求三角形内的相交三角形面积

数学星空

当然，我们也可以得到\(D=E=F\)即\(s=0\)时，\(a,b,c,k_1,k_2,k_3,m,n,p\)的关系式：(注：\(m=\tan(\alpha),n=\tan(\beta),p=\tan(\gamma))\))

\(4a^8k_2^2m^2n^2p^2-8a^6b^2k_2^2m^2n^2p^2+8a^6b^2k_2k_3m^2n^2p^2+8a^6c^2k_1k_2m^2n^2p^2-8a^6c^2k_2^2m^2n^2p^2+4a^4b^4k_2^2m^2n^2p^2-16a^4b^4k_2k_3m^2n^2p^2+4a^4b^4k_3^2m^2n^2p^2-16a^4b^2c^2k_1k_2m^2n^2p^2+8a^4b^2c^2k_1k_3m^2n^2p^2-8a^4b^2c^2k_2^2m^2n^2p^2-16a^4b^2c^2k_2k_3m^2n^2p^2+4a^4c^4k_1^2m^2n^2p^2-16a^4c^4k_1k_2m^2n^2p^2+4a^4c^4k_2^2m^2n^2p^2+8a^2b^6k_2k_3m^2n^2p^2-8a^2b^6k_3^2m^2n^2p^2+8a^2b^4c^2k_1k_2m^2n^2p^2-16a^2b^4c^2k_1k_3m^2n^2p^2-16a^2b^4c^2k_2k_3m^2n^2p^2-8a^2b^4c^2k_3^2m^2n^2p^2-8a^2b^2c^4k_1^2m^2n^2p^2-16a^2b^2c^4k_1k_2m^2n^2p^2-16a^2b^2c^4k_1k_3m^2n^2p^2+8a^2b^2c^4k_2k_3m^2n^2p^2-8a^2c^6k_1^2m^2n^2p^2+8a^2c^6k_1k_2m^2n^2p^2+4b^8k_3^2m^2n^2p^2+8b^6c^2k_1k_3m^2n^2p^2-8b^6c^2k_3^2m^2n^2p^2+4b^4c^4k_1^2m^2n^2p^2-16b^4c^4k_1k_3m^2n^2p^2+4b^4c^4k_3^2m^2n^2p^2-8b^2c^6k_1^2m^2n^2p^2+8b^2c^6k_1k_3m^2n^2p^2+4c^8k_1^2m^2n^2p^2-4a^8k_2m^2n^2p^2+4a^6b^2k_2m^2n^2p^2-4a^6b^2k_3m^2n^2p^2-4a^6c^2k_1m^2n^2p^2+4a^6c^2k_2m^2n^2p^2+4a^4b^4k_2m^2n^2p^2+4a^4b^4k_3m^2n^2p^2+4a^4b^2c^2k_1m^2n^2p^2+24a^4b^2c^2k_2m^2n^2p^2+4a^4b^2c^2k_3m^2n^2p^2+4a^4c^4k_1m^2n^2p^2+4a^4c^4k_2m^2n^2p^2-4a^2b^6k_2m^2n^2p^2+4a^2b^6k_3m^2n^2p^2+4a^2b^4c^2k_1m^2n^2p^2+4a^2b^4c^2k_2m^2n^2p^2+24a^2b^4c^2k_3m^2n^2p^2+24a^2b^2c^4k_1m^2n^2p^2+4a^2b^2c^4k_2m^2n^2p^2+4a^2b^2c^4k_3m^2n^2p^2+4a^2c^6k_1m^2n^2p^2-4a^2c^6k_2m^2n^2p^2-4b^8k_3m^2n^2p^2-4b^6c^2k_1m^2n^2p^2+4b^6c^2k_3m^2n^2p^2+4b^4c^4k_1m^2n^2p^2+4b^4c^4k_3m^2n^2p^2+4b^2c^6k_1m^2n^2p^2-4b^2c^6k_3m^2n^2p^2-4c^8k_1m^2n^2p^2+4a^8k_2^2m^2n^2+4a^8k_2^2n^2p^2+a^8m^2n^2p^2-8a^6b^2k_2^2m^2n^2-8a^6b^2k_2^2n^2p^2+8a^6b^2k_2k_3m^2np+8a^6b^2k_2k_3n^2p^2+8a^6c^2k_1k_2m^2n^2+8a^6c^2k_1k_2mnp^2-8a^6c^2k_2^2m^2n^2-8a^6c^2k_2^2n^2p^2+4a^4b^4k_2^2m^2n^2+4a^4b^4k_2^2n^2p^2-16a^4b^4k_2k