AoPS Academy
[/quote]
, do not answer with "
is a solution" only. Either you post any kind of proof or at least something unexpected (like "
is the smallest solution). Someone that does not see that is a solution of the above without your post is completely wrong here, this is an IMO-level forum.
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.
has an incenter 
Suppose 
. Let be the midpoint of . Suppose that 
. 
meets 
again at point 
. Let points and be such that is the midpoint of 
and is the midpoint of 
. Point lies on the plane such that 
is a parallelogram, and suppose the angle bisectors of 
and 
concur on 
.
and 
meet at 
and 
. Prove that 
.
with another positive integer 
, as long as 
is a perfect square. For example, if the number on the blackboard is 
, Carmela can replace it with 
, because 
, then replace it with 
, because 
. If the number on the blackboard is initially , determine all integers that Carmela can write on the blackboard after finitely many operations.
with circumcenter 
and orthocenter 
, let 
be an arbitrary point on the circumcircle of triangle such that does not lie on line 
and that line 
is not parallel to line . Let be the point on the circumcircle of triangle such that 
is perpendicular to , and let be the point on line such that 
. Let and be the points on the circumcircle of triangle such that 
is a diameter, and 
and 
are parallel. Let be the midpoint of 
.
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)
be a cyclic quadrilateral. A circle centered at passes through and and meets lines 
and again at points and (distinct from 
). Let denote the orthocenter of triangle 
Prove that if lines 
are concurrent, then triangle and 
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. for which it is always possible to color crosses in two colors such that any balanced set has crosses of both colors.
and 
be the centers of the squares 
and 
Then 
and 
are congruent by SAS criterion, due to 
and 
Therefore, 
and 
If 
then quadrilateral 
is cyclic 
is right. Lines 
and 
are respectively parallel to 
and 
since 
are midpoints of 
Hence 
is isosceles right at 
and 
, and 
, so 
is congruent to 
, so 
. Also we know 
and 
, so 
.
such that 
satisfying 
and 
be the midpoint of the side 
of a triangle 
.
and 
are constructed externally on 
and 
with the right angles at 
is right-angled and isosceles.
,
.
.
.
which is the midpoint of 
.
is right-angled and isosceles 
.