13.2: Inradius of h-triangle
The inradius of any h-triangle is less than \(\dfrac{1}{2} \cdot \ln3\) .
- Proof
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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
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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.