AoPS Academy
, let 
. Point 
lies on segment 
such that 
. Let 
and 
be the circumcenter and incenter of triangle 
, respectively. Let 
and 
be the circumcenter and incenter of triangle 
, respectively. Suppose that lines 
and 
intersect at point 
. Find the length of 
.
be positive reals. Prove that there is a cyclic permutation 
of 
such that the inequality:
holds for all positive real numbers 
and 
.
of positive integers fast-growing if 
for all 
. Also, for a sequence 
of real numbers and a sequence 
of positive integers, let us denote![\[
S(a, b) = \sum_{n=1}^{\infty} \left| a(b_n) + a(b_n + 1) + \cdots + a(b_{n+1} - 1) \right|.
\]](http://latex.artofproblemsolving.com/2/f/c/2fc080e56fa9d8d6b1cf8a0bb2cb404a65d9ec73.png)
, 
such that for every sequence , if all the series![\[
\sum_{n=1}^{\infty} a(n), \quad S(a, b) \quad \text{and} \quad S(a, c)
\]](http://latex.artofproblemsolving.com/0/8/a/08aa112d16a9c772145ff03a74fb512b4d02340d.png)
are convergent, then the series 
is also convergent?, , 
such that for every sequence , if all the series![\[
S(a, b), \quad S(a, c) \quad \text{and} \quad S(a, d)
\]](http://latex.artofproblemsolving.com/5/8/9/589be33c766fde9e227f56d280d02322585e2ba3.png)
are convergent, then the series is also convergent?
be holomorphic on 
, continuous on 
and such that the limit 
exists. Prove that 
attains its maximum on the circle 
.
. Prove that for all polynomial 
on 
, we have
Can I change 1 by a constant 
?
![$E\subseteq \{f:[0,1]\to \mathbb{R}\mid f \textnormal{ is Riemann-integrable}\}$](http://latex.artofproblemsolving.com/e/7/0/e701e0cca0851ee57e951ab823ee407e652574fe.png)
such that 
posseses the following properties:
If 
for 
with 
, then 
;
There exists 
with 
.![$E=\{f:[0,1]\to \mathbb{R}\mid f \textnormal{ is Riemann-integrable}\}$](http://latex.artofproblemsolving.com/9/1/f/91ff4dcf8ff60af8a83d6dd4d9abc45e8620a6c1.png)
.
be an 
real matrix with minimal polynomial 
. Prove that the trace of 
is zero.
![$C[a,b]$](http://latex.artofproblemsolving.com/e/e/6/ee63b22d81304485423eec80b34bbf21b655d75d.png)
is the set of all continuous functions on the closed interval ![$[a,b]$](http://latex.artofproblemsolving.com/8/e/c/8ecbd1ba3da8f2adef66a63f2ab32c47e63fa734.png)
. For, define![$$d(f, g) = \sup_{x \in [a, b]} |f(x) - g(x)|.$$](http://latex.artofproblemsolving.com/a/4/a/a4a62b37713788c94992853b0eaadcbabbca86d6.png)
(b) Consider the closedness and compactness of the following set in the space :![$$B = \{ f \in C[a, b] : |f(x)| \leq 1, \text{ for all } x \in [a, b] \}.$$](http://latex.artofproblemsolving.com/2/e/d/2ed13593e5bad1171aad457042dcdc9149d9d931.png)
be the open unit disk in the complex plane and let 
be a holomorphic function such that 
. Let the Taylor series of be 
. Prove that the number of zeroes of (counted with multiplicities) equals 
.
.
such that satisfies two following statements?
which contains 
.
if 
is an odd natural number, 
if is an even natural number ( is different from 0).
be two complex matrices of the same rank, and let 
be a positive integer. Prove that 
if and only if 
.
, because last four digits would have to be divisible by 
, but in each of this cases numbers 
are not. Similarly, we can see that 
leads to contradiction. is coprime to 
. We know that 
is coprime with 
and 
for all , so function 
is bijective on integers relatively prime to 
(we count modulo ). To see that this function is bijective, we can also see that 
iff 
divides 
, but 
cannot divide neither 
, nor 
if 
is not divisible by , so we would have to have 
. The same argument works modulo 
. Therefore, we can have 
. Obviously, we can also have 
, so our answer is 
.
and 
Prove that
such that for any positive integer 
there exists a positive integer 
are equal to 
