Torricelli points on a circle Points $P_{1}, P_{2}, P_{3}, \ldots, P_{n-1}$ are marked on side $BC$ of a regular triangle $ABC$, so that the rays $A P_{1}, A P_{2}, A P_{3}, \ldots, A P_{n-1}$ split $\angle B A C$ into $n$ equal angles. In this case, triangle $ABC$ is divided into $n$ triangles. The points $T_{1}, T_{2}, T_{3}, \ldots, T_{n}$ are the Torricelli points of these triangles.

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*Prove that the points $T_{1}, T_{2}, T_{3}, \ldots, T_{n}, B, C$ lie on the same circle.

*Find the radius of this circle if $AB=1$.


According to my calculations, the radius of the circle is $R=\frac{1}{2 \sin \frac{\pi}{3 n}}$, but I do not yet know how to prove that the points lie on the same circle. I was told that it is necessary to prove that all angles $\angle B T_{i} C$ are equal to $\pi-\frac{\pi}{3 n}$.
P.S. The angle $A T_{i} P_{i}$ is $120^\circ$, so a circle is described around $A B P_{i} T_{i}$. Therefore, the angle $B T_{i} P_{i}$ is equal to $B A P_{i}$. For similar reasons, the angle $C T_{i} P_{i-1}$ is equal to $C A P_{i-1}$. Together, these angles give $\frac{\pi}{3}-\frac{\pi}{3 n}$.
The angle $B T_{i} C$ consists of the two angles mentioned above and the angle $P_{i-1} T_{i} P_{i}$, which is equal to $\frac{2 \pi}{3}$. Therefore, it is equal to $\pi-\frac{\pi}{3 n}$, whence it is clear that it does not depend on $i$, therefore all Torricelli points lie on the circle.
If $R$ is its radius, then by the sine theorem $2 R \sin \frac{\pi}{3 n}=B C=1$, which allows us to find $R$.
 A: EDIT 1:
Do not know if the following construction is any useful HINT.
All Fermat points $T_i$ have three $120^{\circ}$ vertices that surround them.
The Toricelli-Fermat Points minimize distances to vertices in a geometrical sense for each triangle by an algebraic/differential calculus procedure .
In an  more inclusive physical sense we see three equal forces symmetrically contribute to a static stability when they have equal subtended angles at all points when a prismatic net has soap bubbles at any of $(120^{\circ})$ nodes trying to equally pull the common $T_i$ node point away.I.e., the given regiven was imagined to look somewhat to a triangular sector.
Attempted to realize inside a new airfoil-like boundary (of adequate strength or rigidity). I suppose such a honeycomb structure region would be stable for all $AP_i T_i$ points.
I.e., the given region was imagined to look somewhat similar to a thin concave lens in a triangulated configuration including point A..
Remember the 4- point motorway total path minimizing surface tension problem solution? If polyurethane foam is cast into such a closed mold, each closed cell could form this way (ie filled with a maze of Toricelli points). I tried to hand sketch one.. how a narrower airfoil like contour region would be populated with more such Torricelli trianguled nodes.
Even now I have not been 100% complete with a proper answer... taking support of the Queen of sciences  itself is perhaps still a valid answer recourse or so I thought!


