, 2012

Let $BC = a$, $CA = b$, $AB = c$, $AM = x$.

Triangles $AMN$ and $ACB$ are similar, so
$$\frac{x}{b} = \frac{AN}{c} = \frac{MN}{a}.$$
We obtain
$$AN = \frac{cx}{b}, \quad MN = \frac{ax}{b}. \qquad (1)$$
The relation $BM \cdot CN \cdot BC = MN^3$ is equivalent to
$$a(c-x)\left(b-\frac{cx}{b}\right) = \frac{a^3x^3}{b^3},$$
and hence we get
$$a^2x^3 - b^2cx^2 + b^2(b^2 + c^2)x - cb^4 = 0. \qquad (2)$$
Since $b^2 + c^2 = a^2$, it follows that
$$a^2x^3 - b^2cx^2 + a^2b^2x - cb^4 = 0. \qquad (3)$$
The relation (3) is equivalent to
$$(a^2x - b^2c)x^2 + b^2(a^2x - b^2c) = 0,$$
so we get
$$(a^2x - b^2c)(x^2 + b^2) = 0. \qquad (4)$$
From (4) it follows
$$x = \frac{b^2c}{a^2} \qquad (5)$$
Draw $AD \perp BC$, $DM' \perp AB$, $DN' \perp AC$, where $D \in BC$, $M' \in AB$, $N' \in AC$. We have $AD^2 = AM' \cdot AB$, so
$$AM' = \frac{AD^2}{AB} = \frac{b^2c^2}{a^2c} = \frac{b^2c}{a^2} = x = AM.$$
This implies $M = M'$. Also, from $AD^2 = AN' \cdot AC$, we get
$$AN' = \frac{AD^2}{AC} = \frac{b^2c^2}{a^2b} = \frac{bc^2}{a^2} = \frac{cx}{b} = AN, \text{ hence } N = N'.$$
Therefore $AMDN$ is a rectangle. It follows that the symmetric point of $A$ with respect to the midpoint of segment $MN$ is $D$, which completes the proof.

Consider the coordinates system with the origin at $A$ and $AB$ and $AC$ are the coordinate axes. We have $A(0, 0)$, $B(b, 0)$, $C(0, c)$, $M(m, 0)$, $N(0, n)$.

The relation $BM \cdot CN \cdot BC = MN^3$ is equivalent to
$$(b-m)(c-n)\sqrt{b^2+c^2} = (\sqrt{m^2+n^2})^3 \quad (1)$$
The quadrilateral $CNMB$ is cyclic. Hence, from the power of $A$ with respect to its circumcircle we get $nc = mb$. That is, $n = \frac{b}{c}m$. Replacing in (1) we get
$$(b-m) \left(c - \frac{b}{c}m\right) \sqrt{b^2+c^2} = \left(\sqrt{m^2\left(1+\frac{b^2}{c^2}\right)}\right)^3 \quad (2)$$
But the relation (2) is equivalent to
$$(b-m) \left(c - \frac{b}{c}m\right) a = \frac{a^3}{c^3}m^3. \quad (3)$$
Hence
$$(m^2 + c^2)(a^2m - bc^2) = 0,$$
which gives $m = \frac{bc^2}{a^2}$. The symmetric point $A_1$ of $A$ with respect to the midpoint of segment $MN$ has the coordinates $(m, n)$. But $A_1 \in BC$ if and only if $\frac{m}{b} + \frac{n}{c} = 1$, which is equivalent to $mc + nb = bc$. The last relation is equivalent to
$$mc + \frac{b^2}{c}m = bc, \text{ and hence } m = \frac{bc^2}{b^2 + c^2} = \frac{bc^2}{a^2},$$
which is already proved.