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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.
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.
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