Consider five points in the plane, with no three of them collinear. Every pair of points among them is joined by a line. In how many ways can we color these lines by red or blue, so that no three of the points form a triangle with lines of the same color.

Label the five points as $$\{1,2,3,4,5\}$$. Join every pair of points; the resulting graph is the complete graph $$K_5$$ with $$10$$ edges. We must colour every edge red or blue so that no three vertices form a triangle whose three sides have the same colour.

Step 1: Degree restriction at each vertex
Pick any vertex, say $$1$$. Let $$d_R(1)$$ and $$d_B(1)$$ be the number of red and blue edges incident at $$1$$; clearly $$d_R(1)+d_B(1)=4$$.

Assume $$d_R(1)\ge 3$$. Take three red neighbours of $$1$$, say $$2,3,4$$. Among the three edges $$\{2,3\},\{3,4\},\{4,2\}$$ at least one is red or blue:
- If any of them is red, say $$\{2,3\}$$, then $$1,2,3$$ create a red triangle - forbidden.
- Otherwise all three are blue, giving blue triangle $$2,3,4$$ - also forbidden.
Thus $$d_R(1)\not\ge 3\Rightarrow d_R(1)\le 2$$. The same argument with colours swapped gives $$d_B(1)\le 2$$.

Because $$d_R(1)+d_B(1)=4$$ and both quantities are $$\le 2$$, we must have $$d_R(1)=d_B(1)=2$$. Since the choice of vertex was arbitrary, $$d_R(v)=d_B(v)=2 \quad\text{for every vertex }v.$$

Step 2: Structure of the red (and blue) subgraphs
The red edges form a $$2$$-regular graph on five vertices. A $$2$$-regular graph is a disjoint union of cycles. Because triangles are forbidden, no cycle may have length $$3$$. Hence the only possibility is a single cycle of length $$5$$, i.e. the red subgraph is the $$5$$-cycle $$C_5$$.

The blue edges are the complement of the red edges inside $$K_5$$, so they also form the complement of a $$5$$-cycle. But the complement of $$C_5$$ on five vertices is again $$C_5$$. Thus the blue subgraph is another $$5$$-cycle, automatically free of triangles as required.

Step 3: Counting admissible colourings
Every admissible colouring is obtained by choosing which $$5$$-cycle will be coloured red; the remaining five edges automatically become blue.
Conversely, any red $$5$$-cycle gives a colouring with no monochromatic triangle, as neither colour contains a triangle.

The number of distinct Hamiltonian (length-5) cycles in $$K_5$$ is well known:
For $$K_n$$ it is $$\dfrac{(n-1)!}{2}$$ (the factor $$2$$ divides out the two directions of traversal). Hence for $$n=5$$ we get $$\frac{(5-1)!}{2}=\frac{4!}{2}=12.$$

Step 4: Conclusion
There are exactly $$12$$ ways to colour the $$10$$ edges of $$K_5$$ red or blue so that no monochromatic triangle is formed.

Answer: 12