On a given circle, six points $A$, $B$, $C$, $D$, $E$, and $F$ are chosen at random, independently and uniformly with respect to arc length. Determine the probability that the two triangles $ABC$ and $DEF$ are disjoint, i.e., have no common points.
Problem
Source: USAMO 1983 Problem 1
Tags: AMC, USA(J)MO, USAMO, probability, combinatorics unsolved, combinatorics
professordad
17.08.2011 03:06
Let the vertices be A, B, C, D, E, and F. Basically this turns into a problem where you want to find the number of ways that A, B, and C are together in a circular arrangement of ABCDEF. Fix A; then in the circle, we have three cases: either B and C are both to the left of A, both are to the right, or A is in the middle. For each case we have $2 \cdot 6 = 12$ arrangements, because we can arrange B and C 2! ways, and then the other three 3! ways. So $\tfrac{36}{120} = \boxed{\tfrac{3}{10}}$.
This should be moved to High School Intermediate or so...at least the Combinatorics section...
ahaanomegas
17.08.2011 03:58
professordad wrote: This should be moved to High School Intermediate or so...at least the Combinatorics section... This was put here because it is from a math olympiad and it is unsolved. Also, I have a question: what does 'disjoint' mean here?
GlassBead
17.08.2011 04:07
Disjoint in this case means that any edge of one triangle does not intersect the other.
RedFireTruck
30.04.2021 05:32
We see that the points must have $A$, $B$, and $C$ consecutive.
WLOG assume $A$ is fixed.
We see that there are $5!=120$ ways to arrange the points.
We see that there are $2!3!3=36$ ways that $A$, $B$, and $C$ are consecutive.
We see that out answer is $\frac{36}{120}=\boxed{\frac{3}{10}}$.