I know I am missing something in this problem, but I don't know what:

Let $P_1, P_2, ..., P_{1997}$ be distinct points in the plane. Connect the points with the line segments $P_1P_2, P_2P_3, P_3P_4, ..., P_{1996}P_{1997}, P_{1997}P_{1}$. Can one draw a line that passes through the interior of every one of these segments?

According to how I understand the problem, the solution is easy. It's clearly not possible to have one line pass through every line segment in this picture.

enter image description here

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    $\begingroup$ So you have shown that it is not always possible. could it be sometimes possible? $\endgroup$ Jul 6, 2017 at 18:51
  • $\begingroup$ Are you interpreting the question incorrectly or am I? 'Can one draw a line that passes through the interior of every one of these segments?' - I'd take that to mean the area inside. $\endgroup$
    – Shuri2060
    Jul 6, 2017 at 18:55
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    $\begingroup$ @Shuri2060 I think I was wrong. Possibility should depend on the parity of the number of points, and i don't think it works for an odd number... $\endgroup$ Jul 6, 2017 at 19:02
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    $\begingroup$ @HagenvonEitzen Ah okay, so I guess the problem could be reformulated as "find neccesary and sufficient (or at least sufficient) conditions such that it is possible"? $\endgroup$
    – Ovi
    Jul 6, 2017 at 19:13
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    $\begingroup$ @Ovi the better statement of the problem would be: Can you distribute 1997 points on the plane such that if one connects $P_iP_{i+1}$ and $P_{1997}P_1$ then there exists a line crossing each open (meaning the points excluded) segment. I suggest you start with three points instead of 1997... $\endgroup$
    – Hamed
    Jul 6, 2017 at 20:50

1 Answer 1


Consider such a collection of points ${<}P_i{>}$ as described with $N$ points.

Now let us assume that a line can be drawn crossing each of the line segments ${<}L_i{>}$ where $L_N$ connects $P_N$ and $P_{1}$ and all other $L_i$ connect $P_i$ and $P_{i+1}$.

Now we know that if $P_i$ is one one side of the line, $P_{i+1}$ is on the other side of the line, and $P_1$ and $P_N$ are on opposite sides of the line. We can thus infer that $N$ is even, since all odd points are on the same side of the line as $P_1$.

Thus when such a line can be drawn, $N$ must be even, and by contrapositive, if $N$ is odd, such a line cannot be drawn.


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