4. Given a 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. We rotate triangle $ABP$ around vertex $B$ until vertex $B$ coincides with vertex $C$. Let $Q$ denote the new position of vertex $P$ after the rotation. Since $\angle QBC = \angle PBA$, it follows that $\angle QB\Pi = 90^{\circ}$. Because of this, and since $\overline{BP} = \overline{BQ}$, we have

$$
\angle PQB = 45^{\circ} \text{ and } \overline{PQ}^2 = 2 \overline{BP}^2
$$

From the problem statement, we have $\overline{BP} = 2 \overline{AP}$,

![](https://cdn.mathpix.com/cropped/2024_06_05_4826f81c3e0b08e4acb7g-4.jpg?height=430&width=492&top_left_y=1746&top_left_x=994)
$\overline{CP} = 3 \overline{AP}$, so from this and the previous equality, we get:

$$
\overline{PQ}^2 + \overline{CQ}^2 = 8 \overline{AP}^2 + \overline{AP}^2 = \overline{CP}^2
$$

Since the Pythagorean theorem holds for triangle $CQP$, $\angle CQP = 90^{\circ}$. Finally,

$$
\angle APB = \angle CQB = \angle PQB + \angle CQP = 45^{\circ} + 90^{\circ} = 135^{\circ}
$$

## Third Year