[/quote]
,
such that 
is a perfect square.
,
and 
are in the rays 
and 
,
and 
.
.
be real numbers such that 
.
and 
.
-
,
be 
,
lies on arc 
,
intersects the circumcircle of 
.
and 
,
intersects 
at 
respectively, prove that 
are concurrent.
such that
Prove that
Let 
Prove that![$$ (a+b)(a+1)(b+1) \leq \frac{3}{2}+\sqrt[3]{6}+\sqrt[3]{36}$$](http://latex.artofproblemsolving.com/f/5/e/f5e68204507aa9c3e71c28844bbee3a804546b94.png)
be a positive integer larger than 
.
of subsets of 
having the following properties:
distinct sets of 
and 
be positive integers such that 
divides 
.
.
where 
are nine distinct integers. Prove that there exists an integer 
such that for all integers 
the number 
is divisible by a prime number greater than 20.
be the lengths of the sides of a triangle. Prove that![\[\frac{1}{(a-4b)^2}+\frac{1}{(b-4c)^2}+\frac{1}{(c-4a)^2} \geqslant \frac{1}{ab+bc+ca}.\]](http://latex.artofproblemsolving.com/7/4/b/74b7aa5621bc52977d9acd448d74293dc27a8e00.png)
of points in the plane is balanced if, for any two different points 
in 
in 
.
.
,
points.
let 
be the number whose 
-
-
.
.
from the set ![\[f(n)=\sum_{k=1}^{n} \gcd(k,n),\qquad n\in \mathbb{N}.\]](http://latex.artofproblemsolving.com/2/a/5/2a5a20395450d37d0352ab75e9e4875ab7803805.png)
for every two relatively prime 
.
the equation 
has a solution.
such that the equation 
.
Y by loveGeometrytothemoon
Let 
be a triangle. Point 
moves on the opposite ray of 
such that 
. Let 
be the midpoint of 
. 
cuts 
at 
a) Suppost that
is the circumcenter of 
and 
is the circumcenter of 
, Prove that 
b) Let
be the midpoint of 
and 
be the orthocenter of triangle 
. Prove that when 
varies on the opposite ray of 
, 
go through a fixed point
be a triangle. Point 
moves on the opposite ray of 
such that 
. Let 
be the midpoint of 
. 
cuts 
at 
a) Suppost that
is the circumcenter of 
and 
is the circumcenter of 
, Prove that 
b) Let
be the midpoint of 
and 
be the orthocenter of triangle 
. Prove that when 
varies on the opposite ray of 
, 
go through a fixed pointThis post has been edited 1 time. Last edited by 799786, Mar 4, 2022, 10:31 AM
Reason: typo
Reason: typo
intersects 
at 
and 
,
and 
.
.
and 
are corresponding points. So 
and therefore 
.
be the midpoints of 
,
.
and 
have the same radius, so 
are orthocenter of 
are collinear.
.
be second intersection of 
and 
.
we have 
.
and 
.
are collinear and 
are collinear. We also have 
.
.
or 
.
are midpoints of 
.
is parallelogram. So 
.
then 
and 
.
.
which is a fixed point.
