Thursday, October 14, 2010

Fox 308 - Solution

Bleaug says "a logical follow-up indeed", and solves:


  1. If we dilate PBA and QBC, points A and C will certainly meet on the semicircle, but how do we know they meet at point D and not some other point?

  2. your question makes sense. My solution above took some shortcuts.

    Basically, in this solution the two semi-circles are constructed in reverse starting from an arbitrary point D projected vertically in B. Then taking A' and C' as the intersection of triangle PDQ with the semi-circles we show that A'BC'D is a rectangle of center M and that its A'C' diagonal is tangent to the two semi-circles. Since there is only one such tangent in the upper half plan, A' and C' coincide with the given A and C points.


  3. Thank you, Bleaug; this Fox has been frustrating me for the past week. I did still need to work out in my own mind why A'C' is a tangent (because MA' and MC' are equal to tangent MB) but now I get it.