}A{A}_2=M{O}_a\text{and}M{O}_b\parallel B{B}_1,M{O}_a\parallel A{A}_2$$ Hence $M{O}_a$, $M{O}_b$ are also perpendicular so in fact $△O_{b}M{O}_a$ is an isosceles right triangle. This is also trivially the case for $△AM{O}_c$. Now applying our lemma to $△AM{O}_b$ at vertex $M$ we get $O_b{O}_c$ and $A{O}_a$ are perpendicular which is exactly what we wanted.

Similarly we get $B{O}_b\perp O_a{O}_c$ and $C{O}_c\perp O_a{O}_b$ so lines $A{O}_a$, $B{O}_b$, $C{O}_c$ concur at $H$, the orthocentre of $△O_a{O}_b{O}_c$. As $A$ lies on $\omega$ and on $O_{a}H$ it follows $A$ is the reflection of $H$ in line $O_b{O}_c$.



Claim: $H=B$ or $H=C$

Proof: Assume not. By the previous observations we get $O_{c}H=O_{c}A=O_{c}B$. Hence as $O_c{O}_a\perp BH$ and $B\ne H$ this means $B$ is the reflection of $H$ in $O_c{O}_a$ so $B$ lies on $\omega$.

Similarly, $C$ lies on $\omega$. But then we get: $$\angle ACB=180^{\circ }-\angle B{O}_{c}A=90^{\circ }\text{and}\angle CBA=180^{\circ }-\angle A{O}_{b}C=90^{\circ }$$ 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: $$\angle A{O}_{c}C=\angle A{O}_{c}B=90^{\circ }$$ Also $\angle A{O}_{b}C=90^{\circ }$ 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 $△ABC$.