A really nice problem. A kind of geometric inequality.

The answer is $t \leq 4$.

The answer is $0 < t \le 4$. First, we claim that all $t=4$ (and hence $t<4$) work. Indeed, fix $n$ and let $M > 2 n^3$ be a large integer. Construct a $2M$-gon $A_0 A_1 \dots A_{2M-1}$. Consider the triangles $\triangle A_0 A_1 A_M$, $\triangle A_1 A_2 A_M$, \dots, up to $\triangle A_{n-1} A_n A_M$. [asy][asy] size(6cm); draw(unitcircle); pair A = dir(170); pair B = dir(165); pair C = dir(0); dot("$A_{k-1}$", A, dir(180)); dot("$A_k$", B, dir(150)); dot("$A_M$", C, dir(C)); filldraw(A--B--C--cycle, rgb(.8,.8,.8), black); dot("$A_0$", dir(180), dir(225)); [/asy][/asy] The semi-perimeter of the $k$th triangle ($1 \le k \le n$) is \begin{align*} s_k &= \sin\left( \frac{(M+(k-1))\pi}{2M} \right) + \sin\left( \frac{(M-k)\pi}{2M} \right) + \sin \left( \frac{\pi}{2M} \right) \\ &= \sin\left( \frac{\pi}{2M} \right) + \cos\left( \frac{(k-1)\pi}{2M} \right) + \cos\left( \frac{k\pi}{2M} \right) \\ &= \left( \frac{\pi}{2M} - O(M^{-3}) \right) + \left( 2 - O\left( (kM^{-1})^2 \right) \right) \\ &= 2 + \frac{\pi}{2M} - o(M^{-1}). \end{align*}where we have used Taylor's theorem in the form $\sin x = x - O(x^{-3})$ and $\cos x = 1 - O(x^{-2})$. As $M \to \infty$, we see $s_k > 2$ for all $1 \le k \le n$, and so all the triangles have perimeter greater than $4$. Remark: Many other possible constructions exist for $t=4$. Ankan gives the following one: initially start with a right triangle $ABC$, with hypotenuse $BC$, and $A$ near $B$. The idea is to replace $ABC$ with two triangles $AXC$ and $XBC$, with $X$ chosen on arc $\widehat{AB}$, really close to $B$, so that the perimeter of $AXC$ is greater than $4$. Repeat. On the other hand, suppose $t > 4+\varepsilon$ for some $\varepsilon > 0$. Claim: Any triangle with perimeter at least $t$ has area exceeding $\frac{1}{2} \varepsilon^{3/2}$. Proof. Let $a$, $b$, $c$ be the side lengths and $s$ the semiperimeter. We have $s \ge 2+\varepsilon/2$ and $a,b,c \le 2$, hence $\min(s-a,s-b,s-c) > \varepsilon/2$, so the area is at least $\sqrt{s(s-a)(s-b)(s-c)} > \sqrt{2 \cdot (\varepsilon/2)^3}$ by Heron's formula. $\blacksquare$ Thus there can be at most finitely many such triangles.

Answer $0<t\leq4$

BOBTHEGR8 wrote: Answer $0<t\leq4$

We should have perimetet $\ge 4$

@ above ,its a typo

The answer is all $t\in (0;4].$ Let $P(XYZ)$ denotes the perimeter of $\triangle XYZ.$ Construction. It's enough to consider case $t=4.$ Let $\omega '$ be a cemicircle of $\omega$ with diameter $AB_{n+1}.$ Pick $B_1\in \omega ';$ by the triangle inequality $P(AB_{n+1}B_1)>2\cdot |AB_{n+1}|=4.$ Next, for $i=2,3,\dots ,n$ because of $P(AB_{n+1}B_{i-1})> 4$ we can choose take $B_i$ on arc $B_{i-1}B$ of $\omega '$ (sufficently near to the $B$) for which $P(AB_{i-1}B_i)>4.$ Thus the collection of $\triangle AB_iB_{i+1}$ for $i=1,2,\dots ,n$ is good. Upper bound. Now assume there is a good collection with $N$ triangles, each of perimeter greater than $4+\varepsilon .$ Pick an arbitrary triangle $ABC$ from collection; since each of it's side has length at most $2,$ the shortest side has length at least $\varepsilon.$ Using formula connecting area of triangle and it's circumradius we obtain $\text{area}(ABC)>\frac{\varepsilon^3}{4}.$ Since triangles of collection are disjoint, it follows that $N<\frac{4\pi}{\varepsilon^3}$ and hence the conclusion.

