tag:blogger.com,1999:blog-6500033298667240354.post7663372884918246864..comments2024-02-19T00:34:12.578-08:00Comments on Always Creative Geometry Problems plus Occasionally Annoying Philosophy: Fox 308 - Solution8foxeshttp://www.blogger.com/profile/09567328431908997738noreply@blogger.comBlogger3125tag:blogger.com,1999:blog-6500033298667240354.post-60002405061611275162010-10-17T20:59:26.107-07:002010-10-17T20:59:26.107-07:00Thank you, Bleaug; this Fox has been frustrating m...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.Bob Rydennoreply@blogger.comtag:blogger.com,1999:blog-6500033298667240354.post-16711953314678184442010-10-17T15:49:18.225-07:002010-10-17T15:49:18.225-07:00your question makes sense. My solution above took ...your question makes sense. My solution above took some shortcuts.<br /><br />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.<br /><br />bleaugAnonymousnoreply@blogger.comtag:blogger.com,1999:blog-6500033298667240354.post-91238475903217555112010-10-17T10:11:16.744-07:002010-10-17T10:11:16.744-07:00If we dilate PBA and QBC, points A and C will cert...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?Bob Rydennoreply@blogger.com