What relationship exists between the sides $a, b, c$ of a triangle $(a<b<c)$, if the triangle constructed from the medians is similar to the original one? Choose $c$ such that if $a=47, b=65$. Can a smaller set of natural numbers that satisfy this relationship be found?
I. We denote the vertex opposite to the sides $a, b, c$ as $A$, $B$, and $C$ respectively, and the medians from these vertices as $s_{a}, s_{b}, s_{c}$ respectively.

Among the medians corresponding to two different sides, the one corresponding to the larger side is smaller. For example, if $aS B=\frac{2}{3} s_{b}, \quad s_{a}>s_{b}$,

According to the assumption, $s_{a}>s_{b}>s_{c}$. Therefore, the triangle formed by the medians can only be similar to the original triangle if

$$
a: b: c=s_{c}: s_{b}: s_{a}
$$

The following relationships are known between the sides and the medians:

$$
4 s_{a}^{2}=-a^{2}+2 b^{2}+2 c^{2}, \quad 4 s_{b}^{2}=2 a^{2}-b^{2}+2 c^{2}, \quad 4 s_{c}^{2}=2 a^{2}+2 b^{2}-c^{2}
$$

Squaring the ratio pair $b: c=s_{b}: s_{a}$ derived from (1), we get the following equality:

$$
\frac{b^{2}}{c^{2}}=\frac{s_{b}^{2}}{s_{a}^{2}}=\frac{4 s_{b}^{2}}{4 s_{a}^{2}}=\frac{2 a^{2}-b^{2}+2 c^{2}}{-a^{2}+2 b^{2}+2 c^{2}}
$$

Removing the fractions and rearranging in terms of powers of $b$,

$$
2 b^{4}+\left(3 c^{2}-a^{2}\right) b^{2}-2 c^{2}\left(c^{2}+a^{2}\right)=0
$$

This is a quadratic equation in $b^{2}$, and its discriminant is

$$
\left(3 c^{2}-a^{2}\right)^{2}+16 c^{2}\left(c^{2}+a^{2}\right)=25 c^{4}+10 a^{2} c^{2}+a^{4}=\left(5 c^{2}+a^{2}\right)^{2}
$$

Thus, taking only the positive solution for $b^{2}$,

$$
b^{2}=\frac{a^{2}-3 c^{2}+5 c^{2}+a^{2}}{4}=\frac{a^{2}+c^{2}}{2}
$$

With this root, according to (2),

$$
4 s_{a}^{2}=3 c^{2}, \quad 4 s_{b}^{2}=\frac{3}{2}\left(a^{2}+c^{2}\right)=3 b^{2}, \quad 4 s_{c}^{2}=3 a^{2}
$$

Thus, (1) indeed holds in this case. We have also found that the triangle formed by the medians is a $\sqrt{3} / 2$ scaled-down version of the original triangle.

Now, using (3) with $a=47$ and $b=65$, we get $c=79$, and these can indeed be the sides of a triangle.

II. From (3) and the fact that $a$, and thus $c+b>b+a$. Letting the numerator and denominator of the fraction in its simplest form be $u, v$ respectively, where $u>v$ and $u, v$ are coprime,

$$
(c+b) v=(b+a) u, \quad(b-a) v=(c-b) u
$$

Eliminating $a$ and $c$ from the two equations, we get the ratio of the two sides:

$$
\frac{b}{c}=\frac{u^{2}+v^{2}}{u^{2}+2 u v-v^{2}}, \quad \frac{b}{a}=\frac{u^{2}+v^{2}}{-u^{2}+2 u v+v^{2}}
$$

Thus,

$$
a=2 u v-u^{2}+v^{2}, \quad b=u^{2}+v^{2}, \quad c=2 u v+u^{2}-v^{2}
$$

is an integer solution to the problem and consists of coprime integers if one of $u$ and $v$ is even and the other is odd.

Agota Berecz (Mako, Jozsef A. g. III. o. t.)[^0]


[^0]:    ${ }^{1}$ See, for example, Exercise 752, K. M. L. 25 (1962/11) 149.o.