In triangle $ABC$, let $I, O, H$ be the incenter, circumcenter and orthocenter, respectively. Suppose that $AI = 11$ and $AO = AH = 13$. Find $OH$.

[i]Proposed by Kevin You[/i]
1. Given that \( AH = AO \), we start by using the known relationship between the orthocenter \( H \), circumcenter \( O \), and the circumradius \( R \) of a triangle. Specifically, we have:
   \[
   AH = 2R \cos A
   \]
   Since \( AH = AO = R \), we can set up the equation:
   \[
   2R \cos A = R
   \]
   Dividing both sides by \( R \), we get:
   \[
   2 \cos A = 1 \implies \cos A = \frac{1}{2}
   \]
   Therefore, \( \angle A = 60^\circ \).

2. It is a well-known fact that when \( \angle A = 60^\circ \), the points \( B, H, I, O, C \) are concyclic (i.e., they lie on a common circle). The center of this circle is the midpoint \( M \) of the minor arc \( \widehat{BC} \) of the circumcircle of \( \triangle ABC \).

3. The radius of this circle is the circumradius \( R \) of \( \triangle ABC \), which is given as \( 13 \). Therefore, the distance \( AM \) can be calculated as:
   \[
   AM = AI + IM
   \]
   Since \( IM = R = 13 \) and \( AI = 11 \), we have:
   \[
   AM = 11 + 13 = 24
   \]

4. Since \( AOMH \) forms a rhombus (because \( AO = AH \) and both are equal to \( R \)), we can use the parallelogram law to find \( OH \). The parallelogram law states:
   \[
   OH^2 + AM^2 = 4R^2
   \]
   Substituting the known values:
   \[
   OH^2 + 24^2 = 4 \cdot 13^2
   \]
   Simplifying:
   \[
   OH^2 + 576 = 4 \cdot 169
   \]
   \[
   OH^2 + 576 = 676
   \]
   \[
   OH^2 = 100
   \]
   \[
   OH = 10
   \]

The final answer is \(\boxed{10}\).