回復 1# Superconan 的帖子
填充第3題
這題應是求實數解，\(x,y,z\)同為正或同為負，先假設\(x,y,z>0\)
\(\begin{align}
& xy+yz+zx=1 \\
& x=\tan \frac{A}{2},y=\tan \frac{B}{2},z=\tan \frac{C}{2},A+B+C=\pi \\
& x+\frac{1}{x}=\frac{\sin \frac{A}{2}}{\cos \frac{A}{2}}+\frac{\cos \frac{A}{2}}{\sin \frac{A}{2}}=\frac{1}{\cos \frac{A}{2}\sin \frac{A}{2}}=\frac{2}{\sin A} \\
& y+\frac{1}{y}=\frac{2}{\sin B},z+\frac{1}{z}=\frac{2}{\sin C} \\
& \\
& 20\left( x+\frac{1}{x} \right)=21\left( y+\frac{1}{y} \right)=29\left( z+\frac{1}{z} \right) \\
& \frac{20}{\sin A}=\frac{21}{\sin B}=\frac{29}{\sin C} \\
& {{20}^{2}}+{{21}^{2}}={{29}^{2}} \\
& \sin A=\frac{20}{29},\sin B=\frac{21}{29},\sin C=1 \\
& x=\tan \frac{A}{2}=\frac{\sin A}{1+\cos A}=\frac{\frac{20}{29}}{1+\frac{21}{29}}=\frac{2}{5} \\
& y=\tan \frac{B}{2}=\frac{\sin B}{1+\cos B}=\frac{\frac{21}{29}}{1+\frac{20}{29}}=\frac{3}{7} \\
& z=\tan \frac{C}{2}=1 \\
& \left( x,y,z \right)=\left( \frac{2}{5},\frac{3}{7},1 \right) \\
\end{align}\)
由於\(x,y,z\)可以同為負，還有另一組解\(\left( x,y,z \right)=\left( -\frac{2}{5},-\frac{3}{7},-1 \right)\)
