The answer is yes. By complex numbers, $z_1a+z_2b+z_3c=0$ forms triangles $ABC$ similar to a fixed triangle if $z_1+z_2+z_3=0$, so put some coefficients $z_i$ into a matrix $A$ such that $Ap=0$ implies the vector $p$ has the points satisfying the conditions. Treat all nonzero elements of the matrix as variables. The rank of $A$ is at most $n-2$ so the null space has dimension at least $2$. One element of the null space is $(1,1,\ldots,1)$. To show no two elements of $p$ are equal, WLOG by translation $p_{n-1}=p_n=0$ and show that $p$ cannot be part of the null space, so delete columns $n-1$ and $n$. This forms an $(n-2)\times(n-2)$ matrix $A'$ which we will show has nonzero determinant which is sufficient. Since column $n$ contained at least one $z$ variable, the other $z$ variables in that row have no conditions, so if the determinant of $A'$ is zero for all choices of $z$, the matrix formed by removing that row and column must have determinant $0$ for both choices of $z$. By Hall, there exists $n-2$ nonzero entries sharing no rows or columns. We can repeat this process over and over by picking one of the $n-2$ nonzero entries that are still remaining until we get a $1\times1$ determinant by the condition since it ensures no $k\times k$ submatrix has three nonzero entries in every row. Thus, since the null space has dimension at least $2$ and contains no element with two equal entries for some choice of $z$ variables, there exists a suitable vector $p$ containing all of the points satisfying the conditions.

I wonder if this problem becomes harder (psychologically) to solve if you replace "less than 61 degrees" with "acute"

IAmTheHazard wrote: I wonder if this problem becomes harder (psychologically) to solve if you replace "less than 61 degrees" with "acute" not for my solution

IAmTheHazard wrote: I wonder if this problem becomes harder (psychologically) to solve if you replace "less than 61 degrees" with "acute" People have been locked up for lesser crimes.

Really nice.

Can we use the same method in 2022 CTST P6 to show that we can triple-colour every elements in $\{1,…,n\}$ s.t. for all $\mathcal{A}\in\mathcal{S}, \mathcal{A}$ has three elements in the same color or in different colors?I tried to apply Hall theorem but didn't work out.Anyway cool problem with linear algebra.

basically same as above -- The answer is yes. We induct on $n \ge 3$. Call each integer triple in $\mathcal{S}$ a triangle. The key claim is the following: Claim. There exists a $3$-coloring of $\{1, 2, ..., n\}$ such that each triangle is either monochromatic or properly $3$-colored (i.e., its numbers have pairwise distinct colors), and that all three colors are used at least once. Proof. It suffices to solve $x_i + x_j + x_k = 0$ for each triangle $\{i, j, k\}$ in $\mathbb{F}_3^{n}$. There are $n-2$ equations and $n$ variables, so the dimension of the solution space is at least $2$. Hence we can find a solution not in $\{ c \times (1, 1, ..., 1) : c \in \mathbb{F}_3\}$. $\square$ Let $C_i$ be the set of numbers colored with color $i$. Then $\max (|C_0|, |C_1|, |C_2|) < n$. Let $S_i$ be the set of triangles whose numbers are entirely contained in $C_i$. Then by the condition we have either $|C_i| \ge \left|\cup_{A \in S_i} A \right| \ge |S_i| +2$ or $|S_i|=0$. In both cases, there is a configuration of points $\mathcal{P}_i := \{P_j : j \in C_i \}$ such that any triangle in $S_i$ has angles smaller than $61^{\circ}$. Now fix a large equilateral triangle $A_0A_1A_2$ (say, with side length $10^{10}$). We will "shrink down and translate" those configurations $\mathcal{P}_i$ so that they become sufficiently small (say, each pair of points have distance $\le 10^{-9}$) and that the distance from each point in $\mathcal{P}_i$ to $A_i$ is at most $10^{-9}$. Since each triangle is either monochromatic or properly colored, it is easy to check that this construction works. We're done.

Actually, the idea is the same as in the above post, but with a bit more combinatorial approach. Consider a bipartite graph $G(A,V)$ with vertices $A:=\{A_1,A_2,\dots,A_{n-2}\}$ and $V:=\{1,2,\dots, n\}$, Connect each vertex $A_i\in A$ with the elements of the set $A_i$. We prove that the vertices in $V$ can be colored in three colors such that at lest two colors are present and the following condition holds: For each $a\in A$, the set of its neighbours $N(a)$ is either monochromatic or a rainbow (all three colors are present). Call such a coloring proper. Interpret each color as an element of $\mathbb{Z}_3$. For each coloring $f:V\to\mathbb{Z}_3$ of vertices in $V$ we assign a coloring $g:A\to \mathbb{Z}_3$ of vertices of $A$ by the formula $$g(a):=\sum_{v\in N(a)} f(v), \forall a\in A \qquad(1).$$Note that the number of all colorings $f$ of $V$ is $3^n$ and the number of colorings of $A$ are $3^{n-2}$. This means that there exists at least $9$ distinct colorings $f_0,f_1,\dots,f_8$ that are mapped, using the formula in $(1)$, to the same coloring $g$ (actually, 4 such colorings are enough). By linearity of $(1)$ we get $$0=\sum_{v\in N(a)} (f_i-f_0)(a),\forall a\in A, i=1,2,3 \qquad (2).$$Since $f_i-f_0\not\equiv 0$, at least one of these $3$ functions(colorings), say $f_1-f_0$ is not constant. But the condition $(2)$ means that the coloring $f:=f_1-f_0$ is a proper coloring and not a constant. Further, we partition the indices in $[n]$ into groups of equally colored and we place the corresponding points of the triangles near the vertices of a equilateral triangle with large enough side length. For each of these indices we apply the same procedure - proper coloring and placing them at (near) the vertices of equilateral triangle and so on till each monochromatic group consists of just 1 point. Remark. It's just an attempt to make the solution more combinatorial, though it's a pure linear algebra problem. Any attempt to make it combinatorial, makes it more difficult.