Saudi Arabia Mathematical Competitions 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 $$AN=\frac{cx}{b},MN=\frac{ax}{b}.(1)$$ The relation $BM\cdot CN\cdot BC=M{N}^3$ is equivalent to $$a(c-x)\left(b-\frac{cx}{b}\right)=\frac{a^3{x}^3}{b^3},$$ and hence we get $$a^2{x}^3-b^{2}c{x}^2+b^2(b^2+c^2)x-c{b}^4=0.(2)$$ Since $b^2+c^2=a^2$, it follows that $$a^2{x}^3-b^{2}c{x}^2+a^2{b}^{2}x-c{b}^4=0.(3)$$ The relation (3) is equivalent to $$(a^{2}x-b^{2}c)x^2+b^2(a^{2}x-b^{2}c)=0,$$ so we get $$(a^{2}x-b^{2}c)(x^2+b^2)=0.(4)$$ From (4) it follows $$x=\frac{b^{2}c}{a^2}(5)$$ Draw $AD\perp BC$, $D{M}^{\prime }\perp AB$, $D{N}^{\prime }\perp AC$, where $D\in BC$, $M^{\prime }\in AB$, $N^{\prime }\in AC$. We have $A{D}^2=A{M}^{\prime }\cdot AB$, so $$A{M}^{\prime }=\frac{A{D}^2}{AB}=\frac{b^2{c}^2}{a^{2}c}=\frac{b^{2}c}{a^2}=x=AM.$$ This implies $M=M^{\prime }$. Also, from $A{D}^2=A{N}^{\prime }\cdot AC$, we get $$A{N}^{\prime }=\frac{A{D}^2}{AC}=\frac{b^2{c}^2}{a^{2}b}=\frac{b{c}^2}{a^2}=\frac{cx}{b}=AN,\text{ hence }N=N^{\prime }.$$ 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.

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Alternative solution.

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 $BM\cdot CN\cdot BC=M{N}^3$ is equivalent to $$(b-m)(c-n)\sqrt{b^2+c^2}=(\sqrt{m^2+n^2}{)}^3(1)$$ 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 $$(b-m)\left(c-\frac{b}{c}m\right)\sqrt{b^2+c^2}={\left(\sqrt{m^2\left(1+\frac{b^2}{c^2}\right)}\right)}^3(2)$$ But the relation (2) is equivalent to $$(b-m)\left(c-\frac{b}{c}m\right)a=\frac{a^3}{c^3}{m}^3.(3)$$ Hence $$(m^2+c^2)(a^{2}m-b{c}^2)=0,$$ which gives $m=\frac{b{c}^2}{a^2}$. 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 $$mc+\frac{b^2}{c}m=bc,\text{ and hence }m=\frac{b{c}^2}{b^2+c^2}=\frac{b{c}^2}{a^2},$$ which is already proved.