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How can someone prove in Euclidean geometry that the statement

"A line connecting an interior and an exterior point of a circle should intersect the circle at some point"

follows from the axioms of Hilbert or Birkhoff? I cannot find any relevant information inside Hilbert's book or Birkhoff's paper.

Sumac
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    The first question is what do you mean by an interior/exterior point of a circle. A circle in the plane is different from a circle in three dimensional Euclidean space. – Mark Bennet Mar 22 '19 at 09:30
  • @MarkBennet I am sorry, you are right. Everything is on the same plane. An interior point of a circle with center $O$ and radius $r$ is a point $A$ such that $|OA| < r$. A point $A$ is exterior of that circle if $|OA| > r$. – Sumac Mar 22 '19 at 11:03

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From the definitions you gave in the comments,

Everything is on the same plane. An interior point of a circle with center $O$ and radius $r$ is a point $A$ such that $|OA|<r$. A point $A$ is exterior of that circle if $|OA|>r$.

it is fairly straight forward to see that, if we choose point $A$ such that $|OA|<r$ (inside the circle) and point B such that $|OB|>r$ (outside the circle) then at some point on the line that joins $A$ and $B$ we must have a point $C$ such that $|OC|=r$ and thus the point is on the circle.

Consider this diagram

circle

It is obvious that $|OA|<r$ and $|OB|>r$, matching our definitions.

We draw the purple line $AB$. We can immediately see that this crosses the circle line, now it just remains to prove this fact.

If we plot a graph of $|OP|$ for all points $P$ on $AB$ then we will get something of the following shape.

graph

We can use the Intermediate Value Theorem to say that there must be a point between $A$ and $B$ where $|OP|=r$ and we call this point $C$ and say that it must lie on the circle.

lioness99a
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  • Why is there a point on the line such that $|OC| = r$? From Birkhoff's axioms, for example, what we know is that if $A$ is a point on a line, then for every $\varepsilon > 0$, there is on either side of $A$ a point $B$ such that $|AB| = \varepsilon$. We don't know anything about $OA$ or $OB$. I should also clarify that I would like to avoid using the Pythagorean theorem (unless there is no other way). – Sumac Mar 22 '19 at 11:48
  • You have one point where it is less than $r$ and one point where it is greater than $r$ so there has to be a point between where it is equal to $r$ if the two points are joined together... See Intermediate Value theorem – lioness99a Mar 22 '19 at 11:53
  • Yes, I know about the Intermediate Value theorem. The problem is that what you have proved is that there is a point $C$ such that $|OC| = r$. You haven't shown that $C$ is on the line. – Sumac Mar 22 '19 at 12:03
  • We choose $C$ such that it is on the line $AB$. The circle is defined by all points $P$ where $|OP|=r$. Therefore $C$ is on the circle as $|OC|=r$ – lioness99a Mar 22 '19 at 12:10
  • How do we know, from the axioms, that there is a $C$ on the line that has $|OC| = r$? The fact that there are points on the line $A,B$ with $|OA| < r$ and $|OB| > r$ does not imply that there is a point on the line such that $|OC| = r$. Which axioms of Euclidean geometry are you using to get what we need? – Sumac Mar 22 '19 at 12:35
  • Honestly, I just intuitively 'see' it. I'm not sure how else to write it down to explain it beyond what I have already said, sorry. I can try and draw a picture later if that would help – lioness99a Mar 22 '19 at 12:49
  • @Sumac I have updated my answer with a diagram and a graph. I hope this is helpful – lioness99a Mar 22 '19 at 13:10
  • Thank you for your answer, and I understand all of that. The problem is that you assume, for example, that $|OP|$ is a continuous function, which is not that obvious from the axioms. Everything, of course, is absolutely clear if we use coordinates. The problem is deriving this from the axioms (e.g. Birkhoff's axioms). – Sumac Mar 22 '19 at 13:29
  • Glad I could help a bit. I can't comment on Birkhoff's axioms as I have never come across them, sorry – lioness99a Mar 22 '19 at 13:31
  • In a pure geometric axiomatic system, usually you cannot express the real numbers, because the system is too weak. Therefore, you cannot use any results from analysis, or analytic geometry. – Trebor Mar 22 '19 at 13:44