223. A regular triangle is inscribed in a circle. Find the set of intersection points of the altitudes of all possible triangles inscribed in the same circle, two sides of which are parallel to two sides of the given regular triangle.
223. Let (Fig. 59) $ABC$ be the original equilateral triangle, $A_{1} B_{1} C_{1}$ be an arbitrary triangle ($A_{1} C_{1} \parallel A C, A_{1} B_{1} \parallel A B$),

![](https://cdn.mathpix.com/cropped/2024_05_21_6282cf4a54d2a9633c48g-124.jpg?height=337&width=660&top_left_y=568&top_left_x=290)

Fig. 59.

$O$ is the center of the circle, $O_{1}$ is the orthocenter of $\triangle A_{1} B_{1} C_{1}$. Let $\widehat{B O B_{1}}=\varphi$. Since $O_{1} B_{1} \parallel O B, \widehat{O B_{1} O_{1}}=\varphi$; since $\widehat{C_{1} O_{1} B_{1}} = \widehat{C_{1} O B_{1}} = 120^{\circ}$, the quadrilateral $C_{1} O_{1} O B_{1}$ is inscribed in some circle, and thus $\widehat{O_{1} O C_{1}} = \widehat{O_{1} B_{1} C_{1}} = 30^{\circ} - \varphi$. Therefore, $\widehat{O_{1} O B} = \varphi + 120^{\circ} + 30^{\circ} - \varphi = 150^{\circ}$, i.e., the line $O O_{1}$ is parallel to $C B$. To determine how far the point $O_{1}$ can "move" along this line, note that to determine the position of the point $O_{1}$, one needs to draw a line through the variable point $B_{1}$ parallel to $O B$ until it intersects with the line passing through $O$ parallel to $C B$. The most distant points will obviously be obtained for the ends of the diameter perpendicular to $O B$. Thus, the desired set will be the segment $M N$ of the line parallel to $C B$, with a length of $4 R$ and its midpoint at $O$, and the entire set will consist of three such segments.