Belarusian Mathematical Olympiad

Let the abscissae of the points $A,B,C,D$ be $a,b,c,d$ respectively. Since $AC$ is a diameter of the given circle, $M$ is its center. Let $(p,q)$ be the coordinates of $M$ then the equation of the circle is $(x-p{)}^2+(y-q{)}^2=R^2$ where $R$ denotes its radius.

Coordinates $(x;y)$ of any point $A,B,C,D$ is the solution of the system of equations $$\{\begin{array}{l}(x-p{)}^2+(y-q{)}^2=R^2,\\ y=x^2.\end{array}$$ Therefore, the abscissae $a,b,c,d$ of these points are the solutions of the equation $(x-p{)}^2+(x^2-q{)}^2-R^2=0$ which is equivalent to $x^4+(1-2q)x^2-2px+p^2+q^2-R^2=0$. The left side of this equation is a polynomial of degree 4, for which 4 roots are known: $a,b,c,d$. Write Viète's formulas for it:



$$a+b+c+d=0;(1)$$ $$ab+bc+cd+ad+ad+bd=1-2q;(2)$$ $$abc+bcd+cda+dab=2p;(3)$$ $$abcd=p^2+q^2-R^2.(4)$$ Since $M$ is the midpoint of $AC$, $2p=a+c$, using (1) we get $b+d=-2p$. Further, from (3) it follows that $$2p=(abc+acd)+(dab+bcd)=ac(b+d)+bd(a+c)=2p(bd-ac),$$ whence either $p=0$ or $bd-ac=1$. If $p=0$ then $a=-c$ and $b=-d$, so $A$ and $B$ are symmetric to $C$ and $D$ with respect to the axis $Oy$. Hence the quadrilateral $ABCD$ is self-intersecting which contradicts the problem. Thereby, $bd-ac=1$. Rewrite (1) in the form $a+c=-(b+d)$. Hence $(a+c{)}^2=(b+d{)}^2$ or $a^2+2ac+c^2=b^2+d^2+2bd$, whence, using $2bd=2ac+2$ we get $a^2+c^2=b^2+d^2+2$, i.e. $\frac{a^2+c^2}{2}=\frac{b^2+d^2}{2}+1$ which is equivalent to $\frac{a^2+c^2}{2}-\frac{b^2+d^2}{2}=1$. It remains to note that $\frac{a^2+c^2}{2}$ and $\frac{b^2+d^2}{2}$ are the coordinates of $M$ and $N$, therefore the projection of the segment $MN$ to the axis $Oy$ is equal to $1$.