From baby Hartshorne: showing the exterior of a circle is segment connected

I'm finishing up reading about Hilbert's axioms, and this final problem from Hartshorne's Geometry: Euclid and Beyond is throwing me for a loop.

I'm working in a Hilbert plane, so the axioms of incidence, betweenness and congruence apply. I have the additional Archimedes' axiom, which states that for given line segments $AB$ and $CD$, there is a natural number $n$ such that $n$ copies of $AB$ added together will be greater than $CD$. However, I do not have use of the parallel postulate, nor the circle-circle intersection axiom.*

I want to show that the exterior of a circle is a segment connected set, which means that if $B$ and $C$ are two points outside a circle $\Gamma$, then there exists a third point $D$ such that $BD$ and $DC$ are entirely outside $\Gamma$.

My intuitive argument is straightforward. We can draw line BC. If this is outside $\Gamma$, we are done. Otherwise we can find $l$ to be the perpendicular bisector of $BC$ and take points on $l$, further and further 'out' until we find some $D$ such that $BD$ and $DC$ are outside $\Gamma$.

I don't know how to formalize my idea, or why Archimedes' axiom is necessary for this to work. Hartshorne includes the warning that without it or the parallel postulate, the exterior of $\Gamma$ may not be segment connected. Also, I'm not quite sure how to show that it is indeed eventually outside $\Gamma$, since without the circle-circle intersection axiom, I can't find the center of $\Gamma$ nor can I find tangents to $\Gamma$ from $B$ or $C$. I'd be grateful for any explanation, thank you.

*This states that for two circles, if one contains a point inside the other and a point outside the other, then the two circles intersect in (necessarily) two points.

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If $\Gamma$ is your circle whose exterior you're trying to show is segment-connected, then it automatically comes with a center $O$ and radius $OA$, so you don't need the circle-intersection axiom to find the center and radius. Then triangle inequality, Archimedes' axiom, some estimates on how far out you need to go (and which direction), similarity, congruences, etc. should be enough to formalize your idea, but it'll still be very painful and messy. –  Vladimir Sotirov Mar 8 '11 at 6:31
@Vladimir, I had been toying with and idea like that. Suppose the exterior is not segment connected. Suppose $l$ is a ray originating at $C$ that is entirely outside the circle. Then for any point $D$ on $l$, $\triangle BCD$ must have two legs $BC$ and $BD$ inside $\Gamma$. If we push $D$ along $l$ by a fixed increment $CD$ every time, the line $BD$ is in a sense bounded by the tangent line from $B$ to $\Gamma$ on that side of the circle. (Does this exist, even if we can't construct it?) So by the triangle inequality, $nCD$ will always be less than $BC$ plus the length of the tangent at $B$... –  yunone Mar 8 '11 at 6:39
...contrary to Archimedes' axiom. What do you think? –  yunone Mar 8 '11 at 6:40

... that the exterior of a circle is a segment connected set, which means that if $B$ and $C$ are two points outside a circle $\Gamma$, then there exists a third point $D$ such that $\overline{BD}$ and $\overline{DC}$ are entirely outside $\Gamma$.

is not true in neutral plane geometry (a.k.a. the Hilbert plane). More precisely, it is not provable since it is true in Euclidean geometry and false in hyperbolic geometry, both of which are models of neutral geometry.

However, using the definition of a segment-connected set given in Hartshorne's Geometry page 80:

A subset $W$ of the plane is segment-connected if given any two points $A, B \in W$, there is a finite sequence of points $A = A_1, A_2, \dots , A_n = B$ such that for each $i = 1, 2,..., n-1$, the segment $\overline {A_iA_{i+1}}$ is entirely contained within $W$,

it can be shown that, in neutral geometry, the exterior of a circle is segment-connected.

Proof that the claim is true in Euclidean geometry

Let circle $\Gamma$ have radius $r$ and center $O$. If the line $\ell$ connecting $B$ and $C$ does not pass through $\Gamma$, we are done. Otherwise, let us construct a line $b$ passing through $B$ perpendicular to $\overleftrightarrow{OB}$ and a line $c$ passing through $C$ perpendicular to $\overleftrightarrow{OC}$ If line $\ell$ does not pass through $O$, $B$ and $C$ are on the same side of a line $m$ parallel to $\ell$ and passing through $O$, and the measure of angle $\angle BOC$ is less than $180$. If we consider $\ell$ a transversal cutting lines $b$ and $c$ and consecutive interior angles $\beta$ and $\gamma$ sum to less than $180$. Thus by Euclid's fifth postulate lines $b$ and $c$ intersect at some point $D$ and because all points on $b$ are at least $OB$ from $O$ and all points on $c$ are at least $OC$ from $O$ the segments $\overline{BD}$ and $\overline{DC}$ are entirely outside $\Gamma$.

