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Let´s suppose that there is a line l on which lie exactly 3 points of some triangle. Lets assume that none of the points are vertices of the triangle. Then, since no line intersects a side of a triangle at more than 1 (and less than infinity) of points, each of the three points must belong to a different side. Lets call the points A,B,C (B is between A and C) and sides on which the points lie a,b,c.

No two sides of a triangle are parallel. Therefore there must be an intersection point X1, where sides a and b meet. Further, again since no two sides are parallel X1 does not lie on l. Let´s consider the rays X1,A and X1,B. The only way how C could lie on one of those rays is if X1 lied on l, but it doesnt. Since sides a and b lie on the these two rays, the point C is not part of either the side a and b.

What remains, is to check if C lies on c. But we know that the rays X1,A and X2,B after they go through A and B remain on the same side of l. There is no way how could we draw a line segment between two points on them (after they go through A and B) and get intersection with l.

Does this count as a rigorous proof?

Adam
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  • Your beginning seems a little bit strange to me. You said that if 3 points of a triangle are colinear, then they are on different edges. It should be: if 3 points of a triangle are colinear, then they are on the same edge. – Taladris Nov 27 '13 at 12:26
  • Thats not what I say at all. I say if there is a line on which there are exactly 3 triangle points, then each point belongs to a different side of the triangle. – Adam Nov 27 '13 at 12:44
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    Your argument gets into a bit of a muddle: You say in the first paragraph that point $C$ is on side $c$; then you start your third paragraph off wanting to check whether $C$ can be on $c$ (and then apparently deciding that this is impossible). And, while I think most of your assertions may be correct (if somewhat inelegantly made), the final one "there is no way [to do this thing]" seems unsupported. I'll suggest some adjustments in an answer. – Blue Nov 27 '13 at 12:58
  • While we're at it: You didn't define your notion of "triangle". It seems you mean the union of its edges. – Christian Blatter Nov 27 '13 at 13:01
  • Yes, that´s what I mean. Are you suggesting that my notion of a triangle somehow differs from the standard? – Adam Nov 27 '13 at 13:04
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    It is common to start from vertics $X_1$ etc. (and in fact call these $A,B,C$) instead of concluding their existence from nonparallelity of the edges ... This makes your argument a bit confusing – Hagen von Eitzen Nov 27 '13 at 13:09
  • You should elaborate more on betweenness as the claim is obviously false if we do not require the three points of intersection to lie between the vertices. (And which axiom system are you using in the first place? Euclid? Hilbert?) – Hagen von Eitzen Nov 27 '13 at 13:11
  • :-D, seriously? betweennes doesnt really claim anything, it is there just to let you know where on my picture I put the labels A, B and C.... – Adam Nov 27 '13 at 13:15
  • As far as the axiom system goes, the answer is: I myself dont know... I am not working at such a level of detail. – Adam Nov 27 '13 at 13:22
  • @adam: what is the difference between my first comment and your first comment? edge=side. – Taladris Nov 28 '13 at 04:29
  • Well, both are valid deductions. The second one just came more naturally to me while doing the proof. One difference is that in your A implies B, A can actually be true. In my case the deduction is more hypothetical. – Adam Nov 28 '13 at 14:11
  • At least as far as Euclidean plane is concerned. If you look at Fano plane, you will see that the 3 points in question indeed lie on separate edges of their triangle. As I would say in my more dramatic days: "Behold my prophecy has come true." – Adam Nov 29 '13 at 11:32

3 Answers3

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Too long for a comment ...

Let's try tightening-up your description. I'll note that we can eliminate the need to state (and repeat) "no sides of a triangle are parallel", and/or to introduce $X_1$, by simply identifying the triangle in question.

To begin, I believe this captures your set-up:

Suppose line $\ell$ contains exactly three distinct, non-vertex points of $\triangle PQR$, and call those points $A$, $B$, $C$.

Now, you want to claim that each point lies on a separate edge of the triangle. As you indicate, this is a consequence of $\ell$ containing no vertices:

No two of these points lie on one edge of the triangle (otherwise, $\ell$ would coincide with that edge and contain the vertices at its endpoints, but $\ell$ contains no vertices). We'll say that $A$ lies on the edge opposite $P$; $B$ lies on the edge opposite $Q$, and $C$ lies on the edge opposite $R$.

