17.3: Bolyai's construction
- Page ID
- 23691
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\(\newcommand{\avec}{\mathbf a}\) \(\newcommand{\bvec}{\mathbf b}\) \(\newcommand{\cvec}{\mathbf c}\) \(\newcommand{\dvec}{\mathbf d}\) \(\newcommand{\dtil}{\widetilde{\mathbf d}}\) \(\newcommand{\evec}{\mathbf e}\) \(\newcommand{\fvec}{\mathbf f}\) \(\newcommand{\nvec}{\mathbf n}\) \(\newcommand{\pvec}{\mathbf p}\) \(\newcommand{\qvec}{\mathbf q}\) \(\newcommand{\svec}{\mathbf s}\) \(\newcommand{\tvec}{\mathbf t}\) \(\newcommand{\uvec}{\mathbf u}\) \(\newcommand{\vvec}{\mathbf v}\) \(\newcommand{\wvec}{\mathbf w}\) \(\newcommand{\xvec}{\mathbf x}\) \(\newcommand{\yvec}{\mathbf y}\) \(\newcommand{\zvec}{\mathbf z}\) \(\newcommand{\rvec}{\mathbf r}\) \(\newcommand{\mvec}{\mathbf m}\) \(\newcommand{\zerovec}{\mathbf 0}\) \(\newcommand{\onevec}{\mathbf 1}\) \(\newcommand{\real}{\mathbb R}\) \(\newcommand{\twovec}[2]{\left[\begin{array}{r}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\ctwovec}[2]{\left[\begin{array}{c}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\threevec}[3]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\cthreevec}[3]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\fourvec}[4]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\cfourvec}[4]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\fivevec}[5]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\cfivevec}[5]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\mattwo}[4]{\left[\begin{array}{rr}#1 \amp #2 \\ #3 \amp #4 \\ \end{array}\right]}\) \(\newcommand{\laspan}[1]{\text{Span}\{#1\}}\) \(\newcommand{\bcal}{\cal B}\) \(\newcommand{\ccal}{\cal C}\) \(\newcommand{\scal}{\cal S}\) \(\newcommand{\wcal}{\cal W}\) \(\newcommand{\ecal}{\cal E}\) \(\newcommand{\coords}[2]{\left\{#1\right\}_{#2}}\) \(\newcommand{\gray}[1]{\color{gray}{#1}}\) \(\newcommand{\lgray}[1]{\color{lightgray}{#1}}\) \(\newcommand{\rank}{\operatorname{rank}}\) \(\newcommand{\row}{\text{Row}}\) \(\newcommand{\col}{\text{Col}}\) \(\renewcommand{\row}{\text{Row}}\) \(\newcommand{\nul}{\text{Nul}}\) \(\newcommand{\var}{\text{Var}}\) \(\newcommand{\corr}{\text{corr}}\) \(\newcommand{\len}[1]{\left|#1\right|}\) \(\newcommand{\bbar}{\overline{\bvec}}\) \(\newcommand{\bhat}{\widehat{\bvec}}\) \(\newcommand{\bperp}{\bvec^\perp}\) \(\newcommand{\xhat}{\widehat{\xvec}}\) \(\newcommand{\vhat}{\widehat{\vvec}}\) \(\newcommand{\uhat}{\widehat{\uvec}}\) \(\newcommand{\what}{\widehat{\wvec}}\) \(\newcommand{\Sighat}{\widehat{\Sigma}}\) \(\newcommand{\lt}{<}\) \(\newcommand{\gt}{>}\) \(\newcommand{\amp}{&}\) \(\definecolor{fillinmathshade}{gray}{0.9}\)Assume we need to construct a line thru \(P\) asymptotically parallel to the given line \(\ell\) in the h-plane.
If \(A\) and \(B\) are ideal points of \(\ell\) in the projective model, then we could simply draw the Euclidean line \((PA)\). However the ideal points do not lie in the h-plane; therefore there is no way to use them in the construction.
In the following construction we assume that you know a compass-and-ruler construction of the perpendicular line; see Exercise 5.22.
- Drop a perpendicular from \(P\) to \(\ell\); denote it by \(m\). Let \(Q\) be the foot point of \(P\) on \(\ell\).
- Erect a perpendicular from \(P\) to \(m\); denote it by \(n\).
- Mark by \(R\) a point on \(\ell\) distinct from \(Q\).
- Drop a perpendicular from \(R\) to \(n\); denote it by \(k\).
- Draw the circle \(\Gamma\) with center \(P\) and the radius \(QR\). Mark by \(T\) a point of intersection of \(\Gamma\) with \(k\).
- The line \((PT)_h\) is asymptotically parallel to \(\ell\).
Explain what happens if one performs the Bolyai construction in the Euclidean plane.
- Answer
-
Add texts here. Do not delete this text first.
To prove that Bolyai’s construction gives the asymptotically parallel line in the h-plane, it is sufficient to show the following:
Assume \(P\), \(Q\), \(R\), \(S\), \(T\) be points in h-plane such that
- \(S\in (RT)_h\),
- \((PQ)_h\perp (QR)_h\),
- \((PS)_h\perp(PQ)_h\),
- \((RT)_h\perp (PS)_h\) and
- \((PT)_h\) and \((QR)_h\) are asymptotically parallel.
Then \(QR_h=PT_h\).
- Proof
-
We will use the projective model. Without loss of generality, we may assume that \(P\) is the center of the absolute. As it was noted on page , in this case the corresponding Euclidean lines are also perpendicular; that is, \((PQ)\perp (QR)\), \((PS)\perp(PQ)\), and \((RT) \perp (PS)\).
Let \(A\) be the common ideal point of \((QR)_h\) and \((PT)_h\). Let \(B\) and \(C\) denote the remaining ideal points of \((QR)_h\) and \((PT)_h\) respectively.
Note that the Euclidean lines \((PQ)\), \((TR)\), and \((CB)\) are parallel.
Therefore,
In particular,
\(\dfrac{AC}{AB}=\dfrac{AT}{AR}=\dfrac{AP}{AQ}.\)
It follows that
In particular, \(\dfrac{AT\cdot CP}{TC\cdot PA}=\dfrac{AR\cdot BQ}{RB\cdot QA}\). Applying the formula for h-distance 17.2.1, we get that \(QR_h=PT_h\).