If $\ell$ does pass through $O$, then without loss of generality let us assume $OB \leq OC$. Set $\epsilon = \frac{OB-r}{2}$ and construct a line $n$ perpendicular to $\ell$ at $O$. Find a point $O'$ on $n$ a distance $\epsilon$ from $O$. Let us now consider a second circle $\Gamma'$ with center $O'$ and radius $r+\epsilon$. The circle $\Gamma'$ fully contains $\Gamma$ since, if point $X$ is in or on $\Gamma$, $OX \leq r$ and, by the triangle inequality, $O'X \leq OX + \epsilon \leq r+ \epsilon$. So $X$ is in or on $\Gamma'$. Point $B$ is not in $\Gamma'$ since $m$ is perpendicular to $\ell$ at $O$ and $O'B > OB = r + 2\epsilon > r+ \epsilon$. We get a similar result for point $C$. Hence, we can apply the first part of this proof to $B$, $C$ and the new circle $\Gamma'$ to address this case.

Proof that the claim is false in hyperbolic geometry

Consider any circle $\Gamma$ with center $O$ and radius $r$ and with diameter $\overline{QQ'}$. Let $\ell$ be the line perpendicular to line $\overleftrightarrow{QQ'}$ at $O$, and let $P$ and $P'$ be the points of $\ell$ which intersect the circle. Let $m$ be the line passing through $P$ that is asymptotic to line $\overleftrightarrow{QQ'}$ on the $Q$ side of $O$. Drop a perpendicular from $O$ to the line $m$, and let $F$ be the foot of that perpendicular. The point $F$ will be the closest approach that $m$ comes to $O$ and, since $m$ is not perpendicular to $\ell$ at $P$, $OF < r$. Let $R$ be a point of $\overrightarrow{OF}$ between $F$ and the circle.

Let $n$ be the line through $R$ that is asymptotic to $m$ in the other direction. since $m$ and $n$ are asymptotic, any line passing through one, in the direction of the asymptote, passes through the other. So the line $\ell$ passes through both. Consider any one of the points of $\ell$ between $m$ and $n$ — call it $B$. Any line through $B$ that avoids the circle must pass through $n$ and must also pass through $m$ but on the $Q'$ side of $\ell$. It consequently cannot cross $m$ on the $Q$ side and so cannot cross the ray $\overrightarrow{OQ}$. We can make the same argument on the other side of $\ell$ to find points, $B$, near $P$ just outside the circle such that if a line passes through $B$ but not the circle, it cannot intersect ray $\overrightarrow{OQ'}$. There is, apparently, no line through these points that can avoid the circle and go across line $\overleftrightarrow{QQ'}$. Of course, we can make a similar argument about points, $C$, near $P'$ just outside the circle. However, if we are to find a $D$ for a pair of these $B$'s and $C$'s such that $\overline{BD}$ and $\overline{DC}$ are entirely outside $\Gamma$, one or the other of $\overline{BD}$ or $\overline{DC}$ must cross $\overleftrightarrow{QQ'}$.

We have shown that in hyperbolic geometry there are points in the exterior of any circle that require more than two segments to connect them and stay in the exterior. Fooling around with the Poincaré disk, reveals that the larger the circle, the more segments may be needed.

Proof that, in neutral geometry, the exterior of a circle is segment-connected

Consider any circle $\Gamma$ with center $O$ and radius $r$ and any points $B$ and $C$ in the circle's exterior. Construct a perpendicular line $m_1$ to line $\overleftrightarrow{OB}$ at point $B$. Note that all of line $m_1$ lies outside the circle. Find a point $A_1$ on line $m_1$ to one side of $B$ at a distance of $2r$ from $B$ and let angle $\theta$ be the measure of angle $\angle BOA_1$. If $C$ falls on or in the interior of angle $\angle BOA_1$, then the ray $\overrightarrow{OC}$ will intersect $m_1$ at a point $E$. Connect $B$ to $E$ with a segment that lies on $m_1$ and so lies entirely outside $\Gamma$, and then finish by connecting $E$ to $C$ with a segment which will also lie outside since $E$ ad $C$ both lie outside $\Gamma$. If $C$ does not fall in the interior of angle $\angle BOA_1$, connect B to $A_1$, and then find the next point of our segment connection on ray $\overrightarrow{OA_1}$ at distance $OB$ from $O$ and call the point $B_1$.

