AoPS Academy
be a cyclic quadrilateral. A circle centered at 
passes through 
and 
and meets lines 
and 
again at points 
and 
(distinct from 
). Let 
denote the orthocenter of triangle 
Prove that if lines 
are concurrent, then triangle 
and 
are similar.
+1 w
such that 
, where 
are non-negative reals such that 
.
.
be a triangle with orthocenter and let 
be the second intersection of the circumcircle of triangle 
with the internal bisector of the angle 
. Let 
be the circumcenter of triangle 
and 
the orthocenter of triangle 
. Prove that the length of segment 
is equal to the circumradius of triangle .
such that the following property holds: if we list the divisors of 
in increasing order as 
, then we have![\[ d_2 - d_1 \leq d_3 - d_2 \leq \dots \leq d_k - d_{k-1}. \]](http://latex.artofproblemsolving.com/a/d/c/adc94af88b3579164aabba51bd021442133e364a.png)
that satisfy![\[
f(x^2-y)+2yf(x)=f(f(x))+f(y)
\]](http://latex.artofproblemsolving.com/f/b/b/fbb693770cd90d8a144026608074122e54da5802.png)
for all 
.
be the incircle of a fixed equilateral triangle . Let 
be a variable line that is tangent to and meets the interior of segments and 
at points and 
, respectively. A point 
is chosen such that 
and 
. Find all possible locations of the point , over all choices of .
be positive real numbers such that ![\(a+b+c=4\sqrt[3]{abc}\)](http://latex.artofproblemsolving.com/2/8/2/28214a1814762b71858db1543fae1a50212eaafa.png)
. Prove that ![\[2(ab+bc+ca)+4\min(a^2,b^2,c^2)\ge a^2+b^2+c^2.\]](http://latex.artofproblemsolving.com/7/0/4/7043a170ac5535465a5a1b4daccb1f1de3043280.png)
are multiples of 
or 
but not 
?
![\[222{,}222-22{,}222-2{,}222-222-22-2?\]](http://latex.artofproblemsolving.com/1/5/6/156c3662fe4c874a80e54887f76c302c13b9d14c.png)
So i think that, as an Olympiad problem, this problem deserves a place here, but if it's more convenient to place it in the intermediate forum, it doesn't matter 
so every number of the form 
can be written in that form, and only such numbers do. hence the answer to the first problem is 
minus the last term. so these sums are equal to 
So numbers that can be written in both ways are those numbers of the form 
so the rest is easy.
so every number of the form can be written in that form, and only such numbers do. hence the answer to the first problem is minus the last term. so these sums are equal to So numbers that can be written in both ways are those numbers of the form so the rest is easy.
not , I assume, you missed the case 
so every number of the form can be written in that form, and only such numbers do. hence the answer to the first problem is minus the last term. so these sums are equal to So numbers that can be written in both ways are those numbers of the form so the rest is easy. not , I assume, you missed the case gives 0, which is not a positive integer.
