50. IMAC 2010 seniors, the first day.
by Virgil Nicula, Jul 6, 2010, 3:12 AM
Proposed problem. Let 
be a triangle and let 
be the foot of the 
- altitude. The circle 
with the diameter ![$[AD]$](//latex.artofproblemsolving.com/0/f/3/0f3e4c424371b27673db323ced8ef0777940c0d4.png)
meet again 
, 
in the points
, 
respectively. Denote the meetpoint 
of the tangents to the circle in 
, 
. Prove that 
is the -median in 
(Serbia).
Proof 1. Denote
and the second intersection 
of the circle with the line 
. Is well-known or prove easily that the quadrilateral 
is
harmonically and the ray
is the -symmedian in 
. Observe that 
, i.e. 
what
means the quadrilateral
is cyclically. Since 
and is -symmedian in 
obtain that is -median in 
.
======================================================================================================================================
. Therefore,
. Thus is harmonically and 
is -symmedian
in . In conclusion obtain that 
is an harmonical division and 
.
Proof 2. Since the circles with diameters
and 
are orthogonal to it follows that 
and 
pass through the midpoints 
and 
of and respectively.
Let
and 
. Thus, 
becomes orthocenter of 
. Moreover 
since 
is the polar of the intersection
w.r.t. 
. Since is the midpoint of , then is the midpoint of the line goes through the midpoint of 
.
be a triangle and let 
be the foot of the 
- altitude. The circle 
with the diameter ![$[AD]$](http://latex.artofproblemsolving.com/0/f/3/0f3e4c424371b27673db323ced8ef0777940c0d4.png)
meet again 
, 
in the points
, 
respectively. Denote the meetpoint 
of the tangents to the circle in 
, 
. Prove that 
is the -median in 
(Serbia).Proof 1. Denote
and the second intersection 
of the circle with the line 
. Is well-known or prove easily that the quadrilateral 
isharmonically and the ray
is the -symmedian in 
. Observe that 
, i.e. 
whatmeans the quadrilateral
is cyclically. Since 
and is -symmedian in 
obtain that is -median in 
.======================================================================================================================================
. Therefore,
. Thus is harmonically and 
is -symmedianin
. In conclusion obtain that 
is an harmonical division and 
.Proof 2. Since the circles with diameters
and 
are orthogonal to it follows that 
and 
pass through the midpoints 
and 
of and respectively.Let
and 
. Thus, 
becomes orthocenter of 
. Moreover 
since 
is the polar of the intersection
w.r.t. 
. Since is the midpoint of , then is the midpoint of the line goes through the midpoint of 
.This post has been edited 23 times. Last edited by Virgil Nicula, Nov 23, 2015, 4:41 PM
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