Joshua Foley

Answered

2022-07-04

Consider the system of equations in real numbers $\text{}x,y,z\text{}$ satisfying

$\frac{x}{4\text{}\sqrt{{x}^{2}+1}}\text{}=\text{}\frac{y}{5\text{}\sqrt{{y}^{2}+1}}\text{}=\text{}\frac{z}{6\text{}\sqrt{{z}^{2}+1}}$

and $\text{}x+y+z\text{}=\text{}xyz\text{}$. Find $\text{}x,y,z\text{}.$.

I substituted $\text{}x,y,z\text{}$ as $\text{}\mathrm{tan}({\theta}_{1})\text{},\text{}\mathrm{tan}({\theta}_{2})\text{},\text{}\mathrm{tan}({\theta}_{3})\text{}$, where

${\theta}_{1}\text{}+\text{}{\theta}_{2}\text{}+\text{}{\theta}_{3}\text{}=\text{}180\xba\text{}\text{}.$

$\frac{x}{4\text{}\sqrt{{x}^{2}+1}}\text{}=\text{}\frac{y}{5\text{}\sqrt{{y}^{2}+1}}\text{}=\text{}\frac{z}{6\text{}\sqrt{{z}^{2}+1}}$

and $\text{}x+y+z\text{}=\text{}xyz\text{}$. Find $\text{}x,y,z\text{}.$.

I substituted $\text{}x,y,z\text{}$ as $\text{}\mathrm{tan}({\theta}_{1})\text{},\text{}\mathrm{tan}({\theta}_{2})\text{},\text{}\mathrm{tan}({\theta}_{3})\text{}$, where

${\theta}_{1}\text{}+\text{}{\theta}_{2}\text{}+\text{}{\theta}_{3}\text{}=\text{}180\xba\text{}\text{}.$

Answer & Explanation

toriannucz

Expert

2022-07-05Added 16 answers

to any values $a,b,c$ instead of $\frac{1}{4},\frac{1}{5},\frac{1}{6}$

If we put

$\begin{array}{}\text{(1)}& \varphi (a,b,c)=ab+ac-bc,\text{}\psi (a,b,c)=ab+ac+bc\end{array}$

then LAcarguy’s method leads to the following explicit formulas for the solutions :

$\begin{array}{}\text{(3)}& x=\frac{\epsilon}{\varphi ({a}^{2},{b}^{2},{c}^{2})}\sqrt{G(a,b,c)},\text{}y=\frac{\epsilon}{\varphi ({b}^{2},{c}^{2},{a}^{2})}\sqrt{G(a,b,c)},\text{}z=\frac{\epsilon}{\varphi ({c}^{2},{a}^{2},{b}^{2})}\sqrt{G(a,b,c)}\end{array}$

where $\epsilon $ is $+1$ or $-1$. In your initial question where $a=\frac{1}{4},b=\frac{1}{5},c=\frac{1}{6}$, one finds $G=\frac{7}{921600}$,

$\begin{array}{}\text{(4)}& x=\epsilon \frac{\sqrt{7}}{3},\text{}y=\epsilon \frac{5\sqrt{7}}{9},\text{}z=3\epsilon \sqrt{7}\end{array}$

If we put

$\begin{array}{}\text{(1)}& \varphi (a,b,c)=ab+ac-bc,\text{}\psi (a,b,c)=ab+ac+bc\end{array}$

then LAcarguy’s method leads to the following explicit formulas for the solutions :

$\begin{array}{}\text{(3)}& x=\frac{\epsilon}{\varphi ({a}^{2},{b}^{2},{c}^{2})}\sqrt{G(a,b,c)},\text{}y=\frac{\epsilon}{\varphi ({b}^{2},{c}^{2},{a}^{2})}\sqrt{G(a,b,c)},\text{}z=\frac{\epsilon}{\varphi ({c}^{2},{a}^{2},{b}^{2})}\sqrt{G(a,b,c)}\end{array}$

where $\epsilon $ is $+1$ or $-1$. In your initial question where $a=\frac{1}{4},b=\frac{1}{5},c=\frac{1}{6}$, one finds $G=\frac{7}{921600}$,

$\begin{array}{}\text{(4)}& x=\epsilon \frac{\sqrt{7}}{3},\text{}y=\epsilon \frac{5\sqrt{7}}{9},\text{}z=3\epsilon \sqrt{7}\end{array}$

Ayaan Barr

Expert

2022-07-06Added 6 answers

Starting with $\frac{sint}{4}=\frac{sinu}{5}=\frac{sinw}{6}$, then use the law of sines to get: $\frac{a}{4}=\frac{b}{5}=\frac{c}{6}$. So $a=\frac{2c}{3}$, $b=\frac{5c}{6}$. So apply the law of cosines to have: ${a}^{2}={b}^{2}+{c}^{2}-2bccosA$. Substituting these values of $a$ and $b$ into the above equation we can solve for $cosA$, and then $tanB$, and $tanA$, and $tanC$.

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