AoPS Academy
(i.e., the square on 
has as one of its sides and lies outside ). Show that the lines drawn from the vertices 
, 
, and 
to the centers of the opposite squares are concurrent.
be positive integers. Maximize 
if 
.
![$\{x_n\} \subseteq [0,1]$](http://latex.artofproblemsolving.com/8/e/4/8e4206325387485d73765c9d7070ecb1ce18ff60.png)
, there exists some constant 
, for any 
, among the sequence 
and 
could be chosen to satisfy 
and 
.
such that
diverges for 
(harmonic sequence) but converges for 
because 
, is there a value between and such that 
converges![$f(x)\in R[x]$](http://latex.artofproblemsolving.com/a/1/5/a15dc97c051e81ae9ff0c060eaa4367bc723ccc9.png)
show that 
has at least one root of multiplicity one
of integers satisfying![\[x_0 = 0, |x_n| = |x_{n-1} + 1|\]](http://latex.artofproblemsolving.com/4/8/e/48e7b6f98f2062879377a9a2e507a63eac9269a8.png)
.
.
therefore u can use 1 and -1 alternating coming to the minimum of 0, because there is sequence is periodic with the period 4, and the first 4 besides 
are 0 and they repeat themselves 250 times. also 0 is the smallest number possible, since 
when is positive. In other words, you can't go from 1 to -1.
. 24 more numbers. Then, you go
Prove that
holds inequality 
![\[
\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}
\frac{1 - e^{-2} \cos\left(2\left(u + \tan u\right)\right)}
{1 - 2e^{-2} \cos\left(2\left(u + \tan u\right)\right) + e^{-4}}
\, \mathrm{d}u
\]](http://latex.artofproblemsolving.com/9/1/1/91165e6b531233c56b799ed32ded982d57b6c87a.png)
,
and 
and so forth.
.
and this means that 
or 
since we want the minimum. Then 
,
,
,
.