Skip to main content
\(\newcommand{\id}{\mathrm{id}}\) \( \newcommand{\Span}{\mathrm{span}}\) \( \newcommand{\kernel}{\mathrm{null}\,}\) \( \newcommand{\range}{\mathrm{range}\,}\) \( \newcommand{\RealPart}{\mathrm{Re}}\) \( \newcommand{\ImaginaryPart}{\mathrm{Im}}\) \( \newcommand{\Argument}{\mathrm{Arg}}\) \( \newcommand{\norm}[1]{\| #1 \|}\) \( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\) \( \newcommand{\Span}{\mathrm{span}}\)
Mathematics LibreTexts

13.2: Inradius of h-triangle

  • Page ID
    23662
  • \( \newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} } \) \( \newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}} \)\(\newcommand{\id}{\mathrm{id}}\) \( \newcommand{\Span}{\mathrm{span}}\) \( \newcommand{\kernel}{\mathrm{null}\,}\) \( \newcommand{\range}{\mathrm{range}\,}\) \( \newcommand{\RealPart}{\mathrm{Re}}\) \( \newcommand{\ImaginaryPart}{\mathrm{Im}}\) \( \newcommand{\Argument}{\mathrm{Arg}}\) \( \newcommand{\norm}[1]{\| #1 \|}\) \( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\) \( \newcommand{\Span}{\mathrm{span}}\) \(\newcommand{\id}{\mathrm{id}}\) \( \newcommand{\Span}{\mathrm{span}}\) \( \newcommand{\kernel}{\mathrm{null}\,}\) \( \newcommand{\range}{\mathrm{range}\,}\) \( \newcommand{\RealPart}{\mathrm{Re}}\) \( \newcommand{\ImaginaryPart}{\mathrm{Im}}\) \( \newcommand{\Argument}{\mathrm{Arg}}\) \( \newcommand{\norm}[1]{\| #1 \|}\) \( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\) \( \newcommand{\Span}{\mathrm{span}}\)

    Theorem \(\PageIndex{1}\)

    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

    \(\measuredangle_hXIY+ \measuredangle_hYIZ+ \measuredangle_hZIX=2 \cdot \pi\)

    截屏2021-02-24 下午3.30.51.png

    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

    \(\begin{array} {rcl} {r} & < & {\dfrac{1}{2} \cdot \ln \dfrac{1 + \cos \dfrac{\pi}{3}}{1 - \cos \dfrac{\pi}{3}}} \\ {} & = & {\dfrac{1}{2} \cdot \ln \dfrac{1 + \dfrac{1}{2}}{1 - \dfrac{1}{2}}} \\ {} & = & {\dfrac{1}{2} \cdot \ln 3.} \end{array}\)

     

    Exercise \(\PageIndex{1}\)

    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.