Nine points are given in a square having unit area.
Is it always possible to choose three of these points as the vertices of a triangle, such that the area of the triangle is at most 1/8?
Provide valid reasoning for your answer.
Divide the unit square into four squares each with side 1/2.
With nine points and four squares at least one square will contain three points.
Those three points determine a triangle which has maximum area when each leg is 1/2.
Area triangle <= (1/2)*bh = (1/2)*(1/2)*(1/2) = 1/8
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Posted by xdog
on 2023-02-13 19:40:07 |