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.
has circumcircle 
and circumcenter 
. A circle 
with center 
intersects the segment 
at points 
and 
, such that 
, , , and 
are all different and lie on line in this order. Let 
and 
be the points of intersection of and , such that , , , , and lie on in this order. Let 
be the second point of intersection of the circumcircle of triangle 
and the segment 
. Let 
be the second point of intersection of the circumcircle of triangle 
and the segment 
.
and 
are different and intersect at the point 
. Prove that lies on the line 
.
be a point inside a triangle 
such that![\[ \angle APB - \angle ACB = \angle APC - \angle ABC.
\]](http://latex.artofproblemsolving.com/a/6/e/a6e3dc27d0457682fa22af4f918f58d9cd8403bc.png)
, 
be the incenters of triangles 
, 
, respectively. Show that the lines 
, 
, 
meet at a point.
cups labeled 
, where the 
-th cup has capacity liters. In total, there are liters of water distributed among these cups such that each cup contains an integer amount of water. In each step, we may transfer water from one cup to another. The process continues until either the source cup becomes empty or the destination cup becomes full.
Prove that from any configuration where each cup contains an integer amount of water, it is possible to reach a configuration in which each cup contains exactly 1 liter of water in at most 
steps.
Prove that in at most 
steps, one can go from any configuration with integer water amounts to any other configuration with the same property.
for which 
divides 
.
and 
, 
cannot be a power of 
.
of natural numbers such that for every pair of natural numbers 
and 
, the following inequality holds:![\[
\frac{1}{2} < \frac{\gcd(a_r, a_s)}{\gcd(r, s)} < 2
\]](http://latex.artofproblemsolving.com/1/6/7/167679c1707b957d87311298ea5b72347a9bdc45.png)
such that 
for all 
such that for all real numbers 
and 
,![\[f(xf(y))+f(y)=f(x+y)+f(xy).\]](http://latex.artofproblemsolving.com/d/2/f/d2ff6ed448cf39e7ec9bce6b944965d5e89b9878.png)
table filled with natural numbers, we say it is a divisor table if:
-th row are exactly all the divisors of some natural number 
,
-th column are exactly all the divisors of some natural number 
,
for every 
.
is given. Determine the smallest natural number 
, divisible by , such that there exists an divisor table, or prove that such does not exist., a line 
to tangent , and another circle disjoint from such that and lie on opposite sides of . The tangents to from a variable point on meet at 
and 
. Prove that, as varies over , the circumcircle of 
is tangent to two fixed circles.
be a triangle and let 
be its incircle. Denote by 
and 
the points where is tangent to sides and 
, respectively. Denote by 
and 
the points on sides and , respectively, such that 
and 
, and denote by 
the point of intersection of segments 
and 
. Circle intersects segment at two points, the closer of which to the vertex is denoted by 
. Prove that 
.
![\begin{align*}
\frac{a}{c} &= b-\sqrt{b}+c \\
\sqrt{\frac{a}{c}} &= \sqrt{b}+1 \\
\sqrt[4]{\frac{a}{c}} &=\sqrt[3]{b}-1
\end{align*}](http://latex.artofproblemsolving.com/0/c/d/0cd22d36acedbcb10d76d1b884aa35418273ed52.png)
be positive integers. Prove that the expression![\[\frac{\gcd(a+b,ab)}{\gcd(a,b)}\]](http://latex.artofproblemsolving.com/1/b/7/1b78a2027c3b7156fbb566901d2cfd088cc09f00.png)
is always a positive integer, and determine all possible values it can take.
is continious. WLOG 
. First we prove by induction that 
and 
is increasing:
. By induction hypothesis 
and thus 
which implies 
. Also, 
and 
was increasing, thus 
which means 
.
for all , then it is bounded from above and hence has a limit, say 
. Then 
Now we prove that 
Suppose it isn't, then 
has a limit, say 
. Then 
And so 
But 
, and so 
goes to infinity, a contradiction. Thus, goes to infinity, and so does 
. As 
, we know that 
diverges.
. Take 
to be a big positive integer, and choose to be large enough that 
and 
. Then 
Great, now split the sum 
into sums depending on 
. We will consider one of them because the other are the same. 
because 
for 
. Thus, when 
goes to infinity we get 
Other sums can be considered in the same way. Thus, we get 
But now remember how we were choosing and . We could take to be large enough that 
and to be the floor of 
, and so 
. Then we can bound the sum above 
where 
is a small number bigger than . And so the series 
diverges, which is a contradiction. Thus there exists a positive integer such that 
.
satisfying
Prove
attaining all positive values and positive numbers 
are given. Numbers 
satisfy
for every 
.
.