## Task 2 - 201222

Determine all positive real numbers $k$ for which the numbers

$$
a=\frac{2 k}{k+1}, \quad b=\frac{k+1}{2}, \quad c=\sqrt{k}
$$

are the measures of the (measured with the same unit) side lengths of a right triangle.
For every positive real $k$ it holds that $(k+1)^{2}=(k-1)^{2}+4 k \geq 4 k$, thus $\frac{k+1}{2} \geq \frac{2 k}{k+1}$ and $\frac{k+1}{2} \geq \sqrt{k}$.

If now $k$ is a positive real number for which the numbers

$$
a=\frac{2 k}{k+1} ; \quad b=\frac{k+1}{2} ; \quad c=\sqrt{k}
$$

are the measures of the side lengths of a right-angled triangle, then it follows:

It holds that $b \geq a$ and $b \geq c$, so $b$ is the measure of the hypotenuse length and by the Pythagorean theorem:

$$
\begin{aligned}
\frac{(k+1)^{2}}{4} & =\frac{4 k^{2}}{(k+1)^{2}}+k \\
(k+1)^{4} & =16 k^{2}+4 k(k+1)^{2} \\
(k+1)^{2}\left((k+1)^{2}-4 k\right) & =16 k^{2} \\
(k+1)^{2}(k-1)^{2}-16 k^{2} & =0 \\
\left(k^{2}-4 k-1\right)\left(k^{2}+4 k-1\right) & =0
\end{aligned}
$$

One of the two brackets must be zero, from which the following solutions result:

$$
k= \pm 2 \pm \sqrt{5}
$$

Since $\sqrt{5}>2$, but $k>0$ should hold, the sign in front of the $\sqrt{5}$ must definitely be +, and thus follows

$$
k=\sqrt{5} \pm 2
$$

Conversely, these two values satisfy the given equations; therefore, they have the required property according to the converse of the Pythagorean theorem.

Thus, exactly the two numbers given in (1) are the sought ones.

Solution adopted from [5] and improved by MontyPythagoras

## 2.Solution:

The ratio of the side lengths in a right-angled triangle is defined by the Pythagorean theorem. From the formulas $a(k), b(k), c(k)$ given in the problem text, we get $a^{2}=\frac{4 k^{2}}{k^{2}+2 k+1}$ for $b^{2}=\frac{k^{2}+2 k+1}{4}$ and for $c^{2}=k$.

After rearrangement, it follows that

$$
\frac{1}{a^{2}}=\frac{k^{2}+2 k+1}{4 k^{2}}
$$

and with the ratio

$$
\frac{1}{a^{2}}=\frac{b^{2}}{k^{2}}
$$

the quadratic equation for $k$

$$
\begin{gathered}
k^{2}-k a^{2}-a^{4}=0 \\
k_{1,2}=\frac{a^{2}}{2} \pm \sqrt{\left(\frac{a^{2}}{2}\right)^{2}+a^{4}} \\
k_{1,2}=\frac{a^{2}}{2} \pm \sqrt{\frac{a^{4}}{4}+a^{4}} \\
k_{1}=\frac{a^{2}}{2}(1+\sqrt{5}) \\
k_{2}=\frac{a^{2}}{2}(1-\sqrt{5})
\end{gathered}
$$

Since $k_{2}<0$, this solution is discarded as it does not meet the conditions of the problem statement. By substituting ( $k=k_{1}$ ), we get

$$
a^{2}=\frac{2 k}{1+\sqrt{5}}=\frac{4 k^{2}}{k^{2}+2 k+1}
$$

and after further calculation

$$
\begin{gathered}
k^{2}-2 k \sqrt{5}+1=0 \\
k_{3}=\sqrt{5}+2 \\
k_{4}=\sqrt{5}-2
\end{gathered}
$$

the two numerical values $k_{3}, k_{4}$, which, when checked, are also the solution values of the problem statement.