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.
-excenter of 
is 
and 
. The foot from 
to 
is 
and 
. The intersection of 
and 
is 
. The circumcenter of 
is 
.
is tangent to the -excircle if is on the incircle of .
kawaii if it satisfies 
(
is the number of positive factors of , while 
is the number of integers not more than that are relatively prime with ). Find all that is kawaii.
with 
and 
. Lines and intersect at , and lines 
and 
intersect at 
. Let 
be the midpoints of sides 
, respectively. Let 
and 
be points on segment 
and 
, respectively, so that 
is the angle bisector of 
and 
is the angle bisector of 
. Prove that 
is parallel to 
if and only if 
divides into two triangles with equal area.
for which, for all 
, the following identity holds:![\[
f(x) f(yf(x)) + y f(xy) = \frac{f\left(\frac{x}{y}\right)}{y} + \frac{f\left(\frac{y}{x}\right)}{x}
\]](http://latex.artofproblemsolving.com/9/0/c/90c180110402e1a32b70edb2b0a03a28727457d1.png)
is chosen. Prove that there exists a prime number that divides infinitely many terms of the sequence 
, where![\[
b_k = a^{k^k} \cdot 2^{2^k - k} + 1.
\]](http://latex.artofproblemsolving.com/1/7/5/1751a60482d729a36c71b77ac9c978e724f40da0.png)
be a logarithmic spiral centered at the origin (ie curve satisfying for any point on it, line 
makes a fixed angle with the tangent to at ). Let 
be a rectangular hyperbola centered at the origin, scaled such that it is tangent to the logarithmic spiral at some point. on the spiral, the polar of wrt. is tangent to the spiral.
such that every positive integer appears exactly once and the product of the first terms is a perfect 
power ?
be non-negative real numbers such that ![\[b+c\leqslant a+1,\quad c+a\leqslant b+1,\quad a+b\leqslant c+1.\]](http://latex.artofproblemsolving.com/b/9/e/b9e86e898c0536b7323a03611d5bdbf679caa710.png)
Prove that 
be the set of black cells at the end of the game. Ali and Shayan respectively aim to minimize and maximise the perimeter of the shape by playing optimally. (The perimeter of shape is defined as the total length of the boundary segments between a black and a white cell.), assuming both players play optimally?
called ‘nice’ if and only if 
and 
are both perfect squares. How many positive integers 
such that 
are composites for all nice numbers ?
decide whether there exist pairwise distinct positive integers 
such that for every 
, 
divides 
.
of positive integers with the following property: all of the terms of the sequence 
have a greatest common divisor 
be the set of the first perfect squares of nonzero integers. Suppose that 
for some 
. Here 
stands for the set 
. Prove that 
holds for every 
.
be the set of the first perfect squares of nonzero integers. Suppose that for some . Here stands for the set . Prove that holds for every .
as the set of vertices and exactly one edge (potentially a loop) connecting two elements with sum 
for 
. For every path of length 
the difference of the two ends is the difference of the squares corresponding to the edges. We will count the number of paths of length in two different ways.
there are at most as many ways to write as a difference of squares as there are divisors of since 
is equivalent to 
. But for 
, there are 
divisors for every positive integer . So there are 
paths of length in the graph.
paths of length in the graph where 
is the degree of 
. (Here we use Jensen's on the convex function 
.) paths is in both and 
. So 
must be in 
for every 
.
