Let $P=\{P_1,P_2\cdots P_n\}$ be a set of $n\geq 4$ points in the plane and $P_iP_j$ be the line segment connecting $P_i$ and $P_j$ that satisfy:

$(1)$Any three points of $P$ are not on a line;

$(2)$In the set $\{P_1P_2,P_2P_3,\cdots,P_{n-1}P_n,P_nP_1\}\setminus \{P_iP_{i+1}\}$,$P_iP_{i+1}$ only intersect with $P_{i-1}P_i$ and $P_{i+1}P_{i+2}$ ,$i=1,2,\cdots,n$(where $P_0=P_n,P_{n+1}=P_1$).

I conjecture that there must exist $1\leq i\leq n$ such that there is no point of $P$ in the interior of $\bigtriangleup P_{i-1}P_iP_{i+1}$,where $P_0=P_n,P_{n+1}=P_1$.

Is that right?


1 Answer 1


Yes, the polygon formed by the broken line $P_1\dots P_n$ may be triangulated by $n-3$ inner diagonals and in any triangulation there exists a triangle (at fact, at least two triangles for $n\geq 4$) of the form $P_{i-1}P_iP_{i+1}$.

  • 4
    $\begingroup$ This is known as Meister's two-ears theorem, where those empty triangles are the "ears" of the polygon. $\endgroup$ Dec 10, 2014 at 11:59

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