Your second paragraph discusses a point $X_1$, which we already have: it's $R$. So,

Since $C$ doesn't lie on $\overleftrightarrow{PR}$ or $\overleftrightarrow{QR}$, it certainly doesn't lie on the rays $\overrightarrow{RA}$ or $\overrightarrow{RB}$ within those lines.

And then ... your third paragraph. It's not clear what's going on here. You want to "check if $C$ is on $c$" (although we have already declared that it is), but even so, your unsupported assertion about not being able to find an intersection with $\ell$ doesn't seem to have any bearing on the "check" your doing with line $c$. I'm not sure how to advise fixing this, because I seem to have fallen off of your train of thought.

Incidentally: When you get into these kinds of "intuitively obvious" arguments, you have to be really clear about your assumptions. It's especially important in this case, because your result is false in certain geometries.

Below is a picture of the Fano plane. It's a complete picture of the entire geometry, which consists of only seven points (represented by dots) and seven lines (represented by the segments and the circle). You can check that some fundamental geometric notions apply here: any two points lie on exactly one line; and any two lines meet at exactly one point. (There are no parallels here.) It's a perfectly good geometry ... but ... if we label the outer points $PQR$, the center point $O$, and the rest $A$, $B$, $C$, then "line $ABC$" contains exactly one point from each side of $\triangle PQR$.

enter image description here

Your response is likely to be "That's not what I'm talking about! I'm talking about good ol' regular geometry with infinitely many points and no silly circular lines!" And that's the point. You obviously (and very reasonably) want to rule out weird cases like the Fano plane ... because it probably (and very reasonably) never even occurred to you. But to do so, you have to be exceedingly careful not to argue by intuition: "There is no way how could we draw a line segment between two points on them (after they go through $A$ and $B$) and get intersection with $\ell$" isn't a valid argument to make, because there is a way on the Fano plane. You need to explain what it is about good ol' regular geometry that lets us know we aren't on the Fano plane.

If this nit-picking seems crazy, then welcome to the 19th century! It's around that time that mathematicians started realizing that they'd been making far too many intuitive assumptions about geometry for far too long, and they started establishing super-nit-picky foundations for the subject. Take a look, for instance, at David Hilbert's axioms for geometry; where Euclid thought just five postulates were sufficient to describe good ol' standard geometry, Hilbert realized there should be twenty to properly document our intuition.

(The good news is that the nit-picking only happens at the very-very fundamental level, with results such as yours that get to the very heart of what it means, for instance, for points to be collinear. Once we've essentially agreed on those foundations, we can happily move on without thinking too hard about them any more. After all, you don't see many geometry posts here at M.SE that start out saying "Assuming our geometry is governed by Hilbert's axioms, blah, blah, blah ..."; we just invoke the Pythagorean theorem or the Inscribed Angle Theorem or whatever and get about our business.)

Blue
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  • I understand that assuming something and then checking whether is true seems weird. Its supposed to be one of those proofs where R implies not R therefore not R. Does it make sense? – Adam Nov 27 '13 at 13:47
  • Also, the proposition C lies on c is conditional on there being 3 colinear points on a trinagle. So one can understand the proof structure like this: Assume A, A implies B, B implies C but C cannot be true, therefore not A. – Adam Nov 27 '13 at 13:52
  • I would like to know if you consider this kind of logical structure acceptable. In any case moving on to my "unsupported assertion", we have already established that the point C doesnt lie on sides a or b of the triangle if we manage to prove that it doesnt lie on side c - we will have proven that C is not on the triangle contradicting our assumption that there is a line on which lie exactly three triangle points – Adam Nov 27 '13 at 14:15
  • @Adam: Proof by contradiction is a perfectly valid technique, and I can see that that's what you were attempting. Unfortunately, your final paragraph kinda fell apart for other reasons. – Blue Nov 27 '13 at 14:16
  • On the rays R,B and R,A we can find sides a and b of the triangle. So if we cannot find two points (one from each ray) such that if we connect them with a line segment point C will lie on that line segment, we will have shown that point C does not lie on side C, is that correct? – Adam Nov 27 '13 at 14:21
  • @Adam: "if we manage to prove that [$C$] doesn't lie on side $c$, [then] we will have proven that $C$ is not on the triangle". True enough. Now you just have to manage the "manage to prove" part. ;) – Blue Nov 27 '13 at 14:21
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    @Adam: "If we cannot find two points [...] such that if we connect them with a line segment, point $C$ will lie on that line segment, we will have shown that point $C$ does not lie on side $c$, is that correct?" This is correct. There might be an easier formulation of your overall argument, but this is not an invalid path to take. The question is: How do you rigorously prove that "we cannot find two points [...]"? – Blue Nov 27 '13 at 14:29
  • Well my idea was this: we know that sides a and b start at R and go through A and B (since we want to connect the ENDS of lines a and b to produce line c, we do not have to consider points on the rays R,A and R,B prior to their intersection with line l), after they intersect with line l they will remain on the same side of l therefore you will never be able to connect them and get an intersection with l (if there were such an intersection the two points would have to be on opposite sides of l). What do you think? – Adam Nov 27 '13 at 14:41
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    @Adam: I'm not sure what you're getting at with the "opposite sides" bit, but you'll have to be a good deal more precise about why this fails in regular geometry, because (for instance) it doesn't fail on the Fano plane. (I'm now getting the "avoid extended discussions" note from M.SE. I need to get on to some other work anyway. I'll let you mull this stuff over for a while.) – Blue Nov 27 '13 at 14:48
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    The whole time I think, that Blue must be incredibly thick-headed not to get where I am getting with "opposite sides". Now I realize his comment contains more than a grain of truth. The only reason we can speak of opposite sides in Eucliedean geometry is PSA. In Fano plane, there are no "opposite sides". – Adam Nov 29 '13 at 12:13
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Blue's answer shows that we have to use somehow that everything takes place in the familiar euclidean plane $E^2$. In the following a triangle $T$ is determined by its three non-collinear vertices and consists of the union of the three connecting segments.

