Show there are 1977 non-similar triangles such that$\frac{\sin X+\sin Y+\sin Z}{\cos X+\cos Y+\cos Z}=\frac{12}7$and$\sin X\sin Y\sin Z=\frac{12}{25}$

Show that there are 1977 non-similar triangles whose angles $X$, $Y$, $Z$ satisfy the conditions $$\begin{align} \frac{\sin{X}+ \sin{Y}+ \sin{Z}}{\cos{X}+\cos{Y}+ \cos{Z}}=\frac{12}{7} \tag1\\[4pt] \sin{X}\sin{Y} \sin{Z}=\frac{12}{25} \tag2 \end{align}$$

My attempt: $$\begin{align} \sin{X}+ \sin{Y}+ \sin{Z}=4\cos{\frac{X}{2}} \cos{\frac{Y}{2}} \cos{\frac{Z}{2}} \\ \cos{X}+ \cos{Y}+ \cos{Z}=1+4\sin{\frac{X}{2}} \sin{\frac{Y}{2}} \sin{\frac{Z}{2}}\\ X+Y+Z=\pi \\ \sin{X} \sin{Y}\sin{Z}=8\cos{\frac{X}{2}} \cos{\frac{Y}{2}} \cos{\frac{Z}{2}}\sin{\frac{X}{2}} \sin{\frac{Y}{2}} \sin{\frac{Z}{2}}=\frac{12}{25}\end{align}$$ Here after what to do to find the required results.

How to show such points which satisfy these conditions?

Please, help. Thanks in advance.

Solution 1:

Using standard notation,
given $\triangle ABC$ with angles $\alpha,\beta,\gamma$, side lengths $a,b,c$, semiperimeter $\rho$, radius $r$ of the inscribed circle, and radius $R$ of the circumscribed circle,

\begin{align} \frac{\sin\alpha+ \sin\beta+ \sin\gamma} {\cos\alpha+ \cos\beta+ \cos\gamma} &=\frac{12}{7} \tag{1}\label{1} ,\\ \sin\alpha\sin\beta\sin\gamma &=\frac{12}{25} \tag{2}\label{2} . \end{align}

Using known identities,

\begin{align} \sin\alpha+ \sin\beta+ \sin\gamma &= \frac\rho R=u \tag{3}\label{3} ,\\ \cos\alpha+ \cos\beta+ \cos\gamma &= \frac rR+1=v+1 \tag{4}\label{4} ,\\ \sin\alpha\sin\beta\sin\gamma &= \frac{\rho r}{2R^2}=\tfrac12\,uv \tag{5}\label{5} , \end{align}

we ca rewrite \eqref{1}-\eqref{2} in terms of parameters $u=\rho/R,\,v=r/R$ as

\begin{align} \frac u{v+1}&=\frac{12}{7} \tag{6}\label{6} ,\\ \tfrac12\,uv &=\frac{12}{25} \tag{7}\label{7} . \end{align}

The system \eqref{6}-\eqref{7} has just two solutions,

\begin{align} u &= -\frac{24}{35},\quad v = -\frac75 \tag{8}\label{8} ,\\ u &= \frac{12}5,\quad v =\frac25 \tag{9}\label{9} , \end{align}

and obviously, only positive is valid, so, there is only one type of triangle with given properties.

Solution of the cubic equation \begin{align} x^3-2u\,x^2+(u^2+v^2+4v)\,x-4uv &=0 \tag{10}\label{10} ,\\ x^3-\frac{24}5\,x^2+\frac{188}{25}\,x-\frac{96}{25} &=0 \tag{11}\label{11} \end{align}

gives a unique triplet of side lengths of triangle with $R=1$, which satisfies \eqref{1} and \eqref{2}

\begin{align} a&=\frac65,\quad b=\frac85,\quad c=2 \tag{12}\label{12} . \end{align}

As we can see, this triangle is similar to the famous $3-4-5$ right-angled triangle.

Indeed, we have \begin{align} \sin\alpha&=\frac35,\quad\sin\beta=\frac45,\quad\sin\gamma=1 \tag{13}\label{13} ,\\ \cos\alpha&=\frac45,\quad\cos\beta=\frac35,\quad\cos\gamma=0 \tag{14}\label{14} , \end{align}

\begin{align} \sin\alpha+\sin\beta+\sin\gamma &= \frac{12}5 \tag{15}\label{15} ,\\ \cos\alpha+ \cos\beta+ \cos\gamma &= \frac75 \tag{16}\label{16} ,\\ \frac{\sin\alpha+ \sin\beta+ \sin\gamma} {\cos\alpha+ \cos\beta+ \cos\gamma} &=\frac{12}{7} \tag{17}\label{17} ,\\ \sin\alpha\sin\beta\sin\gamma &= \frac{12}{25} \tag{18}\label{18} . \end{align}