Two cyclists leave simultaneously a point $P$ in a circular runway with constant velocities $v_1, v_2 (v_1 > v_2)$ and in the same sense. A pedestrian leaves $P$ at the same time, moving with velocity $v_3 = \frac{v_1+v_2}{12}$ . If the pedestrian and the cyclists move in opposite directions, the pedestrian meets the second cyclist $91$ seconds after he meets the first. If the pedestrian moves in the same direction as the cyclists, the first cyclist overtakes him $187$ seconds before the second does. Find the point where the first cyclist overtakes the second cyclist the first time.
1. **Define Variables and Initial Conditions:**
   - Let \( v_1 \) and \( v_2 \) be the velocities of the first and second cyclists, respectively, with \( v_1 > v_2 \).
   - Let \( v_3 = \frac{v_1 + v_2}{12} \) be the velocity of the pedestrian.
   - Let \( r \) be the radius of the circular runway.
   - The pedestrian meets the second cyclist 91 seconds after meeting the first cyclist when moving in opposite directions.
   - The first cyclist overtakes the pedestrian 187 seconds before the second cyclist when moving in the same direction.

2. **Pedestrian and Cyclists Moving in Opposite Directions:**
   - Time to meet the first cyclist: 
     \[
     t_1 = \frac{2\pi r}{v_1 + v_3} = \frac{2\pi r}{v_1 + \frac{v_1 + v_2}{12}} = \frac{24\pi r}{13v_1 + v_2}
     \]
   - Time to meet the second cyclist:
     \[
     t_2 = \frac{2\pi r}{v_2 + v_3} = \frac{2\pi r}{v_2 + \frac{v_1 + v_2}{12}} = \frac{24\pi r}{v_1 + 13v_2}
     \]
   - Difference in meeting times:
     \[
     t_2 - t_1 = \frac{24\pi r}{v_1 + 13v_2} - \frac{24\pi r}{13v_1 + v_2} = 91
     \]
   - Simplifying the equation:
     \[
     \frac{24\pi r (13v_1 + v_2) - 24\pi r (v_1 + 13v_2)}{(v_1 + 13v_2)(13v_1 + v_2)} = 91
     \]
     \[
     \frac{288\pi r (v_1 - v_2)}{(v_1 + 13v_2)(13v_1 + v_2)} = 91
     \]

3. **Pedestrian and Cyclists Moving in the Same Direction:**
   - Time for the first cyclist to overtake the pedestrian:
     \[
     t_3 = \frac{2\pi r}{v_1 - v_3} = \frac{2\pi r}{v_1 - \frac{v_1 + v_2}{12}} = \frac{24\pi r}{11v_1 - v_2}
     \]
   - Time for the second cyclist to overtake the pedestrian:
     \[
     t_4 = \frac{2\pi r}{v_2 - v_3} = \frac{2\pi r}{v_2 - \frac{v_1 + v_2}{12}} = \frac{24\pi r}{11v_2 - v_1}
     \]
   - Difference in overtaking times:
     \[
     t_4 - t_3 = \frac{24\pi r}{11v_2 - v_1} - \frac{24\pi r}{11v_1 - v_2} = 187
     \]
   - Simplifying the equation:
     \[
     \frac{24\pi r (11v_1 - v_2) - 24\pi r (11v_2 - v_1)}{(11v_2 - v_1)(11v_1 - v_2)} = 187
     \]
     \[
     \frac{288\pi r (v_1 - v_2)}{(11v_2 - v_1)(11v_1 - v_2)} = 187
     \]

4. **Solving the System of Equations:**
   - Let \( v_1 = 2\pi r w_1 \) and \( v_2 = 2\pi r w_2 \).
   - Substitute into the equations:
     \[
     \frac{144(w_1 - w_2)}{(w_1 + 13w_2)(13w_1 + w_2)} = 91
     \]
     \[
     \frac{144(w_1 - w_2)}{(11w_2 - w_1)(11w_1 - w_2)} = 187
     \]
   - Divide the two equations:
     \[
     \frac{187}{91} = \frac{(w_1 + 13w_2)(13w_1 + w_2)}{(11w_2 - w_1)(11w_1 - w_2)}
     \]
   - Simplify and solve for \( w_1 \) and \( w_2 \):
     \[
     w_1^2 - \frac{34}{15}w_1w_2 + w_2^2 = 0
     \]
     \[
     w_2 = \frac{3}{5} w_1
     \]
   - Substitute back to find \( w_1 \):
     \[
     \frac{144(w_1 - \frac{3}{5}w_1)}{(w_1 + 13 \cdot \frac{3}{5}w_1)(13w_1 + \frac{3}{5}w_1)} = 91
     \]
     \[
     w_1 = \frac{90}{17017}, \quad w_2 = \frac{54}{17017}
     \]

5. **Finding the Point of Overtaking:**
   - Time for the first cyclist to overtake the second cyclist:
     \[
     t_5 = \frac{2\pi r}{v_1 - v_2} = \frac{1}{w_1 - w_2} = \frac{17017}{36}
     \]
   - Distance covered by the first cyclist:
     \[
     2\pi r \cdot \frac{17017}{36} \cdot \frac{90}{17017} = 2.5 \times 2\pi r
     \]
   - Distance covered by the second cyclist:
     \[
     2\pi r \cdot \frac{17017}{36} \cdot \frac{54}{17017} = 1.5 \times 2\pi r
     \]
   - The required point is the opposite of \( P \).

The final answer is \(\boxed{\text{the opposite of } P}\)