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Suppose that we are given 40 points equally spaced around the perimeter of a square, so that four of them are located at the vertices and the remaining points divide each side into ten congruent segments. If P, Q, and R are chosen to be any three of these points which are not collinear, then how many different possible positions are there for the centroid of \(\triangle PQR\)?

 Dec 14, 2018
 #1
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Case 1: We choose two points that are on the same side of the square. There are 38 other points, 8 of which are on the same side. There are 30 points to choose from so that they are not collinear. 

 

Case 2: We choose two points that are not on the same side. At this point, any of the other 38 points work.

 

We also need to consider the 4 vertices, so we subtract our total possible centroids by 4. 

 

30 + 38 -4 = 64, so there are \(64\) possible centroids. 

 Dec 15, 2018
 #2
avatar+701 
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If you think about it, there are 100 possible intersections of points on the square where the centroid could be, but the points cannot be collinear, so the centroid cannot be on the edge of the square. There are 36 points on the edge of the square, 10+10+8+8 or 9*4, so 100 - 36 = 64. There are \(\boxed{64}\) possible centroids.

 

- PM 

PartialMathematician  Dec 15, 2018
 #3
avatar+102459 
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If every possible triangle has its own distinct centroid then I think the answer will be closer to

 

4*3*10C2  +  10*10*10 = 540+1000 = 1540

There might be some double counting here I think but I expect the answer is much bigger than 64

 Dec 15, 2018
edited by Melody  Dec 15, 2018
edited by Melody  Dec 15, 2018

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