CircleBundleProblem

Circle bundle problem

Consider three points {A,B,C} in general position. Let L be a variable line through C intersecting line AB at D. On L and on both sides of D take points {M,N} such that DM²=DN² =DA*DB. The circle (c) through {M,N,A} passes also through another fixed point H (variation of a problem proposed by John Pounios).

Project C on the medial line of segment AB and construct the isosceles FAB, such that FC is parallel to AB. Construct the circle as required, DM²=DN²=DB*DA. Let G the other than A intersection point of this circle with AB and set x=BD and a=AB/2. Then (2a+x)x=DM²=(2a+x)DG implies DG=x. Consequently the middle S of AG has AS=a+x, which implies that SD=a.
Project now C on line AB to V. The right angled triangles {KDS,DCV} are similar and KS/SD=DV/VC. Thus K moves on a fixed line intersecting AB at a point I. But KS is zero precisely when DV is zero, then S coinciding with I , consequently lying at distance a from V.
One verifies easily that H is the symmetric of A with respect to line KI.
As a corollary one obtains a circle passing through points {I,B,H,F,C}.

Remark-1 Points {M,N} are the contact points of two circles tangent to line L=CD and passing through {A,B}.
Remark-2 Analogous result holds for the circles passing through points {B,M,N}. One can even easily see that the two circles {A,M,N} and {B,M,N} are equal and the corresponding fixed point H' on FB is such that BH'=AH.
Remark-3 Note that with changing direction of line CD the contact points {M,N} move on a circular cubic curve passing through the six points {A,B,C,F,H,H'}. The cubic passes also through two other easily constructible points, namely the two points {N',M'} symmetric with respect to V (on line CV) such that VA*VB=VM'²=VN'². Besides the asymptote ot the cubic is parallel to the x-axis.

Circle bundle problem

See Also