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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 a triangle. Points 
and lie on sides and 
, respectively, such that 
. Segments 
and 
meet at . The bisector of angle 
meet segments and at and , respectively. Prove that 
.
and 
are the 
-angle bisector of the 
and 
.So applying angle bisector theorem we get,![\[\frac{QD}{QA}=\frac{CD}{AC} \implies \frac{AD}{QA}=\frac{CD+AC}{AC}=\frac{CD+CE+AE}{AC}.\]](http://latex.artofproblemsolving.com/f/5/c/f5c19abc34178a2dcace41939ab72a769a95aa7d.png)
![\[\frac{EC}{ER}=\frac{BD+EC+CD}{BE}\]](http://latex.artofproblemsolving.com/9/4/f/94ffeba8d2a23e2f2f0915f061cefa410083844e.png)
.Cancelling two common terms from these two equations we get, ![\[\frac{QA}{AC}\cdot \frac{EC}{ER}=\frac{AD}{BE}\]](http://latex.artofproblemsolving.com/0/8/c/08cdc68f9007dffc0c29d7c7737644f857470657.png)
with the transversal 
we get,![\[ \frac{ER}{PR}\cdot \frac{AC}{EC}\cdot \frac{PQ}{QA}= 1 \implies \frac{PQ}{AD}=\frac{PR}{BE}\]](http://latex.artofproblemsolving.com/1/7/6/1766bc54c3dec573f12fd027e9bdafaa97efcd81.png)
And since 
we have:
Now using Menelaus Theorem in 
wrt 
