13.2: Inradius of h-triangle
- Page ID
- 23662
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\(\newcommand{\avec}{\mathbf a}\) \(\newcommand{\bvec}{\mathbf b}\) \(\newcommand{\cvec}{\mathbf c}\) \(\newcommand{\dvec}{\mathbf d}\) \(\newcommand{\dtil}{\widetilde{\mathbf d}}\) \(\newcommand{\evec}{\mathbf e}\) \(\newcommand{\fvec}{\mathbf f}\) \(\newcommand{\nvec}{\mathbf n}\) \(\newcommand{\pvec}{\mathbf p}\) \(\newcommand{\qvec}{\mathbf q}\) \(\newcommand{\svec}{\mathbf s}\) \(\newcommand{\tvec}{\mathbf t}\) \(\newcommand{\uvec}{\mathbf u}\) \(\newcommand{\vvec}{\mathbf v}\) \(\newcommand{\wvec}{\mathbf w}\) \(\newcommand{\xvec}{\mathbf x}\) \(\newcommand{\yvec}{\mathbf y}\) \(\newcommand{\zvec}{\mathbf z}\) \(\newcommand{\rvec}{\mathbf r}\) \(\newcommand{\mvec}{\mathbf m}\) \(\newcommand{\zerovec}{\mathbf 0}\) \(\newcommand{\onevec}{\mathbf 1}\) \(\newcommand{\real}{\mathbb R}\) \(\newcommand{\twovec}[2]{\left[\begin{array}{r}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\ctwovec}[2]{\left[\begin{array}{c}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\threevec}[3]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\cthreevec}[3]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\fourvec}[4]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\cfourvec}[4]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\fivevec}[5]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\cfivevec}[5]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\mattwo}[4]{\left[\begin{array}{rr}#1 \amp #2 \\ #3 \amp #4 \\ \end{array}\right]}\) \(\newcommand{\laspan}[1]{\text{Span}\{#1\}}\) \(\newcommand{\bcal}{\cal B}\) \(\newcommand{\ccal}{\cal C}\) \(\newcommand{\scal}{\cal S}\) \(\newcommand{\wcal}{\cal W}\) \(\newcommand{\ecal}{\cal E}\) \(\newcommand{\coords}[2]{\left\{#1\right\}_{#2}}\) \(\newcommand{\gray}[1]{\color{gray}{#1}}\) \(\newcommand{\lgray}[1]{\color{lightgray}{#1}}\) \(\newcommand{\rank}{\operatorname{rank}}\) \(\newcommand{\row}{\text{Row}}\) \(\newcommand{\col}{\text{Col}}\) \(\renewcommand{\row}{\text{Row}}\) \(\newcommand{\nul}{\text{Nul}}\) \(\newcommand{\var}{\text{Var}}\) \(\newcommand{\corr}{\text{corr}}\) \(\newcommand{\len}[1]{\left|#1\right|}\) \(\newcommand{\bbar}{\overline{\bvec}}\) \(\newcommand{\bhat}{\widehat{\bvec}}\) \(\newcommand{\bperp}{\bvec^\perp}\) \(\newcommand{\xhat}{\widehat{\xvec}}\) \(\newcommand{\vhat}{\widehat{\vvec}}\) \(\newcommand{\uhat}{\widehat{\uvec}}\) \(\newcommand{\what}{\widehat{\wvec}}\) \(\newcommand{\Sighat}{\widehat{\Sigma}}\) \(\newcommand{\lt}{<}\) \(\newcommand{\gt}{>}\) \(\newcommand{\amp}{&}\) \(\definecolor{fillinmathshade}{gray}{0.9}\)The inradius of any h-triangle is less than \(\dfrac{1}{2} \cdot \ln3\).
- Proof
-
Let \(I\) and \(r\) be the h-incenter and h-inradius of \(\triangle_hXYZ\).
Note that the h-angles \(XIY\), \(YIZ\) and \(ZIX\) have the same sign. Without loss of generality, we can assume that all of them are positive and therefore
\(\measuredangle_hXIY+ \measuredangle_hYIZ+ \measuredangle_hZIX=2 \cdot \pi\)
We can assume that \(\measuredangle_hXIY \ge \dfrac{2}{3} \cdot \pi\); if not relabel \(X\), \(Y\), and \(Z\).
Since \(r\) is the h-distance from \(I\) to \((XY)_h\), Proposition 13.1.1 implies that
\(\begin{array} {rcl} {r} & < & {\dfrac{1}{2} \cdot \ln \dfrac{1 + \cos \dfrac{\pi}{3}}{1 - \cos \dfrac{\pi}{3}}} \\ {} & = & {\dfrac{1}{2} \cdot \ln \dfrac{1 + \dfrac{1}{2}}{1 - \dfrac{1}{2}}} \\ {} & = & {\dfrac{1}{2} \cdot \ln 3.} \end{array}\)
Let \(\square_h ABCD\) be a quadrangle in the h-plane such that the h-angles at \(A\), \(B\), and \(C\) are right and \(AB_h=BC_h\). Find the optimal upper bound for \(AB_h\).
- Hint
-
Note that the angle of prarllelism of \(B\) to \((CD)_h\) is bigger than \(\dfrac{\pi}{4}\), and it converge to \(\dfrac{\pi}{4}\) as \(CD_h \to \infty\).
Applying Proposition 13.1.1, we get that
\(BC_h < \dfrac{1}{2} \cdot \ln \dfrac{1 + \dfrac{1}{\sqrt{2}}}{1 - \dfrac{1}{\sqrt{2}}} = \ln (1 + \sqrt{2}).\)
The right hand side is the limit of \(BC_h\) if \(CD_h \to \infty\). Therefore, \(\ln (1 + \sqrt{2})\) is the optimal upper bound.