A "nearly regular" $H_{0}$ triangle has angles of $59.5^{\circ}, 60^{\circ}$, and $60.5^{\circ}$. Consider the following sequence of triangles. Let $H_{1}$ be the pedal triangle of $H_{0}$, $H_{2}$ be the pedal triangle of $H_{1}$, and so on. Which one in the sequence will be the first obtuse triangle?
In triangle $ABC$, let $\alpha=60^{\circ}, \beta=59.5^{\circ}, \gamma=60.5^{\circ}$, and in the pedal triangle $H_{1}=A_{1} B_{1} C_{1}$, $A_{1}$ is the foot of the altitude from $A$, $B_{1}$ is the foot of the altitude from $B$, and $C_{1}$ is the foot of the altitude from $C$, and $M$ is the orthocenter.

![](https://cdn.mathpix.com/cropped/2024_05_02_5c31a9552a1bc106e1e9g-1.jpg?height=466&width=570&top_left_y=263&top_left_x=772)

It is easily seen that the quadrilaterals $A_{1} M C_{1} B, B_{1} M A_{1} C, C_{1} M B_{1} A$ are cyclic, each with two opposite right angles. (We know that triangle $ABC$ is acute and $M$ is inside the triangle, which we will use in the proof.)

From the cyclic quadrilateral $AB_{1}MC_{1}$, it follows that $\angle B_{1}MC = \angle B_{1}AC_{1} = \alpha$, and in the cyclic quadrilateral $B_{1}MA_{1}C$, $\angle B_{1}A_{1}C = \angle B_{1}MC = \alpha$, as they are angles subtended by the same arc; similarly, $\angle C_{1}A_{1}B = \alpha$. This implies that in the pedal triangle $H_{1}$, $\angle B_{1}A_{1}C_{1} = 180^{\circ} - 2\alpha$. Since there is no distinguished vertex or side in the triangle, similarly $\angle C_{1}B_{1}A_{1} = 180^{\circ} - 2\beta$ and $\angle A_{1}C_{1}B_{1} = 180^{\circ} - 2\gamma$.

Thus, for the angles of the pedal triangles, we get the following sequence:

In $H_{1}$: $\alpha_{1}=180^{\circ}-2 \cdot 60^{\circ}=60^{\circ}, \beta_{1}=180^{\circ}-2 \cdot 59.5^{\circ}=61^{\circ}, \gamma_{1}=180^{\circ}-2 \cdot 60.5^{\circ}=59^{\circ}$, and so on, in the pedal triangles $H_{2}, H_{3}, \ldots$, one angle is always $60^{\circ}$, and for the other two angles:

$$
\begin{array}{cc}
H_{2}: & \beta_{2}=58^{\circ}, \gamma_{2}=62^{\circ}, \quad H_{3}: \beta_{3}=64^{\circ}, \gamma_{3}=56^{\circ} \\
H_{4}: & \beta_{4}=52^{\circ}, \gamma_{4}=68^{\circ}, \quad H_{5}: \beta_{5}=76^{\circ}, \gamma_{5}=44^{\circ} \\
H_{6}: & \beta_{6}=28^{\circ}, \gamma_{6}=92^{\circ}
\end{array}
$$

Thus, the sixth pedal triangle is the first obtuse triangle in the sequence.