Open with smile!

yup.....we have to make the assumption that the triangles are those formed by the points as vertices

but in abhirups post there are only 2 monochromatic Δs
one red and one blue

maybe i am missing some of your Δs

how can u miss so many triangles

because there are somany blue lines

there are many blue triangles within the outer hexagon

yes so there is only one blue Δ that can be formed using the points on hexagon i don't know how you are seeing many Δ's

checkk thetriangle that ive shaded

that is also monochromatic isnt it

ya got it now philip

but the figure and the lines are 2D there is no question of 3D

"There are 6 points in a plane" this is from the question

so it is 2d

am i wrong here

I made the assumption that the triangles are those formed by the vertices as otherwise the question becomes too complicated to handle. If, instead (as subash indicates) you want to consider triangles formed from all possible intersections, the first hurdle is : just how many triangles are formed. If the points form a regular hexagon you have a number of pairs of lines that dont interect. In the general case, too its difficult too predict how many triangles are there Thats why if you limit it to just the triangles with these six points as vertices, you are assured of 20 triangles. What is more, no matter what their orientation, I can shuffle the points around to form a hexagon without altering the number of triangles (hence the WLOG). This simplification is what made me use that assumption. Otherwise you will be on a wild goose chase.

thanx evry1!

so finally wats the ans....

2 coz we r considering only those which formed from vertices of the hexagon

yes.......initially i thought all the possible triangles, but then it wud b much difficult 2 find the ans.....so, only the triangles which r formed by vertices.

According to me the answer should be two. I would argue this way:

Start with a situation where all the lines are coloured red. WLOG we can consider a hexagon. Call a monochromatic triangle good, and bad otherwise. Then, to begin with, we have 20 good triangles. Now we change the colours of the lines one by one and see how many triangles turn bad .

First if we take two opposite bases, we turn for each of the bases, 4 triangles bad. So, now we have only 12 good triangles. This way as each of the bases turn colour, a total of 18 triangles turn bad with one side in each such triangle of the wrong colour.

Now only two triangles are left and changing the colour of one of the lines does not change the number of good triangles. This means we reached the case of minimum number of good triangles.

Hence the minimum is 2 (In abhirup's case we have one red and one blue triangle making that two good triangles)

[The assumption here is that we are considering only triangles formed by the six vertices]

yup there are two monochromatic triangles

let's wait for othersl....