tag:blogger.com,1999:blog-6213919025256134910.post882456214640224491..comments2023-05-15T08:32:02.241-07:00Comments on The Math Contest: Answers to 2002.Curmudgeonhttp://www.blogger.com/profile/04323026187622872114noreply@blogger.comBlogger3125tag:blogger.com,1999:blog-6213919025256134910.post-18081232083410697662011-11-03T13:40:03.145-07:002011-11-03T13:40:03.145-07:00FH is the altitude of AFE, FH is perpendicular to ...FH is the altitude of AFE, FH is perpendicular to AE.Curmudgeonhttp://www.blogger.com/profile/04323026187622872114noreply@blogger.comtag:blogger.com,1999:blog-6213919025256134910.post-58556282462700020332011-11-03T08:49:52.109-07:002011-11-03T08:49:52.109-07:00In problem 40, where, exactly is H?In problem 40, where, exactly is H?abelliahttp://abellia.myopenid.com/noreply@blogger.comtag:blogger.com,1999:blog-6213919025256134910.post-14762728933608806582011-04-02T14:23:31.660-07:002011-04-02T14:23:31.660-07:00Interesting. I used the sine rule for the second p...Interesting. I used the sine rule for the second part of problem 38, thinking that it would at least avoid me having to solve a quadratic, but of course triangle ABD is isosceles, so everything cancels very nicely in the second application of the cosine rule. A nice question.Jonnoreply@blogger.com