The points A, B, C are not collinear. There are three ellipses, each pair of which intersects. One has foci A and B, the second has foci B and C and the third has foci C and A. Show that the common chords of each pair intersect.
Solution
We show first that each pair of ellipses intersects in just two points. Clearly, it is possible for them not to intersect at all, but we are obviously meant to assume that each pair intersects in at least two points. We have to show that they do not intersect in more than two points.
Let the point P be a distance a from A, b from B and c from C. The ellipse with foci A and B can be taken to be the set of points P such that a + b = k3 (a constant). Similarly, the ellipse with foci A and C can be taken to be the set of points P such that a + c = k2 (another constant). So if P belongs to both ellipses, then (subtracting) b - c = k3 - k2. But this is the equation of one branch of a hyperbola, which intersects one of the ellipses in at most two points. So the two ellipses intersect in two points.
Now consider the equation (k2 - k3)a2 - k2b2 + k3c2 = k2k3(k2 - k3), (*). If P lies on the ellipse foci A, B and on the ellipse foci A, C, then we have: b = k3 - a, c = k2 - a. So substituting in the lhs of (*) we get: (k2 - k3)a2 - k2(k3 - a)2 + k3(k2 - a)2 = k2k3(k2 - k3) = rhs. So the two points on both ellipses satisfy the equation (*). But suppose P is (x1, x2), A is (a1, a2), B is (b1, b2), and C is (c1, c2). Then we have a2 = (x1 - a1)2 + (x2 - a2)2 etc and hence we may write (*) as: (k2 - k3)( (x1 - a1)2 + (x2 - a2)2) - k2( (x1 - b1)2 + (x2 - b2)2) + k3( (x1 - c1)2 + (x2 - c2)2) = k2k3(k2 - k3). But we see that the terms in x12 and x22 cancel, so that (*) is a linear equation and hence represents a straight line. So it is the equation of the common chord of the two ellipses.
So if we take the third ellipse to be b + c = k1, then the equations of the three common chords are:
(k2 - k3) a2 - k2 b2 + k3 c2 = k2k3(k2 - k3), (1)
k1 a2 + (k3 - k1) b2 - k3 c2 = k3k1(k3 - k1), (2)
-k1 a2 + k2 b2 + (k1 - k2) c2 = k1k2(k1 - k2), (3).
But now we find that the linear combination k1(k2 + k3 - k1) (1) + k2(k3 + k1 - k2) (2) + k3(k1 + k2 - k3) (3) is zero. So the three lines are concurrent.
© John Scholes
jscholes@kalva.demon.co.uk
5 Oct 2002