Wondering if we can prove the following or if it is already proved where can I get the proof.Let $v_1, v_2, v_3, \ldots, v_n$ and $t$ be $n+1$ vertexes in a directed graph. $v_1, v_2, v_3, \ldots, v_n$ form directed acyclic graph. $t$ is connected to each and everyone of $v_1, v_2, v_3, \ldots, v_n$. Now since $v_1, v_2, v_3, \ldots, v_n$ are connected in an acyclic manner, if there is a cycle then it will involve $t$. Can we show that all cycles of length more then 3 will always involve a cycle of length 3? Remember $t$ is connected to every $v_1, v_2, v_3, \ldots, v_n$ and there is no pair wise cycle.Are there any other similar problems with different wordings?ThanksI have probably not worded the question more clearly.Say the acyclic directed graph of vertices $v_1, v_2, v_3, \ldots, v_n$ is $v_1 \rightarrow v_2 \rightarrow v_3 \rightarrow \cdots \rightarrow v_n$. Each $v$ has an edge to $t$. Say there is a cycle $t \rightarrow v_1 \rightarrow v_2 \rightarrow v_3 \rightarrow t$. Such a cycle seems to surely involve a cycle of length 3 i.t either $t \rightarrow v_1 \rightarrow v_2 \rightarrow t$ or $t \rightarrow v_2 \rightarrow v_3 \rightarrow t$. But an not being able to proof this. 
Cycles in a directed graph
ds.algorithms;graph theory;directed acyclic graph