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## Solutions to TRTR Ch. 2 (An ancient theorem and a modern question) 2.1-2.4
- *NOTE**: these are my personal solutions. I have made them as logically sound as I can but they may still contain mistakes. Also there might be alternative ways of solving a problem. I will try to point those out whenever I am aware.
### 2.1 (easy)
Consider three lines $a$, $b$ and $c$ that lie in a plane, where $c$ is a transversal of $a$ and $b$. We rotate $a$ while keeping its intersection with $c$ fixed (this point will serve as point $P$ in Fig. 2.8). As $a$ rotates, the value of the sum of the two interior angles on the same side of $c$ changes. When this sum is smaller than two right angles, $a$ and $b$ will intersect somewhere on that side of $c$. When this sum is larger than two right angles, $a$ and $b$ will intersect somewhere on the other side of $c$. It is only when that sum is exactly equal to two right angles that $a$ and $b$ are parallel. That is to say, there is one unique position of $a$ for it to be parallel with $b$, i.e., Playfair's conclusion holds.
### 2.2 (medium)
Under an arbitrary chice of $C$, we have for areas
$\Delta = C^{-1} [\pi - (\alpha+\beta+\gamma)]\ ,$
and for lengths
$d(\mathrm{A}, \mathrm{B}) = D \log\frac{\mathrm{QA}\cdot\mathrm{PB}}{\mathrm{QB}\cdot\mathrm{PA}}\ ,$
where $D$ is a coefficient of proportionality to be determined. Different choices of $C$ correspond to different units of measurement, under which the numerical values of lengths should change in the same fashion as the square root of the numerical values of areas, i.e., $D\propto C^{-1/2}$. The text also implies $D=1$ when $C=1$, hence $D = C^{-1/2}$.
- *NOTE 1**: Here we are not doing a similarity transformation in the hyperbolic plane (which would be illegal). We are just measuring lengths using different units. If 1km = 1000m, then 1km$^2$ = $1000^2$m$^2$. This is always true regardless of the geometry.
- *NOTE 2**: The lengths $\mathrm{QA}$, $\mathrm{PB}$, etc. apparing in the hyperbolic distance formula refer to the Euclidean length of a *line segment* connecting two points, *not* the distance along a circular arc. See e.g. [Wikipedia: Poincaré disk model](https://en.wikipedia.org/wiki/Poincar%C3%A9_disk_model)
### 2.3 (easy)
If $A$, $B$ and $C$ are three successive points on a hyperbolic straight line, then we have
$d(\mathrm{A}, \mathrm{B}) + d(\mathrm{B}, \mathrm{C}) = \log\frac{\mathrm{QA}\cdot\mathrm{PB}}{\mathrm{QB}\cdot\mathrm{PA}} + \log\frac{\mathrm{QB}\cdot\mathrm{PC}}{\mathrm{QC}\cdot\mathrm{PB}} = \log\frac{\mathrm{QA}\cdot\mathrm{PC}}{\mathrm{QC}\cdot\mathrm{PA}} = d(\mathrm{A}, \mathrm{C})\ .$
### 2.4 (medium)
First, we need to understand how to get the projective representation (like the one illustrated in Fig. 2.16) from the conformal representation (as depicted in Fig. 2.11). From Fig. 2.15, it seems that one does this by simply pushing any point radially outward until it hits the secant which meets the bounding circle at the same points as the *hyperbolic straight line passing through said point in the conformal representation*. However, it is not at all obvious that this method yields a uniquely defined result, as there are an infinite number of hyperbolic straight lines passing through a point. Therefore, we need to establish the uniqueness of such result before proceeding to calculating the expansion factor.
