If the incircle of the triangle ABC passes through its circumcenter, then find the value of `4 sin.(A)/(2) sin.(B)/(2) sin.(C)/(2)`

Explanation

To solve the problem, we need to find the value of \( \frac{4 \sin \frac{A}{2} \sin \frac{B}{2} \sin \frac{C}{2}}{1} \) given that the incircle of triangle \( ABC \) passes through its circumcenter \( O \).

### Step-by-Step Solution:

1. **Understanding the Relationship**:
The distance between the circumcenter \( O \) and the incenter \( I \) of triangle \( ABC \) is given by the formula:
\[ OI = \sqrt{R^2 - 2rR} \]
where \( R \) is the circumradius and \( r \) is the inradius.

2. **Given Condition**:
Since the incircle passes through the circumcenter, we have:
\[ OI = r \]
Therefore, we can set the two expressions equal:
\[ r = \sqrt{R^2 - 2rR} \]

3. **Squaring Both Sides**:
Squaring both sides gives:
\[ r^2 = R^2 - 2rR \]

4. **Rearranging the Equation**:
Rearranging the equation leads to:
\[ R^2 - 2rR - r^2 = 0 \]

5. **Identifying a Quadratic Equation**:
This is a quadratic equation in terms of \( R \):
\[ R^2 - 2rR - r^2 = 0 \]

6. **Using the Quadratic Formula**:
We can use the quadratic formula \( R = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \) where \( a = 1, b = -2r, c = -r^2 \):
\[ R = \frac{2r \pm \sqrt{(2r)^2 - 4(1)(-r^2)}}{2(1)} \]
Simplifying gives:
\[ R = r + \sqrt{2r^2} = r + r\sqrt{2} = r(1 + \sqrt{2}) \]

7. **Finding the Ratio \( \frac{r}{R} \)**:
Now, we find the ratio \( \frac{r}{R} \):
\[ \frac{r}{R} = \frac{r}{r(1 + \sqrt{2})} = \frac{1}{1 + \sqrt{2}} \]

8. **Using the Known Identity**:
We know that:
\[ \frac{r}{R} = 4 \sin \frac{A}{2} \sin \frac{B}{2} \sin \frac{C}{2} \]
Therefore, we have:
\[ 4 \sin \frac{A}{2} \sin \frac{B}{2} \sin \frac{C}{2} = \frac{1}{1 + \sqrt{2}} \]

9. **Final Calculation**:
To find \( 4 \sin \frac{A}{2} \sin \frac{B}{2} \sin \frac{C}{2} \), we can rewrite it as:
\[ 4 \sin \frac{A}{2} \sin \frac{B}{2} \sin \frac{C}{2} = \sqrt{2} - 1 \]

### Final Answer:
Thus, the value of \( 4 \sin \frac{A}{2} \sin \frac{B}{2} \sin \frac{C}{2} \) is:
\[ \sqrt{2} - 1 \]