I would suggest going with “What is the probability that these three points will form a triangle containing the center of the square?” and avoiding all ambiguity, while keeping the elegance of your solution intact.

Jonathan

]]>Three collinear points form a triangle — it just happens to be degenerate. (Consequently, it has a lot in common with my friends.) I think the James and James math dictionary provides a definition of triangle without stipulating that they must not be collinear, allowing for the degenerate case. The book is at my office, so I’ll check it out on Tuesday — but I think it’s James and James; if not, I know that I’ve seen definitions that allow for this very case, and I’ll see if I can find a reference.

I definitely don’t like any rewording of the problem where the answer must be changed to 4/15.

]]>1. Reword

“What is the probability that the center of the square will be contained within the triangle formed by these three points?” to

“What is the probability that the center of the square will be contained within a triangle formed by these three points?” or to

“What is the probability that these three points will form a triangle containing the center of the square?”

2. Modify your solution by multiplying by the probability a triangle is actually formed (15/16). This is not so nice, and yields my quibble-answer.

Jonathan

]]>The solution offered is nice. But see the bottom. I have a quibble.

I considered 4 cases:

3 points on a side P = 1/16

2 points on a side, and one on an adjacent side P = 6/16

2 points on a side, and one on the opposite side P = 3/16

points on 3 sides P = 6/16

The first case (3 on a side) yields no triangles.

The second (2 on a side, one adjacent) never encloses the center.

The third case (2 on a side, one opposite) I used analysis like yours, but found P = length of the segment on one side, integrated, got 1/3.

The fourth, points on 3 sides, I connected the points on opposite sides. The adjacent point will enclose the center either on the left, or on the right, so 1/2

But here’s the problem

3/16 x 0 = 0

6/16 x 1/3 = 1/16

6/16 x 1/2 = 3/16

Adds to 4/16, right? But you’ve asked about the triangle, and in 1/16 of the cases there is no triangle. Looks to me like 4/15 of the triangles enclose the center.

]]>I don’t know the answer to your question, and I don’t easily see a why to extend my argument to the n-dimensional case. I’ll need to give it some thought. I may also send it to my friend Harold Reiter; he and his friend Arthur Holshouser solved a different generalization (see comment #7 above), so they may have thoughts on this one, too.

]]>a colleague of mine and I tried a brute force simulation for the 2 dimensional case and obviously got the 1/4 approximation (we used a Mathematica script together with this piece of knowledge: http://mathworld.wolfram.com/TriangleInterior.html)

I challenged my colleague (who is a much better mathematician than me) to generalise the result and we conjectured that for the n dimensional case the probability of finding the centre of the n-dimensional cube within a n-dimensional pyramid created by n+1 random points on the hyperfaces shoud be 1/2^n.

Quite surprisingly that is the case and a few hundred thousands trials per dimension are showing exactly that.

Could any of the 2-dimensional proofs easily be extended to the n-dimensional case?

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