Difficulties with directed angles

Question

In most geometry books, I've studied with, it did not mention directed angles at all. Then I came across Evan Chen's EGMO which uses directed angles from the start to finish. Although he has written an explanation of the concept, I found directed angles quite cumbersome and confusing to understand. Is there a better way to visualise and understand directed angles so that it can seem more relatable when solving geometry problems because right now it seems more like an unnecessary and an extra step? And is it just me who's having trouble with it?

Answer 1

I had the exact same problem when studying EGMO, ad I wasted a lot of time, but the truth is, I didn't do what I should have done in the first place, that is just sit down and study directed angles till you understand it. It is really not a very difficult topic and I think if you just sit down and study it for some time (Only EGMO is needed, or maybe you could look at this handout by evan) instead of procrastinating (trust me, this is what is happening to you right now) you will soon understand it very well.

And I have to say is it needed very much, in a lot of problems (for instance see ISL 2019/G1) the only acceptable way to do that problem might be using directed angles, or when you will study inversion (ch-8), you will need directed angles a lot, for instance, my proof here is possible only because of directed angles.

But what I would say is you do not need directed right now that much, just do like the first 3 chapters using normal angles if you are not able to understand directed right now, and once you gain some geo experience, then sit down and just study directed angles and really try to understand it for like a couple hours or so, and then resolve some of the angle chasing problems from previous chapters using directed (at least that is what i did to understand directed angles).

Anyways,

All the best in your olympiad journey, and have fun!

Answer 2

In my opinion, the best definition of a direct (or oriented) angle of vector half lines is to establish a bijection with plane rotations, following the comments of @Jean Marie and @David G. Stork.

$\tt Definition\;1\quad$ A vector half line in the 2-dimensional (real) vector space $E\,$ is a subset of $E$ of the form

$$ \Delta := \mathbb R_+\,v = \left\{\lambda v\;:\;\lambda\geq0\right\}\qquad (\text{with } v\neq 0). $$

A direct angle of half lines $\,$(or simply angle)$\;\theta\,$ in $E\,$ is an ordered pair of half lines, which we will denote by

$$ \theta := (\Delta_1,\Delta_2)^\wedge. $$

Two angles $\,\theta=(\Delta_1, \Delta_2)^\wedge,\; \theta'=(\Delta'_1, \Delta'_2)^\wedge$ of an euclidean space $E$ are equal iff, by definition, there is a rotation of $E$, i.e. an element $\varrho$ of the group $\mathbb{SO}(E)$, such that

$$ \varrho(\Delta_1)=\Delta'_1\qquad\text{and}\qquad \varrho(\Delta_2)=\Delta'_2. $$

The set $\;\cal U\;$ of equivalence classes of equal angles is in a bijection with the group $\mathbb{SO}(E)$, so is itself a group, called the group of angles of E. These groups are therefore isomorphic, and are denoted additively the first and multiplicatively the second. They are also commutative because dim(E)=2. The isomorphism

$$ ρ\;:\;\mathcal U\;\longrightarrow\;\mathbb{SO}(E) $$

associates to each class $[\theta]$ of equal angles the corresponding rotation.

It is interesting to point out that, despite its name, an oriented angle does not depends on any orientation of $E$: it can be viewed as a simple function $\;E\longrightarrow E\;$ with the constraint of being a rotation carrying a given half line onto another given half line.

But, what about the common notion of an angle as a set of points (or vectors in our case) between two half lines? After all, we are fond of this vision of angle. To do this, however, the plane must be oriented, otherwise there would be ambiguity in the choice of the point/vectors to be considered. In fact we can describe $\theta=(\Delta_1, \Delta_2)^\wedge$ by "going" from $\Delta_1$ to $\Delta_2$ in one direction or the opposite, thus obtaining explementary angles. We can proceed in the following way.

$\tt Definition\; 2\quad$ An orientation over the vector space $\,E\;$ is a non zero alternating bilinear form

$$ \Psi\;:\;E\times E \;\longrightarrow\; \mathbb R. $$

This orientation form distinguishes positively and negatively oriented bases of $\,E\;$ according to the sign it takes on the given bases. The space $\,E\;$ equipped with an orientation form $\Psi$ is called an oriented vector space.

$\tt Definition\;3\quad$ Let $\Delta_1=\mathbb R_+v_1,\;\; \Delta_2=\mathbb R_+v_2, \;$ with $v_1, v_2\neq0$.
The open angular sector associated with the angle $\theta=(\Delta_1, \Delta_2)^\wedge$ of the oriented vector space $\,E\;$ is the set

$$ S^°(\theta) = S^°(\Delta_1, \Delta_2) := \left\{v\in E\raise.3ex\smallsetminus\{0\}\;:\;\rm{pos}\Big(\Psi(v_1, v), \Psi(v, v_2), \Psi(v_2, v_1)\Big)\geq2\right\}, $$

where $\;\rm{pos}(\alpha, \beta, \gamma)\;$ denote the numbers of strictly positive elements contained into the round parenthesis.

It isn't difficult to check, by some pictures for example, that this definition corresponds to our intuitive idea of angle as a set of points, and that it depends, obviously, on chosen orientation.

Affine case$\quad$ What has been said so far is easily transposed to affine planes, which is the natural setting of classical plane geometry.

Let $\mathcal A(E)$ be an affine real plane with $E$ as the vector space of translations. Given three points $V, A_1, A_2$ of $\mathcal A(E)$ with $A_1$ and $A_2\neq V$, the angle of vertex $V$ passing through $A_1$ and $A_2$ is a triple $(V;\Delta_1, \Delta_2)^\wedge$, where, for $i=1,2$:

$$ \Delta_i=\big\{P\in \mathcal A(E)\;:\;P-V=\lambda(A_i-V) \;\text{ for some } \lambda\geq0\big\}= $$

$$ =V+\mathbb R_+(A_i-V). $$

The space $\mathcal A(E)$ is oriented if such is $E$. The open angular sector associated with the angle $(V;A_1,A_2)^\wedge$ of the oriented affine space $\mathcal A(E)$ is the set

$$ S^°(V;A_1,A_2) := \hskip8cm $$ $$ :=\Big\{P\in \mathcal A(E)\;:\;\rm{pos}\Big(\Psi(A_1-V, P-V), \Psi(P-V, A_2-V), \Psi(A_2-V, A_1-V)\Big)\geq 2\Big\}. $$

As can be seen, the affine case reduces to the vector case, where the vectors all apply to the vertex $V$.