Solution

Let the explosion occurred at time $t_0$, then the time of signal registration at point $A$ is equal to:

$$t_A = t_0 + \frac{L + x}{c}, \, (1)$$

where $x = BO$. Similar to point $B$

$$t_B = t_0 + \frac{x}{c}, \,(2)$$

For point $C$

$$t_C = t_0 + \frac{L - x}{c}, \,(3)$$

Let's subtract the second equation from the first equation:

$$t_A – t_B = \frac{L}{c}. \,(4)$$

And from the first equation we subtract the third equation

$$t_A – t_C = \frac{2x}{c}. \,(5)$$

From equation $(4)$ let's express

$$c = \frac{L}{t_A – t_B},$$

and from $(5)$

$$x = \frac{t_A – t_C}{2} \cdot c.$$

Then the required distance

$$AO = L + x = L + \frac{t_A – t_C}{2}\frac{L}{t_A – t_B}$$

After the transformation

$$AO = \frac{3t_A – 2t_B – t_C}{2(t_A – t_B)} \cdot L. \, (6)$$

To determine the moment of time at which the explosion occurred, we substitute in the expression

$$t_A = t_0 + \frac{L + x}{c}, \; c = \frac{L}{t_A – t_B}$$ $$\frac{t_A – t_C}{2} = \frac{x}{c}$$

after transformations

$$t_0 = t_B - \frac{1}{2} \cdot (t_A – t_C)$$