Problem 1313

1. Identify the given information and the goal:
- Given: $ID = 2$, $IA = 3$, $IX = 4$
- Goal: Compute the inradius $r$ of triangle $ABC$.

2. Understand the geometric setup:
- $I$ is the incenter of $\triangle ABC$.
- $D$ is the foot of the perpendicular from $A$ to $BC$.
- $X$ is a point such that $AX$ is a diameter of the circumcircle of $\triangle ABC$.

3. **Use the given distances to find $AX$:**
- Since $AX$ is a diameter of the circumcircle, $AX = 2R$, where $R$ is the circumradius.
- Using the given distances, we can set up the following relation:
$$\frac{IA}{ID} = \frac{XA}{XI}$$
Substituting the given values:
$$\frac{3}{2} = \frac{XA}{4}$$
Solving for $XA$:
$$XA = \frac{3 \cdot 4}{2} = 6$$

4. **Use the Pythagorean theorem in $\triangle AIX$:**
- We know $IA = 3$, $IX = 4$, and $AX = 6$.
- By the Pythagorean theorem in $\triangle AIX$:
$$AI^2 + IX^2 = AX^2$$
Substituting the known values:
$$3^2 + 4^2 = 6^2$$
$$9 + 16 = 36$$
$$25 = 36$$
This confirms that the distances are consistent.

5. **Find the length $IS$:**
- Using the Pythagorean theorem in $\triangle ASI$ and $\triangle ASX$:
$$3^2 - IS^2 = 6^2 - (IS + 4)^2$$
Simplifying:
$$9 - IS^2 = 36 - (IS + 4)^2$$
$$9 - IS^2 = 36 - (IS^2 + 8IS + 16)$$
$$9 - IS^2 = 36 - IS^2 - 8IS - 16$$
$$9 = 20 - 8IS$$
$$8IS = 11$$
$$IS = \frac{11}{8}$$

6. **Use the similarity of triangles $\triangle DIA_1$ and $\triangle AIS$:**
- Since $\triangle DIA_1 \sim \triangle AIS$ with a ratio of similitude $\frac{ID}{IA} = \frac{2}{3}$:
$$\frac{ID}{IA} = \frac{2}{3}$$
Therefore:
$$r = IA_1 = \frac{2}{3} IS = \frac{2}{3} \cdot \frac{11}{8} = \frac{22}{24} = \frac{11}{12}$$

The final answer is $\boxed{\frac{11}{12}}$.