[/quote]
,
be a cyclic quadrilateral. A circle centered at 
passes through 
and 
and meets lines 
and 
again at points 
and 
(distinct from 
)
denote the orthocenter of triangle 
Prove that if lines 
are concurrent, then triangle 
and 
are similar.
satisfying 
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.
,
onto 
past 
.
be on the perpendicular bisector of 
such that 
are on the same side of ![\[\frac12\angle ALE=1.4\angle ABE+3.4\angle ACE-558^\circ\]](http://latex.artofproblemsolving.com/c/6/3/c63eeb053f1fa8a29bef93516814ae121382219e.png)
Let the reflection of 
and 
be 
and 
,
and 
such that 
.
and 
intersect the circumcircles of 
and 
at 
and 
,
and 
intersect 
and 
at 
and 
.
intersect 
.
such that 
![\[
C(p-1, k-1) \equiv (-1)^{k-1} \pmod{p}
\]](http://latex.artofproblemsolving.com/4/0/7/4072235afbaa1f2d8d4c9ca044615a113a4306ce.png)
for 
,
is the binomial coefficient 
choose 
.
satisfy that: 
and 
.
intersect at points 
and 
and 
.
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 function such that for every regular 
-
we have 
.
for all reals 
.
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Here it is. 
I wrote this article here sometime ago, and made the PDF file today.
Please post your comments and let's discuss about it.
Here is the final version (version 6) of this article in PDF:
There was a solution for IMO ShortList 2007, Problem N5 in older versions of article, but now there isn't. I removed the solution of this problem because it had a long solution with [almost] no use of LTE. But I attached the solution of this problem for those who are interested.
I wrote this article here sometime ago, and made the PDF file today.
Please post your comments and let's discuss about it.
Here is the final version (version 6) of this article in PDF:
There was a solution for IMO ShortList 2007, Problem N5 in older versions of article, but now there isn't. I removed the solution of this problem because it had a long solution with [almost] no use of LTE. But I attached the solution of this problem for those who are interested.
This post has been edited 12 times. Last edited by Amir Hossein, Jan 3, 2025, 8:59 PM
for prime 
has many properties in common with logarithms. Examples:
is odd, then 
.
would be 
,
.
such that 
.
since 
changes this to 
or 
.
for some integer 
and 
for an integer 
is the 
by 
.
and 
show. You should change this to be coherent with a lot of other texts on this subject.
and 
.
and 
for some integer 
.
.
.
must be a pure power of 
.
or equivalently 
.
.
and the RHS is 
.
.
.
,
.
,
separately. First 
.
or 
.
and 
both powers of 
.
.
and 
.
which is true. This is the only solution to this case.
.
.
.
.
.
is the only solution.
,
notation. 
,
contradicts the statement.
where 
be a prime divisor of 
.
to be the order of 
is the minimal odd divisor of 
.
.
for some 
and 
.
.
.
.
,
for some odd 
.
and you'll understand it by yourself.
,
