Mathematics

Mathematics

Problem

If two curves, whose equations are ax2+2hxy+by2+2gx+2fy+c=0ax^2 + 2hxy + by^2 + 2gx + 2fy + c = 0 and a′x2+2h′xy+b′y2+2g′x+2f′y+c′=0a'x^2 + 2h'xy + b'y^2 + 2g'x + 2f'y + c' = 0 intersect in four concyclic points, prove that a−bh=a′−b′h′\frac{a-b}{h} = \frac{a'-b'}{h'}
Intuitive explanation

The key insight is that four concyclic points lie on a circle, which has a special algebraic form: the coefficient of xyxy must be zero. When two conics intersect at four concyclic points, their difference (a linear combination) must represent that circle. This forces a relationship between how each conic "twists"—measured by the ratio of the quadratic terms' asymmetry to their coupling—to be identical for both curves.

Intuitive explanation

The key insight is that four concyclic points lie on a circle, which has a special algebraic form: the coefficient of xyxy must be zero. When two conics intersect at four concyclic points, their difference (a linear combination) must represent that circle. This forces a relationship between how each conic "twists"—measured by the ratio of the quadratic terms' asymmetry to their coupling—to be identical for both curves.

Browse the problem library