3
$\begingroup$

By orientation I mean roughly speaking whether we rotate clockwise or anti-clockwise. Formally I want to define relation $\sim$ between triangles (contained in the same plane) such that $\triangle abc\sim\triangle pqr$ iff enumerations of these triangles are both clockwise or both anticlockwise. Then I want to prove this relation is an equivalence relation and has exactly two equivalence classes.

I suppose all this can be done within Hilbert's axioms of incidence and order (no congruence, continuity or parallel axioms).

$\endgroup$
  • $\begingroup$ Define a vector as an ordered pair of points. Using the Parallel Postulate, one can say what it means for a vector $ (A,B) $ to be similar to a vector $ (P,Q) $. Let $ (A,B,C) $ and $ (P,Q,R) $ be ordered triples of non-collinear points. Pick $ D $ so that $ \overline{BD} $ bisects $ \angle ABC $, and pick $ S $ so that $ \overline{QS} $ bisects $ \angle PQR $. $\endgroup$ – Transcendental Sep 6 at 3:43
  • $\begingroup$ Then we say that $ \triangle ABC $ and $ \triangle PQR $ are orientation congruent if and only if (i) $ \triangle ABC \equiv \triangle PQR $ in the usual sense, and (ii) if $ (Q,A') $, $ (Q,C') $, and $ (Q,D') $ are, respectively, similar to $ (B,A) $, $ (B,C) $, and $ (B,D) $, then $ \angle A'QP \equiv \angle C'QR \equiv \angle D'QS $. $\endgroup$ – Transcendental Sep 6 at 3:47
  • $\begingroup$ What is the source of this definition? Is this your own idea? Can you prove (in euclidean geometry) that the fact that $\triangle ABC$ and $\triangle PQR$ are orientation congruent implies that $\mathcal{O}(A,B,C)=\mathcal{O}(P,Q,R)$ according to my definition? $\endgroup$ – Kulisty Sep 7 at 17:15
  • $\begingroup$ Hi Kulisty. Yes, the definitions above are mine. Given points $ A,B,C,P,Q,R $, it’s true in Euclidean geometry that if $ (A,B,C) $ is orientation congruent to $ (P,Q,R) $, then $ \mathcal{O}(A,B,C) = \mathcal{O}(P,Q,R) $, and I’m sure that you already know that the converse doesn’t hold in general (if $ \mathcal{O}(A,B,C) = \mathcal{O}(P,Q,R) $, then all we can say is that $ \triangle ABC $ and $ \triangle PQR $ are similarly oriented). Anyway, I believe your definition of orientation below to be absolutely correct. Good work! :) $\endgroup$ – Transcendental Sep 8 at 2:34
1
$\begingroup$

Some time ago I came up with my own idea and I'm posting it now. Everything is done in an ordered plane i.e. within axioms of connection and order (no congruence, continuity or parallel axioms).

Points will be denoted by small letters $a,b,\ldots$. Sets like lines, rays, halfplanes etc. will be denoted by capital letters $A,B,K,L,M,N,\ldots$. If $a\neq b$, then the only line passing through $a,b$ will be denoted by $\overleftrightarrow{ab}$. If $A$ is a ray, then we use the symbol $o(A)$ to denote the origin of $A$ and $L(A)$ to denote the line, in which $A$ is contained. A ray complementary to $A$ will be denoted by $A^*$. If $M$ is a half-plane, the complementary half-plane will be denoted by $M^*$.

Then we introduce the notion that rays $A$ and $B$ have the same orientation (or direction) and the notion of orientation (direction) of the line. Roughly speaking rays $A$ and $B$ have the same orientation (direction) iff they are contained in the same line and are directed similarly. The orientation (direction) of a line is a set of all similarly directed rays contained in this line (formally an equivalence class). For each line there are exactly two directions associated with this line. If $\mathcal{D}$ is an orientation of a line, we denote the opposite orientation by $\mathcal{D}^*$. For the detailed definition see the book "Foundations of geometry" by Borsuk and Szmielew (1960) (pages 37,38).

