(b) WLOG the line is horizontal. Let $S$ be the set of all segments, $A$ be the $n$ with the leftmost left coordinate (ties broken by rightmost right coordinate), $B$ be the $n$ with rightmost right coordinate (ties broken by leftmost left coordinate), and $I$ be the remaining segments in $S$. Observe that the rightmost left endpoint in $A$ is still left of the leftmost right endpoint in $A$; otherwise the condition would necessitate that there is a segment in $S\setminus A$ with a leftmore left endpoint (contradicting the definition of $A$). A similar claim holds for $B$. It follows that the segments in $A$ are pairwise intersecting, and likewise that the segments in $B$ are pairwise intersecting. Thus I f there were a segment in both $A$ and $B$, then it would satisfy the desired property, and we'd be done. Suppose not; that $A$ and $B$ are disjoint, and that in particular there is a unique $i\in I$. Consider the $a\in A$ with leftmost right endpoint (break ties yada yada...). By hypothesis, it must intersect $i$ or some $b'\in B$ (or both). If the latter is the case, then again we'd be done, so suppose not. Analogously consider the corresponding $b\in B$; by a similar argument we can assume that it intersects $i$. But then $i$ itself must have the desired property, and the result follows $\blacksquare$

(a) Let's there are $a,b,c,d\ in\mathbb{Z}$ suchthat $ x^4+3x^3+6x^2+9x+12=(x^2+ax+b)(x^2+cx+d)$ $\implies$\begin{align} a+c=3\\ bd=12\\ b+d+ac=6\\ bc+ad=9 \end{align}$(2)+(3)+(4):(a+c+1)(b+d)=27$ $\implies b+d=\frac{27}{4}$ So its is contradic with $a,b\in\mathbb{Z}$ Done$\ddot\smile$

a) Just apply Eisenstein Criterion with $p=3$.

(b) A different point of view makes the problem almost trivial. We may view the situation as a timeline for the following activity. We have a room with entrance and exit. Everyone of the $2n+1$ people enters the room, stays there some time and then exits. Each person meets inside the room at least $n$ others. Prove that there is a person who meets all other people inside the room. Proof. Consider the $n+1$-th person, who enters the room. The moment he/she enters, all $n$ persons, that have entered before, are still in. Indeed, if someone had gone away, he/she would not comply to the requirement. If all those $n+1$ persons had exited before the last person entered, that last person would also break the rule. So, there is a person $P$ among the first $n+1$ have entered, who is still inside the moment the last person enters. Hence, $P$ has met all the others inside the room.

FERZIO wrote: (a) Let's there are $a,b,c,d\ in\mathbb{Z}$ suchthat $ x^4+3x^3+6x^2+9x+12=(x^2+ax+b)(x^2+cx+d)$ $\implies$\begin{align} a+c=3\\ bd=12\\ b+d+ac=6\\ bc+ad=9 \end{align}$(2)+(3)+(4):(a+c+1)(b+d)=27$ $\implies b+d=\frac{27}{4}$ So its is contradic with $a,b\in\mathbb{Z}$ Done$\ddot\smile$ $(a+c+1)(b+d)=ab+(bc+b+ad+dc+d)=ab+27$