tag:blogger.com,1999:blog-40658289389239287.post2590163351247124391..comments2024-01-13T15:24:59.486-08:00Comments on ΜΑΘΗΜΑΤΙΚΑ: PAUL PRIZEΑνωπολίτηςhttp://www.blogger.com/profile/11021660591505869543noreply@blogger.comBlogger1125tag:blogger.com,1999:blog-40658289389239287.post-73141009424071790412012-01-19T01:33:39.499-08:002012-01-19T01:33:39.499-08:00Suppose, $ H_a,H_b,H_c,H $ are the orthocenters of...Suppose, $ H_a,H_b,H_c,H $ are the orthocenters of $ \Delta A"BC,\Delta B"CA,\Delta C"AB, \Delta ABC $.<br />Note that, those nine-point centers are the midpoints of $ OH,OH_a,OH_b,OH_c $. So its enough to prove that $ <br />HH_aH_bH_c $ is cyclic. <br />Now note that $ AHH_aA" $ is a parallelogram. So $ HH_a\parallel AA" $ and $ HH_a=AA" $ and similar for others.<br />Suppose, $ A_2,B_2,C_2 $ are the diametrically opposite points of $ A",B",C" $ wrt $ (O) $. Note that $ AA_2,BB_2,CC_2 $ are concurrent at the reflection of $ D $ on $ O $.<br />Now suppose, $ A_1,B_1,C_1 $ are the foot of the perpendiculars from $ O $ to $ AA_2,BB_2,CC_2 $. Note that $ OA_1\parallel AA" $ and $ OA_1=\frac {AA"}{2} $. Similar for others.<br />So the configuration $ HH_aH_bH_c $v is homothetic to $ OA_1B_1C_1 $.<br />Clearly $ O,A_1,B_1,C_1 $ lie on the circle with diameter $ OD' $ where $ D' $ is reflection of $ D $ on $ O $.<br />So $ HH_aH_bH_c $ is cyclic. So done.Chandanhttps://www.blogger.com/profile/14305444043101312742noreply@blogger.com