Consider all good triangles containing a certain vertex $A$. The other two vertices of any such triangle lie on the circle $\omega_{A}$ with unit radius and center $A$. Since $P$ is convex, all these vertices lie on an arc of angle less than $180^{\circ}$. Let $L_{A} R_{A}$ be the shortest such arc, oriented clockwise (see Figure 1). Each of segments $A L_{A}$ and $A R_{A}$ belongs to a unique good triangle. We say that the good triangle with side $A L_{A}$ is assigned counterclockwise to $A$, and the second one, with side $A R_{A}$, is assigned clockwise to $A$. In those cases when there is a single good triangle containing vertex $A$, this triangle is assigned to $A$ twice. There are at most two assignments to each vertex of the polygon. (Vertices which do not belong to any good triangle have no assignment.) So the number of assignments is at most $2 n$. Consider an arbitrary good triangle $A B C$, with vertices arranged clockwise. We prove that $A B C$ is assigned to its vertices at least three times. Then, denoting the number of good triangles by $t$, we obtain that the number $K$ of all assignments is at most $2 n$, while it is not less than $3 t$. Then $3 t \leq K \leq 2 n$, as required. Actually, we prove that triangle $A B C$ is assigned either counterclockwise to $C$ or clockwise to $B$. Then, by the cyclic symmetry of the vertices, we obtain that triangle $A B C$ is assigned either counterclockwise to $A$ or clockwise to $C$, and either counterclockwise to $B$ or clockwise to $A$, providing the claim. ! Figure 1 ! Figure 2 Assume, to the contrary, that $L_{C} \neq A$ and $R_{B} \neq A$. Denote by $A^{\prime}, B^{\prime}, C^{\prime}$ the intersection points of circles $\omega_{A}, \omega_{B}$ and $\omega_{C}$, distinct from $A, B, C$ (see Figure 2). Let $C L_{C} L_{C}^{\prime}$ be the good triangle containing $C L_{C}$. Observe that the angle of $\operatorname{arc} L_{C} A$ is less than $120^{\circ}$. Then one of the points $L_{C}$ and $L_{C}^{\prime}$ belongs to $\operatorname{arc} B^{\prime} A$ of $\omega_{C}$; let this point be $X$. In the case when $L_{C}=B^{\prime}$ and $L_{C}^{\prime}=A$, choose $X=B^{\prime}$. Analogously, considering the good triangle $B R_{B}^{\prime} R_{B}$ which contains $B R_{B}$ as an edge, we see that one of the points $R_{B}$ and $R_{B}^{\prime}$ lies on arc $A C^{\prime}$ of $\omega_{B}$. Denote this point by $Y, Y \neq A$. Then angles $X A Y, Y A B, B A C$ and $C A X$ (oriented clockwise) are not greater than $180^{\circ}$. Hence, point $A$ lies in quadrilateral $X Y B C$ (either in its interior or on segment $X Y$ ). This is impossible, since all these five points are vertices of $P$. Hence, each good triangle has at least three assignments, and the statement is proved. Comment 1. Considering a diameter $A B$ of the polygon, one can prove that every good triangle containing either $A$ or $B$ has at least four assignments. This observation leads to $t \leq\left\lfloor\frac{2}{3}(n-1)\right\rfloor$. Comment 2. The result $t \leq\left\lfloor\frac{2}{3}(n-1)\right\rfloor$ is sharp. To construct a polygon with $n=3 k+1$ vertices and $t=2 k$ triangles, take a rhombus $A B_{1} C_{1} D_{1}$ with unit side length and $\angle B_{1}=60^{\circ}$. Then rotate it around $A$ by small angles obtaining rhombi $A B_{2} C_{2} D_{2}, \ldots, A B_{k} C_{k} D_{k}$ (see Figure 3). The polygon $A B_{1} \ldots B_{k} C_{1} \ldots C_{k} D_{1} \ldots D_{k}$ has $3 k+1$ vertices and contains $2 k$ good triangles. The construction for $n=3 k$ and $n=3 k-1$ can be obtained by deleting vertices $D_{n}$ and $D_{n-1}$. ! Figure 3