# Equations

#### sbhatnagar

##### Active member
Find $$x, \ y, \ z \ \in \mathbb{R}$$ satisfying

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

#### Opalg

##### MHB Oldtimer
Staff member
Find $$x, \ y, \ z \ \in \mathbb{R}$$ satisfying

$\frac{4\sqrt{x^2+1}}{x}=\frac{5\sqrt{y^2+1}}{y}= \frac{6\sqrt{z^2+1}}{z} \ \ \ \text{and} \ \ \ xyz=x+y+z$
Numerically, there is a solution somewhere close to the point given by $$x=0.882,\quad y=1.470,\quad z=7.935.$$ But I don't see any analytic way to get an exact answer, and I doubt whether this problem has a neat solution.

#### mfb

##### New member
With x=sinh(a), y=sinh(b), z=sinh(c), you can get rid of the roots:

tanh(a)/4 = tanh(b)/5 = tanh(c)/6
sinh(a) sinh(b) sinh(c) = sinh(a) + sinh(b) + sinh(c)

However, I do not know how to solve this.

and I doubt whether this problem has a neat solution.
Me too.

#### sbhatnagar

##### Active member
The problem can be solved analytically and the answers are

$x=\frac{\sqrt{7}}{3}, \ y= \frac{5\sqrt{7}}{9} , \ z=3\sqrt{7}$

#### sbhatnagar

##### Active member
Hint

Put $$x=\tan \alpha , \ \ y= \tan \beta , \ \ z=\tan \gamma$$, $$\dfrac{-\pi}{2} < \alpha , \ \beta , \ \gamma < \dfrac{+\pi}{2}$$.

#### sbhatnagar

##### Active member
Solution

Let $$x=\tan \alpha , \ y=\tan \beta , \ z= \tan \gamma$$ such that $$\dfrac{-\pi}{2}<\alpha , \beta , \gamma < \dfrac{+\pi}{2}$$

$\frac{4\sqrt{\tan^2 \alpha + 1}}{\tan \alpha} = \frac{5\sqrt{\tan^2 \beta + 1}}{\tan \beta} = \frac{6\sqrt{\tan^2 \gamma + 1}}{\tan \gamma}$

$\Rightarrow \ \frac{4}{\sin \alpha}= \frac{5}{\sin \beta}=\frac{6}{\sin \gamma}$

Again

\begin{aligned} \tan \alpha \tan \beta \tan \gamma &=\tan \alpha + \tan \beta +\tan \gamma \\ \tan \alpha (\tan \beta \tan \gamma -1) &=\tan\beta + \tan \gamma \\ -\tan \alpha &= \frac{\tan\beta + \tan \gamma }{1- \tan\beta \tan \gamma } \\ -\tan \alpha &= \tan(\beta +\gamma) \\ \tan(k\pi - \alpha) &= \tan(\beta + \gamma) \\ \alpha +\beta +\gamma &=k\pi\end{aligned}

Taking $$k=1$$ we get $$\alpha +\beta +\gamma =\pi$$ which implies that there exists a triangle whose angles are $$\alpha , \ \beta , \ \gamma$$ and whose sides opposite to these angles are proportional to 4,5 and 6 respectively.

Let the sides of such a triangle be $$4k, \ 5k, \ 6k$$.

If $$s$$ is the semiperimeter of the triangle then

$s=\frac{15k}{2}$

$\tan \frac{\alpha}{2}=\sqrt{\frac{(s-4k)(s-6k)}{s(s-5k)}}=\sqrt{\frac{\dfrac{5k}{2} \times \dfrac{3k}{2}}{\dfrac{15k}{2} \times \dfrac{7k}{2}}}=\sqrt{\frac{1}{7}}$

$x = \tan \alpha=\frac{2\tan \dfrac{\alpha}{2}}{1-\tan^2 \dfrac{\alpha}{2}}=\dfrac{2\sqrt{\dfrac{1}{7}}}{1-\dfrac{1}{7}}=\frac{\sqrt{7}}{3}$

Similarly, $$y=\dfrac{5\sqrt{7}}{9}$$ and $$z=3\sqrt{7}$$.

#### Sudharaka

##### Well-known member
MHB Math Helper
The problem can be solved analytically and the answers are

$x=\frac{\sqrt{7}}{3}, \ y= \frac{5\sqrt{7}}{9} , \ z=3\sqrt{7}$
Hi sbhatnagar,

Note that, $$x=-\dfrac{\sqrt{7}}{3},\,y=-\dfrac{5\sqrt{7}}{9}\mbox{ and }z=-3\sqrt{7}$$ is another solution.

Kind Regards,
Sudharaka.