A survey of participants was conducted at the Olympiad. $ 90\%$ of the participants liked the first round, $60\%$ of the participants liked the second round, $90\%$ of the participants liked the opening of the Olympiad. Each participant was known to enjoy at least two of these three events. Determine the percentage of participants who rated all three events positively.
1. Let \( P \) be the total number of participants. Define the following:
   - \( a \): Number of participants who liked both the first and second rounds.
   - \( b \): Number of participants who liked both the first round and the opening.
   - \( c \): Number of participants who liked both the second round and the opening.
   - \( d \): Number of participants who liked all three events.

2. Given:
   - \( \frac{a + b + d}{P} = 0.9 \) (90% liked the first round)
   - \( \frac{a + c + d}{P} = 0.6 \) (60% liked the second round)
   - \( \frac{b + c + d}{P} = 0.9 \) (90% liked the opening)

3. We need to determine the percentage of participants who liked all three events, i.e., \( \frac{d}{P} \).

4. Adding the given equations:
   \[
   \frac{a + b + d}{P} + \frac{a + c + d}{P} + \frac{b + c + d}{P} = 0.9 + 0.6 + 0.9
   \]
   \[
   \frac{(a + b + d) + (a + c + d) + (b + c + d)}{P} = 2.4
   \]
   \[
   \frac{2a + 2b + 2c + 3d}{P} = 2.4
   \]

5. Simplifying the equation:
   \[
   \frac{2(a + b + c) + 3d}{P} = 2.4
   \]
   \[
   2(a + b + c) + 3d = 2.4P
   \]

6. Since each participant liked at least two events, we have:
   \[
   a + b + c + d = P
   \]

7. Substitute \( a + b + c = P - d \) into the equation:
   \[
   2(P - d) + 3d = 2.4P
   \]
   \[
   2P - 2d + 3d = 2.4P
   \]
   \[
   2P + d = 2.4P
   \]
   \[
   d = 2.4P - 2P
   \]
   \[
   d = 0.4P
   \]

8. Therefore, the percentage of participants who liked all three events is:
   \[
   \frac{d}{P} = \frac{0.4P}{P} = 0.4 = 40\%
   \]

The final answer is \(\boxed{40\%}\).