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Selection and Training Session

Belarus geometry

Problem

and are two parallel chords of a parabola. Circle passing through points , intersects circle passing through , at points , .

Prove that if belongs to the parabola, then also belongs to the parabola.
Solution
First note that all parabolas are similar, thus we may consider the parabola . Let , , , , be the abscissae of the points , , , , respectively. We use the following easy lemmas.

Lemma 1. The chords and of the parabola are parallel if and only if .

Lemma 2. The points , , , of the parabola (at least three of them let be different) are concyclic if and only if .

From Lemma 2 it follows that in addition to , , the circle has one more common point with the parabola, the abscissa of being . Similarly, has common point with the parabola, its abscissa . Since , we have , so . This means that belongs to the parabola.

Techniques

Cartesian coordinatesConstructions and loci