Problem 1486

1. Given: Point $M$ is chosen in triangle $ABC$ such that the radii of the circumcircles of triangles $\triangle AMC$, $\triangle BMC$, and $\triangle BMA$ are no smaller than the radius of the circumcircle of $\triangle ABC$.

2. To Prove: All four radii are equal.

3. Step 1: Let $R$ be the circumradius of $\triangle ABC$, and let $R_1$, $R_2$, and $R_3$ be the circumradii of $\triangle AMC$, $\triangle BMC$, and $\triangle BMA$ respectively. We are given that:
$$R_1 \geq R, \quad R_2 \geq R, \quad R_3 \geq R$$

4. Step 2: Recall the formula for the circumradius $R$ of a triangle $\triangle ABC$:
$$R = \frac{a}{2 \sin A} = \frac{b}{2 \sin B} = \frac{c}{2 \sin C}$$
where $a, b, c$ are the sides opposite to angles $A, B, C$ respectively.

5. Step 3: For the circumradius of $\triangle AMC$ to be at least $R$, we must have:
$$\angle AMC \geq 180^\circ - \angle ABC$$
This is because the circumradius $R_1$ of $\triangle AMC$ is given by:
$$R_1 = \frac{AM \cdot CM}{2 \sin \angle AMC}$$
For $R_1 \geq R$, we need $\sin \angle AMC \leq \sin (180^\circ - \angle ABC) = \sin \angle ABC$, which implies $\angle AMC \geq 180^\circ - \angle ABC$.

6. Step 4: Similarly, for the circumradius of $\triangle BMC$ to be at least $R$, we must have:
$$\angle BMC \geq 180^\circ - \angle BAC$$

7. Step 5: For the circumradius of $\triangle BMA$ to be at least $R$, we must have:
$$\angle BMA \geq 180^\circ - \angle ACB$$

8. Step 6: Adding these inequalities, we get:
$$\angle AMC + \angle BMC + \angle BMA \geq (180^\circ - \angle ABC) + (180^\circ - \angle BAC) + (180^\circ - \angle ACB)$$
Simplifying, we have:
$$\angle AMC + \angle BMC + \angle BMA \geq 540^\circ - (\angle ABC + \angle BAC + \angle ACB)$$
Since $\angle ABC + \angle BAC + \angle ACB = 180^\circ$ (sum of angles in $\triangle ABC$), we get:
$$\angle AMC + \angle BMC + \angle BMA \geq 540^\circ - 180^\circ = 360^\circ$$

9. Step 7: However, the sum of the angles around point $M$ in the plane is exactly $360^\circ$. Therefore, equality must hold in all the inequalities:
$$\angle AMC = 180^\circ - \angle ABC, \quad \angle BMC = 180^\circ - \angle BAC, \quad \angle BMA = 180^\circ - \angle ACB$$

10. Step 8: Since equality holds, the circumradii $R_1, R_2, R_3$ must all be equal to $R$.

$\blacksquare$