Points A, B, C, and D (no three of which are collinear) lie in a plane. If point D lies randomly within ΔABC's anticomplementary triangle, then what is the probability that 4gon ABCD is (concave, convex, reflex)?
What would be the answer if ΔABC is equilateral and "ΔABC's anticomplementary triangle" is replaced with "ΔABC's circumcircle"?
NOTE: The anticomplementary triangle of a given triangle
is formed by three lines. Each line passes through
a vertex of the given triangle and is parallel to
the opposite side.
NOTE: The circumcircle of a given triangle is the unique
circle which passes through each of its vertices.
If the anticomplementary triangle is labeled A'B'C', where A' is the vertex farthest from A, etc., then you need to find the areas of triangles ABC, ABC', BCA' and ACB', as well as of A'B'C'.
If D falls within ABC' or BCA', the quadrilateral ABCD would be reflex. If D falls within ABC, the quadrilateral ABCD would be concave, and if within ACB', it would be convex. So the probabilities would be the areas involved divided by the area of A'B'C'. The area of ABC is easily found by Heron's formula, but to get the others, you'd need to know the dimensions, in terms of the known triangle ABC.
Part II:
For the equilateral triangle and circumcircle, since the size doesn't matter, only the relative sizes of the pieces, let's take the radius of the circle as 1, so the circle has area pi.
The base of the triangle is then sqrt(3), and its height is 3/2, so its area is 3 sqrt(3)/4. This leaves pi  3sqrt(3)/4 for the rest of the circle, divided among three segments, each of which thus has area (pi  3sqrt(3)/4)/3 = pi/3  sqrt(3)/4. So the probability of a reflex quadrilateral would be 2(pi/3  sqrt(3)/4) / pi; the probability of a concave quadrilateral would be 3 sqrt(3) / (4 pi); and the probability of a convex quadrilateral would be (pi/3  sqrt(3)/4) / pi. Numerically these are approximated by .391002218955772, .413496671566344 and .195501109477886 respectively, which do add to 1, allowing for rounding error in the last position.

Posted by Charlie
on 20061103 09:55:11 