The transversal number τ (H) of a hypergraph H is the minimum cardinality of a set of vertices that intersects all edges of H. The matching number α (H) of H is the size of a largest matching in H, where a matching is a set of pairwise disjoint edges in H. A hypergraph is intersecting if each pair of edges has a nonempty intersection. Equivalently, H is an intersecting hypergraph if and only if α (H) = 1. We observe that τ (H) ≤ r for an intersecting hypergraph H of rank r. For an intersecting hypergraph H of rank r without isolated vertex, we call H 1-special if τ (H) = r; H is maximal 1-special if H is 1special and adding any missing r-edge to H increases the matching number. Furthermore, H is called 1-edge-critical if for any e ∈ E(H) and any v ∈ e, v-shrinking e increases the matching number; H is called 1-vertex-critical if for every vertex v in H deleting the vertex v and v-shrinking all edges incident with v increases the matching number. The intersecting hypergraphs, as defined above, are said to be matching critical in the sense that the matching number increases under above definitions of criticality [M.A. Henning and A.Yeo, Quaest. Math. 37 (2014) 127-138]. Let n i (r) (i = 2, 3, 4, 5) denote the maximum order of a hypergraph in each class of matching critical intersecting hypergraphs. In this paper we study the extremal behavior of matching critical intersecting hypergraphs. We show that n 2 (r) = n 3 (r) and n 4 (r) = n 5 (r) for all r ≥ 2, which answers an open problem on matching critical intersecting hypergraphs posed by Henning and Yeo. We also give a strengthening of the result n 4 (r) = n 5 (r) for intersecting r-uniform hypergraphs. arXiv:1512.02871v1 [math.CO] 9 Dec 2015 Lemma 3.5. For r 2, every hypergraph H in H 5 min (r) with |V (H)| = n 5 (r) is r-uniform. Proof. Suppose, to the contrary, that H contains a t-edge e such that t < r. Since H is minimal 1-vertex-critical, there exists a vertex u in e such that qd H−e (u) < 2 by Lemma 3.4, hence there exists an edge f of H whose intersecting with e is only u, i.e., e ∩ f = {u}. Clearly, f \ {u} is a transversal in H − e. Note that H has rank r, without loss of generality, we may assume that f is an r-edge of H. Let H be the hypergraph obtained from H by adding a new vertex v and a new edge (f \ {u}) ∪ {v} to H and replacing e by e ∪ {v}. By the construction of H , V (H ) = V (H) ∪ {v} and E(H ) = (E(H) \ {e}) ∪ {e ∪ {v}, (f \ {u}) ∪ {v}}. Hence |V (H )| = n 5 (r) + 1 and H has rank r. Note that f \ {u} is a transversal of H − e, so (f \ {u}) ∪ {v} meets with all edges of H . Hence H is a intersecting hypergraph with rank r. On the other hand, we note that d(v) = qd(v) = 2 and qd(u) 2 for all u ∈ H . Thus H is still 1-vertex-critical, i.e.,H ∈ H 5 (r). But then |V (H )| ≤ n 5 (r), which is a contradiction.By Lemma 3.5 and Theorem 3.1, we obtain a strengthening of Theorem 3.1.Theorem 3.2. For any integers r ≥ 2, n 4 r = n 4 (r) = n 5 (r) = n 5 r .