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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

$$\measuredangle_hXIY+ \measuredangle_hYIZ+ \measuredangle_hZIX=2 \cdot \pi$$

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.