_3m^2np-16a^4b^4k_2k_3n^2p^2+4a^4b^4k_3^2m^2p^2+4a^4b^4k_3^2n^2p^2-2a^4b^4m^2n^2p^2-16a^4b^2c^2k_1k_2m^2np-16a^4b^2c^2k_1k_2mn^2p+16a^4b^2c^2k_1k_3m^2np-8a^4b^2c^2k_1k_3m^2p^2-8a^4b^2c^2k_1k_3mn^2p+16a^4b^2c^2k_1k_3mnp^2+8a^4b^2c^2k_2^2m^2n^2-32a^4b^2c^2k_2^2mn^2p+8a^4b^2c^2k_2^2n^2p^2-16a^4b^2c^2k_2k_3mn^2p-16a^4b^2c^2k_2k_3mnp^2-8a^4b^2c^2m^2n^2p^2+4a^4c^4k_1^2m^2n^2+4a^4c^4k_1^2m^2p^2-16a^4c^4k_1k_2m^2n^2-16a^4c^4k_1k_2mnp^2+4a^4c^4k_2^2m^2n^2+4a^4c^4k_2^2n^2p^2-2a^4c^4m^2n^2p^2+8a^2b^6k_2k_3m^2np+8a^2b^6k_2k_3n^2p^2-8a^2b^6k_3^2m^2p^2-8a^2b^6k_3^2n^2p^2-8a^2b^4c^2k_1k_2m^2n^2+16a^2b^4c^2k_1k_2m^2np+16a^2b^4c^2k_1k_2mn^2p-8a^2b^4c^2k_1k_2mnp^2-16a^2b^4c^2k_1k_3m^2np-16a^2b^4c^2k_1k_3mnp^2-16a^2b^4c^2k_2k_3mn^2p-16a^2b^4c^2k_2k_3mnp^2+8a^2b^4c^2k_3^2m^2p^2-32a^2b^4c^2k_3^2mnp^2+8a^2b^4c^2k_3^2n^2p^2-8a^2b^4c^2m^2n^2p^2+8a^2b^2c^4k_1^2m^2n^2-32a^2b^2c^4k_1^2m^2np+8a^2b^2c^4k_1^2m^2p^2-16a^2b^2c^4k_1k_2m^2np-16a^2b^2c^4k_1k_2mn^2p-16a^2b^2c^4k_1k_3m^2np-16a^2b^2c^4k_1k_3mnp^2-8a^2b^2c^4k_2k_3m^2np+16a^2b^2c^4k_2k_3mn^2p+16a^2b^2c^4k_2k_3mnp^2-8a^2b^2c^4k_2k_3n^2p^2-8a^2b^2c^4m^2n^2p^2-8a^2c^6k_1^2m^2n^2-8a^2c^6k_1^2m^2p^2+8a^2c^6k_1k_2m^2n^2+8a^2c^6k_1k_2mnp^2+4b^8k_3^2m^2p^2+4b^8k_3^2n^2p^2+b^8m^2n^2p^2+8b^6c^2k_1k_3m^2p^2+8b^6c^2k_1k_3mn^2p-8b^6c^2k_3^2m^2p^2-8b^6c^2k_3^2n^2p^2+4b^4c^4k_1^2m^2n^2+4b^4c^4k_1^2m^2p^2-16b^4c^4k_1k_3m^2p^2-16b^4c^4k_1k_3mn^2p+4b^4c^4k_3^2m^2p^2+4b^4c^4k_3^2n^2p^2-2b^4c^4m^2n^2p^2-8b^2c^6k_1^2m^2n^2-8b^2c^6k_1^2m^2p^2+8b^2c^6k_1k_3m^2p^2+8b^2c^6k_1k_3mn^2p+4c^8k_1^2m^2n^2+4c^8k_1^2m^2p^2+c^8m^2n^2p^2-4a^8k_2m^2n^2-4a^8k_2n^2p^2+12a^6b^2k_2m^2n^2-8a^6b^2k_2m^2np+4a^6b^2k_2n^2p^2-4a^6b^2k_3m^2p^2-4a^6b^2k_3n^2p^2-4a^6c^2k_1m^2n^2+4a^6c^2k_1m^2p^2-8a^6c^2k_1mnp^2+4a^6c^2k_2m^2n^2+4a^6c^2k_2n^2p^2-12a^4b^4k_2m^2n^2+16a^4b^4k_2m^2np+4a^4b^4k_2n^2p^2+4a^4b^4k_3m^2p^2+4a^4b^4k_3n^2p^2+4a^4b^2c^2k_1m