By parallel lines we mean disjoint or equal lines.

From here definitions are invented by me. For example the definition of parallel rays is not the same as usual definition of this notion (sometimes called limiting parallel). As a sidenote I'll say that these definitions are equivalent under the assumption of the parallel postulate (which would be Playfair's axiom in this setting) but I don't assume it.

We say that rays $A,B$ are parallel, what we denote $A\parallel B$, iff lines $L(A),L(B)$ are parallel and $A,B$ have the same direction if $L(A)=L(B)$ or $A$ and $B$ lie on the same side of $\overleftrightarrow{o(A)o(B)}$ if $L(A)\cap L(B)=\emptyset$

We say that directions $\mathcal{D}$, $\mathcal{D}_1$ of some lines are parallel, what we denote by $\mathcal{D}\parallel\mathcal{D}_1$, iff for all rays $D\in\mathcal{D},D_1\in\mathcal{D}_1$, rays $D,D_1$ are parallel.

Let's head to the actual orientation definition:

Consider the family $\mathcal{R}$ of all triples of the form $(L,\mathcal{D},M)$, where $L$ is a line, $\mathcal{D}$ is an orientation of $L$ and $M$ is a halfplane with boundary $L$. We index this family the following way: $\mathcal{R}=\{O_A\}_{A\in\mathcal{A}}$, where $O_A=(L_A,\mathcal{D}_A,M_A)$ for all $A\in\mathcal{A}$. We define the relation $\sim$ on the set $\mathcal{R}$ the following way: Fix $A,B\in\mathcal{A}$ and consider cases:

  1. $L_A=L_B$

If $\mathcal{D}_A=\mathcal{D}_B$, then $O_A\sim O_B :\iff M_A=M_B$.

If $\mathcal{D}_A=\mathcal{D}_B^*$, then $O_A\sim O_B :\iff M_A=M_B^*$

  1. $L_A\cap L_B=\emptyset$.

If $\mathcal{D}_A\parallel \mathcal{D}_B$, then $$O_A\sim O_B :\iff\left( L_A\subset M_B\wedge L_B\subset M_A^*\right)\vee\left( L_A\subset M_B^*\wedge L_B\subset M_A\right)$$

If $\mathcal{D}_A\parallel \mathcal{D}_B^*$, then $$O_A\sim O_B :\iff\left( L_A\subset M_B\wedge L_B\subset M_A\right)\vee\left( L_A\subset M_B^*\wedge L_B\subset M_A^*\right)$$

  1. $L_A\cap L_B=\{o\}$

    Then $$O_A\sim O_B :\iff\left(P_A\subset M_B \wedge P_B\subset M_A^*\right)\vee\left(P_A\subset M_B^* \wedge P_B\subset M_A\right)$$, where $P_A\in\mathcal{D}_A,o(P_A)=o, P_B\in\mathcal{D}_B,o(P_B)=o$.

I managed to prove that $\sim$ is an equivalence relation with exactly two equivalence classes. I suppose my proof is correct although it's extremely long and tedious. We may call the equivalence classes orientations.

As for the original problem with relation between triangles we can easily introduce the orientation determined by three noncollinear points in the following way:

$$\mathcal{O}(a,b,c):=[(\overleftrightarrow{ab},[\overrightarrow{ab}],M(\overleftrightarrow{ab},c))]_{\sim}$$

where $M(\overleftrightarrow{ab},c)$ is a halfplane with boundary $\overleftrightarrow{ab}$ having point c.

$\endgroup$
  • $\begingroup$ This is an excellent definition of orientation in neutral geometry. In layman terms, it enables one to consistently distinguish the left and right sides of a directed line. $\endgroup$ – Transcendental Sep 8 at 2:50

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.