Thursday, July 1, 2010

Fox 296 - Solutions

Among similar ones, Bleaug finds a contradiction for Fox 296: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, therefore
OB^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:

3 comments:

  1. Good work!
    296 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!

    ReplyDelete
  2. A new solution is added to this post.
    Thank you Giannno.

    ReplyDelete
  3. Do not use all of these Private Money Lender here.They are located in Nigeria, Ghana Turkey, France and Israel.My name is Mrs.Ramirez Cecilia, I am from Philippines. Have you been looking for a loan?Do you need an urgent personal or business loan?contact Fast Legitimate Loan Approval he help me with a loan of $78.000 some days ago after been scammed of $19,000 from a woman claiming to be a loan lender from Nigeria but i thank God today that i got my loan worth $78.000.Feel free to contact the company for a genuine financial call/whats-App Contact Number +918929509036 Email:(fastloanoffer34@gmail.com)

    ReplyDelete