Problem:
Three distinct segments are chosen at random among the segments whose endpoints are the vertices of a regular 12-gon. What is the probability that the lengths of these three segments are the three side lengths of a triangle with positive area?
Answer Choices:
A. 715553β
B. 572443β
C. 143111β
D. 10481β
E. 286223β Solution:
Assume without loss of generality that the regular 12-gon is inscribed in a circle of radius 1 . Every segment with endpoints in the 12-gon subtends an angle of 12360βk=30k degrees for some 1β€kβ€6. Let dkβ be the length of those segments that subtend an angle of 30k degrees. There are 12 such segments of length dkβ for every 1β€kβ€5 and 6 segments of length d6β. Because dkβ=2sin(15kβ), it follows that d2β=2sin(30β)=1,d3β=2sin(45β)=2β, d4β=2sin(60β)=3β,d6β=2sin(90β)=2,
d1β=2sin(15β)=2sin(45ββ30β)=2sin(45β)cos(30β)β2sin(30β)cos(45β)=26ββ2ββ, and d5β=2sin(75β)=2sin(45β+30β)=2sin(45β)cos(30β)+2sin(30β)cos(45β)=26β+2ββ.β
If aβ€bβ€c, then daββ€dbββ€dcβ and the segments with lengths daβ,dbβ, and dcβ do not form a triangle with positive area if and only if dcββ₯daβ+dbβ. Because d2β=1<6ββ2β=2d1β<2β=d3β, it follows that for (a,b,c)β{(1,1,3),(1,1,4),(1,1,5),(1,1,6)}, the segments of lengths daβ,dbβ,dcβ do not form a triangle with positive area. Similarly,
d3β=2β<26ββ2ββ+1=d1β+d2β<3β=d4βd4β<d5β=26β+2ββ=26ββ2ββ+2β=d1β+d3β, and d5β<d6β=2=1+1=2d2ββ
so for (a,b,c)β{(1,2,4),(1,2,5),(1,2,6),(1,3,5),(1,3,6),(2,2,6)}, the segments of lengths daβ,dbβ,dcβ do not form a triangle with positive area. Finally, if aβ₯2 and bβ₯3, then daβ+dbββ₯d2β+d3β=1+2β>2β₯dcβ, and also if aβ₯3, then daβ+dbββ₯2d3β=22β>2=dcβ. Therefore the complete list of forbidden triples (daβ,dbβ,dcβ) is given by (a,b,c)β{(1,1,3),(1,1,4), (1,1,5),(1,1,6),(1,2,4),(1,2,5),(1,2,6),(1,3,5),(1,3,6),(2,2,6)}.
For each (a,b,c)β{(1,1,3),(1,1,4),(1,1,5)}, there are (122β) pairs of segments of length daβ and 12 segments of length dcβ. For each (a,b,c)β{(1,1,6),(2,2,6}}, there are (122β) pairs of segments of length daβ and 6 segments of length dcβ. For each (a,b,c)β{(1,2,4),(1,2,5),(1,3,5)}, there are 123 triples of segments with lengths daβ,dbβ, and dcβ. Finally, for each (a,b,c)β{(1,2,6),(1,3,6)}, there are 122 pairs of segments with lengths daβ and dbβ, and 6 segments of length dcβ. Because the total number of triples of segments equals βββ(122β)3ββ ββ=(663β), the required probability equals