Baltic Way 2025 · Problem 15

Let $O$ and $\omega$ denote the circumcentre and circumcircle of $ABC$. Without loss of generality, suppose $O,A_b,A_c$ are collinear in this order. Then

$$\angle OA_bA=90^\circ-\angle ABC=\angle AA_cO,$$

so

$$\triangle A_bAO\sim\triangle AA_cO.$$

Therefore

$$\frac{OA}{OA_b}=\frac{OA_c}{OA},$$

or equivalently

$$OA^2=OA_b\cdot OA_c.$$

Thus $A_b$ and $A_c$ are interchanged by inversion in $\omega$. By symmetry, the same is true for $B_c,B_a$ and for $C_a,C_b$. Consequently the circumcircles of $A_bB_cC_a$ and $A_cB_aC_b$ are interchanged by inversion in $\omega$. The centres of two inverse circles and the centre of inversion are collinear, so their circumcentres are collinear with $O$, as required.