[/quote]
,
.
meet at point 
which is not on the segment 
.
meet 
respectively. Let 
be the midpoint of 
lie on a circle.
red cards, 
white cards, and 
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 
,
.
,
such that 
satisfying 
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 
,
be an arbitrary point on the circumcircle of triangle 
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)
crosses are placed. A set of 2025 cells we will call balanced if no two of them are in the same row or column. It is known that any balanced set has at least 
crosses.
Y by Adventure10, Mango247
In the acute-angled triangle 
, the foot of the perpendicular from 
to 
is 
. Let ![$[i]l[/i]$](//latex.artofproblemsolving.com/6/9/7/697de0b0060ad98f93ed92ee626d8bdcb57a1ba8.png)
be the tangent to the circle at . The foot of the perpendicular from 
to is 
, Prove that 
is parallel to 
.
See here: http://www.bmoc.maths.org/home/bmo1-2014.pdf
, the foot of the perpendicular from 
to 
is 
. Let ![$[i]l[/i]$](http://latex.artofproblemsolving.com/6/9/7/697de0b0060ad98f93ed92ee626d8bdcb57a1ba8.png)
be the tangent to the circle at . The foot of the perpendicular from 
to is 
, Prove that 
is parallel to 
.See here: http://www.bmoc.maths.org/home/bmo1-2014.pdf
This post has been edited 1 time. Last edited by AndrewTom, Nov 30, 2013, 3:39 PM
,
and 
.
is tangent to the circumcircle of 
at 
.
,
is cyclic. Now by equal inscribed angles of circle 
.
,
we get 
(the last equation follows from the angles of triangle 
)
,