^2n^2-16a^4b^2c^2k_1m^2np+4a^4b^2c^2k_1m^2p^2+16a^4b^2c^2k_1mn^2p-8a^4b^2c^2k_2m^2n^2+48a^4b^2c^2k_2mn^2p+16a^4b^2c^2k_2mnp^2-8a^4b^2c^2k_2n^2p^2-16a^4b^2c^2k_3m^2np+4a^4b^2c^2k_3m^2p^2+8a^4b^2c^2k_3mn^2p+16a^4b^2c^2k_3mnp^2-4a^4b^2c^2k_3n^2p^2+4a^4c^4k_1m^2n^2-12a^4c^4k_1m^2p^2+16a^4c^4k_1mnp^2+4a^4c^4k_2m^2n^2+4a^4c^4k_2n^2p^2+4a^2b^6k_2m^2n^2-8a^2b^6k_2m^2np-4a^2b^6k_2n^2p^2+4a^2b^6k_3m^2p^2+4a^2b^6k_3n^2p^2+4a^2b^4c^2k_1m^2n^2+16a^2b^4c^2k_1m^2np-4a^2b^4c^2k_1m^2p^2-16a^2b^4c^2k_1mn^2p+8a^2b^4c^2k_1mnp^2+4a^2b^4c^2k_2m^2n^2-16a^2b^4c^2k_2mn^2p+16a^2b^4c^2k_2mnp^2+4a^2b^4c^2k_2n^2p^2+16a^2b^4c^2k_3m^2np-8a^2b^4c^2k_3m^2p^2+48a^2b^4c^2k_3mnp^2-8a^2b^4c^2k_3n^2p^2-8a^2b^2c^4k_1m^2n^2+48a^2b^2c^4k_1m^2np-8a^2b^2c^4k_1m^2p^2+16a^2b^2c^4k_1mn^2p-4a^2b^2c^4k_2m^2n^2+8a^2b^2c^4k_2m^2np+16a^2b^2c^4k_2mn^2p-16a^2b^2c^4k_2mnp^2+4a^2b^2c^4k_2n^2p^2+16a^2b^2c^4k_3m^2np+4a^2b^2c^4k_3m^2p^2-16a^2b^2c^4k_3mnp^2+4a^2b^2c^4k_3n^2p^2+4a^2c^6k_1m^2n^2+12a^2c^6k_1m^2p^2-8a^2c^6k_1mnp^2-4a^2c^6k_2m^2n^2-4a^2c^6k_2n^2p^2-4b^8k_3m^2p^2-4b^8k_3n^2p^2-4b^6c^2k_1m^2n^2-4b^6c^2k_1m^2p^2+4b^6c^2k_3m^2p^2-8b^6c^2k_3mn^2p+12b^6c^2k_3n^2p^2+4b^4c^4k_1m^2n^2+4b^4c^4k_1m^2p^2+4b^4c^4k_3m^2p^2+16b^4c^4k_3mn^2p-12b^4c^4k_3n^2p^2+4b^2c^6k_1m^2n^2+4b^2c^6k_1m^2p^2-4b^2c^6k_3m^2p^2-8b^2c^6k_3mn^2p+4b^2c^6k_3n^2p^2-4c^8k_1m^2n^2-4c^8k_1m^2p^2+4a^8k_2^2n^2+a^8m^2n^2+a^8m^2p^2+a^8n^2p^2-8a^6b^2k_2^2n^2+8a^6b^2k_2k_3np-4a^6b^2m^2n^2+8a^6c^2k_1k_2mn-8a^6c^2k_2^2n^2-4a^6c^2m^2p^2+4a^4b^4k_2^2n^2-16a^4b^4k_2k_3np+4a^4b^4k_3^2p^2+6a^4b^4m^2n^2-2a^4b^4m^2p^2-2a^4b^4n^2p^2-16a^4b^2c^2k_1k_2mn+8a^4b^2c^2k_1k_3mp-8a^4b^2c^2k_2^2n^2-16a^4b^2c^2k_2k_3np+16a^4b^2c^2m^2np-16a^4b^2c^2mn^2p-16a^4b^2c^2mnp^2+4a^4b^2c^2n^2p^2+4a^4c^4k_1^2m^2-16a^4c^4k_1k_2mn+4a^4c^4k_2^2n^2-2a^4c^4m^2n^2+6a^4c^4m^2p^2-2a^4c^4n^2p^2+8a^2b^6k_2k_3np-8a^2b^6k_3^2p^2-4a^2b^6m^2n^2+8a^2b^4c^2k_1k_