1 and 4. Given an acute triangle $ABC$, determine for which points in its interior the following inequalities are satisfied:

$$
1 \leq \frac{\angle APB}{\angle ACB} \leq 2, \quad 1 \leq \frac{\angle BPC}{\angle BAC} \leq 2 \quad \text{and} \quad 1 \leq \frac{\angle CPA}{\angle CBA} \leq 2
$$
Solution:

Let $O$ be the circumcenter of triangle $ABC$. The value of angle $ACB$, being inscribed in the circle, is half of angle $AOB$. For any point $P$ on the arc $AOB$, we have $\angle APB = \angle AOB = 2 \angle ACB$. Therefore, this arc divides the interior of the triangle into two parts: for points $Q$ located on one side of the arc $AOB$, the value of angle $\angle AQB$ is greater than $\angle AOB$, and for those located on the other side of the arc, the angle $\angle AQB$ is less than $\angle AOB$. Thus, the points inside the triangle that lie on the arc $AOB$ or outside the circular segment $AOC$ are the ones that satisfy the first condition $1 \leq \frac{\angle APB}{\angle ACB} \leq 2$.

![](https://cdn.mathpix.com/cropped/2024_04_26_46c0569b5329ed838c66g-1.jpg?height=680&width=651&top_left_y=1602&top_left_x=702)

The same reasoning for the chords $BC$ and $CA$ leads us to the conclusion that point $O$ is the only one that can satisfy all three conditions.

![](https://cdn.mathpix.com/cropped/2024_04_26_46c0569b5329ed838c66g-2.jpg?height=676&width=671&top_left_y=233&top_left_x=698)