Given is an acute angled triangle $ ABC$. Determine all points $ P$ inside the triangle with
\[1\leq\frac{\angle APB}{\angle ACB},\frac{\angle BPC}{\angle BAC},\frac{\angle CPA}{\angle CBA}\leq2\]
To solve the problem, we need to find all points \( P \) inside an acute-angled triangle \( ABC \) such that the following inequalities hold:
\[ 1 \leq \frac{\angle APB}{\angle ACB}, \frac{\angle BPC}{\angle BAC}, \frac{\angle CPA}{\angle CBA} \leq 2. \]

1. **Understanding the Angles:**
   - Let \( \angle ACB = \gamma \).
   - Let \( \angle BAC = \alpha \).
   - Let \( \angle CBA = \beta \).
   - We need to find \( P \) such that:
     \[ 1 \leq \frac{\angle APB}{\gamma} \leq 2, \]
     \[ 1 \leq \frac{\angle BPC}{\alpha} \leq 2, \]
     \[ 1 \leq \frac{\angle CPA}{\beta} \leq 2. \]

2. **Using the Circumcenter \( O \):**
   - Consider the circumcircles \( \omega_1 \) of \( \triangle AOB \), \( \omega_2 \) of \( \triangle BOC \), and \( \omega_3 \) of \( \triangle COA \), where \( O \) is the circumcenter of \( \triangle ABC \).
   - The angles at \( O \) are:
     \[ \angle AOB = 2\gamma, \]
     \[ \angle BOC = 2\alpha, \]
     \[ \angle COA = 2\beta. \]

3. **Analyzing the Angles at \( P \):**
   - For \( P \) to satisfy the given inequalities, \( P \) must lie in such a way that:
     \[ \gamma \leq \angle APB \leq 2\gamma, \]
     \[ \alpha \leq \angle BPC \leq 2\alpha, \]
     \[ \beta \leq \angle CPA \leq 2\beta. \]

4. **Considering the Regions:**
   - The point \( P \) must lie in the region where the angles formed by \( P \) with the vertices of \( \triangle ABC \) are constrained by the circumcircles \( \omega_1, \omega_2, \omega_3 \).
   - Since \( \angle APB, \angle BPC, \angle CPA \) are constrained by the circumcircles, \( P \) must be the circumcenter \( O \) itself. This is because \( O \) is the only point where the angles subtended by the sides of the triangle at \( O \) are exactly twice the angles at the vertices.

5. **Conclusion:**
   - Therefore, the only point \( P \) that satisfies all the given inequalities is the circumcenter \( O \) of \( \triangle ABC \).

\(\blacksquare\)

The final answer is \( \boxed{ O } \) (the circumcenter of \( \triangle ABC \)).