be 
.
meet at point 
be 
meet 
respectively. Let 
be the midpoint of 
lie on a circle.
red cards, 
white cards, and 
blue cards. A player plays all the cards one at a time. With each play he accumulates a penalty. If he plays a blue card, then he is charged a penalty which is the number of white cards still in his hand. If he plays a white card, then he is charged a penalty which is twice the number of red cards still in his hand. If he plays a red card, then he is charged a penalty which is three times the number of blue cards still in his hand. Find, as a function of 
and 
the minimal total penalty a player can amass and all the ways in which this minimum can be achieved.
and a set 
consisting of 
disting positive integers smaller than 
are given. Prove that there exists a positive integer 
that can be written in the form 
,
in at least 
different ways.
lies on side 
lies on side 
.
and 
be points on segments 
and 
,
is parallel to 
be a point on line 
,
.
be the point on line 
,
.
,
with 
.
lie on 
respectively, such that 
.
to meet the circumcircle of 
respectively. Let the circumcircles of 
meet at points 
.
to 
,![$[ABC]$](http://latex.artofproblemsolving.com/d/3/3/d33cc80fa8f093e155c5be46d2e5d9da3d7e1ef5.png)
.
,
.
to meet the circumcircle of 
at 
,
.
and 
.
Equality holds when 
Equality holds when 
Equality holds when 
Y by Adventure10, Mango247
Let 
be a convex quadrilateral and let 
be a circle tangent to the lines 
and 
at points 
and 
, respectively. intersects the line segments 
and 
again at 
and 
, respectively, which are both different from 
. Let 
be the point of intersection of the lines 
and 
. Given 
, prove that the line is tangent to th circumcircle of the triangle 
.
be a convex quadrilateral and let 
be a circle tangent to the lines 
and 
at points 
and 
, respectively. intersects the line segments 
and 
again at 
and 
, respectively, which are both different from 
. Let 
be the point of intersection of the lines 
and 
. Given 
, prove that the line is tangent to th circumcircle of the triangle 
.This post has been edited 1 time. Last edited by Eray, Jun 24, 2016, 1:08 PM
,
and let 
,
,
give us 
give us 
give us 
give 
give 
and so 
as we want
thus 
are collinear.let 
the intersection of 
through 
and so 
.
tangent to 
which means that 
is tangent to the circumcircle of ABG.
gives that ![\[\angle BAC+\angle FCE=\angle GDC+\angle ECA+\angle FAD=\angle BGC+\angle GCA=180-\angle CAG-\angle AGB=\angle BAC+\angle GBA\]](http://latex.artofproblemsolving.com/1/3/8/138f8fb42a32ccdb0c1c916452c1d4c5f25bcb16.png)
Thus 
as desired
