Is there a right triangle in which the legs $a$ and $b$, and the height $m$ corresponding to the hypotenuse, satisfy the following relationship:

$$
m=\frac{1}{5} \sqrt{9 b^{2}-16 a^{2}}
$$
I. solution. To find the answer to the question, we add to the given equality the equalities that hold for every right-angled triangle:

$$
\begin{aligned}
& a^{2}+b^{2}=c^{2} \\
& a b=c m=2 t
\end{aligned}
$$

(where $c$ is the hypotenuse and $t$ is the area of the triangle), and solve the resulting system of 4 unknowns. If there is a solution, then the answer is yes. It is foreseeable, however, that we cannot obtain a definite solution, since we can only eliminate 2 unknowns. It is expected, however, that we can determine the ratio of the unknowns, the shape of the triangle. Indeed, in (1)-(3) there is no term without an unknown, and after squaring (1), every term in our equations has the same degree (in the usual expression: each equation is homogeneous). From (1)

$$
25 m^{2}=9 b^{2}-16 a^{2}
$$

Let's eliminate $m$ and $c$ from the system. From (3), considering (2),

$$
m^{2}=\frac{a^{2} b^{2}}{c^{2}}=\frac{a^{2} b^{2}}{a^{2}+b^{2}}
$$

and thus from $\left(1^{\prime}\right)$, by cross-multiplying,

$$
\begin{gathered}
25 a^{2} b^{2}=\left(a^{2}+b^{2}\right)\left(9 b^{2}-16 a^{2}\right)=-16 a^{4}-7 a^{2} b^{2}+9 b^{4} \\
16 a^{4}+32 a^{2} b^{2}-9 b^{4}=0
\end{gathered}
$$

Adding $25 b^{4}$ to both sides, the left side becomes a perfect square, specifically a positive value squared: $\left(4 a^{2}+4 b^{2}\right)^{2}=25 b^{4}$, so $4 a^{2}+4 b^{2}=5 b^{2}$, and from this $4 a^{2}=b^{2}$, and since $a$ and $b$ are both positive, $2 a=b, b: a=2$.

Accordingly, there is a triangle with the desired property.

Zsuzsa Torda (Budapest, Berzsenyi D. High School I. o. t.)

II. solution. We can characterize the triangle in a different form if we add to (1) the equations that are essentially equivalent to (2) and (3):

$$
m^{2}=c_{1} c_{2}, \quad a^{2}=c c_{1}, \quad b^{2}=c c_{2}, \quad c_{1}+c_{2}=c
$$

Here $c_{1}$ and $c_{2}$ are the projections of $a$ and $b$ on the hypotenuse. Thus, through more equations, but more simply, just by solving a quadratic equation, we arrive at the result. Based on our first three auxiliary equations, we eliminate the quadratic terms in (1'), and then $c^{2}$ based on the fourth:

$$
\begin{aligned}
& 25 c_{1} c_{2}=9 c c_{2}-16 c c_{1} \\
& 25 c_{1}\left(c-c_{1}\right)=9 c\left(c-c_{1}\right)-16 c c_{1} \\
& 25 c_{1}^{2}-50 c c_{1}+9 c^{2}=0, \quad \text { from which } \quad\left(5 c_{1}-5 c\right)^{2}=4 c^{2}
\end{aligned}
$$

and since $c_{1}<c$, then $-5 c_{1}+5 c=4 c, c_{1}=c / 5$. Then $c_{2}=4 c / 5$, so (1) holds if $m$ divides the hypotenuse into two parts in the ratio $1: 4$.

László Tihanyi (Budapest, Petőfi S. High School I. o. t.)