Assume that a line $\ell\subset E^2$ is given that intersects $T$ in a finite set $S\ne\emptyset$.

The set $S$ can contain at most one vertex $v$ of $T$. In this case it does not contain any further points of the two edges meeting at $v$, and it may or may not contain a point $P$ in the interior of the third edge. It follows that $|S|\in\{1,2\}$ in this case.

In the following we assume that $S$ contains no vertex of $T$. Then it contains $1$, $2$ or $3$ points in the interior of the three edges, but no two of them on the same edge. It is claimed that the case $|S|=3$ is impossible. (The case $|S|=1$ is impossible as well, but this is another story.)

We may assume that the vertices of $T$ are $(0,0)$, $(1,0)$ and $(1,1)$. When the line $\ell$ passes through the points $(p,0)$ with $0<p<1$ and $(0,q)$ with $0<q<1$ then its equation is $${x\over p}+{y\over q}=1\ .\tag{1}$$ On the other hand, the points $(x,y)$ on the third edge of $T$ simultaneously satisfy $$x>0,\quad y>0, \quad x+y=1\ .$$ But $x>0$, $y>0$ together with $(1)$ implies $$x+y<{x\over p}+{y\over q}=1\ ,$$ so that there cannot be a point of $\ell$ on the third edge.

Christian Blatter
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Assuming our geometry is governed by Hilbert´s axioms,

we can choose a line l that contains exactly three distinct non-vertex points of a triangle PQR and call them A,B,C.

Each of those points lie on a separate edge of the triangle. (if two of them lied on the same edge, then the line l would intersect the same edge exactly twice, which is impossible)

We will say that R is the vertex of the triangle where the sides going through A and B meet. Since no two edges of a triangle are parallel, the point R does not lie on l. Let´s consider the rays R,A and R,B. None of them go through C (if they did they would intersect line l at two distinct points A,(B) and C which is impossible.) Therefore the point C does not lie on either of the two edges of the triangle which intersect at point R.

Does it lie on the edge that connects these two? Up until now the proof is valid even in Fano geometry, but here comes the point of departure. The reason for the departure is the Plane seperation axiom which states that if a point A is not on the same side as point B and point A is not on the same side as C, then points B and C are on the same side. This axiom is actually false in the Fano geometry. But lets return to our Euclidean case.

We know that the edges of our triangle that begin at R go through A,(B) but do not stop there (because if they did, A and B would be vertices- and they are not). Consequently P and R are on opposite sides of l and Q and R are on opposite sides of l. By the Plane separation axiom, P and Q are on the same side of l. Consequently the point C does not lie between the vertices P and Q (by definition of "on the same side").

Therefore the point C doesn´t lie on the triangle PQR and our assumption that there exists a line on which there are exactly three points of some triangle must have been wrong.

Adam
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