双曲直角三角形 Heron公式验证

hbghlyj

$a,b,c$是任意正数，$\cosh c=\cosh a\cosh b$，设$s={\tfrac {1}{2}}(a+b+c)$，求证
$$\tanh\frac a2 \tanh\frac b2=\frac{2\sqrt{\tanh {\frac {s}{2}}\tanh {\frac {s-a}{2}}\tanh {\frac {s-b}{2}}\tanh {\frac {s-c}{2}}}}{1-\tanh {\frac {s}{2}}\tanh {\frac {s-a}{2}}\tanh {\frac {s-b}{2}}\tanh {\frac {s-c}{2}}}$$

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Mathematica验证：

1. a=RandomReal[{0,10}];
2. b=RandomReal[{0,10}];
3. s=(a+b+c)/2;
4. c=ArcCosh[Cosh[a]Cosh[b]];
5. Tanh[a/2]Tanh[b/2]-(2Sqrt[Tanh[s/2]Tanh[(s-a)/2]Tanh[(s-b)/2]Tanh[(s-c)/2]])/(1-Tanh[s/2]Tanh[(s-a)/2]Tanh[(s-b)/2]Tanh[(s-c)/2])//Chop

复制代码

输出0

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来源：
在Pythagoras in the hyperbolic plane提到了$\cosh a\cosh b = \cosh c.$
在Area of a right angled hyperbolic triangle as function of side lengths提到了$\tan\frac S2 = \tanh\frac a2 \tanh\frac b2$
在Heron formulae in non-Euclidean geometries提到了$\tan ^{2}{\frac {S}{4}}=\tanh {\frac {s}{2}}\tanh {\frac {s-a}{2}}\tanh {\frac {s-b}{2}}\tanh {\frac {s-c}{2}}.$
由正切半角公式$\tan\frac {S}{2}=\frac{2\tan(\frac {S}{4})}{1-\tan^2(\frac {S}{4})}$把上面两式消去$S$就得到这个问题。

