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

The Pythagorean theorem and Hilbert axiomsWhy is Euclidean geometry scale-invariant?What accounts for the special relationship between Euclidean geometry and other branches of math?Connecting Coordinate Geometry and Plane GeometryIn Neutral Geometry, prove that the opposite sides of a rectangle are congruent.Without using angle measure how do I prove two lines are parallel to the same line are parallel to each other?An exersise of Euclidean geometryCounterexample to Secant Tangent Theorem?Can one state and prove that Euclidean space has genus $0$ in Hilbert's geometry?Proving Hilbert's Axioms as Theorems in $ℝ^n$

0

$begingroup$

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.

euclidean-geometry

share|cite|improve this question

asked Mar 22 at 9:18

SumacSumac

392213

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  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.
  $endgroup$
  – Mark Bennet
  Mar 22 at 9:30
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @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 $|OA| > r$.
  $endgroup$
  – Sumac
  Mar 22 at 11:03
  

  
  
  
  

add a comment | 

0

$begingroup$

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.

euclidean-geometry

share|cite|improve this question

asked Mar 22 at 9:18

SumacSumac

392213

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  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.
  $endgroup$
  – Mark Bennet
  Mar 22 at 9:30
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @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 $|OA| > r$.
  $endgroup$
  – Sumac
  Mar 22 at 11:03
  

  
  
  
  

add a comment | 

$begingroup$

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.

euclidean-geometry

share|cite|improve this question

asked Mar 22 at 9:18

SumacSumac

392213

$endgroup$

share|cite|improve this question

asked Mar 22 at 9:18

SumacSumac

392213

share|cite|improve this question

asked Mar 22 at 9:18

SumacSumac

392213

asked Mar 22 at 9:18

SumacSumac

392213

asked Mar 22 at 9:18

SumacSumac

392213

1 Answer
1

active

oldest

votes

0

$begingroup$

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

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.

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.

share|cite|improve this answer

edited Mar 22 at 13:27

answered Mar 22 at 11:34

lioness99alioness99a

3,9012727

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  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).
  $endgroup$
  – Sumac
  Mar 22 at 11:48
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  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
  $endgroup$
  – lioness99a
  Mar 22 at 11:53
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  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.
  $endgroup$
  – Sumac
  Mar 22 at 12:03
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  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 $|OC|=r$
  $endgroup$
  – lioness99a
  Mar 22 at 12:10
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  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?
  $endgroup$
  – Sumac
  Mar 22 at 12:35
  

  
  
  
  

 | 
show 5 more comments

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0

$begingroup$

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

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.

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.

share|cite|improve this answer

edited Mar 22 at 13:27

answered Mar 22 at 11:34

lioness99alioness99a

3,9012727

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  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).
  $endgroup$
  – Sumac
  Mar 22 at 11:48
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  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
  $endgroup$
  – lioness99a
  Mar 22 at 11:53
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  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.
  $endgroup$
  – Sumac
  Mar 22 at 12:03
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  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 $|OC|=r$
  $endgroup$
  – lioness99a
  Mar 22 at 12:10
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  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?
  $endgroup$
  – Sumac
  Mar 22 at 12:35
  

  
  
  
  

 | 
show 5 more comments

0

$begingroup$

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

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.

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.

share|cite|improve this answer

edited Mar 22 at 13:27

answered Mar 22 at 11:34

lioness99alioness99a

3,9012727

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  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).
  $endgroup$
  – Sumac
  Mar 22 at 11:48
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  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
  $endgroup$
  – lioness99a
  Mar 22 at 11:53
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  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.
  $endgroup$
  – Sumac
  Mar 22 at 12:03
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  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 $|OC|=r$
  $endgroup$
  – lioness99a
  Mar 22 at 12:10
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  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?
  $endgroup$
  – Sumac
  Mar 22 at 12:35
  

  
  
  
  

 | 
show 5 more comments

$begingroup$

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

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.

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.

share|cite|improve this answer

edited Mar 22 at 13:27

answered Mar 22 at 11:34

lioness99alioness99a

3,9012727

$endgroup$

share|cite|improve this answer

edited Mar 22 at 13:27

answered Mar 22 at 11:34

lioness99alioness99a

3,9012727

answered Mar 22 at 11:34

lioness99alioness99a

3,9012727

answered Mar 22 at 11:34

lioness99alioness99a

3,9012727