China Southeastern Mathematical Olympiad

Let the set of 12 red points be $A=\{A_1,A_2,\dots ,A_{12}\}$. From $A_1$, one can get 11 chords with red points, but every triangle with vertex $A_1$ has two such chords, and so the 11 chords should be in at least 6 triangles with $A_1$ being an endpoint. The same applies to $A_i$ ($i=2,3,\dots ,12$); we need

$12\times 6=72$ triangles, and every triangle has 3 points. So $n\ge \frac{72}{3}=24$.

On the other hand, the following example shows that $n$ can be 24.

Consider a circle whose perimeter is 12, and the 12 red points are on the circle with equal distance.

The number of chords with red endpoint is $\left(\genfrac{}{}{0ex}{}{12}{2}\right)=66$. If the length of the minor arc to a chord is $k$, then say “the chord belongs to $k$.” So we have only six kinds of chords, and the number of chords belonging to 1, 2, ..., 5 is 12, and that belonging to 6 is 6.



If three chords belonging to $a,b,c$ ($a\le b\le c$) can be the sides of a triangle, then $a+b=c$ or $a+b+c=12$. So the triangles $(a,b,c)\in \{(1,1,2),(2,2,4),(3,3,6),(2,5,5),(1,2,3),(1,3,4),(1,4,5),(1,5,6),(2,3,5),(2,4,6),(3,4,5),(4,4,4)\}$.

Now we give an example:

The number of (1, 2, 3) triangles is 6, whose vertices are $\{2,3,5\},\{4,5,7\},\{6,7,9\},\{8,9,11\},\{10,11,1\},\{12,1,3\}$.

The number of (1, 5, 6) triangles is 6, whose vertices are $\{1,2,7\},\{3,4,9\},\{5,6,11\},\{7,8,1\},\{9,10,3\},\{11,12,5\}$.

The number of (2, 3, 5) triangles is 6, whose vertices are $\{2,4,11\},\{4,6,1\},\{6,8,3\},\{8,10,5\},\{10,12,7\},\{12,2,9\}$.

The number of (4, 4, 4) triangles is 3, whose vertices are $\{1,5,9\},\{2,6,10\},\{3,7,11\}$.

The number of (2, 4, 6) triangles is 3, whose vertices are $\{4,6,12\},\{8,10,4\},\{12,2,8\}$.

So the minimum $n$ is 24.