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be 
. The tangents to at 
meet at point 
. For a point 
on line 
which is not on the segment , let the midpoint of 
be 
. Lines 
meet again at points 
respectively. Let 
be the midpoint of . Prove that the points 
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 
, for 
in at least 
different ways.
, point 
lies on side and point 
lies on side 
. Let 
and 
be points on segments 
and 
, respectively, such that 
is parallel to 
. Let 
be a point on line 
, such that lies strictly between and , and 
. Similarly, let 
be the point on line 
, such that lies strictly between and , and 
.
, and are concyclic.
with 
. The points 
lie on 
respectively, such that 
. Extend 
to meet the circumcircle of at 
respectively. Let the circumcircles of 
meet at points 
. Extend 
to at , given 
, find ![$[ABC]$](http://latex.artofproblemsolving.com/d/3/3/d33cc80fa8f093e155c5be46d2e5d9da3d7e1ef5.png)
.
with 
, denote the incenter of the triangle as 
. Extend 
to meet the circumcircle of 
at 
, find the length of 
.
and 
. Prove that
Equality holds when 
Equality holds when 
Equality holds when 
be an acute-angled triangle and 
the midpoint of . Let be a point on segment 
and its projection on . If 
and are the projections of on and then the interior angule bisectors of 
and 
are parallel.
, 
, 
, and 
, the angle bisector of is parallel to that of 
. Now if 
are such that 
, and 
, we require the angle bisectors of 
and to coincide. This, however is equivalent to 
. Denoting the first by 
and the second by 
, from sine law applied on 
, we compute:![\[\frac{\sin x}{AE}\cdot\frac{EM}{\cos x} = \frac{\sin\angle EAN}{NE}\cdot\frac{NE}{\sin\beta}=\frac{\sin\angle EAN}{\sin\beta} = \frac{\sin\angle EAP}{\sin\gamma} =\frac{\sin\angle EAP}{PE}\cdot \frac{PE}{\sin\gamma} =\frac{\sin y}{AE}\cdot\frac{EM}{\cos y} \]](http://latex.artofproblemsolving.com/3/d/3/3d3f0113ffc02f81b68ba64e71cff80f65311154.png)
Therefore 
, and we're done.
