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$Figure 3 shows a graph structure centered around a block B_s. The boundary of B_s is a cycle containing vertices a_1, w_{s-1}, x_1, x_2, w_s, b_1, b_{q-1}, b_q, and z_c. Inside B_s, there is a region T. Two paths, L_1 and L_2, connect the boundary. L_1 connects a_1 to w_{s-1} and x_1. L_2 connects b_1 to b_{q-1} and x_2. The region T is bounded by L_1, L_2, and a path F' connecting y_1 and y_2. The path F' is shown in red, and L_1 and L_2 are shown in blue. The vertices z_1, z_2, z_3, \dots, z_{c-1} are also marked on the boundary of B_s.$

Figure 3 shows a graph structure centered around a block B_s. The boundary of B_s is a cycle containing vertices a_1, w_{s-1}, x_1, x_2, w_s, b_1, b_{q-1}, b_q, and z_c. Inside B_s, there is a region T. Two paths, L_1 and L_2, connect the boundary. L_1 connects a_1 to w_{s-1} and x_1. L_2 connects b_1 to b_{q-1} and x_2. The region T is bounded by L_1, L_2, and a path F' connecting y_1 and y_2. The path F' is shown in red, and L_1 and L_2 are shown in blue. The vertices z_1, z_2, z_3, \dots, z_{c-1} are also marked on the boundary of B_s.

Figure 3: Some notations on $B_s$ .

a contradiction. If $|G_2|\ge t$ , then $|B_r|=|G_2|\ge t$ and $m(B_r) . Since B_r is 2-connected and$

$m(B_r)\le m(G)-1<\frac{n-(t-1)}{3t-7}-1=\frac{n-(3t-7)-(t-1)}{3t-7}<\frac{|B_r|-(t-1)}{3t-7},$

contradicting the choice of $G$ . □

4 Proof of Theorem 1.5

The proof proceeds by induction. If $k^{\log_2 3}\le n\le 36k^{\log_2 3}$ , then

$e(G)\le 3n-6\le 3n-6-\frac{n}{36k^{1+\log_2 3}}+\frac{k^3}{4},$

the result holds. So, we assume that

$n>36k^{\log_2 3}.$ (2)

Let $G$ be a $2C_k$ -free plane graph with $e(G)=\operatorname{ex}_P(n, 2C_k)$ . It is clear that $G$ is connected; otherwise we can add an edge between two components of $G$ to make it remain $2C_k$ -free, a contradiction. If $G$ is $C_k$ -free, then by Theorem 1.3, $e(G)\le 3n-6-\frac{n}{4k^{\log_2 3}}\le 3n-6-\frac{n}{36k^{1+\log_2 3}}+\frac{k^2}{2}$ , the result holds. So, we assume that $G$ contains a $C_k$ .

Case 1. $G$ is 2-connected.

In that case, the boundary of each face in $G$ is a cycle. Assume that $c$ is the smallest number of vertices that need to be removed from $G$ to make it $C_k$ -free, and let $v_1, v_2, \dots, v_c$ be one such set of vertices. Then $c\le k$ , since we can remove vertices of a $k$ -cycle to make it $C_k$ -free. Moreover, for any $i, j\in[c]$ , there exists a $k$ -cycle in $G$ such that $v_i$ is contained in the $k$ -cycle, but $v_j$ not.

Case 2.1. $c=1$ .

Since $G$ is 2-connected, it follows that $G'=G-v_1$ is connected and is $C_k$ -free. Assume that $G'$ has $r$ blocks $B_1, B_2, \dots, B_r$ , where $|B_i|\ge k^{\log_2 3}$ for $i\in[q]$ and $|B_i| for q+1\le i\le r . For each i\in[q], since B_i is a C_k -free 2-connected plane graph, it follows that$

$m(B_i)\ge\frac{|B_i|-(k^{\log_2 3}-1)}{3k^{\log_2 3}-7};$