Consider nine points in space, no four of which lie in a plane. We connect each pair of points with an edge (i.e., a straight line segment), and color each such edge red, blue, or leave it uncolored. Determine the smallest value of $n$ such that whenever the number of colored edges is exactly $n$, the set of colored edges necessarily contains a triangle whose edges are all the same color.
The task is actually a graph theory question, which can be rephrased as follows: What is the smallest $n$ such that if a 9-point graph has $n$ edges, then no matter how we color these edges red or blue, the graph will always contain a monochromatic triangle?

To solve this, we will use the following theorem: if a graph has at least 6 points, then the graph or its complement will definitely contain a triangle, or in other words: if the edges of a complete graph with at least 6 points are colored with two colors, the graph will definitely contain a monochromatic triangle.

A 9-point complete graph has $\binom{9}{2}=36$ edges. We will prove that if the graph has at least 33 edges (i.e., at most 3 edges are missing), then it will always contain a monochromatic triangle. Additionally, we will provide an example of a colored graph with 32 edges that does not contain a monochromatic triangle. From this, it follows that $n=33$ is the desired number.

![](https://cdn.mathpix.com/cropped/2024_05_02_a4f7f9267076ca0c30dcg-1.jpg?height=693&width=661&top_left_y=632&top_left_x=713)

If the graph is missing at most 3 edges, then select one endpoint of each missing edge, and remove these at most 3 selected points from the graph. This results in a subgraph of the original graph that has at least 6 points and no missing edges (i.e., the subgraph is complete). According to the theorem mentioned earlier, this subgraph must contain a monochromatic triangle. The triangle is also part of the original graph. This proves that if the graph has at least 33 edges, it will contain a monochromatic triangle.

Now, we will provide an example of a 9-point, 32-edge graph whose edges can be colored with two colors such that it does not contain a monochromatic triangle (Figure 1). Note that among the four missing edges, no two can share a common endpoint, because otherwise, we could remove three points (one being the common endpoint of the two edges) such that the remaining 6-point graph is complete, which would definitely contain a monochromatic triangle.

This condition already determines the structure of the graph; we only need to specify the coloring.

![](https://cdn.mathpix.com/cropped/2024_05_02_a4f7f9267076ca0c30dcg-1.jpg?height=683&width=596&top_left_y=1838&top_left_x=751)

For better clarity, we represent the red edges with solid lines and do not draw the blue edges. The missing $36-32=4$ edges are indicated by dashed lines. From Figure 2, it is clear that the graph contains neither a complete triangle nor an empty triangle.

Remark. The auxiliary theorem can be proven as follows. Let $A$ be any point in the graph. Since at least 5 edges originate from $A$, among these edges, there are three of the same color, say red. Let the endpoints of these edges be $B, C$, and $D$.

If any two of the points $B, C, D$ are connected by a red edge, then these two points together with $A$ form a red triangle. If none of the edges of the triangle $BCD$ are red, then $BCD$ is a blue triangle.