Circumcenter lies on line parallel to BC
by math_pi_rate, Aug 20, 2018, 12:14 PM
Here is a problem I made a few days ago (which I'll later be using as a lemma).
LEMMA Let
be an acute-angled triangle with circumcircle 
. Let 
be the midpoint of arc 
(not containing 
), and let 
be the antipode of in . Let 
. Finally let 
be the circumcenter of 
. Then 
is tangent to .
Anyway, here's a problem that uses this lemma.
Problem (Polish MO 2018 P5): Point is a center of circumcircle of acute triangle , bisector of angle 
cuts side 
in point . Let 
be a point such that, 
and 
. Lines 
and 
intersect in point 
. Show that circle of center in point passing through a point is tangent to line .
LEMMA Let
be an acute-angled triangle with circumcircle 
. Let 
be the midpoint of arc 
(not containing 
), and let 
be the antipode of in . Let 
. Finally let 
be the circumcenter of 
. Then 
is tangent to .
,
,
.
,
is cyclic. As we can calculate both 
and 
in terms of 
,
.
is a point on 
such that 
.
.
is tangent to 
.
are concyclic. Thus as 
, ![\[\angle ODC=\angle OTB = \angle ATB = \angle OAD.\]](http://latex.artofproblemsolving.com/a/b/a/aba9c6b5c99c6aba44a2c470f2a67d4a9989328b.png)
Proved.Anyway, here's a problem that uses this lemma.
Problem (Polish MO 2018 P5): Point
is a center of circumcircle of acute triangle , bisector of angle 
cuts side 
in point . Let 
be a point such that, 
and 
. Lines 
and 
intersect in point 
. Show that circle of center in point passing through a point is tangent to line .
passes through 
at 
.
it suffices to show that 
.
,
.
is the external angle bisector of 
.
such that 
at 
is cyclic.
be the circumcenter of 
.
.
and 
is right-angled at 
,This post has been edited 6 times. Last edited by math_pi_rate, Sep 27, 2018, 12:58 PM
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