66th Belarusian Mathematical Olympiad

Let $A(a,1/a)$, $B(b,1/b)$, $C(c,1/c)$, $D(d,1/d)$, $E(e,1/e)$, $F(f,1/f)$, and positions of all points on the hyperbola are shown in the Figure. It is easy to write the equations of the straight lines $AB$, $DC$, and $EF$: $$AB(l):y=-\frac{1}{ab}x+\frac{1}{a}+\frac{1}{b},EF(l_1):y=-\frac{1}{cf}x+\frac{1}{c}+\frac{1}{f},$$ $$DC(l_2):y=-\frac{1}{cd}x+\frac{1}{c}+\frac{1}{d}.$$ Since all these lines are parallel, their slopes coincide. Therefore, $$ab=cd=ef.(1)$$ Let $M$, $N$, $K$ be the midpoints of the segments $AB$, $EF$, $DC$, respectively. Then $M((a+b)/2;(1/a+1/b)/2)$, $N((e+f)/2;(1/e+1/f)/2)$, $K((c+d)/2;(1/c+1/d)/2)$. Let $O$ be the origin. It is evident that the equation of the straight lines $MO$, $ON$ and $OK$ respectively have the forms $$y=\frac{1}{ab}x,y=\frac{1}{ef}x,y=\frac{1}{cd}x.$$ From (1) it follows that all these lines coincide, so all points $O$, $M$, $N$, $K$ belong to the same straight line $MK$ with the equation $y=\beta x$, where $\beta =1/(ab)=1/(cd)=1/(ef)$. It is easy to see that in any trapezoid the segment connecting the midpoints of its bases passes through the midpoint of any segment which is parallel to the trapezoid bases. Consider the trapezoid $ABCD$ and the segment $GH$, $GH\parallel BC\parallel AD$, such that $E$ and $F$ lie on this segment. Since the line $MK$ passes through the midpoints of $AB$ and $DC$, the line $MK$ passes through the midpoint of the segment $GH$, too. Since $MK$ also passes through the midpoint of the segment $EF$, therefore the midpoints of $GH$ and $EF$ coincide since these segments lie on the same straight line. Thus, $GE=HF$ as required.