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.
, 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.
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.
such that 
where 
.
cannot be a perfect cube of a positive integer.
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.
be an acute triangle with 
, and let be the projection from 
onto . Let be a point on the extension of 
past such that 
. Let 
be on the perpendicular bisector of 
such that and 
are on the same side of and![\[\frac12\angle ALE=1.4\angle ABE+3.4\angle ACE-558^\circ\]](http://latex.artofproblemsolving.com/c/6/3/c63eeb053f1fa8a29bef93516814ae121382219e.png)
Let the reflection of across and be 
and , respectively. Let 
and 
such that 
. Let 
and 
intersect the circumcircles of 
and 
at 
and 
, respectively. Let 
and 
intersect 
and 
at and . Let intersect at . Prove that 
.
touches the sides 
at 
respectively. A circle through and encloses and intersects the line at points and . Prove that the midpoint of lies on the circumircle of 
.
touches the sides at respectively. A circle through and encloses and intersects the line at points and . Prove that the midpoint of lies on the circumircle of .
-centered):
touches the sides at 
respectively. Suppose that is a point on 
and let 
Prove that 
and animate on line 
Define 
Consider the following series of projective maps:
to through 
(
) to through the point at infinity along the pencil of lines perpendicular to 
(
) to 
through any point. (
) swapping and 
(
)
to through a well-known lemma. In order to show that this composition of maps is the identity map, it suffices to check that 
for 3 cases of 
Pick 
the first two are easy to check by the Iran Lemma, and the last is trivial.
. As 
, it suffices to show that 
. Let 
and 
; then because 
, 
which can be checked through algebraic expansion.
and now since 
we have by McLaurin that 
where is the midpoint of , hence 
is cyclic thus we are done 
satisfying 
and 
a
,
