1. **Problem statement:** We have a circle $\omega$ with radius 1. A collection $T$ of triangles is called good if: - Each triangle in $T$ is inscribed in $\omega$. - No two triangles in $T$ share an interior point. We want to determine all positive real numbers $t$ such that for every positive integer $n$, there exists a good collection of $n$ triangles each with perimeter greater than $t$. 2. **Key observations and formulas:** - The circle $\omega$ has radius 1, so its circumference is $2\pi$. - Each triangle is inscribed in $\omega$, so its vertices lie on the circle. - The perimeter of a triangle inscribed in a unit circle depends on the arc lengths between its vertices. 3. **Maximum perimeter of an inscribed triangle:** - The largest perimeter triangle inscribed in a unit circle is an equilateral triangle with side length equal to the chord length subtending $120^\circ$. - The chord length for angle $\theta$ in a unit circle is $2\sin(\theta/2)$. - For an equilateral triangle, each side subtends $120^\circ$, so side length is $2\sin(60^\circ) = 2 \times \frac{\sqrt{3}}{2} = \sqrt{3}$. - Thus, the maximum perimeter of an inscribed triangle is $3 \times \sqrt{3} = 3\sqrt{3} \approx 5.196$. 4. **Packing triangles without overlapping interiors:** - To have $n$ triangles with no common interior points, their interiors must be disjoint. - Since all vertices lie on the circle, the triangles are formed by chords connecting points on the circumference. - The union of all triangles must fit inside the circle without overlapping interiors. 5. **Constructing good collections:** - For any $n$, we can place $3n$ points evenly spaced on the circle. - Connect every consecutive triple of points to form $n$ triangles. - Each triangle will be isosceles or nearly equilateral if points are evenly spaced. - The perimeter of each triangle depends on the arc lengths between its vertices. 6. **Perimeter lower bound for large $n$:** - As $n$ grows, the arcs between points become smaller. - The perimeter of each triangle approaches 6 (the circumference) divided by $n$ times 3, but since the arcs shrink, the perimeter decreases. - To have perimeter greater than $t$, $t$ must be less than or equal to the maximum perimeter possible for such triangles. 7. **Conclusion:** - Since the maximum perimeter of any inscribed triangle is $3\sqrt{3}$, for any $t \geq 3\sqrt{3}$, it is impossible to have triangles with perimeter greater than $t$. - For any $t < 3\sqrt{3}$, we can choose $n$ large enough and construct a good collection of $n$ triangles each with perimeter greater than $t$. **Final answer:** $$\boxed{\text{All positive real numbers } t \text{ such that } 0 < t < 3\sqrt{3}}$$