# Prove a line cannot intersect all 3 of the sides of a triangle at interior points

Here, a story is told of a competitor in the Russian math olympiad of 1945. The boy did not solve a single problem, but received a prize for writing:

I spent much time trying to prove that a straight line can't intersect three sides of a triangle in their interior points but failed for, to my consternation, I realized that I have no notion of what a straight line is.

Assume the standard Euclidean geometry of the plane. What is the simplest way to prove this statement? Can it be done without analytic geometry?

As noted in the quote, a lot of this has to do with how to rigorously define. A related question is: how is Euclidean geometry rigorously defined today, and in particular, what is a line? What is a point? A triangle?

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I'm not sure I believe that story... –  user641 Nov 8 '11 at 23:30
Interesting question. –  Emmad Kareem Nov 9 '11 at 0:17
What's a straight line? –  Jon Beardsley Nov 9 '11 at 1:49
What do you mean by standard Euclidean geometry of the plane? What axioms are you assuming? In Hilbert's axiomatization of Euclidean geometry, one of his betweenness axioms was Pasch's axiom. This stated that for three noncollinear points $A$, $B$, $C$, if a line $\ell$, not containing any of them, intersected the side of $\triangle ABC$ between say $A$ and $B$, then it intersected one and only one other side of $\triangle ABC$, but not both. –  yunone Nov 9 '11 at 3:16
I'm not sure. What is the standard formulation of Euclidean geometry these days? Also, I've heard that Euclid's was insufficient (not rigorous in some way). Why is that? –  Potato Nov 9 '11 at 4:04

Yunone is right that this fact is basically equivalent to Pasch's postulate, which can't be proved from Euclid's axioms. There are many such gaps in the axioms corresponding to incidence results that are obviously true if you draw a picture and hence were left unstated, like that the two circles in his first proposition intersect, and that is why Euclid's axioms are said to be insufficient.

So, one answer to your question is that the result in question is true as a consequence of Pasch's postulate if you are using an axiom system that includes it. But there is no geometric axiom system that can be considered the standard formulation of Euclidean geometry today. Rather, the standard formulation nowadays is in terms of the Cartesian plane. Points are ordered pairs, lines are solution sets of linear equations, distance is given by the usual formula, etc.

In this formulation Pasch's postulate is a theorem that can be proven.

A straightforward way to see it is as a consequence of Menelaus' theorem, which can be proven purely algebraically. If a line intersected the interiors of the three sides, this theorem would give us the product of three positive numbers as -1.

But there's a more direct though ugly way to see it. Given a line intersecting the interiors of all sides of a triangle, you can use coordinates to transform this statement into a series of equations and inequalities between real numbers and show they cannot be satisfied. Here are the unattractive details:

Translations preserve lines and betweenness, so translate one vertex to the origin. Linear transformations also preserve lines and betweennes, so transform the two other vertices to (0,1) and (1,0). So, without loss of generality we can assume our triangle has vertices (0,0), (0,1), and (1,0). Now, suppose a line intersects all three sides. If it were vertical, it would be parallel to one side and thus couldn't intersect all three. Thus, the line is the graph of a function f(x). Suppose it intersects the interior of the two edges meeting the origin. Then

$0 < f(0) < 1$ and there is an r in $(0,1)$ such that $f(r) = 0$. Thus $f(x)$ is decreasing so it can be written as $f(x) = mx + b$ where$m < 0$ and $0 < b < 1$, and also $f(1)=m+b<f(r)=0$ from the fact it is decreasing.

$1-x$ is the equation of the third side. Let's look at the intersection between this side and the line defined by $f(x)$. If $1-x = mx+b$, then $(m+1)x=1-b$. If it intersects the interior of this side, $0 < x < 1$. Thus $m+1 > 0$, so $1-b<m+1$, so $m+b <0$. But $m+b > 0$ was proved above. So it's impossible for any line to intersect the interiors of all three sides.

This proof is ugly, but it demonstrates that Pasch's postulate is a theorem that you can prove in the Cartesian model of Euclidean geometry.

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Excellent answer, and thank you for emphasizing the connection to foundations. –  Potato Nov 11 '11 at 17:27
But you haven't defined anything. What is a "line"? What is a "triangle"? –  goblin Dec 7 '14 at 3:26

In a rather hand-waving way and assuming (a) the straight line cannot intersect a side more than once, (b) intersecting a side at an interior point means passing from inside to outside the triangle or from outside to inside, and (c) the straight line starts and ends outside the triangle:

A straight line starts outside the triangle. If it intersects a side of the triangle at an interior point, it passes inside the triangle. If it intersects another side of the triangle at an interior point, it passes outside the triangle. If it intersects a third side of the triangle at an interior point, it passes inside the triangle, but that is impossible as it ends outside the triangle and there are no more sides to intersect.

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(a) just transforms the problem, I think. –  Stefan Kendall Nov 9 '11 at 3:46
@StefanKendall I would disagree - isn't (a) (two lines intersect in at most one point) just one of Euclid's axioms? –  Steven Stadnicki Nov 9 '11 at 7:23

Let $ABC$ be the triangle in question and suppose the line $\ell$ intersects the relative interior of the three sides. In particular, $\ell$ is not parallel to any of the sides and on one of the open half-spaces in which it cuts the plane there is exactly one of the three vertices —we can assume this is $A$.

Now start translating $\ell$ parallel to itself in the direction normal to itself towards $A$. By hypothesis, $\ell$ intersects the interior of three of the sides of the triangle, and this will not change when we start sliding until we reach $A$. After going through $A$, the number of segments whose interior the line intesects will have dropped by two, and never change again. But we can move the line far away from the triangle, so after going through $A$, the number of sides of the triangle whose interior the line intersects is zero.

This is absurd.

More generally, if $\mathcal A$ is a finite set distinct lines in the plane such that no three are incident to a point and no two are parallel, then the parity of the number of bounded elements of the face poset $\mathcal L(\mathcal A)$ which a line $\ell$ parallel to none of the lines in $\mathcal A$ and which does not pass through an intersection point of two elements of $\mathcal A$, is the same as the parity of the number of elements of $\mathcal A$.

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A line l divides the plane into two halves. Out of three points of the triangle A, B, C, two of them will definitely lie on one half (by the Pigeonhole principle and noting that none of them lie on the line itself because of the initial problem condition). Let these two points be A, B. Then the line l does not intersect the line AB of the triangle.

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