AMC12 2013 A

Answer (E): 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 $\frac{360}{12}k = 30k$ degrees for some $1 \le k \le 6$. Let $d_k$ be the length of those segments that subtend an angle of $30k$ degrees. There are 12 such segments of length $d_k$ for every $1 \le k \le 5$ and 6 segments of length $d_6$. Because $d_k = 2\sin(15k^\circ)$, it follows that $d_2 = 2\sin(30^\circ)=1$, $d_3 = 2\sin(45^\circ)=\sqrt2$, $d_4 = 2\sin(60^\circ)=\sqrt3$, $d_6 = 2\sin(90^\circ)=2$, $d_1 = 2\sin(15^\circ)=2\sin(45^\circ-30^\circ)$ $=2\sin(45^\circ)\cos(30^\circ)-2\sin(30^\circ)\cos(45^\circ)=\frac{\sqrt6-\sqrt2}{2},$ and $d_5 = 2\sin(75^\circ)=2\sin(45^\circ+30^\circ)$ $=2\sin(45^\circ)\cos(30^\circ)+2\sin(30^\circ)\cos(45^\circ)=\frac{\sqrt6+\sqrt2}{2}.$ If $a \le b \le c$, then $d_a \le d_b \le d_c$ and the segments with lengths $d_a$, $d_b$, and $d_c$ do not form a triangle with positive area if and only if $d_c \ge d_a + d_b$. Because $d_2=1<\sqrt6-\sqrt2=2d_1<\sqrt2=d_3$, it follows that for $(a,b,c)\in\{(1,1,3),(1,1,4),(1,1,5),(1,1,6)\}$, the segments of lengths $d_a,d_b,d_c$ do not form a triangle with positive area. Similarly, $d_3=\sqrt2<\frac{\sqrt6-\sqrt2}{2}+1=d_1+d_2<\sqrt3=d_4,$ $d_4<d_5=\frac{\sqrt6+\sqrt2}{2}=\frac{\sqrt6-\sqrt2}{2}+\sqrt2=d_1+d_3,$ and $d_5<d_6=2=1+1=2d_2,$ so for $(a,b,c)\in\{(1,2,4),(1,2,5),(1,2,6),(1,3,5),(1,3,6),(2,2,6)\}$, the segments of lengths $d_a,d_b,d_c$ do not form a triangle with positive area. Finally, if $a\ge2$ and $b\ge3$, then $d_a+d_b\ge d_2+d_3=1+\sqrt2>2\ge d_c$, and also if $a\ge3$, then $d_a+d_b\ge 2d_3=2\sqrt2>2=d_c$. Therefore the complete list of forbidden triples $(d_a,d_b,d_c)$ is given by $(a,b,c)\in\{(1,1,3),(1,1,4),(1,1,5),(1,1,6),(1,2,4),(1,2,5),(1,2,6),(1,3,5),(1,3,6),(2,2,6)\}$. For each $(a,b,c)\in\{(1,1,3),(1,1,4),(1,1,5)\}$, there are $\binom{12}{2}$ pairs of segments of length $d_a$ and 12 segments of length $d_c$. For each $(a,b,c)\in\{(1,1,6),(2,2,6)\}$, there are $\binom{12}{2}$ pairs of segments of length $d_a$ and 6 segments of length $d_c$. For each $(a,b,c)\in\{(1,2,4),(1,2,5),(1,3,5)\}$, there are $12^3$ triples of segments with lengths $d_a,d_b,$ and $d_c$. Finally, for each $(a,b,c)\in\{(1,2,6),(1,3,6)\}$, there are $12^2$ pairs of segments with lengths $d_a$ and $d_b$, and 6 segments of length $d_c$. Because the total number of triples of segments equals $\binom{\binom{12}{2}}{3}=\binom{66}{3}$, the required probability equals $1-\frac{3\cdot 12\cdot \binom{12}{2}+2\cdot 6\cdot \binom{12}{2}+3\cdot 12^3+2\cdot 12^2\cdot 6}{\binom{66}{3}}$ $=1-\frac{63}{286}=\frac{223}{286}.$