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 an acute triangle incribed in a circle 
.Let 
be the midpoint of 
.Let 
and 
be altitudes from 
and 
of triangle , respectively, and let them intersect at 
.Let 
be the intersection point of tangents to the circle at points 
.Prove that 
and 
are concurrent.
and 
. Prove that:
+4 w
of acute triangle is 
. The ray drawn from through the circumcenter 
intersects at 
. Let the midpoint of 
be 
. Define 
, 
, 
, 
, 
, 
similarly. Prove that 
, 
, 
are concurrent.
to 
(including ). He writes some of the numbers on the numerator, and writes 
(multiplication) between each number. Then he writes the rest of the numbers in the denominator and also writes between each number. There is at least one number both in numerator and denominator. The teacher ensures that the fraction is equal to the smallest possible integer possible.
. Prove that ![\[\sqrt{a^2+1}+\sqrt{b^2+1}+\sqrt{c^2+1}\le \sqrt{2}(a+b+c)\]](http://latex.artofproblemsolving.com/6/e/d/6ed654c1d3616a8274ff0838e7bb13dfa0a11b8a.png)
light bulbs are operated by some switches. Each switch works on a subset of the light bulbs. When we use a switch, all the light bulbs in the subset change their state: bulbs that were on turn off, and bulbs that were off turn on. We know that every light bulb is operated by at least one of the switches. Initially, all lamps were off. Find the biggest number 
for which we can surely turn on at least light bulbs.
denote the infinite parallel ruler with the parallel edges being at distance 
from each other. The following construction steps are allowed using ruler 
:
parallel to a given line 
at distance (there are two such lines, both of which can be constructed using this step); and 
with 
two parallel lines at distance such that one of them passes through , and the other one passes through (if 
, there exists two such pairs of parallel lines, and both can be constructed using this step).
be a given positive integer. Prove that based on the three points and there exists 
such that for any 
it is possible to construct points using only on one of the excircles of the triangle.
and 
and 
is non-decelerating if 
holds for all positive integers . We say that a strictly increasing sequence 
is convergence-inducing, if the following statement is true for all real sequences 
: if subsequence 
is convergent and tends to 
for all positive integers 
, then sequence is also convergent and tends to . Prove that a non-decelerating sequence is convergence-inducing if and only if sequence 
, 
, 
is bounded from above.
light bulbs are arranged around a circle, and are consecutively numbered with
. Each bulb can be in one of two states: either it is on or off. In the initial configuration, days we change the current on/off configuration as
, on the -th day we start from the -th bulb and moving in clockwise direction-th day, coincides with the initial one.
let 
and 
denote the feet of the altitudes from 
and , respectively. Let line 
intersect circumcircle at points 
. Similarly, let line 
intersect circumcircle at points 
. Prove that the radical axis of circles 
and 
passes through the orthocenter of triangle
, 
, and![\[
a_{2k+1} = 2 + 2a_k, \quad a_{2k+2} = 2 + a_k + a_{k+1},
\]](http://latex.artofproblemsolving.com/4/e/c/4ec892df19b4ec97af9bb1e5d1954e6c26be3172.png)
for all integers 
. Determine all positive integers 
such that![\[
\frac{a_n}{n}
\]](http://latex.artofproblemsolving.com/3/d/1/3d1d012e7c41e1e09abd14626a79b1fb1cbab34d.png)
is an integer.
where 
?
?
?
divisible by 
prime numbers prime
is a prime number, show that 
is composite.
or 
.
,
,
which is impossible, so 
.
.
such that 
is a perfect cube.
,
,
.
where p is a prime,then 
where 
we have 
,
If for an odd prime 
we have 
,
which is also impossible(it is also clear that k=0 or k=1).Then 
.
then 
,
.
or 
It is trivial that 
.
.
has residue 
mod 
,
.
,
.
doesn't divide 
,
so 
,
.
take 
take 
take 
a succession of positive integers such that for every couple of positive integers 
we have 
. Prove that there exists a positive integer 
such that 
.
is obvious. Let 
be called a good pair if 
for some 
.
are 
is a good pair iff 
.
,
then let 
,
to both sides gives 
so the claim is proven.
,
for 
and 
.
are 
pairwise good integers. First we show that 
are pairwise good. This follows from 
by induction and our definition of 
.
is a good pair. Note that 
as desired, so we have 
.
and 
such that 
and 
we have![\[
\text{gcd}(x+y,mn)>1
\]](http://latex.artofproblemsolving.com/0/e/3/0e3bae9e964367fac00f7f728b1db4fecf514ba3.png)
be the subset of primes dividing 
the subset of primes dividing 
to be the set of overlap in 
prime factors. If none exist take it to be 
.
hence 
,
.
such for any 
:
.
and 
are coprime and 
,
-
.
.
t
.
