In $\triangle ABC$, $AB = 40$, $BC = 60$, and $CA = 50$. The angle bisector of $\angle A$ intersects the circumcircle of $\triangle ABC$ at $A$ and $P$. Find $BP$.

[i]Proposed by Eugene Chen[/i]
1. **Scale the Triangle:**
   Given the sides of $\triangle ABC$ are $AB = 40$, $BC = 60$, and $CA = 50$, we can scale the triangle down to a simpler ratio. Let's scale it down by a factor of 10, resulting in a triangle with sides $AB = 4$, $BC = 6$, and $CA = 5$.

2. **Angle Bisector Theorem:**
   Let $D$ be the point where the angle bisector of $\angle A$ intersects $BC$. By the Angle Bisector Theorem, we have:
   \[
   \frac{BD}{DC} = \frac{AB}{AC} = \frac{4}{5}
   \]
   Let $BD = x$ and $DC = y$. Then:
   \[
   x + y = 6 \quad \text{and} \quad \frac{x}{y} = \frac{4}{5}
   \]
   Solving these equations, we get:
   \[
   x = \frac{4}{9} \cdot 6 = \frac{24}{9} = \frac{8}{3}
   \]
   \[
   y = \frac{5}{9} \cdot 6 = \frac{30}{9} = \frac{10}{3}
   \]
   Thus, $BD = \frac{8}{3}$ and $DC = \frac{10}{3}$.

3. **Stewart's Theorem:**
   To find $AD$, we use Stewart's Theorem, which states:
   \[
   AB^2 \cdot DC + AC^2 \cdot BD = AD^2 \cdot BC + BD \cdot DC \cdot BC
   \]
   Plugging in the values:
   \[
   4^2 \cdot \frac{10}{3} + 5^2 \cdot \frac{8}{3} = AD^2 \cdot 6 + \frac{8}{3} \cdot \frac{10}{3} \cdot 6
   \]
   \[
   16 \cdot \frac{10}{3} + 25 \cdot \frac{8}{3} = 6AD^2 + \frac{480}{9}
   \]
   \[
   \frac{160}{3} + \frac{200}{3} = 6AD^2 + \frac{160}{3}
   \]
   \[
   \frac{360}{3} = 6AD^2
   \]
   \[
   120 = 6AD^2
   \]
   \[
   AD^2 = 20
   \]
   \[
   AD = \sqrt{20} = 2\sqrt{5}
   \]

4. **Power of a Point:**
   By the Power of a Point theorem, we have:
   \[
   AD \cdot DP = BD \cdot DC
   \]
   Let $DP = x$. Then:
   \[
   2\sqrt{5} \cdot x = \frac{8}{3} \cdot \frac{10}{3}
   \]
   \[
   2\sqrt{5} \cdot x = \frac{80}{9}
   \]
   \[
   x = \frac{80}{9 \cdot 2\sqrt{5}} = \frac{40}{9\sqrt{5}} = \frac{40\sqrt{5}}{45} = \frac{8\sqrt{5}}{9}
   \]

5. **Similarity of Triangles:**
   Since $\triangle BDP$ and $\triangle ADC$ are isosceles and similar, we have:
   \[
   \frac{BP}{BD} = \frac{AD}{DC}
   \]
   \[
   \frac{BP}{\frac{8}{3}} = \frac{2\sqrt{5}}{\frac{10}{3}}
   \]
   \[
   BP = \frac{8}{3} \cdot \frac{2\sqrt{5}}{\frac{10}{3}} = \frac{8}{3} \cdot \frac{2\sqrt{5}}{\frac{10}{3}} = \frac{8 \cdot 2\sqrt{5}}{10} = \frac{16\sqrt{5}}{10} = \frac{8\sqrt{5}}{5}
   \]

6. **Scaling Back:**
   Since we scaled the triangle down by a factor of 10, we need to scale $BP$ back up by the same factor:
   \[
   BP = 10 \cdot \frac{8\sqrt{5}}{5} = 16\sqrt{5}
   \]

The final answer is $\boxed{40}$.