[/quote]
,
such that 
,
,
with center 
and radius 
intersects the circle 
be a point on the circle 
and 
.
.
.
and from 
and 
as the reflections of 
and 
be the circumcircle of 
.
the second intersection of line 
with 
the intersection of ray 
with 
is the circumcenter of 
can be inscribed within a circle.
![\[\sqrt{a+h_B}+\sqrt{b+h_C}+\sqrt{c+h_A}=\sqrt{a+h_C}+\sqrt{b+h_A}+\sqrt{c+h_B}\]](http://latex.artofproblemsolving.com/5/d/6/5d6881be29327bcdc3e7bd76c221968578c7f50b.png)
is isosceles.
be all the divisors of 
.
are all divisors for some 
(
)
be a sequence of positive real numbers that satisfies 
and 
for every natural number 
for every natural number 
and 
.
?
be positive real numbers such that : 
. Prove that : 
be the set of positive integers. Find all functions 
such that![\[ m^2 + f(n) \mid mf(m) +n \]](http://latex.artofproblemsolving.com/2/f/4/2f409d1de993f1af8fd839bb8e9f87a57e1b8608.png)
and 
(ans- 
)
satisfying 
Prove that 
When does equality hold?
Y by Adventure10, Mango247
On the circle 
six points 
are on the circle counterclockwise. 
is the diameter of the circle and it is perpendicular to 
Also, lines 
is concurrent. Let 
be the foot of altitude from 
to 
and let 
be the foot of altitude from 
to 
Prove that the area of 
is less than half the area of 
six points 
are on the circle counterclockwise. 
is the diameter of the circle and it is perpendicular to 
Also, lines 
is concurrent. Let 
be the foot of altitude from 
to 
and let 
be the foot of altitude from 
to 
Prove that the area of 
is less than half the area of 
![[asy]
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[/asy]](http://latex.artofproblemsolving.com/7/b/7/7b72d490296f469b91d9a2b5e9ded68b15dc84aa.png)
be the intersection of 
and 
and 
are cyclic and it follows that 
and 
.
as well (how to prove this? And what next?)
be 
,
and 
Therefore, we know that 
and so there is a homthety at 
sending 
to 
Hence, we have that ![$[\triangle CMN] = \frac{CX^2}{CY^2} [\triangle CAE].$](http://latex.artofproblemsolving.com/d/1/3/d134b5079b65a8232414ef3dfc12f243d1a7ba3e.png)
Also, observe that ![$[\triangle CAE] = \frac{CZ}{CF} \cdot [CAFE]$](http://latex.artofproblemsolving.com/7/d/6/7d6ebb329680f6f22bf728b81fd694cb598db3d7.png)
.
,
comes into play. Notice that 
,
Observe that 
.
,
and 
we have that 
Now, we get that 
,
