If this is correct, the answer should be $k = 2024$. Generally, if the number of points are $3n$, $n$ for each color, then the minimum number of bolivian triangles over all configuration satisfying the property of the problem is $n$. First, for each colored point $T$, we move $T$ such that the direction $\overrightarrow{OT}$ is preserved and the length $|OT|$ is normalized to $1$. This won't change the condition of the problem, since $O$ is in the interior of $\triangle XYZ$ before moving the points if and only if $O$ is in the interior of $\triangle XYZ$ after moving the points. Therefore, now all the colored points are located on a unit circle centered at $O$. The construction for $2024$ bolivian triangles as follows. [asy][asy]import graph;size(6cm); pair O=(0,0); pair R1 = dir(30), R2 = dir(150), R11 = 1.1 * dir(210), R22 = 1.1 * dir(330); pair B1 = dir(225), B2 = dir(315), B3 = dir(120), B4 = 1.1 * dir(300); draw(circle(O,1)); dot(O); dot(R1, red); dot(R2, red); dot(B1, blue); dot(B2, blue); dot(B3, blue); draw(O--R1, red, EndArrow(5)); draw(O--R2, red, EndArrow(5)); draw(O--B1, blue, EndArrow(5)); draw(O--B2, blue, EndArrow(5)); draw(O--B3, blue, EndArrow(5)); draw(O--R11, red + dashed); draw(O--R22, red + dashed); draw(O--B4, blue + dashed); pen darkgreen = rgb(0, 0.5, 0); filldraw(arc(O, dir(230), dir(295))--O--dir(230)--cycle, darkgreen); MP("O",O,S); MP("R_1",R1,NE, red); MP("R_2",R2,NW, red); MP("B_1",B1,SW, blue); MP("B_2",B2,SE, blue); MP("B_3",B3,NW, blue); [/asy][/asy] Put two red points on $R_1$ and $R_2$, three blue points on $B_1$, $B_2$, and $B_3$, and the rest of the colored points on the arc of the green sector. Since the convex hull of the blue points and the red points each contains $O$, so this configuration satisfy the property that the convex hull of the points from any two colors contains $O$. It should be easy to verify that the only bolivian triangles are the triangles formed by $R_1$, $B_3$, and each one of the green points. Therefore, the number of bolivian triangles is $2024$. (note: the dashed segments are the lines opposite to the ray connecting $O$ with $R_1$, $R_2$, and $B_3$) Next, we prove that there always exist $n = 2024$ bolivian triangles in any configuration. For a point $X$, denote $X’$ as its antipode w.r.t. the unit circle. For a point $X$, call two green points which are the closest to $X’$ in the clockwise and counter-clockwise directions as the green splitting points of $X$. (Define the red and blue splitting points similarly). For two points $P$, $Q$ on the circumference, we define $d(P, Q)$ as the angle needed to rotate the ray $\overrightarrow{OP}$ to $\overrightarrow{OQ}$ counter-clockwise, and the arc $\widehat{PQ}$ as the arc of the circle from $P$ to $Q$ moving counter-clockwise. Take a red point $R$ and its green splitting points $G_1$ and $G_2$. If there is a blue point not strictly lying on the arc $\widehat{G_1’G_2’}$, then there is a bolivian triangle with $R$ as one of the vertices. Note that the points $G_1’, G_2’, \dots G_n’$ partition the circumference into $n$ arcs (only intersect at the green points). If not all the blue points lie strictly on one of the partitions, then for each red point, we could find a bolivian triangle with this red point as one of the vertices, meaning that there are at least $n$ bolivian triangles. So, now we assume that all the blue points $B_1, B_2, \dots, B_n$ strictly lie on one of the partitions, let’s say $G_1’G_2’$. So, for each blue point $B$, $G_1$ and $G_2$ are its green splitting points. By repeating the argument from the previous paragraph, either for each blue point there is a bolivian triangle with that point as one of the vertices, creating $n$ bolivian triangles, or we must have all the red points also strictly lie on arc $G_1’G_2’$. Now, we assume the latter holds. If $\alpha = d(G_1, G_2) < 180^{\circ}$, this means the blue and red points can be covered by a sector with angle less than $180^{\circ}$, implying that the convex hull of the blue and red points doesn’t contain $O$, a contradiction. So, $\alpha \ge 180^{\circ}$. [asy][asy]import graph;size(6cm); pair O=(0,0); pair G1 = dir(30), G2 = dir(150), G11 = dir(210), G22 = dir(330); pair R1 = dir(200), R2 = dir(355), R11 = dir(20), R22 = dir(175); pen darkgreen = rgb(0, 0.5, 0); draw(circle(O,1)); dot(O); dot(G1, darkgreen); dot(G2, darkgreen); dot(G11, purple); dot(G22, purple); dot(R11, orange); dot(R22, orange); dot(R1, red); dot(R2, red); draw(O--G1, darkgreen, EndArrow(5)); draw(O--G2, darkgreen, EndArrow(5)); draw(O--R1, red, EndArrow(5)); draw(O--R2, red, EndArrow(5)); draw(O--(1.1*G11), purple+dashed); draw(O--(1.1*G22), purple+ dashed); draw(O--(1.1*R11), orange+dashed); draw(O--(1.1*R22), orange+ dashed); MP("O",O,S); MP("G_2",G1,NE, darkgreen); MP("G_1",G2,NW, darkgreen); MP("R_1",R1,SW, red); MP("R_2",R2,SE, red); MP("G_2'",G11,E, purple); MP("G_1'",G22,W, purple); MP("R_1'",R11,E, orange); MP("R_2'",R22,W, orange); path alpha = anglemark(G2, O, G1); draw(alpha, darkgreen); label("$\alpha$", alpha, 9 * dir(160), darkgreen); markscalefactor=0.02; path beta = anglemark(R1, O, R2); draw(beta, red); label("$\beta$", beta, S, red); [/asy][/asy] If all the red points are strictly on arc $\widehat{G_1'G_1}$, then the convex hull of red and green points doesn’t contain $O$, so there must exist a red point in arc $\widehat{G_1G_2’}$. Similarly, there exists a red point on arc $\widehat{G_1'G_2}$. Take the red points $R_1$ and $R_2$ on both of them which is the furthest from $G_1$ and $G_2$, respectively. We must have $\beta = d(R_1, R_2) \le 180^{\circ}$ (or $d(R_2, R_1) \ge 180^{\circ}$), because the convex hull of the green and red points contains $O$. Note that for each green point $G$, the points $R_1$ and $R_2$ are their red splitting points. Therefore, if there exists a blue point outside the arc $R_1’R_2’$, there must exist at least $n$ bolivian triangles, one for each green point. So, all the blue points lie strictly on arc $R_1’R_2’$. But this means the convex hull of the green and blue points doesn’t contain $O$, a contradiction. Therefore, there are at least $n$ bolivian triangles.