13yup.....we have to make the assumption that the triangles are those formed by the points as vertices
There are 6 points in a plane such that no 3 are collinear. Each point is joined to another by a coloured line which is either blue or red. Find the minimum no. of monochromatic triangles.
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1but in abhirups post there are only 2 monochromatic Δs
one red and one blue
maybe i am missing some of your Δs
11how can u miss so many triangles
because there are somany blue lines
there are many blue triangles within the outer hexagon
1yes 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
1[11] but we have to take Δs formed by the points on the hexagon only !!
see prophet sir's post wat he's written in the end
i mean the minimum will be found using that assumption
11ya 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
341I 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.
11no phil question was
There are 6 points in a plane such that no 3 are collinear. Each point is joined to another by a coloured line which is either blue or red. Find the minimum no. of monochromatic triangles.
so it is one plane so no need of meeting in space it is one plane
this from prophet sir's solution
The assumption here is that we are considering only triangles formed by the six vertices
but is this given in the question
12 coz we r considering only those which formed from vertices of the hexagon
341Ppl are awaiting one clarification from you, aditya. The triangles in question are just the ones formed from the 6 points as vertices isnt it?
1yes.......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.
1no subhash you dont know which of those lines do really intersect in space
11in abhirups figure i can see more than 2 same coloured triangle infact many of them in the central blue region
341According 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]
11maybe 35
my method minimum monochromaticity wil be when out of 6C2 total lines 8are of one colour and 7 are of another colour and from the 7 same colour lines 7C3 triangles can be made
please tell me ur views on this soln
1sry prateek... i dunno the ans....i wildly guessed 20..but not sure abt it