Solve the equation

$$
x^{2}+\left(\frac{5 x}{x+5}\right)^{2}=1
$$
I. Solution. The fraction on the left side of the equation can be transformed so that the unknown appears only once:

$$
\frac{5 x}{x+5}=\frac{5}{1+\frac{5}{x}}
$$

(Clearly, $x=0$ is not a solution.) Of course, $x$ can also be written in a similar form:

$$
x=\frac{5}{\frac{5}{x}}
$$

If we now apply the substitution $y=0.5+\frac{5}{x}$, the equation becomes symmetric:

$$
\frac{25}{(y-0.5)^{2}}+\frac{25}{(y+0.5)^{2}}=1
$$

This equation, when multiplied by the common denominator $\left(y^{2}-0.25\right)^{2}$, becomes quadratic in $y^{2}$ (while in the original equation, the odd-degree terms remain after multiplication):

$$
50\left(y^{2}+0.25\right)=y^{4}-0.5 y^{2}+0.25^{2}
$$

From this, we arrive at the equation

$$
16 y^{4}-808 y^{2}-199=0
$$

whose positive root is

$$
y^{2}=\frac{101+20 \sqrt{26}}{4}
$$

From this, we get two solutions:

$$
x_{1}=\frac{10}{\sqrt{101+20 \sqrt{26}}-1} \approx 0.7549 \text { and } x_{2}=\frac{10}{-\sqrt{101+20 \sqrt{26}}-1} \approx-0.6559
$$

The transformations are reversible throughout, so these values are indeed solutions.

II. Solution. We will solve the more general equation $x^{2}+\left(\frac{a x}{x+a}\right)^{2}=1$ for the case $|a| \geq 1$. (The case $|a|<1$ is a bit more complicated, but we don't need it: in the problem, $a=5$.)

Consider the function $x \mapsto x^{2}+\left(\frac{a x}{x+a}\right)^{2}$; for $|x| \geq 1$, it takes values greater than one, while at $x=0$ its value is zero. Thus, the equation has at least one root in the intervals $(-1,0)$ and $(0,1)$, and no roots elsewhere. (We used the fact that our function is continuous). Our equation precisely means that for some $\alpha$, $x=\sin \alpha$ and $\frac{a x}{x+a}=\cos \alpha$. Substituting $x=\sin \alpha$ into the fraction $\frac{a x}{x+a}$ and rearranging, we get the equation

$$
a(-\cos \alpha+\sin \alpha)=\sin \alpha \cos \beta
$$

Squaring this and using the identity $\sin ^{2} \alpha+\cos ^{2} \alpha=1$, we get a quadratic equation in $u=\sin \alpha \cos \alpha$: $u^{2}+2 u a^{2}-a^{2}=0$, i.e., $\sin \alpha \cos \alpha=u=-a^{2} \pm \sqrt{a^{4}+a^{2}}$. It is known that

$$
|\sin \alpha \cos \alpha|=\left|\frac{\sin 2 \alpha}{2}\right| \leq \frac{1}{2}
$$

Moreover,

$$
-a^{2}-\sqrt{a^{4}+a^{2}}<-2 a^{2} \leq-2, \quad \text { if } \quad|a| \geq 1
$$

Therefore, only the root $u=-a^{2}+\sqrt{a^{4}+a^{2}}$ provides a solution for $\alpha$. In the following, we calculate the values of $\sin \alpha+\cos \alpha$ and $\sin \alpha-\cos \alpha$. According to (1),

$$
-\cos \alpha+\sin \alpha=-a+e \sqrt{1+a^{2}} \quad\left(e=\frac{|a|}{a}=1 \quad \text { or } \quad-1\right) .
$$

On the other hand, $(\cos \alpha+\sin \alpha)^{2}=1+2 u=1-2 a^{2}+2|a| \sqrt{1+a^{2}}$, so

$$
\cos \alpha+\sin \alpha= \pm \sqrt{\left(1-2 a^{2}+2|a| \sqrt{1+a^{2}}\right)}
$$

Adding the two equations and dividing by two:

$$
x=\sin \alpha=\frac{e \sqrt{1+a^{2}}-a \pm \sqrt{1-2 a^{2}+2|a| \sqrt{1+a^{2}}}}{2}
$$

(The numbers under the square roots are positive.) At most, these two values can be roots. On the other hand, we have seen that the equation has at least two roots, so the above two solutions are the roots of the equation and there are no more.

$$
x_{1,2}=\frac{\sqrt{26}-5 \pm \sqrt{10 \sqrt{26}-49}}{2}
$$

András Boncz (Zalaegerszeg, Zrínyi M. Gymnasium, 2nd grade)

Remarks: 1. Of course, the idea of the first solution can also be applied to the general case.

2. The two solutions give different forms of the result, but it can be directly verified that these are equal. Indeed,

$$
10 \sqrt{26}-49=2-(\sqrt{26}-5)^{2}=\frac{2(\sqrt{26}+5)^{2}-1}{(\sqrt{26}-5)^{2}}
$$

Thus, for example,

$$
\begin{gathered}
x_{1}=\frac{\sqrt{26}-5+\sqrt{10 \sqrt{26}-49}}{2}=\frac{1+\sqrt{2(\sqrt{26}+5)^{2}-1}}{2(\sqrt{26}+5)}= \\
=\frac{1+\sqrt{101+20 \sqrt{26}}}{2(\sqrt{26}+5)}=\frac{100+20 \sqrt{26}}{2(\sqrt{26}+5)} \cdot \frac{1}{\sqrt{101+20 \sqrt{26}}-1}=\frac{10}{\sqrt{101+20 \sqrt{26}-1}}
\end{gathered}
$$