Problem 1531

1. Given Conditions and Setup:
We are given a triangle $ABC$ with external points $X, Y, Z$ such that:
$$\angle BAZ = \angle CAY, \quad \angle CBX = \angle ABZ, \quad \text{and} \quad \angle ACY = \angle BCX.$$
We need to prove that $AX, BY, CZ$ are concurrent.

2. Using Law of Sines:
Consider the triangles $\triangle ABX$ and $\triangle ACX$. By the Law of Sines, we have:
$$\frac{\sin \angle BAX}{BX} = \frac{\sin (\angle ABC + \angle XBC)}{AX}$$
and
$$\frac{\sin \angle CAX}{CX} = \frac{\sin (\angle ACB + \angle XCB)}{AX}.$$

3. Combining Ratios:
Now, consider $\triangle BCX$. By the Law of Sines, we have:
$$\frac{\sin \angle XBC}{CX} = \frac{\sin \angle BCX}{BX}.$$
Combining these ratios, we get:
$$\frac{\sin \angle BAX}{\sin \angle CAX} = \frac{BX \cdot \sin (\angle ABC + \angle XBC)}{CX \cdot \sin (\angle ACB + \angle XCB)} = \frac{\sin \angle XCB \cdot \sin (\angle ABC + \angle XBC)}{\sin \angle XBC \cdot \sin (\angle ACB + \angle XCB)}.$$

4. Applying Similar Steps for Other Vertices:
We can achieve similar equations for the other vertices $B$ and $C$. For example, for $\triangle BCY$ and $\triangle ABY$, and for $\triangle CAZ$ and $\triangle CBZ$.

5. Multiplying the Ratios:
Multiplying the ratios obtained for all three vertices, we get:
$$\left( \frac{\sin \angle BAX}{\sin \angle CAX} \right) \cdot \left( \frac{\sin \angle CBY}{\sin \angle ABY} \right) \cdot \left( \frac{\sin \angle ACZ}{\sin \angle BCZ} \right) = 1.$$
This is a direct application of the Trigonometric form of Ceva's Theorem, which states that for concurrent cevians $AX, BY, CZ$ in $\triangle ABC$, the product of these ratios must equal 1.

6. Conclusion:
By Trigonometric Ceva's Theorem, since the product of the ratios equals 1, the cevians $AX, BY, CZ$ are concurrent.

$\blacksquare$