2mn-16a^2b^4c^2k_1k_3mp-16a^2b^4c^2k_2k_3np-8a^2b^4c^2k_3^2p^2-16a^2b^4c^2m^2np+4a^2b^4c^2m^2p^2+16a^2b^4c^2mn^2p-16a^2b^4c^2mnp^2-8a^2b^2c^4k_1^2m^2-16a^2b^2c^4k_1k_2mn-16a^2b^2c^4k_1k_3mp+8a^2b^2c^4k_2k_3np+4a^2b^2c^4m^2n^2-16a^2b^2c^4m^2np-16a^2b^2c^4mn^2p+16a^2b^2c^4mnp^2-8a^2c^6k_1^2m^2+8a^2c^6k_1k_2mn-4a^2c^6m^2p^2+4b^8k_3^2p^2+b^8m^2n^2+b^8m^2p^2+b^8n^2p^2+8b^6c^2k_1k_3mp-8b^6c^2k_3^2p^2-4b^6c^2n^2p^2+4b^4c^4k_1^2m^2-16b^4c^4k_1k_3mp+4b^4c^4k_3^2p^2-2b^4c^4m^2n^2-2b^4c^4m^2p^2+6b^4c^4n^2p^2-8b^2c^6k_1^2m^2+8b^2c^6k_1k_3mp-4b^2c^6n^2p^2+4c^8k_1^2m^2+c^8m^2n^2+c^8m^2p^2+c^8n^2p^2-4a^8k_2n^2+12a^6b^2k_2n^2-8a^6b^2k_2np-4a^6b^2k_3p^2+4a^6c^2k_1m^2-8a^6c^2k_1mn+4a^6c^2k_2n^2-12a^4b^4k_2n^2+16a^4b^4k_2np+4a^4b^4k_3p^2-12a^4b^2c^2k_1m^2+16a^4b^2c^2k_1mn+8a^4b^2c^2k_2n^2+16a^4b^2c^2k_2np-8a^4b^2c^2k_3mp+12a^4b^2c^2k_3p^2-12a^4c^4k_1m^2+16a^4c^4k_1mn+4a^4c^4k_2n^2+4a^2b^6k_2n^2-8a^2b^6k_2np+4a^2b^6k_3p^2+12a^2b^4c^2k_1m^2-8a^2b^4c^2k_1mn-12a^2b^4c^2k_2n^2+16a^2b^4c^2k_2np+16a^2b^4c^2k_3mp+8a^2b^4c^2k_3p^2+8a^2b^2c^4k_1m^2+16a^2b^2c^4k_1mn+12a^2b^2c^4k_2n^2-8a^2b^2c^4k_2np+16a^2b^2c^4k_3mp-12a^2b^2c^4k_3p^2+12a^2c^6k_1m^2-8a^2c^6k_1mn-4a^2c^6k_2n^2-4b^8k_3p^2-4b^6c^2k_1m^2-8b^6c^2k_3mp+12b^6c^2k_3p^2+4b^4c^4k_1m^2+16b^4c^4k_3mp-12b^4c^4k_3p^2+4b^2c^6k_1m^2-8b^2c^6k_3mp+4b^2c^6k_3p^2-4c^8k_1m^2+a^8m^2+a^8n^2+a^8p^2-4a^6b^2m^2-4a^6b^2n^2-4a^6c^2m^2-4a^6c^2p^2+6a^4b^4m^2+6a^4b^4n^2-2a^4b^4p^2+8a^4b^2c^2m^2-4a^4b^2c^2n^2-4a^4b^2c^2p^2+6a^4c^4m^2-2a^4c^4n^2+6a^4c^4p^2-4a^2b^6m^2-4a^2b^6n^2-4a^2b^4c^2m^2+8a^2b^4c^2n^2-4a^2b^4c^2p^2-4a^2b^2c^4m^2-4a^2b^2c^4n^2+8a^2b^2c^4p^2-4a^2c^6m^2-4a^2c^6p^2+b^8m^2+b^8n^2+b^8p^2-4b^6c^2n^2-4b^6c^2p^2-2b^4c^4m^2+6b^4c^4n^2+6b^4c^4p^2-4b^2c^6n^2-4b^2c^6p^2+c^8m^2+c^8n^2+c^8p^2+a^8-4a^6b^2-4a^6c^2+6a^4b^4+4a^4b^2c^2+6a^4c^4-4a^2b^6+4a^2b^4c^2+4a^2b^2c^4-4a^2c^6+b^8-4b^6c^2+6b^4c^4-4b^2c^6+c^8=0\)

