1. **Problem Statement:** Is it always possible to find a square such that three arbitrary points lie on its boundary? 2. **Analysis:** A square has 4 vertices and 4 sides. To have three arbitrary points on the boundary, they can be on vertices or edges. 3. **Key Insight:** Not all triples of points can be placed on the boundary of a single square. For example, if the points are collinear and spaced irregularly, no square can contain all three on its boundary. 4. **Conclusion:** The statement is **false**. It is not always possible to find such a square for any three points. --- 1. **Problem Statement:** Is it always possible to find an equilateral triangle such that three arbitrary points lie on its boundary? 2. **Key Fact:** Any three points define a unique triangle (unless collinear). We want an equilateral triangle containing all three points on its boundary. 3. **Important Rule:** Given any three points, there exists a unique circle passing through them (circumcircle). An equilateral triangle can be inscribed in a circle. 4. **Result:** By rotating and scaling the equilateral triangle inscribed in the circumcircle, we can position it so that all three points lie on its boundary. 5. **Conclusion:** The statement is **true**. For any three points, an equilateral triangle can be found with all points on its boundary. --- 1. **Problem Statement:** For which integers $n \geq 3$ is it always possible to find a regular $n$-gon such that three arbitrary points lie on its boundary? 2. **Known Cases:** - $n=3$: equilateral triangle, true as above. - $n=4$: square, false as above. 3. **Reasoning:** A regular $n$-gon is inscribed in a circle. Three points define a unique circle. To have all three points on the boundary of the $n$-gon, the points must coincide with vertices or lie on edges. 4. **Key Insight:** For $n=3$, always true. For $n>3$, generally false because the polygon's vertices are fixed by the circle and $n$, and arbitrary points won't align with vertices or edges. 5. **Conclusion:** The statement is true only for $n=3$. **Final answers:** - Part 1: False - Part 2: True - Part 3: True only for $n=3$