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Three Points in a Square (Posted on 2008-03-17) Difficulty: 3 of 5
Three points are chosen at random inside a square. Each point is chosen by choosing a random x-coordinate and a random y-coordinate.

A triangle is drawn with the three random points as the vertices. What is the probability that the center of the square is inside the triangle?

No Solution Yet Submitted by Brian Smith    
Rating: 3.2000 (5 votes)

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An Alembic of Alephantine Animadversions | Comment 15 of 18 |

I hold that it is discreet to prefer discrete solutions (on the principle that Discretion is the better part of palaver), so I decided to transmogrify our problem into a (near-)equivalent, using rectangles instead of triangles.

 

Let there be a square grid of 2n x 2n points. Define its "center" as the intersection of the diagonals of the innermost square.  Consider the set of all rectangles which can be formed on these points (since there are an even numbers of rows and columns, the center will never be ON one of these rectangles, to avoid that issue with triangles).  For any rectangle, define a HIT as when the center is within that rectangle.

 

For any N we can compute the ratio (odds, probability) of hits to the number of possible rectangles.  If these converge as N increases, we have a solution to our version of the problem.  I suggest this version is a fair simulacrum of the original problem (and much easier to compute than with triangles), since any triangle is a bisected parallelogram.

 

The ratios do indeed converge to 0.25 (1/4, 1:4) as N inceases:

 

4 x 4 -- 16 of 36 = 0.4444

 

10 x 10 -- 625 of 2025 = 0.3086

 

20 x 20 -- 10000 of 36100 = 0.2770

 

50 x 50 -- 390625 of 1500625 = 0.2603  u.s.w.

 


  Posted by ed bottemiller on 2008-03-19 10:50:32
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