13.2: Inradius of h-triangle

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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.

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