.
intersects 
at 
,
at 
at 
respectively. Suppose 
and 
intersect again at 
,
is the angle bisector of 
.
and 
Prove that
Let 
Prove that
specifies the greatest common divisor of 
and 
.
![\[ \begin{tabular}{c c c c}
{} & \textbf{Adams} & \textbf{Baker} & \textbf{Adams and Baker} \\
\textbf{Boys:} & 71 & 81 & 79 \\
\textbf{Girls:} & 76 & 90 & ? \\
\textbf{Boys and Girls:} & 74 & 84 & \\
\end{tabular}
\]](http://latex.artofproblemsolving.com/8/5/c/85c75770c3f90af49df0db600dead2a29e6674b3.png)
Y by jhu08, Adventure10, Mango247
An acute-angled 
is inscribed into a circle 
. Let 
be the centroid of 
, and let 
be an altitude of this triangle. A ray 
meets at 
. Prove that the circumcircle of the triangle 
is tangent to 
. (A.I. Golovanov , A.Yakubov)
is inscribed into a circle 
. Let 
be the centroid of 
, and let 
be an altitude of this triangle. A ray 
meets at 
. Prove that the circumcircle of the triangle 
is tangent to 
. (A.I. Golovanov , A.Yakubov)
and 
the projection of 
onto 
midpoint of 
,
are collinear. Now it is easy to see 
,
be point on 
.
be midpoint of 
.
we have that 
and 
share centroid 
be midpoint of 
.
are collinear. Let 
be circumcenter of 
.
are collinear so 
and 
and 
thus 
thus 
so 
(where 
)
.
thus 
so 
and we are done
.
,
,
lie on the unit circle in the complex plane.
.
be the point where ray 
meets the unit circle.
be the point on ray 
such that 
= 
.
lies on the unit circle.
is an isosceles trapezium.
Circumcircle of triangle 
such that 
.
at 
.
are collinear. Note that 
is an isosceles trapezium.![\[\measuredangle BA'H = \measuredangle BA'K = \measuredangle BCK = \measuredangle ABC = \measuredangle ABH\quad\blacksquare\]](http://latex.artofproblemsolving.com/a/4/d/a4d43c9322897f1da51e8e78be1383762c4bc462.png)
is inscribed into a circle 
be the centroid of 
,
be an altitude of this triangle. Ray 
meets 
is tangent to 
.
is an isosceles trapezoid. Then, by an easy homothety argument, we can see that 
.
followed by reflection in the angle bisector of 
.
such that 
is tangent to 
.
,
.
lies on 
,
.
be the foot from 
in 
.
,
is cyclic. Let 
be reflection of 
,
and let 
.
.
antipode of 
,
.
.
lies on the major arc of 
be the second intersection of line 
with 
.
in the perpendicular bisector of 
.![\[\measuredangle BA'H = \measuredangle BA'A'' = \measuredangle BAA'' = \measuredangle ABC\]](http://latex.artofproblemsolving.com/d/e/7/de745f6fce1675dbf38947201fe4768ad68b269d.png)
with 
,
.
.
,
doesn't alter the question condition ??
represents angles taken modulo 
![[asy]
import olympiad;
size(230);
pair A = (-1 , 2) , B = (-3 , -2) , C = (4,-2);
draw(A--B--C--cycle , green);
draw(circumcircle(A,B,C) , cyan);
pair D = (0.5 , -2);
pair E = (1.5 , 0);
pair F = (-2 , 0);
pair G = centroid(A,B,C);
pair H = foot(A,B,C);
pair A_1 = (-1 , -4.5);
draw(circumcircle(A_1, H , B) , orange);
draw(A--H);
dot("$A$" , A , N);
dot("$B$" , B , W);
dot("$C$" , C , NE);
dot("$D$" , D , NE);
dot("$E$" , E , NE);
dot("$F$" , F , W);
dot("$G$" , G , SW);
dot("$H$" , H , SW);
dot("$A_1$" , A_1 , S);
[/asy]](http://latex.artofproblemsolving.com/6/0/d/60de6cf6ac9772a4e88672c6bfad1ada7c471a66.png)
such that 
and so we have 
.
is an isosceles trapezium. Thus we get the following 
easy to conclude, that 
is an isosceles trepezoid.
,
where 
is a cyclic isosceles trapezoid 
.
.
and we are done 
.
with 
is an isosceles trapezoid (for instance, use coordinates to prove the converse), so 
as desired.
then it’s well known that 
is an isosceles trapezium, so from there 
,
and let 
invert at 
is tangent to 
,
and let 
be the reflection of 
and let 
.
.
,
bisects 
is the midpoint of 
lies on perpendicular bisector of 
is perpendicular bisector of 
,
is tangent to 