The answer is $t\le 4$. Note that for each triangle with lengths $a$, $b$, and $c$ we have $a,b,c\le 2$ so $a,b,c\ge t-4$ so if $t>4$ then note that \[A=\frac{abc}{4R}=\frac{abc}{4}\ge \frac{(t-4)^3}{4}\]which means that enough triangles will cover the circle so there can't be infinite. If $t< 4$ then we first start by noting that any right triangle works, so let $AB$ be diameter and $C_1$ be any point on the circle and let it be close to $A$. We will split this into $AC_2B$ and $C_2C_1B$. Let $d$ be the perimeter minus $4$. Then, let $AC_2=\frac{d}{2}$ and we get $C_1C_2\ge BC_1-\tfrac{d}2$ and $AC_2\ge AC-\tfrac{d}2$ so $AC_2+C_1C_2+AC_1\ge AC+AC_1+BC_1-d=4$ so we're done.

We claim the answer is $t \le 4$. If $t>4$ we have the area of a triangle is $\sqrt{s(s-a)(s-b)(s-c)} \ge \sqrt{\frac{t(t-4)^3)}{16}} > 0$, so if $t >4$ we get that the triangles will have an area that is bounded below therefore if we pick $n$ large enough the sum of the areas of the triangles will be larger than the area of the circle and we will have a contradiction. Now if $t \le 4$, we just need to provide a construction for $t = 4$. Let $\overline{AB}$ be a diameter. Pick $C$ on the circle close to $B$. Note that the perimeter of this triangle must be greater than $4$ since $p = AC + BC + AB > 2AB = 4$. Now we want to show that we can divide this triangle into two triangles with perimeter at least $4$. To do this we simply pick $C'$ between $B$ and $C$ such that $BC' > \frac{p-4}{2}$ which obviously exists. Now realize that $CC' > BC - \frac{p-4}{2}$ and $AC' > AB - \frac{p-4}{2}$ so summing gives $AC + C'C + AC' > AC + BC - \frac{p-4}{2} + AB - \frac{p-4}{2} = p - (p-4) = 4$. So we can infinitely split triangles of perimeter greater than $4$ so we are done.

The answer is $\boxed{(0,4]}$ Firstly, we will prove that for all $t \leq 4$ such a collection exists. Let $AB$ be the diameter of a circle. Delete the semicircle below $AB$ Now we will introduce a function $f: P \to P$, where $P$ is a set of all points on the remaining semicircle Let $f(X)$ be such a point $Y$ on arc $BX$, that $BX+XY+YB=4$ (if there are multiple such points, choose the closest to $X$). Note that such a point doesn't exist for all $X$, so when it doesn't exist, function is just not defined Now note that our function is continious and $f(A)=A$. Consider $g(X)=f^n(X)$. It is also continious and also $g(A)=A$. Then for any point $T$ close enough to $A$ there exists such a $M$, that $f(M)=T$. But then take triangles $BMf(M),Bf(M)f(f(M)),\ldots,Bf^{n-1}(M)f^n(M)$ For $t>4$, note that by Heron's formula $S=\sqrt{p(p-a)(p-b)(p-c)} \geq \sqrt{p(p-2)(p-2)(p-2)} > \sqrt{\frac{t}{2}(\frac{t}{2}-2)^3}=C>0$, and so not more than $\frac{\pi}{C}$ triangles can be in the collection