$K_{2^p+1}$ is not a union of $p$ bipartite graphs What I want to show is that among $2^p+1$ points in the plane there are three that determine an angle of size at least $\pi(1-1/p)$.
I was told I have to start with showing for $n=2^p$ that the graph $K_{n+1}$ is not the union of $p$ bipartite graphs but $K_n$ is. 
I have no idea how to prove this. My first ideas were induction or proof by contradition but neither was helpful. It could also have something to do with the fact that every bipartite grpah is 2-colourable.
 A: It seems the following.
Proposition. The graph $K_{2^p+1}$ is not a union of $p$ bipartite graphs. 
Proof. Assume the converse: there exists a family $\{\Gamma_i:i\in [p]=\{1,\dots,p\}\}$ of bipartite subgraphs of a graph $\Gamma=K_{2^p+1}$ such that $\Gamma$ is the union of the family $\{\Gamma_i\}$. Let $(V_i ,W_i)$ be the bipartition of vertices of the graph $\Gamma_i$.  Let $\Delta_i$ be a full bipartite graph with the bipartition $V_i\cup (V\setminus V_i)$ of the vertices, where $V$ is the set of all of vertices of the graph $\Gamma$. Then $\Gamma_i$ is a subgraph of  $\Delta_i$ and $\Delta_i$ is a subgraph of $\Gamma$. Then  $\Gamma$ is the union of the family $\{\Delta_i\}$. Now define a map $f:V\to\{0,1\}^p$ as follows. Let $v$ be a vertex of the graph $\Gamma$. For each index $i$ put $f_i(v)=0$ if $v\in V_i$ and $f_i(v)=1$, otherwise. Put $f=\prod_{i\in [p]} f$, that is $f(v)=(f_1(v),\dots, f_p(v)).$  Since $|V|=2^p+1>2^p=|\{0,1\}^p |$, there exist different vertices $v_1,v_2\in V$  such that $f(v_1)=f(v_2)$. But then the edge between $v_1$ and $v_2$ does not belong to each of the graphs $\Delta_i$, a contradiction.
