with![\[2x^3zf(z)+yf(y) \ge 3yz^2f(x)\]](http://latex.artofproblemsolving.com/a/b/f/abf10d7e679f6f04db556424a0a0eda26f371fd4.png)
for all 
.
intersects 
at 
,
.
intersect again at 
. If 
is the tangent point of the 
be a function taking in positive integers and outputting nonnegative integers, defined as follows:
is the number of positive integers 
with 
such that the equation 
has an integer solution 
.
such tha
Adit Aggarwal, India.
Y by Adventure10, Mango247
The internal angle bisectors of 
, 
, and 
in 
concur at 
and intersect the circumcircle of at 
, 
, and 
, respectively. The circle with diameter 
intersects 
at 
and 
; the circle with diameter 
intersects 
at 
and 
; the circle with diameter 
intersects 
at 
and 
. Show that , , , , , and are concyclic.
, 
, and 
in 
concur at 
and intersect the circumcircle of at 
, 
, and 
, respectively. The circle with diameter 
intersects 
at 
and 
; the circle with diameter 
intersects 
at 
and 
; the circle with diameter 
intersects 
at 
and 
. Show that , , , , , and are concyclic.
,
and 
concur at 
(simple angle chasing). Let 
be common point of 
and 
.
is a radical axis of these two circles. Therefore 
) lets have three different circles. Well, radical axis of every pair is one of sides of triangle. But they are not neither parallel, nor concurrent (they should be, because they are radical axes of pairs of circles in triplet of circles) and that is contradiction.
,
and 
.
![\[\angle PLB = \angle NLB = \frac{\angle C}{2}\]](http://latex.artofproblemsolving.com/7/0/e/70e677c92242863f4bc618a1cdf531853cdf470a.png)
and ![\[\angle LBP = \angle LBM = \angle LBC + \angle CBM = \frac{\angle A + \angle B}{2}\]](http://latex.artofproblemsolving.com/7/3/1/7316e7c468e62e9d4fefd8987df4d168152fafda.png)
then ![\[\angle PLB + \angle LBP = \frac{\angle A + \angle B + \angle C}{2}= 90^{\circ} \Rightarrow NL \perp BM\]](http://latex.artofproblemsolving.com/9/c/b/9cb30a52cc02a35c0f06c690fa817473bc80485b.png)
Thus P lies on 
and lies on 
and 
.![\[Pow (\odot(ID), B) = BD \cdot BE = BP \cdot BI = BH \cdot BJ = Pow(\odot(IN), B)\]](http://latex.artofproblemsolving.com/b/8/1/b81a310f40649e2da5267da5659a35fee72025f1.png)
so 
is cyclic. Analogously 
and 
are both cyclic.
,
and 
not concur at only point (nor are they parallel) the follows that 
