Let $\triangle ABC$ be a triangle with $AB=5, BC=6, CA=7$. Suppose $P$ is a point inside $\triangle ABC$ such that $\triangle BPA\sim \triangle APC$. If $AP$ intersects $BC$ at $X$, find $\frac{BX}{CX}$.

[i]Proposed by Nathan Ramesh
1. Given that $\triangle BPA \sim \triangle APC$, we know that the corresponding sides of these triangles are proportional. This implies that:
   \[
   \frac{BP}{AP} = \frac{AP}{CP}
   \]
   Let $k$ be the ratio such that:
   \[
   \frac{BP}{AP} = k \quad \text{and} \quad \frac{AP}{CP} = k
   \]
   Therefore, we have:
   \[
   BP = k \cdot AP \quad \text{and} \quad AP = k \cdot CP
   \]
   This implies:
   \[
   BP = k^2 \cdot CP
   \]

2. Since $P$ is inside $\triangle ABC$ and $AP$ intersects $BC$ at $X$, we need to find the ratio $\frac{BX}{CX}$. Using the properties of similar triangles, we know that the areas of $\triangle APB$ and $\triangle APC$ are proportional to the squares of the corresponding sides:
   \[
   \frac{[APB]}{[APC]} = \left(\frac{AB}{AC}\right)^2 = \left(\frac{5}{7}\right)^2 = \frac{25}{49}
   \]

3. The ratio of the areas of $\triangle APB$ and $\triangle APC$ is also equal to the ratio of the heights from $B$ and $C$ to $AX$, because they share the common base $AP$. Therefore:
   \[
   \frac{[APB]}{[APC]} = \frac{\text{height from } B \text{ to } AX}{\text{height from } C \text{ to } AX} = \frac{BX}{CX}
   \]

4. Thus, we have:
   \[
   \frac{BX}{CX} = \frac{25}{49}
   \]

The final answer is $\boxed{\frac{25}{49}}$