CircumconicGeneration

1. A circle generation

Start with an equilateral triangle ABC and the diametral point D of A on its circumcircle. For every line through D consider the intersection points {E, F, G} with sides {BC, AB, AC} correspondingly. Then their fourth harmonic point coincides with the second intersection point H of the line with the circumcircle. And inversely, joining every point H of the circumcircle to D and intersecting with line DH the sides of the triangle we find points such that (F,G,E,H) = -1.

The proof follows at once by observing that at point H the angles BHA and AHC are equal, each being 60 degrees. Thus lines {HA, HD} are bisectors of the angle BHC hence H(B,C,E,I) is a harmonic bundle consequently A(F,G,E,H) is also a harmonic bundle, proving the claim. The inverse

2. Circumconic generation

Given a triangle ABC and a circumconic (c) with perspector P. Let D be the other intersection point of AP with c. For every line through D let {B',C',A'} be the intersection points with sides {AB, AC, BC} respectively. Then the fourth harmonic A'' of these points i.e. point such that : (B',C',A',A'') = -1 is on the conic.
Inversely, given a triangle ABC and a point D not on the side-lines of the triangle, consider all lines through D and the corresponding intersections with these lines {A',B',C'}. Let then A'' be the fourth harmonic of these three points: A'' = A'(B',C') (i.e (B',C',A',A'') = -1). Then point A'' describes a conic circumscribing the triangle ABC and passing through D. Further the pivot P of the conic with respect to the triangle is on line AD.

The proof follows from the previous section by applying to the equilateral an appropriate projectivity transforming it to the general triangle ABC and the center of the equilateral to the perspector P of the conic c.

3. And a specialization

Given two parallel lines {b, c} and a third line (a) intersecting them, consider all lines through a fixed point D not lying on the given three lines {a, b, c}. For each line through D define the intersection points {A, B, C} with lines {a, b, c} correspondingly. The fourth harmonic E on each line i.e. point such that (B,C,A,E) = -1 describes a hyperbola.

This property follows from the previous section by letting point A of the triangle ABC shown there go to infinity. When A coincides with the intersection M of line (a) with the middle parallel (m) of lines {b,c} then E goes to infinity, from which it follows that line DM is parallel to an asymptote. The other asymptote is obviously parallel to lines {b,c}. When A obtains the position of M' (parallel to b projection of D on a) points {B,C,E} go to the same point at infinity. Besides the points {B₀, C₀, D} from which passes the hyperbola we know also point M', which is the projection of D on line (m) parallel to line (a). This position is obtained by E when A goes to infinity on a and DA becomes parallel to line (a).
An easily constructible point is point D' on the line through D which is orthogonal to line (a). Then the conic can be identified with the unique conic passing through the five points {D, C₀, B₀, M'', D'}.

4. Problems

[1] Find the hyperbola passing through the vertices of a trapezium and having known directions of asymptotes. (Could use the property shown in the previous paragraph, but this handles a special case of trapezia, such that the parallel to one asymptote from the middle of a side B₀C₀ passes through another vertex of the trapezium.)
[2] The same problem for general quadrangle instead of hyperbola.
[3] Find the structure of all hyperbolas passing through the vertices of a trapezium.
[4] Same problem for a general quadrangle.

1. A circle generation

2. Circumconic generation

3. And a specialization

4. Problems

See Also