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We consider the following iterative construction of a random planar triangulation.Start with a triangle embedded in the plane. In each step, choose a bounded face uniformly at random, add a vertex inside that face and join it to the vertices of the face. After n -3 steps, we obtain a random triangulated plane graph with n vertices, which is called a Random Apollonian Network (RAN). We show that asymptotically almost surely (a.a.s.) a longest path in a RAN has length o(n), refuting a conjecture of Frieze and Tsourakakis. We also show that a RAN always has a cycle (and thus a path) of length (2n − 5) log 2/ log 3 , and that the expected length of its longest cycles (and paths) is n 0.88 . Finally, we prove that a.a.s. the diameter of a RAN is asymptotic to c log n, where c ≈ 1.668 is the solution of an explicit equation.and we say a.a.s. X = o(f (n)) if for every fixed ε > 0, lim n→∞ P X ≤ εf (n) = 1.The authors of [9] conjecture in their concluding remarks that a.a.s. a RAN has a path of length (n). We refute this conjecture by showing the following theorem. Let L m be a random variable denoting the number of vertices in a longest path in a RAN with m faces. Theorem 1.1. A.a.s. we have L m = o(m).Recall that a RAN on n vertices has 2n − 5 faces, so Theorem 1.1 implies that a.a.s. a RAN does not have a path of length (n).We also prove lower bounds for L m deterministically, and its expected value in a RAN. In fact, we will prove lower bounds for C m and E [C m ], where C m denotes the number of Random Structures and Algorithms

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