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Belarus — Selection and Training Session

Geometry Difficulty 5.2 AIME, harder Prove it Belarus

$AB$ and $CD$ are two parallel chords of a parabola. Circle $S_1$ passing through points $A$ , $B$ intersects circle $S_2$ passing through $C$ , $D$ at points $E$ , $F$ .

Prove that if $E$ belongs to the parabola, then $F$ also belongs to the parabola.

Solution

First note that all parabolas are similar, thus we may consider the parabola $y = x^2$ . Let $a$ , $b$ , $c$ , $d$ , $e$ be the abscissae of the points $A$ , $B$ , $C$ , $D$ , $E$ respectively. We use the following easy lemmas.

Lemma 1. The chords $AB$ and $CD$ of the parabola are parallel if and only if $a + b = c + d$ .

Lemma 2. The points $A$ , $B$ , $C$ , $D$ of the parabola (at least three of them let be different) are concyclic if and only if $a + b + c + d = 0$ .

From Lemma 2 it follows that in addition to $A$ , $B$ , $E$ the circle $S_1$ has one more common point $F_1$ with the parabola, the abscissa of $F_1$ being $f_1 = -a - b - e$ . Similarly, $S_2$ has common point $F_2$ with the parabola, its abscissa $f_2 = -c - d - e$ . Since $AB \parallel CD$ , we have $a + b = c + d$ , so $f_1 = f_2$ . This means that $F_1 = F_2 = F$ belongs to the parabola.

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