145. Spain MO P5 - Point on median.
by Virgil Nicula, Oct 5, 2010, 5:28 PM
Let 
be any point on the angle bisector of 
of 
. Let 
, 
be the feet of the perpendiculars drawn from
to
and 
respectively. Show that the point 
for which 
belongs to the 
-median of .
Proof. Denote
, 
so that 
. Observe that 
, 
are cyclically. Therefore,
is -isosceles and 
, i.e. 
belongs to the -median.
Remark. The following properties are wellknown or we can prove similarly, but more easily :
The incircle 
of touches , in 
and 
. Denote 
for which 
. Then 
belongs to -median.
Let 
be a triangle with incircle and circumcircle 
. Denote the midpoint 
of ![$[BC]$](//latex.artofproblemsolving.com/e/a/1/ea1d44f3905940ec53e7eebd2aa5e491eb9e3732.png)
, the intersection 
and the projections
, 
of 
to , . Then 
(you can prove directly or apply Simpson's theorem in this particular case).
Now can choose one from points
or and through the homothety by vertex we"ll obtain the conclusion of proposed problem.
I came up to this idea because we know a simple fact that locus of the midpoint of![$[XY]$](//latex.artofproblemsolving.com/b/d/5/bd5db5e85aa6daea3eebecaea5d26721edd15203.png)
is -median, when , so that 
.
Remark. If , so that 
is cyclically (in this case we say that is antiparallel line/direction to 
), then the locus
of the midpoint of is -symmedian in . Generally, the locus of a point which belongs to the interior of 
and for which
(constant) is a line which pass through . I used notation 
- distance from the point 
to the line 
. Good luck !
be any point on the angle bisector of 
of 
. Let 
, 
be the feet of the perpendiculars drawn from to
and 
respectively. Show that the point 
for which 
belongs to the 
-median of .Proof. Denote
, 
so that 
. Observe that 
, 
are cyclically. Therefore,
is -isosceles and 
, i.e. 
belongs to the -median.Remark. The following properties are wellknown or we can prove similarly, but more easily :
The incircle 
of touches , in 
and 
. Denote 
for which 
. Then 
belongs to -median. Let 
be a triangle with incircle and circumcircle 
. Denote the midpoint 
of ![$[BC]$](http://latex.artofproblemsolving.com/e/a/1/ea1d44f3905940ec53e7eebd2aa5e491eb9e3732.png)
, the intersection 
and the projections
, 
of 
to , . Then 
(you can prove directly or apply Simpson's theorem in this particular case).Now can choose one from points
or and through the homothety by vertex we"ll obtain the conclusion of proposed problem.I came up to this idea because we know a simple fact that locus of the midpoint of
![$[XY]$](http://latex.artofproblemsolving.com/b/d/5/bd5db5e85aa6daea3eebecaea5d26721edd15203.png)
is -median, when , so that 
.Remark. If
, so that 
is cyclically (in this case we say that is antiparallel line/direction to 
), then the locusof the midpoint of
is -symmedian in . Generally, the locus of a point which belongs to the interior of 
and for which
(constant) is a line which pass through . I used notation 
- distance from the point 
to the line 
. Good luck !This post has been edited 8 times. Last edited by Virgil Nicula, Dec 1, 2015, 11:03 AM
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