Where to place x_i's?

Because the diagonals from one vertex divide a convex $61$ -gon into $59$ disjoint triangles, at least $59$ points are needed. It is actually possible to choose exactly $59$ points so that the condition is satisfied.

Let the vertices of the $61$ -gon be $P_1, P_2, ..., P_{61}$ . Mark one point on each region bounded by the diagonals $\overline{P_1P_n}$ , $\overline{P_nP_{61}}$ and $\overline{P_{n-1}P_{n+1}}$ for all $n \in \{ 2,3,...,60 \}$ . Notice that the triangle $P_aP_nP_b$ if $1\leq a<n<b\leq61$ contains the region bounded by the diagonals $\overline{P_1P_n}$ , $\overline{P_nP_{61}}$ and $\overline{P_{n-1}P_{n+1}}$ , which has a marked point in it. Hence, the given construction with $59$ points satisfies the condition.

Fix 1 vertex. The diagonals from this vertex to all the other vertices, along with the perimeter of A, would determine 59 disjoint triangles. We have to place a point in each of these triangles, and any point in a triangle would not be in any of the other disjoint triangles. Hence, we need at least 59 points.

We will show that 59 points are sufficient. Label the vertices of $A$ as $\{A_i\}_{i=1}^{61}$ in order. Let $w_i$ be any point contained in the triangle bounded by the diagonals $A_1 A_i, A_{i-1}A_{i+1}, A_iA_{61}$ for $i=2$ to $60$ . This gives us a set of 59 points. Now, given any vertex triangle $A_pA_qA_r$ with $1\leq p < q < r \leq 61$ , notice that it contains the triangle bounded by $A_1 A_q, A_{q-1}A_{q+1}, A_qA_{61}$ , hence contains $w_{q}$ . Thus, $n=59$ is the minimum.