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?
The reasoning in my earlier comment admits to an easy generalisation: The probability of encompassing the centre inside the triangle remains 1/4, so long as the circumscribing shape is a regular polygon with an even number of sides (hexagon, octagon, etc.)
To see this we only need to examine the triangles in pairs. Let A and B be two vertices of the inscribed triangle and let Z be the centre of the polygon. Connect both A and B to Z. Then we have already seen that the centre will lie in ABC so long as C is in the sector encompassed by the extensions of AZ and BZ.
Now consider the situation if instead of having chosen B, we had chosen the point B' (lying on the extension of BZ the same distance from Z as B). Then the triangle AB'C' will contain the Z iff C' is in the sector encompassed by the extensions of AZ and B'Z.
But the two sectors, taken together encompass exactly one half thearea within the circumscribing polygon. (They are delimited by a straight line passing through Z). Hence, there is a 50% probability that Z will lie in one of these two triangles.
This works with circumscribing regular polygons with an even number of sides because 1) B' will be inside the polygon whenever B is in the polygon. (This would not be true, for instance, if the circumscribing polygon were to be an equilateral triangle), and 2) Because any line drawn through Z divides an even sided regular polygon into sectors with the same area (again, not true for a circumscribing equilateral triangle).
Edited on March 25, 2008, 4:42 pm
Posted by FrankM
on 2008-03-25 16:40:18