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Suppose one given rectangle exist, how can we use the Archimedian property to lengthen or shorten the sides and obtain a rectangle with sides of any prescribed length in neutral geometry (that is, without assuming the parallel postulate, but assuming e.g. the rest of Hilbert's axioms)?

I found an older related question here, but the question is non-satisfactory answered.

EDIT:

The Kulisty answer shows how we can construct an arbitrarily large rectangle. Can we do better? Can we construct a rectangle with sides of any prescribed length, not longer than some given sides, but with sides congruent to two other given segments?

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  • $\begingroup$ Some general thoughts: (1) The how is clear - extend the rectangle in one direction, then the other. It's proving that the result is a rectangle which takes some work. (2) We shouldn't need Archimedes to obtain a rectangle with sides of a prescribed length: that can be done by copying segments with III.1 on this list. Archimedes is necessary to say that we can make an arbitrarily large rectangle: e.g., one containing an arbitrary point. (3) I agree that my question does not have a satisfactory answer. $\endgroup$ – Misha Lavrov May 7 at 23:17
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Notation. $B$ is a betweenness relation. $B(abc)$ means that $b$ lies between $a$ and $c$ (on the same line).

The following theorem holds in neutral geometry (it also holds in any Hilbert plane with Archimedes' axiom, which is slightly weaker theory):

Theorem. Given segment $pq$, for any rectangle, there exists a rectangle $\square abcd$ such that and $bc>pq$ and $ab$ is congruent to the given rectangle's side.

Proof. Let $\square b_0a_0a_1b_1$ be a rectangle with exists by an assumption. I'll define a sequence of points $(a_n)_{n\in\mathbb{N}}$ by recursion:

  • $a_1$ coincides with $a_1$ with is already defined.
  • Assuming that $a_n$ has been defined , $a_{n+1}$ will be the unique point such that $B(a_0a_na_{n+1})$ and $a_na_{n+1}\equiv a_0a_1$

Similarly a sequence of points $(b_n)_{n\in\mathbb{N}}$ is defined by recursion:

  • $b_1$ coincides with $b_1$ with is already defined.
  • Assuming that $b_n$ has been defined , $b_{n+1}$ will be the unique point such that $B(b_0b_nb_{n+1})$ and $b_nb_{n+1}\equiv b_0b_1$

Claim. For any $n\in\mathbb{N}$, $\square b_0a_0a_nb_n$ is a rectangle.

Lemma 1. Given a rectangle $\square abcd$, lines $\overleftrightarrow{ab},\overleftrightarrow{cd}$ are parallel and lines $\overleftrightarrow{bc},\overleftrightarrow{da}$ are parallel.

Lemma 2. Assume we are given four points $a,b,c,d$ such that $c,d$ on the same side of $\overleftrightarrow{ab}$ and $b,c$ on the same side of $\overleftrightarrow{ad}$ and $a,d$ on the same side of $\overleftrightarrow{bc}$. Then $\square abcd$ is a convex quadrilateral.

Lemma 3. Given a rectangle $\square abcd$, $ab\equiv cd$ and $bc\equiv da$.

Lemma 4. Two Saccheri quadrilaterals with equal bases and legs are congruent i.e. summit angles and the summits are equal.

Proof of the claim by induction:

  • $n=1$ holds by the assumption.
  • Fix $n\in\mathbb{N}$ and assume that $\square b_0a_0a_nb_n$ is a rectangle.

    From lemma 1 lines $\overleftrightarrow{a_0a_n}=\overleftrightarrow{a_na_{n+1}}=\overleftrightarrow{a_0a_{n+1}}$ and $\overleftrightarrow{b_0b_n}=\overleftrightarrow{b_nb_{n+1}}=\overleftrightarrow{b_0b_{n+1}}$ are parallel. Let's denote $L:=\overleftrightarrow{a_nb_n}$. From the same lemma lines $\overleftrightarrow{a_0b_0}$ and $\overleftrightarrow{a_nb_n}=L$ are parallel, so $a_0$ and $b_0$ lie on the same side of $L$. Next $a_0$ and $a_{n+1}$ lie on opposite sides of $L$ and similarly $b_0$ and $b_{n+1}$ lie on opposite sides of $L$. Combining last three we get that $a_{n+1}$ and $b_{n+1}$ lie on the same side of $L$. Assumptions of lemma 2 for $\square b_na_na_{n+1}b_{n+1}$ are satisfied so $\square b_na_na_{n+1}b_{n+1}$ is a convex quadrilateral. Also $a_n,b_n$ lie on the same side of $\overleftrightarrow{a_0b_0}$ and $a_n,a_{n+1}$ lie on the same side of $\overleftrightarrow{a_0b_0}$ and $b_n,b_{n+1}$ lie on the same side of $\overleftrightarrow{a_0b_0}$. Combining last three $a_{n+1},b_{n+1}$ lie on the same side of $\overleftrightarrow{a_0b_0}$. Assumptions of lemma 2 for $\square b_0a_0a_{n+1}b_{n+1}$ are satisfied so $\square b_0a_0a_{n+1}b_{n+1}$ is a convex quadrilateral.

