AMC 12 Mock Exam 3 Problem 21

The three points cut the circle into three arcs; let their measures, as fractions of the full circle, be $a,b,c$ with $a+b+c=1$. The angle of the triangle opposite an arc is half that arc's central angle, i.e., $180^\circ\cdot(\text{fraction})$. So the triangle is acute iff $a,b,c<\frac12$. The side opposite an arc of central angle $\phi<180^\circ$ has length $2\sin\frac\phi2$, which exceeds $1$ iff $\phi>60^\circ$. So we need $\frac16<a,b,c<\frac12$.

Rotate so the first point is at $0$, and let the other two be at $u,v\in[0,1)$ (uniform on the unit square). By symmetry take $u<v$ (the triangle $0<u<v<1$ has area $\frac12$), with $a=u$, $b=v-u$, $c=1-v$; the map $(u,v)\mapsto(a,b)$ preserves area. Write $a=\frac16+a'$, etc., with $a'+b'+c'=\frac12$: the region $a',b',c'\ge0$ is a right triangle with legs $\frac12$, area $\frac18$. We must remove the parts where some $a'>\frac13$ (i.e., an arc exceeds $\frac12$): each is a triangle with legs $\frac16$, area $\frac1{72}$, and they are disjoint. The favorable area is $\frac18-\frac3{72}=\frac1{12}$, so the probability is $\frac{1/12}{1/2}=\boxed{\tfrac16}$.