Problem 1747

1. Restate the given condition: The points $A_1, B_1, C_1$ lie on the sides $BC, CA,$ and $AB$ of the triangle $ABC$ respectively. We are given that $AB_1 - AC_1 = CA_1 - CB_1 = BC_1 - BA_1$.

2. Reformulate the condition: The given condition can be replaced with $AB_1 + BA_1 = AB$, $CA_1 + AC_1 = CA$, and $BC_1 + CB_1 = BC$.

3. **Consider the point $A_1$ on $BC$**: Let $A_2$ and $C_2$ be the points on $CA$ such that $CA_1 = CA_2$ and $AC_1 = AC_2$. Since $CA_1 + AC_1 = CA$, we have $A_2 = C_2$.

4. **Define $C_1$**: Take the circle with center $C$ and radius $CA_1$. Let this circle meet the segment $CA$ at $K$. The circle with center $A$ and radius $AK$ meets the segment $AB$ at $C_1$. Hence, $C_1$ is uniquely determined by $A_1$, and the same goes for $B_1$.

5. **Consider the incentre $I$ of triangle $ABC$**: Consider the circle with center $I$ and radius $IA_1$. If this circle is tangent to $BC$, then $A_1$ is the foot of the perpendicular from $I$ to $BC$. Thus, this circle is tangent to the sides $CA$ and $AB$ too. In this case, $O_A, O_B, O_C$ are the midpoints of $IA, IB, IC$ respectively, and the problem follows since $\triangle O_AO_BO_C$ is an enlargement of $\triangle ABC$ with scale factor $\frac{1}{2}$.

6. **Otherwise, the circle is not tangent to any sides of triangle $ABC$**: Let it cross $BC, BC, CA, CA, AB, AB$ in $A_1, A_2, B_0, B_2, C_0, C_2$ (such that the 6 points appear in that order on the circle). Remember that $A_1$ uniquely determines $B_1$ and $C_1$. We will prove that $B_0 = B_1$ and $C_0 = C_1$.

7. **Let $D, E$ be the feet of the perpendiculars from $I$ to $BC, CA$ respectively**: Then $ID \perp A_1A_2$ so $D$ is the midpoint of $A_1A_2$. Similarly, $E$ is the midpoint of $B_2B_0$. Clearly, $CD = CE$. Let $DA_1 = DA_2 = x$ and $EB_2 = EB_0 = y$. Then the power of $C$ with respect to the circle is $CA_2 \cdot CA_1 = CB_2 \cdot CB_0 \implies (CD - x)(CD + x) = (CE - y)(CE + y) \implies CD^2 - CE^2 = x^2 - y^2$ so $x = y$. We conclude that $CA_2 = CB_0$ and $CA_1 = CB_2$.

8. **Reconstruct $C_1$**: Recalling the construction of $C_1$ at the start, we see that $B_2$ is the point $K$, i.e., $AC_1 = AB_2$. By a symmetric argument, we can prove that $AC_1 = AB_2$ and $AC_2 = AB_0$, and so we conclude $C_0 = C_2$. Similarly, $B_0 = B_2$.

9. Equal lengths: Note that $x = y \implies A_1A_2 = B_1B_2$ and symmetrically this length must also be equal to $C_1C_2$.

10. Cyclic quadrilaterals: Note that $\angle IA_1A_2 = \angle IA_2A_1 = 180^\circ - \angle IA_2C = 180^\circ - \angle IB_1C$ due to isosceles triangles. Thus, $IB_1CA_1$ is cyclic (with center $O_C$). Similarly, $O_A$ is the center of $(AB_1IC_1)$. So $IB_1$ is the radical axis of the circles $AB_1C_1$ and $A_1B_1C$. Thus, by symmetry, $\angle IO_CO_A = \frac{1}{2} \angle IO_CB_1$. Now $\angle IO_CB_1 = 2 \angle ICB_1 = C$. So $\angle IO_CO_A = \frac{1}{2} C$. Similarly, $\angle IO_3O_2 = \frac{1}{2} C$ and so $IO_C$ bisects $\angle O_AO_CO_A$. Similarly, $IO_A$ bisects $\angle O_CO_AO_B$ and $IO_B$ bisects $\angle O_AO_BO_C$ and therefore $I$ is the incentre of triangle $O_AO_BO_C$.

$\blacksquare$