In a non-isosceles triangle with side lengths $a, b, c$ being integers, the following relationship holds. What is the smallest height of the triangle?

$$
\frac{a^{2}}{c}-(a-c)^{2}=\frac{b^{2}}{c}-(b-c)^{2}
$$
Let's transform the given relationship as follows:

$$
a^{2}-b^{2}=c\left[(a-c)^{2}-(b-c)^{2}\right]=c(a-b)(a+b-2 c)
$$

Given the assumption $a \neq b$, dividing by $a-b$ yields

$$
a+b=c(a+b)-2 c^{2}
$$

and from this

$$
a+b=\frac{2 c^{2}}{c-1}=\frac{2[(c-1)+1]^{2}}{c-1}=2(c-1)+4+\frac{2}{c-1}=2 c+2+\frac{2}{c-1}
$$

This equality can only hold with integer values of $a, b, c$ if $c-1$ is a divisor of 2, meaning $c-1$ is either 1 or 2.

In the first case, $c=2$ and from (1) $a+b=8$. On the other hand, from the triangle inequality $|a-b|<c$, and since $a+b=8$, we have $a-b<8-2=6$ and $a-b>2-8=-6$. Therefore, $|a-b|<6$.

In the second case, $c=3$ and from (1) $a+b=12$. Similarly, from the triangle inequality $|a-b|<c$, and since $a+b=12$, we have $a-b<12-3=9$ and $a-b>3-12=-9$. Therefore, $|a-b|<9$.

Thus, the triangle is uniquely determined. It is well-known that a triangle with sides of 3, 4, and 5 units is a right triangle (an "Egyptian triangle"), with an area of 6 square units from the legs $c=3$ and $b=4$, and thus the smallest height corresponding to the hypotenuse $a=5$ is $12 / 5=2.4$ units.

Sebestyén Zoltán (Celldömölk, Berzsenyi D. Gymnasium II. grade)

Note. The possible values for the side $c$ can also be obtained from (1) in the form $a+b=2 c^{2} /(c-1)$. Indeed, $c^{2}=c c$ and $c-1$ are relatively prime, so $c-1$ must be a divisor of 2.

Farkas Zoltán (Hódmezővásárhely, Bethlen G. Gymnasium II. grade)