### Third Millennium International Mathematical Olympiad 2009

Grade 10-12

Problem 6

In a Cartesian system of coordinates, a circle of radius r with center at (p, q) meets the parabola with the equation |

|Up| |Contact| |Front page| |Contents|

Copyright © 1996-2018 Alexander Bogomolny

The equation of a circle with center (p, q) and radius r is (x - p)² + (y - q)² = r². We may eliminate x² by multiplying this equation by a and subtracting the result from the equation of the parabola,

(*) | (2pa + b)x = ay² + (1 - 2aq)y + a (p² + q² - r²). |

Dividing by (2pa + b) we indeed get an equation of a parabola

x = Ay² + By + C, |

where A = a / k, B = (1 - 2aq) / k, C = a (p² + q² - r²) / k, k = 2pa + b. However, let's not forget that division is not always possible: _{1, 2} are two solutions of the equations, _{1}_{ 2}

For that problem, Sadik Shahidain received the score of 5 (out of 7) for the following observation:

There is not necessarily another parabola passing through the same four points. In the case of the circle centered at |

|Up| |Contact| |Front page| |Contents|

Copyright © 1996-2018 Alexander Bogomolny

66388831