[/quote]
,
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 
,
.
,
.
meet at point 
.
meet 
respectively. Let 
lie on a circle.
such that 
satisfying 
where 
.
cannot be a perfect cube of a positive integer.
has an incenter 
Suppose 
.
.
meets 
again at point 
.
and 
.
is a parallelogram, and suppose the angle bisectors of 
and 
concur on 
.
and 
meet 
and 
.
.
and 
with another positive integer 
,
is a perfect square. For example, if the number on the blackboard is 
,
,
,
,
.
and orthocenter 
,
and that line 
is not parallel to line 
is perpendicular to 
.
is a diameter, and 
and 
are parallel. Let 
.
and 
are perpendicular.
and 
are perpendicular.
be positive integers and 
be an odd prime such that:![\[
a^p \equiv 1 \pmod{p^n}.
\]](http://latex.artofproblemsolving.com/3/4/8/34815b8211ded2f37b0653525dc4b436356e6eee.png)
Prove that:![\[
a \equiv 1 \pmod{p^{n-1}}.
\]](http://latex.artofproblemsolving.com/0/7/a/07a4ed2ab3c94f41028080f7789cfb6e2973955c.png)
and 
)
Prove that if lines 
are similar.
satisfying![\[\begin{aligned}
\begin{cases}
a+b+c+d=0,\\
a^2+b^2+c^2+d^2=12,\\
abcd=-3.\\
\end{cases}
\end{aligned}\]](http://latex.artofproblemsolving.com/8/b/1/8b1923e0303c0aa79332ee4540ee1885a9cf3f08.png)
Y by itslumi, GeoKing, Rounak_iitr, ItsBesi
We are given an acute triangle 
with 
. Let 
be a point of 
such that 
is tangent to the circumcircle of . Let 
and 
be the circumcenters of triangles 
and 
, respectively, and let 
be the midpoints 
. Prove that the line tangent to the circumcircle of 
through is also tangent to the circumcircle of .
Proposed by Patrik Bak, Slovakia
with 
. Let 
be a point of 
such that 
is tangent to the circumcircle of . Let 
and 
be the circumcenters of triangles 
and 
, respectively, and let 
be the midpoints 
. Prove that the line tangent to the circumcircle of 
through is also tangent to the circumcircle of .Proposed by Patrik Bak, Slovakia
This post has been edited 1 time. Last edited by Tafi_ak, Dec 23, 2022, 10:13 AM
rest is easy chase 
be the tangent from 
.
,
is tangent to 
iff 
bisects 
.
is isosceles, so we need 
,
is sufficient. We have that 
,
.
of 
,
,
and 
)
be the other tangent line through 
,
.
for the obvious reason. Similarly 
.
is a parallelogram. So 
which bisects 
.
.
and 
be the point of 
distinct from 
such that 
is tangent to 
)
.
be circumcenters of 
,
Proof: Note 
,
.
It follows
which implies 
,
![[asy]
size(250);
pair B=dir(-165),C=dir(-15),A=dir(120),O=(0,0),N=1/2*(B+C),S=2*foot(A,B,C)-A,K=IP(N--S,unitcircle),T=2*foot(A,N,O)-A,D=2*A*K/(A+K);
pair E=circumcenter(A,B,D),F=circumcenter(A,D,C),M=1/2*(E+F);
fill(A--D--M--A--cycle,cyan+white+white);
fill(A--S--N--cycle,green+white+white);
draw(A--D--K,brown);
draw(E--F,magenta);
draw(D--C,red);
draw(S--T);
draw(A--M--D,purple);
draw(S--A--N,purple);
draw(A--B--S--C--A,orange+linewidth(0.8));
draw(A--E--D--F--A,fuchsia+linewidth(0.8));
draw(unitcircle,royalblue);
draw(circumcircle(A,O,K),grey);
dot("$A$",A,dir(90));
dot("$B$",B,dir(-110));
dot("$C$",C,dir(C));
dot("$O$",O,dir(60));
dot("$N$",N,dir(-60));
dot("$S$",S,dir(S));
dot("$T$",T,dir(T));
dot("$K$",K,dir(-70));
dot("$D$",D,dir(D));
dot("$E$",E,dir(E));
dot("$F$",F,dir(F));
dot("$M$",M,dir(-140));
[/asy]](http://latex.artofproblemsolving.com/9/d/7/9d7f7c0a68e2cc33c573077b45a33a898384c044.png)
be the spiral similarity centered at 
and 
.
