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there are nine points on or inside the unit square.prove that u will always find three points, which consist a triangle with area not greater than (1/8)

Geometry
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what i did: since the nine points need to be farthest from each other to get the max value of area, one possible configuration is shown in diagram. then it can easily be shown that the are of that 45-45-90 triangle is 1/2(1/2)(1/2) = 1/8 which is max. but i want some alternative general approach , if any. |dw:1352030299200:dw|
should we choose non-collinear points? (the three points) can't we choose the triangle like below? or it will mean that I choose four points? :D |dw:1352030543732:dw|
the triangle formed should not contain any other points on r inside it.

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Other answers:

it's said that the points randomly marked on or inside the square, can I put the points like this? :D |dw:1352030942810:dw|
i am sorry! the exact question is 'there are nine points on or inside the unit square.prove that u will always find three points, which consist a triangle with area not greater than (1/8)' and is edited now......
does this change anything ?
it does. it gets easy now as now we can simply apply pigeonhole principle
idk what is pigeonhole principle ?
divide the unit square into 8 equal triangles first
|dw:1352031901246:dw|
think of them as pigion holes. 8 pigeon holes
we need to assign 9 points to 8 triangles. the points are pogeons
interesting!
9 pigeons, 8 holes. by pigeonhole principle, one pigeon must go into an already occupied hole
that proves, atleast one triangle will have area less than 1/8 becoz, each triangle area is 1/8
wiki has very simple and good explanaiton of pigeonhole principle http://en.wikipedia.org/wiki/Pigeonhole_principle
nice!
whoaa..., I never thought that this problem can be solved using pigeonhole principle :D
also implicitly weve used below to conclude fully : For a given square with any 3 points inside making a triangle, the maximum area of triangle is half the area of square

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