TSC999

第 7 楼的那个最终公式，对否？我用下列数字验了一下，有些出入。
设B点位于坐标原点即B(0,0)，C点是(1,0), A点是(u,v).  u = 0.69639765; v = 0.59005161; k1 = 0.379938648; k2 = 0.272727567; k3 = 0.267042601;
Alpha = 40.68272°; Beta =  24.54524°; Gamma = 48.01613°; 算出△DEF面积是0.0384587，画图，从图上测量计算，面积是0.0385.
而用第 7 楼的公式，结果是0.0347322。二者相差约10%，不知何故？

我用 mathematica 得到的公式是

数学星空

楼上的结论太复杂，并且不太对称，正确的可能性不大

将楼上的参数代入我的模型得到相关参数：

\(a = 1, b = .6635776438, c = .9127598750, h = .5900516100, k1 = .379938648, k2 = .272727567, k3 = .267042601, m = .8596111095, n = .4566801724, p = 1.113193654, u = .6963976500, v = .3036023500\)

\(-612.1304944s^2-255.3815588s+10.69700904=0\) 求解得到：\(s=0.03835943372\)

mathematica

问题虽然复杂,但是不能引起别人的兴趣

creasson

记号:  $\theta  = \angle CDZ,\varphi  = \angle AEX,\phi  = \angle BFY$，并记\[\left\{ \begin{array}{l}
x = \frac{{DC\sin \left( {C - \varphi } \right) + CE\sin \varphi }}{{BC\sin \left( {C + \theta  - \varphi } \right)}}{e^{i\theta }} \\
y = \frac{{EA\sin \left( {A - \phi } \right) + AF\sin \phi }}{{CA\sin \left( {A + \varphi  - \phi } \right)}}{e^{i\varphi }} \\
z = \frac{{FB\sin \left( {B - \theta } \right) + BD\sin \theta }}{{AB\sin \left( {B + \phi  - \theta } \right)}}{e^{i\phi }} \\
\end{array} \right.\]
\[\left\{ \begin{array}{l}
{w_1} = \left( {x - \frac{{DC}}{{BC}}} \right){e^{iC}} + \left( {y + \frac{{CE}}{{CA}}} \right) \\
{w_2} = \left( {y - \frac{{EA}}{{CA}}} \right) + \left( {z + \frac{{AF}}{{AB}}} \right){e^{ - iA}} \\
\end{array} \right.\]
那么$XYZ$的面积\[{S_{XYZ}} = \frac{{A{C^2}}}{2}|{w_1}| \times |{w_2}| \times |\sin \left( {Arg{w_1} - Arg{w_2}} \right)|\]
此由我最近的几何理论得到，暂不发表。此式可做进一步化简，诸位可选一些特例验证此式的正确性。

数学星空

对于15＃的结果，我们可以取下列特例检验(结论不正确)

例一：

　　取\(\angle A=\alpha=\frac{\pi}{6},\angle B=\beta=\frac{\pi}{3},\angle C=\gamma=\frac{\pi}{2},b=\sqrt{3}a,c=2a\)

        根据15＃公式得：\(w_1=\frac{1-\sqrt{3}}{2}+\frac{\sqrt{3}}{6}Ｉ,w_2=\frac{2\sqrt{3}-3}{12}Ｉ+\frac{\sqrt{3}-1}{2}\)

