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The problem:

There are five points on a plane. There is no line that passes through exactly two points. Prove that these five points are collinear.

My attempt: The method I am trying to use is proof by contradiction. Let $P_1$, $P_2$, $P_3$, $P_4$, and $P_5$ be the points. Let's assume that only points $P_1$, $P_2$, $P_3$, and $P_4$ lie on a same line. Let $P_5$ be a distinct point. The line that passes through $P_1\ldots P_4$ has a slope $k_1$. Since $P_5$ is a distinct point, we can define another line with different slope that passes trough points $P_1$ and $P_5$. That is a line that passes through two points. That is a contradiction, since such a line should not exist. So, point $P_5$ has to lie on a same line that passes through other points. So, all of the five points are collinear. $\square$

Is my proof correct?

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3 Answers 3

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If there are five points, and no line goes through exactly two of them, and the five points are not collinear, then exactly three or four points must be collinear.

First, suppose exactly three points are collinear, as $A$, $B$, $C$ below. Then $AD$, $AE$, $BD$, $BE$, $CD$, $CE$, and $DE$ pass through exactly two points, which contradicts the supposition.

And even if two sets of three points are collinear, as $F$, $G$, $H$ and $F$, $J$, $K$, still $GJ$, $GK$, $HJ$, and $HK$ pass through exactly two points, against the supposition.5 collinear points

Finally, if exactly four points are collinear, as $L$, $M$, $N$, $O$, then $LP$, $MP$, $NP$, and $OP$ pass through exactly two points, again a contradiction.

Therefore, the five points must be collinear.

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  • $\begingroup$ Thank you for this! This seems to make perfect sense. I am just a little worried about the sketchs you drew. Is it enough to use just drawings? As the points can be anywhere on a plane. Nevertheless this seems just awesome work so thank you! $\endgroup$
    – Mixer
    Dec 18, 2018 at 17:42
  • $\begingroup$ Good question. Drawings must suitably reflect the given conditions. In the first one, consider $D$ and $E$ as any two points not on straight line $ABC$. In the second, $G$, $H$, and $J$, $K$, are two pairs of random points on two lines crossing at a fifth point $F$. In the third, $P$ is any point not lying on the straight line through $L$, $M$, $N$, $O$. I agree, to be trustworthy, a drawing has to be interpreted and understood, not just looked at. $\endgroup$
    – Edward Porcella
    Dec 18, 2018 at 18:04
  • $\begingroup$ Thank you for reply. In my attempt I was trying to look for a proof that would work without drawings as someone could ask if I have tested all the points. $\endgroup$
    – Mixer
    Dec 19, 2018 at 15:45
  • $\begingroup$ Do you mean in general, that geometric proofs should work without drawings? $\endgroup$
    – Edward Porcella
    Dec 20, 2018 at 16:28
  • $\begingroup$ I was taught to be very cautious with drawings. Drawings can be used to sketch a proof certainly but can they be used as a complete proof in this kind of situation? If the proof deals with triangles for example then drawings are fine but I think they need to be backed up with something else. I am not against your proof but this kind of consept that drawing is sufficient is foreign to me. $\endgroup$
    – Mixer
    Dec 21, 2018 at 17:01
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If not all five points are collinear, then some three of the points must form a triangle. Each edge of this triangle defines a line on which are two of the points. In order to satisfy the three-point-per-line requirement, the other two points must be distributed upon each of these three edge-lines - not possible! Each pair of lines shares one point, so they cannot also share a second, distinct point, so each of the two remaining points must be placed on only one of the three lines, but this can only be done for two of the lines, leaving the third passing through only two points.

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Assume first that three points are collinear. Then, two of the five points may be connected with a distinct line, a contradiction. (break into subcases here) Next, assume that four points are collinear. Then, the remaining point may be connected with one of the four collinear points, a contradiction. The remaining possibility is that all five are collinear. I don't think you even need to mention different slopes. Just that in Euclidean plane, between any two points there is a unique line connecting them.

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  • $\begingroup$ Not quite. The line through the two remaining points might be collinear with one of the three you started with.You've a little more work to do in that case. $\endgroup$
    – Ethan Bolker
    Dec 16, 2018 at 18:30
  • $\begingroup$ Good point Ethan. Maybe the OP can fix this part. $\endgroup$
    – D.B.
    Dec 16, 2018 at 18:33

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