.
.
In particular, this gives
are collinear.
is symmedian (in 
)
.
with 
,
.
lie on 
,
lie on circle with diameter 
It follows 
.
is perpendicular bisector of 
is perpendicular bisector of 
and 
we have that 
is cyclic, where 
is a parallelogram
and 
.
.
but 
at 
.
.
,
but 
thus 
is tangent to the circumcircle of 
Since 
Similarly 
Now we wanna prove that 
is tangent to 
.
too. Then, we make the following key claim.
Then,
.
so, 
.
and 
are collinear. Thus,
as well. Establishing the claim.
and 
.
and 
implying 
.
So, 
too. So points 
are concylic with 
which implies 
which gives our desired tangency.
.
are collinear.
armonic quadrilateral, so 
,
,
are collinear.
,
are collinear.
are concyclic.
,
,
so 
,
is tangent to 
,
,
.
.![\[2\measuredangle EDB = \measuredangle DEB = 2\measuredangle DAB = 2\measuredangle ACB = 90+ \measuredangle ACB\]](http://latex.artofproblemsolving.com/5/4/1/54188efc667ef6f3b3d327c05790aacfcd426aa5.png)
Thus, 
.![\[2\measuredangle BDF = \measuredangle CFD = 2\measuredangle CAD - 180\]](http://latex.artofproblemsolving.com/7/5/1/7514bceea2bbed2bb0bf584cc0bd679ec50eed38.png)
Then, we have that 
.![\[\measuredangle EDF = \measuredangle EDB + \measuredangle BDF = 90 + \measuredangle ACB + \measuredangle CAD - 90 = \measuredangle CAB\]](http://latex.artofproblemsolving.com/1/c/e/1ce477155bbb68f134a836765bd3b817c4cf36c5.png)
Thus,![\[\measuredangle FOE = \measuredangle EDF\]](http://latex.artofproblemsolving.com/f/d/9/fd945770bb77604e6deb05ebdd4f09cfe6c5c4e5.png)
Further note that,![\[\measuredangle OED = 360 - ( \measuredangle BDE +90 + \measuredangle ABD) = 180 + \measuredangle BCA + CBA = 2\measuredangle CBA + \measuredangle BAC\]](http://latex.artofproblemsolving.com/2/3/1/23147fbfff8e2b47b69cfa1277b542697133e1eb.png)
with a similar angle chase, we obtain that 
.![\[\measuredangle OEF = \measuredangle DFO\]](http://latex.artofproblemsolving.com/8/0/8/80804b307a10ebded250233c502f1cd80fb16dad.png)
as well. This implies the required result.
is a cyclic quadrilateral.
.
!
.![\[\measuredangle MDK = \measuredangle ADK + \measuredangle KDM = \measuredangle ADM = \measuredangle MAD\]](http://latex.artofproblemsolving.com/2/2/9/2292c50e6cfa266f493d9f01e36a3498908b252d.png)
This means that the line tangent to 
in 
invert we have that from easy angle chase that 
is cyclic, redefine 
which implies the tangency.
,
to 
is the same as 
by the midpoint, and by radax we have 
colinear so by thales we have 
as desired thus we are done 
.
and 
.
.
.
and 
,
to 
and 
[using EM 
]
,
and we are done. 
.
is tangent to 
.
are collinear.
is a parallelogram. This implies that 
such that 
.
where the last equality comes from symmetry.
so 
.
bisects 
so 
as desired.
is a parallelogram
and 
thus by Salmon Theorem we get that 
hence we get 
we get that 
hence the quadrilateral 
is a parallelogram 
are collinear and 
and 
we get that 
lies on the perpendicular bisector of 
lies on the perpendicular bisector of 
so 
and 
lies on the perpendicular bisector of 
lies on the perpendicular bisector of 
hence 
is the side bisector of 
(because they are tangent to 