        最终得到：\(S_{XYZ}=\frac{b^2}{2}|w_1w_2|\sin(Argw_1-Argw_2)=\frac{(15-7\sqrt{3})a^2}{16}\),

　　但是根据几何知识，我们可以得到：\(S_{XYZ}=\frac{S_{ABC}}{4}=\frac{\sqrt{3}a^2}{8}\)

数学星空

例二：我们取角度为弧度值

　\(A = 79.1066053466134, B = 59.9999999967723, C = 40.8933946469309, R = 1.527525232,a = 3., \alpha = 18.11851636, b = 2.64575131106459, \beta = 18.11851636, c = 2., \gamma = 18.11851636, h = 1.73205080756888, k_1 = .333333333333333, k_2 = .333333333333333, k_3 = .333333333333333, m = 1.73205080756888, n = 1.73205080756888, p = 1.73205080756888, u = 1., v = 2., x_1 = .166666666666667, x_2 = 0.833333333333333e-1, x_3 = .222222222222222, y_1 = .288675134594814, y_2 = .144337567297407, y_3 = .192450089729875\)

    我们可以算得\(S_{XYZ}=s\)满足方程：\(-2.383200502*10^{11}s-8.255648591*10^{10}s^2+1.91637680*10^9＝0\) 即\(s=0.008018914771\)

   而15＃计算的结果为：\(w_1 =0 .744590952277280+.230756975136404Ｉ, w_2 = .375264792217011-0.720464304922004\),\(S_{XYZ}=2.18066159923799\)

creasson

对16的验证暂不做论述，可能某个环节的确有问题，现给出化简后的结果：
记号\[\left\{ \begin{array}{l}
{R_1} = \frac{{DC \times \sin \left( {C - \varphi } \right) + CE \times \sin \varphi }}{{BC \times \sin \left( {C + \theta  - \varphi } \right)}} \\
{R_2} = \frac{{EA \times \sin \left( {A - \phi } \right) + AF \times \sin \phi }}{{CA \times \sin \left( {A + \varphi  - \phi } \right)}} \\
{R_3} = \frac{{FB \times \sin \left( {B - \theta } \right) + BD \times \sin \theta }}{{AB \times \sin \left( {B + \phi  - \theta } \right)}} \\
\end{array} \right.\]
那么有
\[\begin{array}{l}
{2*S_{XYZ}} = {R_2} \times {R_3} \times CA \times AB \times \sin \left( {A + \varphi  - \phi } \right) + {R_2} \times CA \times AF \times \sin \left( {A + \varphi } \right) + {R_3} \times CE \times AB \times \sin \left( {A - \phi } \right) + CE \times AF \times \sin A \\
     + {R_3} \times {R_1} \times BC \times AB \times \sin \left( {B + \phi  - \theta } \right) - {R_3} \times DC \times AB \times \sin \left( {B + \phi } \right) + {R_1} \times BC \times AF \times \sin \left( {B - \theta } \right) - DC \times AF \times \sin B \\
      + {R_1} \times {R_2} \times BC \times CA \times \sin \left( {C + \theta  - \varphi } \right) - {R_1} \times BC \times EA \times \sin \left( {C + \theta } \right) - {R_2} \times DC \times CA \times \sin \left( {C - \varphi } \right) + DC \times EA \times \sin C \\
      - {R_2} \times CA \times CA \times \sin \varphi  \\
\end{array}\]

creasson

以正三角形，取各边中点相连，上式验证OK，以直角三角形(三边分别为$1$,$sqrt(3)$,$2$)，取各边中点相连，上式验证也OK。在一般情形下，我相信它也能通过检验

creasson

如图，取$AB \to 2,AF \to 1,BC \to 2,BD \to 1,CA \to 2,CE \to 1,\varphi  \to \theta ,\phi  \to \theta$
上式恰好也给出$2S_{XYZ} = 2\sqrt 3 \cos ^2\theta $

取$AB \to 8,AF \to 3,BC \to 9,BD \to 4,CA \to 10,CE \to 5,\theta  \to \frac{\pi }{6},\varphi  \to \frac{\pi }{3},\phi  \to \frac{\pi }{2}$
得到\[2S_{XYZ} = \frac{{675\sqrt 3 \left( { - 155891 + 26623\sqrt {77} } \right)}}{{6450964}} = 14.086393777737645288\]