    $\angle b_na_na_{n+1}$ is a supplementary angle to $\angle b_na_na_0$ which is a right angle (by induction hypethesis). So $\angle b_na_na_{n+1}$ is a right angle and similarly $\angle a_nb_nb_{n+1}$ is a right angle. By lemma 3 $a_0a_1\equiv b_0b_1$, so $a_na_{n+1}\equiv a_0a_1\equiv b_0b_1\equiv b_nb_{n+1}$. Also by lemma 3 $a_0b_0\equiv a_nb_n$. So $\square b_na_na_{n+1}b_{n+1}$ and $\square b_0a_0a_1b_1$ are Saccheri quadrilaterals with equal bases and legs. By lemma 4 $\angle a_na_{n+1}b_{n+1}\equiv \angle a_0a_1b_1$ and the latter is a right angle. Similarly $\angle b_nb_{n+1}a_{n+1}$ is a right angle. Finally you can easily observe that all angles of $\square b_0a_0a_{n+1}b_{n+1}$ are right angles. The claim is proved.

To finish the proof, we simply apply Archimedes' axiom and we get that there exists $n\in\mathbb{N}$ such that $a_0a_n>pq$ and we put $\square abcd:=\square b_0a_0a_nb_n$ which we know is a rectangle.

If you assume that there exists a rectangle you can apply this theorem twice and you will get an arbitrarily large rectangle.

Update. As Evgeny Kuznetsov mentioned in the comment this result allows us to construct a rectangle of any prescribed lenght. The next step in proving that existence of a rectangle implies that Euclid's postulate holds, is proving that the sum of measures of angles of any triangle is equal two right angles (I know the proof from the Polish edition of Foundations of Geometry by K.Borsuk and W.Szmielew, not sure where to find it in English). Assume that we know this result. Then we can prove the following theorem in neutral geometry:

Theorem. If there exists a rectangle, then for any segments $pq$ and $rs$ there exists a rectangle $\square abcd$ such that $ab\equiv pq,bc\equiv rs$.

Proof. There exists a right angle. Call its vertex $b$ and find points $a,c$ on angle's rays such that $ab\equiv pq, bc\equiv rs$ (you can do this by one of the axioms). In the halfplane with boundary $\overleftrightarrow{ac}$ complementary to the one with point $b$ there exists a point $d$ such that $\triangle abc\equiv\triangle cda$ (this can be proved with laying off an angle and SAS).

An important step now is to prove that segment $bd$ cuts line $\overleftrightarrow{ac}$ in point $o$ such that $B(aoc)$. Suppose that $o$ does not lie on halfline $\overrightarrow{ac}$ (i.e. $o=a$ or $B(oac)$). Then $$\angle ocb=\angle acb\equiv\angle cad$$ $$\angle ocd=\angle acd\equiv \angle cab$$ ,halfline $\overrightarrow{co}$ lies in the interior of $\angle bcd$ (because $B(bod)$) and $\overrightarrow{ab},\overrightarrow{ad}$ on opposite sides of $\overleftrightarrow{ac}$. All these properties altogether imply that $\overrightarrow{ac}$ lies in the interior of $\angle bad$ which means that segment $bd$ cuts $\overrightarrow{ac}$ so $o$ lies on halfline $\overrightarrow{ac}$. Contradiction. So $o$ lies on the halfline $\overrightarrow{ac}$ and similarly on halfline $\overrightarrow{ca}$ i.e. $B(aoc)$.

Now we know that $\square abcd$ is a convex quadrilateral (because diagonals intersect and this is a sufficient condition) and from this we know that $$|\angle bad|=|\angle bac|+|\angle cad|=|\angle bac|+|\angle acb|=2R-|\angle abc|=2R-R=R$$ whre $R$ is the measure of the right angle. Similarly $\angle bcd$ is a right angle.

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  • $\begingroup$ I can not see how to prove your Lemma 3 without use of Euclid's prop. 29. Since the prop. 29 needs the 5th postulate to be proven, the proof I see isn't neutral. $\endgroup$ – Evgeny Kuznetsov May 10 at 19:43
  • $\begingroup$ Oh I see now. Proof goes by contradiction. If one is larger then measure shorter one on the long one, then new quadrilateral is a Saccheri quadrilateral and summit angles are equal. Assumption of inequality of sides contradicts to Euclid's prop. 16 $\endgroup$ – Evgeny Kuznetsov May 10 at 19:59
  • $\begingroup$ So, we constructed an arbitrarily large rectangle. Can we construct a rectangle with sides of any prescribed length, not longer than some given sides, but with sides congruent to two given segments? $\endgroup$ – Evgeny Kuznetsov May 10 at 20:40
  • $\begingroup$ Actually using the Kulisty answer we can show that angle sum of every triangle is $180^\circ$ using this we can construct a triangle with altitude of the prescribed length. Then we can construct the triangle I was asking for. $\endgroup$ – Evgeny Kuznetsov May 10 at 21:09

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