Now repeat the process with $B_1$. Construct a perpendicular line $m_2$ to line $\overleftrightarrow{OB_1}$ at point $B_1$, and find a point $A_2$ on line $m_2$ to the side of $B_1$ opposite $B$ at a distance of $2r$ from $B_1$. By SAS, the triangle $\triangle OBA_1$ is congruent to $\triangle OB_1A_2$ and so the measure of the angle $\angle B_1OA_2$ is also $\theta$. Again, if $C$ falls on or in the interior of angle $\angle B_1OA_2$, then we can find, as before, a point $E$ on $m_2$ and connect $B_1$ to $E$ to $C$ with segments. If not, connect to $A_2$ and find the next point of our segment connection on ray $\overrightarrow{OA_2}$ at distance $OB$ from $O$ and call the point $B_2$.

We can repeat the process until, by Archimedian property of angle measures, $C$ lies within $\angle B_iOA_{i+1}$ by which time we will have constructed a segment connection between $B$ and $C$ entirely in the exterior of $\Gamma$.

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Thank you Eric for this thoughtful answer. However, I don't see how this directly addresses the question. First, this problem in Hartshorne wants to find a single point $D$ such that $BD$ and $DC$ are completely outside $\Gamma$, not a sequence $A_1, A_2, \dots, A_{i+1}$ of more than one point. Secondly, the Archimedian property given by Hartshorne deals with line segments, not angles. Are they equivalent somehow? And is there a way to cut the sequence down to one single point? –  yunone Mar 9 '11 at 8:45
There is no way to cut the sequence down to one point. Since the claim is not true in hyperbolic geometry, it is not true in neutral geometry (the Hilbert plane). I do not have access to Hartshorne so I cannot tell what the problem is. But you are trying to prove something that is not true. –  Eric Nitardy Mar 9 '11 at 8:55
I have posted screenshots of the problem, as well as problem 11.1 which 12.6 tells me to confer with. I believe I was misled by the suggestion in 11.1 above, so the definition you found on pg. 80 is the correct one. Sorry for the misunderstanding on my part. However, I still don't understand how you have used the Archimidean axiom (A) applied to angles, when it is about segments. Could you explain that point when you get the chance? Thank you. –  yunone Mar 9 '11 at 9:04
Again the only access I have to Hartshorne is the Google excerpts, but I know that Hilbert created an arithmetic of congruent segments (Hilbert's Foundations of Geometry see pages 15-16 and 23-36). I assume he applied that arithmetic to angle measures using Archimedes' approach for successively better estimations of arc-length. –  Eric Nitardy Mar 9 '11 at 9:39
Ok, thank you for bearing with me. Thanks also for this brilliant post! –  yunone Mar 9 '11 at 18:22

hmm... I accidentally stumbled upon this post and I was thinking:

Proof that the claim is true in Euclidean geometry

Let circle Γ have radius r and center O. If the line ℓ connecting B and C does not pass through Γ, we are done. Otherwise, let us construct a line b passing through B perpendicular to OB←→ and a line c passing through C perpendicular to OC←→ If line ℓ does not pass through O, B and C are on the same side of a line m parallel to ℓ and passing through O,

My proof is similar up to here, then I took a different approach.

Assume b // c

Then extend the line OB s.t. it intersects c. Let E be that intersection. Then angle OBG, where G belongs to b, would congruent to angle OEC. Then you have angle OEC and OCE both being right angle.

The contradiction (Greenberg Proposition 4.5 I think...) should yield the point D needed for this proof. Then use Greenberg Propostion 4.6 to show all pts on BD & DC are in the exterior of the ciricle.

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I know it is not an answer and maybe discouraging, but from my modern geometry class I remember the Hilbert axioms are seriously limited, ambiguous, with some conflating definitions. Maybe you should turn to another book.

For your approach, I feel what you meant by "further and further 'out'.." etc is essentially magnifying the line segment. Without the Archimedes' axiom you are not guaranteed you can pass over the circle. I know it is kind of silly but you need it to work. Similarly you may run into other problems with similar approaches without appropriate axioms. For the fifth postulate maybe you can try the Poincare model or Klein model. I am quite sure without it you cannot find $D$.

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