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# Nanairo Batake
## Solutions to TRTR Ch. 2 (An ancient theorem and a modern question) 2.5-2.6
### 2.5 (easy)
The conformality of Beltrami's geometry is a direct result of the conformality of the conformal representation and the conformality of the stereographic projection.
The fact that *hyperbolic* striaght lines are representated by vertical semicircles is apparent from the relation between the hemisphere representation and the projective representation (i.e., vertical projection as depicted in Fig. 2.17(a)).
- *NOTE**: Vertically projecting the projection representation onto the northern hemisphere, and sterepgraphically projecting the conformal representation onto the same hemisphere, result in the same picture. This is easily checked by referring to the expansion factor derived in Exercise 2.4.
### 2.6 (hard)
> *Note to myself:* I was unable to figure out the circle part, because the phrasing of the hint was rather undecipherable, so I looked it up in the original solutions website [here](https://web.archive.org/web/20070403130049if_/http://www.roadsolutions.ox.ac.uk:80/solutions/Solution3.jpg).
- *The angle part**: In Fig. II, we depicted the sphere $S$, centered at $\mathrm{O}$, and an arbitrary point $\mathrm{A}$ inside the equator that is different from $\mathrm{O}$ (the case when $\mathrm{A}$ and $\mathrm{O}$ are the same point is trivial). The plane passing $\mathrm{O}$, $\mathrm{A}$ and the north pole is denoted by $\Pi$, whereas the equatorial plane is denoted by $\mathrm{E}$.
Consider a small triangle $\mathrm{ABC}$ where $\mathrm{B}$ lies on the ray $\mathrm{OA}$ and $\mathrm{C}$ is arbitrary. The projection of this triangle onto the northern hemisphere is denoted by $\mathrm{A'B'C'}$. Draw straight lines $\mathrm{B''A'}$ and $\mathrm{C''A'}$ that are tangent to $\mathrm{B'A'}$ and $\mathrm{C'A'}$ respectively. In order to prove conformality, we need to prove $\angle \mathrm{B''A'C''} = \angle \mathrm{BAC}$.
By basic middle school geometry, we know that $\angle \mathrm{B''A'A} = \angle \mathrm{BAA'}$. Draw the perpendicular bisector $l$ of the line segment $\mathrm{AA'}$, we see that $\mathrm{A'B''}$ and $\mathrm{AB}$ are symmetric (within plane $\Pi$) wrt. $l$.
Consider the plane $\Delta$ that meets $\Pi$ at $l$ and is perpendicular to $\Pi$, and two other planes, plane $\mathrm{B''A'C''}$ (i.e., the tangent plane to sphere $S$ at $\mathrm{A'}$) and plane $\mathrm{BAC}$ (i.e., the equatorial plane $\mathrm{E}$). These two planes are also perpendicular to $\Pi$, and by the conclusion of last paragraph, these two planes are symmetric wrt. plane $\Delta$.
On the other hand, plane $\Delta$ is perpendicular to line $\mathrm{AA'}$, where the two planes $\mathrm{A'AB}$ and $\mathrm{A'AC}$ meet to form a dihedral angle. $\angle \mathrm{B''A'C''}$ and $\angle \mathrm{BAC}$ are exactly what we get when we intersect this dihedral angle with the two planes $\mathrm{B''A'C''}$ and $\mathrm{BAC}$, and these planes have been shown to be symmetric wrt. $\Delta$. Therefore $\angle \mathrm{B''A'C''} = \angle \mathrm{BAC}$.
- *The circle part**: In Fig. III, we again depicted sphere $S$, centered at $\mathrm{O}$, and an arbitrary circle $\Gamma$ *on the northern hemisphere*. The plane passing through $\mathrm{O}$ and the center of $\Gamma$ and the north pole is denoted $\Pi$. We form a cone $\mathrm{X}$ by connecting every point on $\Gamma$ with the south pole $\mathrm{P}$. We need to show that the intersection of $\mathrm{X}$ and the equatorial plane $\mathrm{E}$ is a circle.
The circle $\Gamma$ intersects the plane $\Pi$ at two points $\mathrm{M}$ and $\mathrm{N}$. Straight lines $\mathrm{MP}$ and $\mathrm{NP}$ meet the equatorial plane $\mathrm{E}$ at points $\mathrm{M'}$ and $\mathrm{N'}$ respectively. By basic middle school geometry, the angles $\angle \mathrm{N'M'P}$ and $\angle\mathrm{MNP}$ are equal. Therefore, it is not hard to see that there exists some straight line $t$ on plane $\Pi$ which meets $\mathrm{MP}$ and $\mathrm{NP}$ at equal angles wrt. to which $\mathrm{NM}$ and $\mathrm{M'N'}$ are symmetric.
Consider plane $\Sigma$ that meets $\Pi$ at $t$ and is perpendicular to $\Pi$, as well as two other planes: the plane of circle $\Gamma$ and the equatorial plane $\mathrm{E}$. Clearly, these two planes are symmetric wrt. $\mathrm{\Sigma}$.
Now comes the key question: what shape is the intersection of $\Sigma$ and $\mathrm{X}$? Intuitively, the answer is an ellipse. We can justify this by saying that since $\mathrm{X}$ is a quadratic surface, its intersection with a plane must be a quadratic curve. This curve must also be closed, therefore only ellipses are possible. Moreover, this ellipse must be symmetric wrt. plane $\Pi$ because the whole picture is symmetric wrt. $\Pi$. As a result, $t$ must be one of the two axes of the ellipse, and the other axis (which we denote by $w$) must be perpendicular to $\Pi$.
We now see that the cone $\mathrm{X}$ is symmetric wrt. the plane passing $\mathrm{P}$ and $w$. The plane of circle $\Gamma$ and the equatorial plane $\mathrm{E}$ form the same angle with this plane of symmetry. Therefore the shapes of the intersections of $\mathrm{X}$ and these two planes must be similar to each other, i.e., the intersection of $\mathrm{X}$ and $\mathrm{E}$ is a circle.
- *NOTE**: Stereographic projection does not necessarily map the center of a circle to the center of its image.