In an acute-angled triangle $ABC$, $A_1$ and $B_1$ are the feet of the altitudes from $A$ and $B$ respectively, and $M$ is the midpoint of $AB$. a) Prove that $MA_1$ is tangent to the circumcircle of triangle $A_1B_1C$. b) Prove that the circumcircles of triangles $A_1B_1C,BMA_1$, and $AMB_1$ have a common point.
Problem
Source: French TST 2002
Tags: geometry, circumcircle, geometric transformation, reflection, geometry proposed
mahanmath
17.06.2011 17:17
WakeUp wrote: b) Prove that the circumcircles of triangles $A_1B_1C_1,BMA_1$, and $AMB_1$ have a common point. http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2173258&sid=ac8699c6e7e185bebbcd95fa354f4c05#p2173258 $P$ in above thread is that common point .
WakeUp
17.09.2011 19:52
For a) note that since $A_1,B_1$ lie on the circle with diameter $AB$ (and centre $M$), $MA=MA_1$ so $\angle HA_1M=\angle HAC_1=\angle HCA_1$ so $MA_1$ is tangent to $(HA_1C)$. For b) I think there is a misprint in the problem statement (i copied it from imomath). I think the first circle should be $A_1B_1C$ so I have edited it to reflect this. regardless, it is an easy problem, as mahanmath stated, take P to be the projection of H onto CM then easy angle chasing shows that P lies on all three circles.
vanu1996
20.12.2013 08:16
$M$ be the circumcenter of $AB_1A_1B$,so $AM=MA_1$,enough to prove $\angle MA_1A=\angle HB_1A_1=\angle HCA_1$($H$=orthocenter,$B_1HA_1C$ is cyclic),but $\angle HAB=\angle HCA_1$,hence $\angle HA_1M=\angle HCA_1$,hence done. second part-well known miquel point.