hbghlyj

设$\text{ta}=\tanh\frac a2,\text{tb}=\tanh\frac b2,\text{tc}=\tanh\frac c2,\text{ts}=\tanh\frac s2$，则$\text{ta},\text{tb},\text{tc},\text{ts}\in(0,1).$
由$\cosh x=\frac{\tanh^2\frac x2+1}{1-\tanh^2\frac x2}$得
$$\cosh a\cosh b=\cosh c\implies\frac{\left(\text{ta}^2+1\right) \left(\text{tb}^2+1\right)}{\left(1-\text{ta}^2\right) \left(1-\text{tb}^2\right)}=\frac{\text{tc}^2+1}{1-\text{tc}^2}\implies\text{tc}=\frac{\sqrt{\text{ta}^2+\text{tb}^2}}{\sqrt{\text{ta}^2 \text{tb}^2+1}}$$
由tanh加法公式
$$\tanh s=\tanh(\frac a2+\frac b2+\frac c2)=\frac{\frac{\text{ta}+\text{tb}}{\text{ta} \text{tb}+1}+\text{tc}}{\frac{\text{tc} (\text{ta}+\text{tb})}{\text{ta} \text{tb}+1}+1}=\frac{\sqrt{\text{ta}^2+\text{tb}^2} \sqrt{\text{ta}^2 \text{tb}^2+1}+\text{ta} \text{tb} (\text{ta}+\text{tb})+\text{ta}+\text{tb}}{\text{ta}^2\text{tb}^2+ \left(\text{ta}+\text{tb}\right)^2+1}$$
由tanh半角公式$\tanh(\frac{\tanh^{-1}x}2)=\frac{1-\sqrt{1-x^2}}{x}$
$$\text{ts}=\tanh(\frac{s}2)=\frac{\left(\sqrt{\text{ta}^2+\text{tb}^2}+\text{ta}+\text{tb}\right) \left(-\sqrt{\text{ta}^2 \text{tb}^2+1}+\text{ta} \text{tb}+1\right)}{2 \text{ta} \text{tb}}$$
化简
\begin{align*}
\frac{(\text{ts}-\text{ta}) (\text{ts}-\text{tb})}{(1-\text{ta} \text{ts}) (1-\text{tb} \text{ts})}&=\frac{\text{ta} \text{tb} \left(\sqrt{\text{ta}^2 \text{tb}^2+1}+\text{ta} \text{tb}-1\right)-\sqrt{\text{ta}^2 \text{tb}^2+1}+1}{\text{ta} \text{tb}}\\
\frac{\text{ts} (\text{ts}-\text{tc})}{1-\text{tc} \text{ts}}&=\frac{\text{ta} \text{tb} \left(-\sqrt{\text{ta}^2 \text{tb}^2+1}+\text{ta} \text{tb}+1\right)-\sqrt{\text{ta}^2 \text{tb}^2+1}+1}{\text{ta} \text{tb}}
\end{align*}
两式相乘
$$\frac{\text{ts} (\text{ts}-\text{ta}) (\text{ts}-\text{tb}) (\text{ts}-\text{tc})}{(1-\text{ta} \text{ts}) (1-\text{tb} \text{ts}) (1-\text{tc} \text{ts})}=\frac{\text{ta}^2 \text{tb}^2-2 \sqrt{\text{ta}^2 \text{tb}^2+1}+2}{\text{ta}^2 \text{tb}^2}=\left(\frac{\sqrt{\text{ta}^2 \text{tb}^2+1}-1}{\text{ta} \text{tb}}\right)^2$$
开方
$$\sqrt{\frac{\text{ts} (\text{ts}-\text{ta}) (\text{ts}-\text{tb}) (\text{ts}-\text{tc})}{(1-\text{ta} \text{ts}) (1-\text{tb} \text{ts}) (1-\text{tc} \text{ts})}}=\frac{\sqrt{\text{ta}^2 \text{tb}^2+1}-1}{\text{ta} \text{tb}}$$
由tan半角公式$\tan(\frac{\tan^{-1}x}2)=\frac{\sqrt{x^2+1}-1}{x}$得
$$\tan^{-1}\sqrt{\frac{\text{ts} (\text{ts}-\text{ta}) (\text{ts}-\text{tb}) (\text{ts}-\text{tc})}{(1-\text{ta} \text{ts}) (1-\text{tb} \text{ts}) (1-\text{tc} \text{ts})}}=\frac{\tan^{-1}(\text{ta} \text{tb})}2$$
两边×2
$$2\tan^{-1}\sqrt{\frac{\text{ts} (\text{ts}-\text{ta}) (\text{ts}-\text{tb}) (\text{ts}-\text{tc})}{(1-\text{ta} \text{ts}) (1-\text{tb} \text{ts}) (1-\text{tc} \text{ts})}}=\tan^{-1}(\text{ta} \text{tb})$$
两边取tan
$$\frac{2\sqrt{\frac{\text{ts} (\text{ts}-\text{ta}) (\text{ts}-\text{tb}) (\text{ts}-\text{tc})}{(1-\text{ta} \text{ts}) (1-\text{tb} \text{ts}) (1-\text{tc} \text{ts})}}}{1-\frac{\text{ts} (\text{ts}-\text{ta}) (\text{ts}-\text{tb}) (\text{ts}-\text{tc})}{(1-\text{ta} \text{ts}) (1-\text{tb} \text{ts}) (1-\text{tc} \text{ts})}}=\text{ta} \text{tb}$$
由tanh加法公式$\tanh {\frac {s-a}{2}}=\frac{\text{ts}-\text{ta}}{1-\text{ta}\text{ts}}$，上式化为
$$\frac{2\sqrt{\tanh {\frac {s}{2}}\tanh {\frac {s-a}{2}}\tanh {\frac {s-b}{2}}\tanh {\frac {s-c}{2}}}}{1-\tanh {\frac {s}{2}}\tanh {\frac {s-a}{2}}\tanh {\frac {s-b}{2}}\tanh {\frac {s-c}{2}}}=\tanh\frac a2 \tanh\frac b2$$
证毕！

hbghlyj

上面的计算太繁瑣了，有没有捷徑？

搜了一些资料，可能有用：
hyperbolic area - heron's formula
Hyperbolic plane geometry revisited, Ákos G.Horváth
The Formulas of Möbius-Bretschneider and Möbius-Cagnoli in the Poincaré Disc Model of Hyperbolic Geometry
Non-Euclidean versions of some classical triangle inequalities