TMIMO 2009, Grade 10, Solution to Problem 6

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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 y = ax² + bx + c at four distinct points. Prove that there is another parabola passing through the same four points and find its equation.

Solution

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, y = ax² + bx + c. After simplifications we get

(*)	(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: (2pa + b) = 0 constitute an exceptional case. This happens exactly when p = -b / 2a, i.e., when the center of the circle lies on the axis of the parabola. In this case, the equation (*) is reduced to a quadratic equation in y whilst the parabola degenerates into two parallel lines. If y_{1, 2} are two solutions of the equations, y = y₁ and y = y₂ are the two parallel lines.

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 (0, 0) with radius 4, the parabola 4x² - 16 intersects at 4 points. If one finds a parabola tilted at an angle with three of them, the fourth one will be inside the gap in the parabola. Also, if it is completely tilted 90 degrees, then it is impossible because of the mean-value theorem for derivatives.

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Third Millennium International Mathematical Olympiad 2009 Grade 10-12 Problem 6

Third Millennium International Mathematical Olympiad 2009
Grade 10-12
Problem 6