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 Three Points in a Square (Posted on 2008-03-17)
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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 re: Faster, simpler, better and now FURTHER Comment 18 of 18 |
(In reply to Faster, simpler, better and now FURTHER by FrankM)

Even Further?

I agree with you Frank about extending this to regular polygons with an even number of sides and retaining the probability of 1/4. However, I suggest it can be extended to any closed figure with point symmetry so that parallelograms, ellipses etc would be OK.

I was thinking like this.. (using the words radius, diameter and sector loosely and using a shape with unit area and centre Z):

For a fixed vertex A, anywhere within the shape, let the area of the sector formed by the radii through A and B be 'a', so that 'da' (delta a) gives the probability of a second vertex B lying in an elemental sector of area 'da'. The probability of C giving a triangle ABC enclosing Z is given by the area of the sector between the extensions of AZ and BZ. Provided there is point symmetry about Z, this sector is congruent to the sector between ZA and ZB and so the probability is 'a'.

Multiplying these probabilities and integrating for B anywhere on one side of the diameter AZ, gives a probability of integral of a*da = (a^2)/2 between a=0 and a=1/2. Thus the integral is 1/8 and allowing B to lie on the other side of the diameter doubles this to 1/4.

I'm now wondering whether it's possible to have a shape with vertically opposite sectors being equal in area, but which doesn't have point symmetry so that we could extend again.

Edited on March 19, 2009, 2:53 pm
 Posted by Harry on 2009-03-18 15:41:15

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