In the square $A B C D$, the point $P$ was taken inside such that $A P: B P: C P=1: 2: 3$. What is the measure of the angle $A P B$?
I. solution. Rotate the $B P C$ triangle $90^{\circ}$ around $B$ so that the image of $C$ is $A$. Let $P^{\prime}$ be the rotated image of $P$ (1st figure). Then the $P^{\prime} B P$ triangle is isosceles and the angle at $B$ is a right angle. Therefore, $P^{\prime} P B \varangle=45^{\circ}$, and by the Pythagorean theorem, $P^{\prime} P^{2}=P^{\prime} B^{2}+P B^{2}=2 \cdot P B^{2}$. Choose the length of the $A P$ segment as the unit. Then, according to the condition, $P B=2$ and $P C=P^{\prime} A=3$. Thus, $P^{\prime} A^{2}=9=8+1=P^{\prime} P^{2}+P A^{2}$, so by the converse of the Pythagorean theorem, the angle $P^{\prime} P A \varangle$ is a right angle.

## Therefore

$$
A P B \varangle=A P P^{\prime} \varangle+P^{\prime} P B \varangle=90^{\circ}+45^{\circ}=135^{\circ} .
$$

Based on the work of Puskás Anna (Fazekas M. Fóv. Gyak. Gimn., 10th grade)

The above solution is short and elegant, but it is difficult to come up with. The following solution is not as beautiful, but it is easier to find.

II. solution. Again, choose the length of the $A P$ segment as the unit. Let the side length of the square $A B C D$ be $a$. Take a Cartesian coordinate system such that $A(0,0), B(0, a)$, and $C(a, a)$ (2nd figure). If the coordinates of $P$ are $x$ and $y$, then by the distance formula,

(1) $\quad P A^{2}=1=x^{2}+y^{2}$

(2) $\quad P B^{2}=4=x^{2}+(y-a)^{2}$,

(3) $\quad P C^{2}=9=(x-a)^{2}+(y-a)^{2}$.

Subtracting (1) from (2) and (2) from (3), we get

$$
\begin{aligned}
& 3=a^{2}-2 a y \\
& 5=a^{2}-2 a x
\end{aligned}
$$

From these, $y=\frac{a^{2}-3}{2 a}$ and $x=\frac{a^{2}-5}{2 a}$, which, when substituted back into (1), gives:

$$
1=\frac{a^{4}-10 a^{2}+25}{4 a^{2}}+\frac{a^{4}-6 a^{2}+9}{4 a^{2}}
$$

Rearranging this equation in terms of $a^{2}$, we get the quadratic equation $a^{4}-10 a^{2}+17=0$, from which $a^{2}=5 \pm \sqrt{8}$. Since $P$ is an interior point of the square $A B C D$, it follows that $x>0$, so $a^{2}>5$. Therefore, we only need to consider the solution $a^{2}>5$.

Write the cosine rule for the side $A B$ of the triangle $A P B$. If the angle we are looking for is $A P B \varangle=\varphi$, then $A B^{2}=A P^{2}+$ $B P^{2}-2 \cdot A P \cdot B P \cdot \cos \varphi$, that is,

$$
5+\sqrt{8}=1^{2}+2^{2}-2 \cdot 1 \cdot 2 \cdot \cos \varphi
$$

From this, we get $\cos \varphi=-\frac{\sqrt{2}}{2}$, so $\varphi=135^{\circ}$.