2. In $\triangle A B C$, points $A_{0}, B_{0}$, and $C_{0}$ are the feet of the altitudes from vertices $A, B$, and $C$, respectively, and it is given that $\triangle A B C \sim \triangle A_{0} B_{0} C_{0}$.

a) If it is known that $\triangle A B C$ is acute, calculate its angles.

b) If it is known that $\triangle A B C$ is obtuse, calculate its angles.
2. a) Let the angles of the observed triangle be denoted by $\alpha$, $\beta$, and $\gamma$, and assume without loss of generality that $\alpha \geqslant \beta \geqslant \gamma$. Let $H$ be the orthocenter of $\triangle ABC$. The circle with diameter $BH$ contains points $A_{0}$ and $C_{0}$ (due to the right angles at $A_{0}$ and $C_{0}$), and similarly, the circle with diameter $CH$ contains points $A_{0}$ and $B_{0}$. From the properties of inscribed angles, we get

$$
\begin{aligned}
\measuredangle B_{0} A_{0} C_{0} & =\measuredangle B_{0} A_{0} H+\measuredangle H A_{0} C_{0}=\measuredangle B_{0} C H+\measuredangle H B C_{0} \\
& =\left(\frac{\pi}{2}-\alpha\right)+\left(\frac{\pi}{2}-\alpha\right)=\pi-2 \alpha
\end{aligned}
$$

Similarly, the other two angles in $\triangle A_{0} B_{0} C_{0}$ are $\pi-2 \beta$ and $\pi-2 \gamma$, and for these three angles, the order is $\pi-2 \gamma \geqslant \pi-2 \beta \geqslant \pi-2 \alpha$. From the similarity given in the problem, we have $\alpha=\pi-2 \gamma$, $\beta=\pi-2 \beta$, and $\gamma=\pi-2 \alpha$. From the second equation, we immediately get $\beta=\frac{\pi}{3}$. Subtracting the third equation from the first, we get $\alpha-\gamma=2(\alpha-\gamma)$, i.e., $\alpha=\gamma$, so we have $\alpha=\gamma=\frac{\pi-\frac{\pi}{3}}{2}=\frac{\pi}{3}$.
![](https://cdn.mathpix.com/cropped/2024_06_04_f254ba4f25ce7f8d9df1g-59.jpg?height=640&width=1326&top_left_y=720&top_left_x=192)

Dr 2017 2B 2

b) Now suppose $\alpha \geqslant \beta \geqslant \gamma$, but $\alpha>\frac{\pi}{2}$. Similarly to part (a), we calculate

$$
\begin{aligned}
\measuredangle B_{0} A_{0} C_{0} & =\measuredangle B_{0} A_{0} A+\measuredangle A A_{0} C_{0}=\measuredangle B_{0} B A+\measuredangle A C C_{0} \\
& =\left(\frac{\pi}{2}-(\pi-\alpha)\right)+\left(\frac{\pi}{2}-(\pi-\alpha)\right)=2 \alpha-\pi
\end{aligned}
$$

$\measuredangle A_{0} B_{0} C_{0}=\measuredangle A_{0} B_{0} A+\measuredangle A B_{0} C_{0}=\measuredangle A_{0} B A+\measuredangle C B C_{0}=2 \beta$ and, similarly, $\measuredangle A_{0} C_{0} B_{0}=$ $2 \gamma$. Since $2 \beta \geqslant 2 \gamma$, the largest angle in $\triangle A_{0} B_{0} C_{0}$ is either $2 \alpha-\pi$ or $2 \beta$. In the first case, from the given similarity, we have $\alpha=2 \alpha-\pi$, i.e., $\alpha=\pi$, which is impossible. Therefore, $\alpha=2 \beta$. Since it is impossible that $\gamma=2 \gamma$, it remains that $\gamma=2 \alpha-\pi$ and $\beta=2 \gamma$. Thus, we have $\gamma=2 \alpha-\pi=4 \beta-\pi=8 \gamma-\pi$, i.e., $\gamma=\frac{\pi}{7}$, and then $\beta=\frac{2 \pi}{7}$ and $\alpha=\frac{4 \pi}{7}$.