We use Fig. I to facilitate our proof. Denote by $\Gamma$ the bounding circle of our conformal representation, centered at point $\mathrm{O}$. Let $\mathrm{A}$ be an arbitrary point inside $\Gamma$ other than $\mathrm{O}$. Consider the *unique* hyperbolic straight line $\mathrm{PQ}$ through $\mathrm{A}$ that is *symmetric* wrt. $\mathrm{OA}$. Denote the Euclidean circle which $\mathrm{PQ}$ is part of by $\Gamma_1$, whose center is $\mathrm{W}$. Also, draw an *arbitrary* hyperbolic straight line $\mathrm{RS}$ through $\mathrm{A}$, which is part of another circle $\Gamma_2$.
Let $\mathrm{M}_1$ be the intersection of $\Gamma_1$ and the Euclidean straight line $\mathrm{OA}$. Similarly let $\mathrm{M}_2$ be the intersection of $\Gamma_2$ and $\mathrm{OA}$. We are going to prove that $\mathrm{M_1}$ and $\mathrm{M_2}$ are the same point. Note that $\mathrm{OP}$ and $\mathrm{OR}$ are tangent to $\Gamma_1$ and $\Gamma_2$ respectively, so by basic middle school geometry,
$\mathrm{OP}\cdot\mathrm{OP} = \mathrm{OA}\cdot\mathrm{OM_1}\ ,$
$\mathrm{OR}\cdot\mathrm{OR} = \mathrm{OA}\cdot\mathrm{OM_2}\ .$
Since $\mathrm{OP}=\mathrm{OS}=$ radius of $\Gamma$, we immediately have $\mathrm{OM_1} = \mathrm{OM_2}$. Hence $\mathrm{M_1}$ and $\mathrm{M_2}$ are the same point and we denote it by $\mathrm{M}$ from now on.
Let's recap: There are two circles $\Gamma_1$ and $\Gamma_2$, and an Euclidean straight line $\mathrm{OA}$. These three objects all intersect at $\mathrm{A}$ and $\mathrm{M}$. What we now need to show is that the three *Euclidean* straight lines $\mathrm{PQ}$, $\mathrm{RS}$ and $\mathrm{OA}$ intersect at the same point.
Imagine $\Gamma$, $\Gamma_1$ and $\Gamma_2$ as equators of three Euclidean spheres that are cut in half by the plane which we have been looking at. Then the *Euclidean* line segments $\mathrm{PQ}$, $\mathrm{RS}$ and $\mathrm{AM}$ are just what we see when we look at the intersections (circles) of pairs of spheres. These three line segments must meet at the same point which is the intersection of all three spheres above the equator plane. We denote this point (or more precisely, the image of this point on the plane we are looking at) by $\mathrm{A}'$, and our proof is complete.
Clearly, $\mathrm{A'}$ is the point in the projective representation corresponding to $\mathrm{A}$. It is also very easy to see that $\mathrm{W}$ lies on the Euclidean line $\mathrm{OA}$ by the very construction of $\Gamma_1$. The calculation of the expansion factor (hinted in the caption of Fig. 2.15) is simple. Let $R$ and $r$ be the radii of circles $\Gamma$ and $\Gamma_1$ respectively, and let $r_c$ be the length of $\mathrm{OA}$. From the Euclidean right triangle $\mathrm{POW}$ we have
$(r+r_c^2) = R^2 + r^2\ ,$
which gives us $r = (R^2-r_c^2) / (2r_c)$.
From the similarity of the Euclidean right triangles $\mathrm{POW}$ and $\mathrm{A'OP}$ we have $\mathrm{OA'} = R^2/(r+r_c)$, hence the expansion factor
$\frac{\mathrm{OA'}}{\mathrm{OA}} = \frac{R^2}{r_c(r+r_c)} = \frac{2R^2}{R^2+r_c^2}\ .$
- *NOTE**: Another simpler solution using Beltrami's geometry (as hinted) can be found at the original (now closed down) solutions site [here](https://web.archive.org/web/20070403130704if_/http://www.roadsolutions.ox.ac.uk:80/solutions/Solution2.jpg)