2. Given the square $A B C D$ and a point $P$ inside the square such that $\overline{A P}: \overline{B P}: \overline{C P}=1: 2: 3$. Calculate the angle $\measuredangle A P B$.
Solution. If $\triangle A B P$ is rotated around vertex $B$ until vertex $A$ coincides with vertex $C$ (a rotation with center at $B$ and an angle of $90^{\circ}$). Let $Q$ denote the new position of vertex $P$ after the rotation. Since $\angle Q B C = \angle P B A$, it follows that $\angle Q B P = 90^{\circ}$. Therefore, since $\overline{B P} = \overline{B Q}$, we have

$$
\angle P Q B = 45^{\circ} \text { and } \overline{P Q}^{2} = 2 \overline{B P}^{2} .
$$

![](https://cdn.mathpix.com/cropped/2024_06_05_1695add2a02277f6808fg-2.jpg?height=445&width=546&top_left_y=268&top_left_x=938)

From the condition of the problem, we have $\overline{B P} = 2 \overline{A P}$, $\overline{C P} = 3 \overline{A P}$, so from this and the previous equality, we get:

$$
\overline{P Q}^{2} + \overline{C Q}^{2} = 8 \overline{A P}^{2} + \overline{A P}^{2} = \overline{C P}^{2}
$$

Since Pythagoras' theorem holds for $\triangle C Q P$, $\angle C Q P = 90^{\circ}$. Finally,

$$
\angle A P B = \angle C Q B = \angle P Q B + \angle C Q P = 45^{\circ} + 90^{\circ} = 135^{\circ}
$$