2. Inside the right triangle $ABC\left(\varangle ACB=90^{\circ}\right)$, a point $P$ is observed such that $AP=4, BP=2$, and $CP=1$. The point $Q$, which is symmetric to point $P$ with respect to $AC$, lies on the circumcircle of triangle $ABC$. Determine the angles of triangle $ABC$.


2. Let $\alpha=\angle BAC$. Let $R$ be a point such that triangles $ARB$ and $APC$ are similar and similarly oriented. Since $\frac{AR}{AP}=\frac{AB}{AC}$ and $\angle RAP=\angle BAC$, it follows that $\triangle ARP \sim \triangle ABC$. We obtain $BR=CP \cdot \frac{AB}{AC}=\frac{1}{\cos \alpha}, RP=AP$. $\frac{AC}{BC}=4 \tan \alpha$ and $\angle BRP=\angle ARB - \angle ARP=\angle AQC - \angle ABC=2\alpha$. Now, using the cosine rule in

![](https://cdn.mathpix.com/cropped/2024_06_04_a7c4af2042dd38c4829cg-2.jpg?height=408&width=642&top_left_y=1212&top_left_x=1047)
$\triangle BPR$ we get the equation in $\alpha$:

$4=BP^2=BR^2+RP^2-2BR \cdot RP \cos 2\alpha=\frac{1+16 \sin^2 \alpha - 8 \sin \alpha \cos 2\alpha}{\cos^2 \alpha}$, which simplifies to $16x^3 + 20x^2 - 8x - 3 = 0$ for $x=\sin \alpha$. The solutions to this equation are $x \in \left\{\frac{1}{2}, \frac{3}{2}, -\frac{1}{4}\right\}$, with only $x=\frac{1}{2}$ being meaningful. Therefore, $\alpha=30^\circ$. The angles of triangle $ABC$ are $30^\circ, 60^\circ$, and $90^\circ$.