A circle $\omega$ of radius 1 is given. A collection $T$ of triangles is called good, if the following conditions hold:  (i) each triangle from $T$ is inscribed in $\omega$;  (ii) no two triangles from $T$ have a common interior point.  Determine all positive real numbers $t$ such that, for each positive integer $n$, there exists a good collection of $n$ triangles, each of perimeter greater than $t$.  (South Africa)

A circle $\omega$ of radius 1 is given. A collection $T$ of triangles is called good, if the following conditions hold:  (i) each triangle from $T$ is inscribed in $\omega$;  (ii) no two triangles from $T$ have a common interior point.  Determine all positive real numbers $t$ such that, for each positive integer $n$, there exists a good collection of $n$ triangles, each of perimeter greater than $t$.  (South Africa)
First, we show how to construct a good collection of $n$ triangles, each of perimeter greater than 4. This will show that all $t \leqslant 4$ satisfy the required conditions. Construct inductively an $(n+2)$-gon $B A_{1} A_{2} \ldots A_{n} C$ inscribed in $\omega$ such that $B C$ is a diameter, and $B A_{1} A_{2}, B A_{2} A_{3}, \ldots, B A_{n-1} A_{n}, B A_{n} C$ is a good collection of $n$ triangles. For $n=1$, take any triangle $B A_{1} C$ inscribed in $\omega$ such that $B C$ is a diameter; its perimeter is greater than $2 B C=4$. To perform the inductive step, assume that the $(n+2)$-gon $B A_{1} A_{2} \ldots A_{n} C$ is already constructed. Since $A_{n} B + A_{n} C + B C > 4$, one can choose a point $A_{n+1}$ on the small arc $\widehat{C A_{n}}$, close enough to $C$, so that $A_{n} B + A_{n} A_{n+1} + B A_{n+1}$ is still greater than 4. Thus each of these new triangles $B A_{n} A_{n+1}$ and $B A_{n+1} C$ has perimeter greater than 4, which completes the induction step. ![](https://cdn.mathpix.com/cropped/2024_04_17_6da854f21230b13cd732g-42.jpg?height=366&width=691&top_left_y=1353&top_left_x=682)

We proceed by showing that no $t > 4$ satisfies the conditions of the problem. To this end, we assume that there exists a good collection $T$ of $n$ triangles, each of perimeter greater than $t$, and then bound $n$ from above.

Take $\varepsilon > 0$ such that $t = 4 + 2 \varepsilon$.

Claim. There exists a positive constant $\sigma = \sigma(\varepsilon)$ such that any triangle $\Delta$ with perimeter $2s \geqslant 4 + 2 \varepsilon$, inscribed in $\omega$, has area $S(\Delta)$ at least $\sigma$.

Proof. Let $a, b, c$ be the side lengths of $\Delta$. Since $\Delta$ is inscribed in $\omega$, each side has length at most 2. Therefore, $s - a \geqslant (2 + \varepsilon) - 2 = \varepsilon$. Similarly, $s - b \geqslant \varepsilon$ and $s - c \geqslant \varepsilon$. By Heron's formula, $S(\Delta) = \sqrt{s(s-a)(s-b)(s-c)} \geqslant \sqrt{(2 + \varepsilon) \varepsilon^3}$. Thus we can set $\sigma(\varepsilon) = \sqrt{(2 + \varepsilon) \varepsilon^3}$.

Now we see that the total area $S$ of all triangles from $T$ is at least $n \sigma(\varepsilon)$. On the other hand, $S$ does not exceed the area of the disk bounded by $\omega$. Thus $n \sigma(\varepsilon) \leqslant \pi$, which means that $n$ is bounded from above.

Comment 1. One may prove the Claim using the formula $S = \frac{abc}{4R}$ instead of Heron's formula.

Comment 2. In the statement of the problem condition $(i)$ could be replaced by a weaker one: each triangle from $T$ lies within $\omega$. This does not affect the solution above, but reduces the number of ways to prove the Claim.

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