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.
that are twice differentiable and satisfy 
Find limit of A(n) when n tend to +oo
.
.
1 (mod 7)
are of order 2. 
is of order 7? No elements of order 4 in in the Sylow 7-subgroup.
.
1 (mod 2)
, does this mean that there are no elements of order 2 in the Sylow-2 subgroup, but only elements of order 4. in Dih(28)
is a continuous function on ![$[0,1]$](http://latex.artofproblemsolving.com/e/8/6/e861e10e1c19918756b9c8b7717684593c63aeb8.png)
such that 
, then there exists a 
such that:![\[
F(c) = \frac{1}{c} \int_0^c F(x) \,dx
\]](http://latex.artofproblemsolving.com/8/f/e/8fea4d84bc98b6f4953e19976cfa554945a37ea6.png)
be an integer. Show that, if 
is an integer, then it is a perfect square.[/quote]
. If is an integer, then 
is at least rational, so that 
must be a perfect square then. Using Conway's topograph method, I have found out that the smallest non-negative pairs 
for which this happens are 
and 
, and that, for every such pair 
, the "next" such pair can be calculated as
The eigenvalues of that matrix are irrational, however, so that any calculation which uses powers of that matrix is a little cumbersome. There must be an easier way, but I cannot find it. Can you?
interesting if is composite and for every divisor 
of at least one of 
and 
is also a divisor of with 
, such that 
and 
.
such that 
is prime, 
with 
and ?
be an acute triangle with orthocenter 
and circumcenter 
. Let 
and be points on the line segments 
and 
respectively such that 
is a parallelogram. Prove that 
.
that satisfy 
for all 
and 
satisfies the following recursion formulas: 
, 
and 
. Find 
and 
be a triangle, and let 
and 
be points on segment symmetric with respect to the midpoint of . Let 
denote the circle passing through and tangent to line at 
. Similarly, let 
denote the circle passing through and tangent to line 
at 
. Let the circles and intersect again at point 
(
). Prove that 
.
houses on a straight street are one or more boys. At what point should all the boys meet so that the sum of the distances that they walk is as small as possible?
. Let 
the meeting point.
, notice that if is not on 
, we can say 
or 
. If is on , 
. We conclude that 
, for all 
.
. To achieve the equality note that, by definition, 
for 
. must be on 
.
, notice that if is not on 
, we can say 
or 
. If is on , 
. We conclude that 
, for all . Moreover, 
.
. To achieve the equality note that, by definition, 
for and 
if and only if 
.
is on 
, so must be on .
