Take an arbitrary AOC triangle such that OC=3OA. Build B as intersection of triangle's circumscribed circle and AOC internal bisector. B is such that ABC is isoceles, i.e. AB=BC. Symmetry in OABI implies AB=BI, hence IBC is isoceles. Orthogonal projection of B on OC coincides with H such that IH= HC=OA. OHB is rectangle, thereforeOB^2 = (2OA)^2 + BH^2 => OB > 2OA iff angle(AOC) > 0.
Strictly speaking, if angle(AOC)=0 there is a solution if we admit that a straight line is a circle with center at infinity (e.g. in projective plane).
And Giannno solves too:
Good work!
ReplyDelete296 is a strong statement. See it this way: angle(AOC) can take any values between 0 and 180. Also note, for any value of angle(AOC), we can rotate the construction around point O, by 180-(AOC) degrees. Yet, we won't be able to find a construction satisfying "x, 2x, 3x" requirement.
Interesting!
A new solution is added to this post.
ReplyDeleteThank you Giannno.