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.
(where 
are the digits and 
) we define the \textit{permutation sum} as the value 
For example the permutation sum of 20253 is 
Let 
and 
be two fivedigit integers with the same permutation sum.
.
be a triangle. Let 
be a points on the segment 
such that 
. Let 
be the mid-point of 
. Let 
intersect 
in 
and 
in 
respectively. Determine 
.
, Its incircle 
and 
escircle 
are tangent to at 
and 
respectively. 
intersects line at 
. Line intersects at , and line intersects at 
. Line 
intersects at 
. Prove that 
.
such that:
be a triangle with circumcenter 
. Let 
be the midpoints of 
and 
respectively and let 
be the projection of on 
. Let be the projection of 
on 
. Let 
intersect 
at points 
and 
. Let 
intersct at points 
. Let 
intersect 
at 
and 
intersect at 
. Then show that 
are concurrent.
and 
respectively be the incentre, circumcentre and circumcircle of triangle . Points 
are chosen on , such that 
, and point 
is the foot of the altitude from 
to 
. If 
are similarly defined, prove that lines 
and 
concurr on 
.
be a sequence that consists of an initial block of 
positive distinct integers that then repeat periodically. This means that 
are distinct positive integers and 
for every positive integer . The terms of the sequence are not known and the goal is to find the period . To do this, at each move it possible to reveal the value of a term of the sequence at your choice. is one of the first 
prime numbers, find for which values of there exist a strategy that allows to find revealing at most 
terms of the sequence.
such that, for all real numbers 
with 
, 
of positive integers for which there exist positive integers 
and 
such that
holds for all integers 
(Note that 
denotes the greatest common divisor of integers 
and 
)
is defined by 
and, for each 
, the number 
is the smallest positive integer greater than 
that has the following property: for each 
, the number 
is not a perfect square. Prove that, for all 
, it holds that 
.
. Now suppose that 
for 
where 
. Let's prove that 
. Consider 
and 
. and have no common element. Assume that 
is not in . 
and 
hence there must exists some 
such that both 
and 
are perfect squares for 
.![\[\sqrt{(N-2)^2+l+m+a_i}\geq \sqrt{(N-2)^2+l+a_i}+1\]](http://latex.artofproblemsolving.com/3/c/0/3c082456e591ba185d3e199ecc3f963fa3dbfc10.png)
![\[(N-2)^2+l+m+a_i\geq (N-2)^2+l+a_i+1+2\sqrt{(N-2)^2+l+a_i}\]](http://latex.artofproblemsolving.com/9/3/b/93b4df6d3e72787bdfab8d1efa423db8a5f5a348.png)
![\[2N-4>m\geq 1+2\sqrt{(N-2)^2+l+a_i}\geq 1+2(N-1)\geq 2N-1\]](http://latex.artofproblemsolving.com/5/d/a/5da118493cf2a5703b1b6d374b0d9bd9f6930052.png)
Which is impossible as desired.
. It comes down to a single lemma:
for wich 
, for each 
.
we would have: 
.
, but that's a contradiction because 
. 
such 
(
to be precise) there must be at least one 
for which there is no with 
and that means , as we wanted.
Prove that
and let 
and 
and 
respectively. Let the circle with centre 
and 
.
,
and 
,
.
and 
be feet of perpendiculars from 
and 
respectively. Prove that 
,
and 
are concurrent.
s
