Let $\triangle ABC$ be a triangle with circumcenter $O$ satisfying $AB=13$, $BC = 15$, and $AC = 14$.  Suppose there is a point $P$ such that $PB \perp BC$ and $PA \perp AB$.  Let $X$ be a point on $AC$ such that $BX \perp OP$. What is the ratio $AX/XC$?

[i]Proposed by Thomas Lam[/i]
1. **Identify the key observation**: The problem states that $BX$ is the $B$-symmedian of $\triangle ABC$. A symmedian in a triangle is a line that passes through a vertex and is the reflection of the median with respect to the angle bisector at that vertex.

2. **Define the circle $\omega$**: Let $\omega$ be the circle with center $P$ and radius $PB$. Since $PB \perp BC$ and $PA \perp AB$, $P$ is the orthocenter of $\triangle ABP$ and $\triangle BCP$.

3. **Radical axis concept**: The line $BX$ is the radical axis of the circle $\omega$ and the circumcircle $\odot(ABC)$. The radical axis of two circles is the locus of points that have equal power with respect to both circles.

4. **Define the power function**: Define the function
   \[
   f(T) := \operatorname{pow}_\omega(T) - \operatorname{pow}_{\odot(ABC)}(T),
   \]
   where $\operatorname{pow}_\omega(T)$ is the power of point $T$ with respect to circle $\omega$ and $\operatorname{pow}_{\odot(ABC)}(T)$ is the power of point $T$ with respect to the circumcircle of $\triangle ABC$.

5. **Compute the power of points $A$ and $C$**:
   \[
   \operatorname{pow}_\omega(A) = PA^2 - PB^2 = -AB^2,
   \]
   because $PA \perp AB$ and $PB \perp BC$.
   \[
   \operatorname{pow}_\omega(C) = PC^2 - PB^2 = BC^2,
   \]
   because $PC \perp BC$ and $PB \perp BC$.

6. **Affine map property**: The function $f$ is an affine map, meaning it is a linear transformation followed by a translation. For $f(X)$ to equal zero, $X$ must lie on the radical axis, which implies:
   \[
   f(X) = 0.
   \]

7. **Determine the ratio $AX/XC$**: Since $f$ is an affine map, the ratio of the distances $AX$ and $XC$ is determined by the values of $f(A)$ and $f(C)$. Specifically,
   \[
   \frac{AX}{XC} = \frac{\operatorname{pow}_\omega(A) - \operatorname{pow}_{\odot(ABC)}(A)}{\operatorname{pow}_\omega(C) - \operatorname{pow}_{\odot(ABC)}(C)} = \frac{-AB^2}{BC^2}.
   \]

8. **Substitute the given side lengths**:
   \[
   \frac{AX}{XC} = \frac{AB^2}{BC^2} = \frac{13^2}{15^2} = \frac{169}{225} = \frac{169}{225}.
   \]

The final answer is $\boxed{\frac{169}{225}}$.