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.
different point-like cities were founded one by one; from the second onwards, each newly founded city and the nearest existing city (the older one, if there were two) were declared sister cities. The surviving map of the empire shows the cities and the distances between them, but not the order in which they were founded. Historians have tried to deduce from the map that each city had at most 41 sister cities.
, give a map from which this deduction can be made.
, this conclusion cannot be drawn from any map.
is divided into cyclic quadrilaterals with pairwise disjoint interiors. None of the vertices of the cyclic quadrilaterals in the decomposition is an interior point of a side of any cyclic quadrilateral in the decomposition or of a side of the quadrilateral . Prove that is also a cyclic quadrilateral.
and a positive integer 
, we denote the -fold composition of by 
. A polynomial with real coefficients is called perfect if for each integer there is a positive integer so that 
is an integer. Is it true that for each perfect polynomial , there exists a positive 
so that for each integer there is 
for which is an integer?
be a positive integer, with divisors 
. Prove that 
is always less than 
, and determine when it is a divisor of .
and 
be sequences of real numbers for which 
and
for all positive integers . Prove that is eventually increasing (that is, there exists a positive integer 
for which 
for all 
).
be an integer. Let 
, 
, ..., 
be positive real numbers such that ![\[n^2 + 1 > \left( t_1 + t_2 + \cdots + t_n \right) \left( \frac{1}{t_1} + \frac{1}{t_2} + \cdots + \frac{1}{t_n} \right).\]](http://latex.artofproblemsolving.com/1/a/b/1ab410ebc17d9f6c38a3e7ae0f6efa955ffddfd3.png)
Show that 
, 
, 
are side lengths of a triangle for all 
, 
, with 
.
be an acute triangle with incenter 
and circumcenter 
. The incircle touches sides 
and 
at 
and 
respectively, and 
is the reflection of 
over . The circumcircles of 
and 
meet at 
, and the circumcircles of 
and meet at a point 
, where 
is the midpoint of 
. Prove that if 
and meet at 
, then 
.
and 
Prove that
such that divides 
but 
does not divide 
.
with right angles at and 
has an inscribed circle with center and radius 
. Let the lengths of the parallel sides 
and 
be 
and 
, and the intersection of the diagonals 
and 
be 
. to one of the trapezoid sides has the length . and as a function of and .
be a point on the cicumcircle of . Given a point 
, denote by 
the intersection points of the lines 
with the circumcircle. Prove that the intersection points of the lines 
and 
, 
and 
, 
and are collinear.
are two arbitrary points in the plane of with circumcircle 
cut 
again at 
and 
cut again at 
cut 
at 
Then are collinear.
are collinear.
and 
.![\[P=\frac{1}{5-2a}+\frac{1}{5-2b}+\frac{1}{5-2c}.\]](http://latex.artofproblemsolving.com/c/3/7/c3708768a84c2bbb1e4fae502a9bb1e92cb13d02.png)
.
implies 
collinear. Similarly 
collinear. Thus we get the conclusion.