Belorusija, 2012

Geometry Difficulty 5.2 AIME, harder Prove it Belarus

Two distinct points $A$ and $B$ are marked on the left half of the parabola $y = x^2$ . Consider any pair of parallel lines which pass through $A$ and $B$ and intersect the right half of the parabola at points $C$ and $D$ . Let $K$ be the intersection point of the diagonals $AC$ and $BD$ of the obtained trapezoid $ABCD$ . Let $M, N$ be the midpoints of the bases of $ABCD$ .
Prove that the difference $KM - KN$ depends only on the choice of points $A$ and $B$ but does not depend on the pair of parallel lines described above.
(I. Voronovich)

Solution

Hint. Show that if the given points have coordinates $A(a; a^2)$ , $B(b; b^2)$ , then $KM - KN = (b - a)^2$ .

Want a route through all this instead of an archive? The track puts 2,604 problems in a working order, from Junior Challenge level to the IMO shortlist.

Source: MathNet, licensed CC-BY-4.0. Statement reproduced verbatim; metadata (topic, difficulty) added by this project.