[/quote]
,
such that 
.
be an interior point for it such that 
.
When does equality hold?
be positive reals. Prove that there is a cyclic permutation 
of 
such that the inequality:![\[
\frac{a}{xa + yb + zc} + \frac{b}{xb + yc + za} + \frac{c}{xc + ya + zb} \geq \frac{3}{x + y + z}
\]](http://latex.artofproblemsolving.com/5/5/3/553527fc40a1b7d29dbdfb4b5e63635308f46c85.png)
and 
.
be real numbers such that 
.![\[
3 + (2 - \sqrt{3}) \cdot \frac{(b-c)^2}{b+(\sqrt{3}-1)c} \leq a+b+c
\]](http://latex.artofproblemsolving.com/5/a/8/5a83eaa7470bcca6546599ff4e296e4efec69b16.png)
and determine all the cases when the equality occurs.
have the same incenter 
is concyclic iff 
is concurrent
be the nonnegative integers in 
where 
.
.
.
such that 
where 
,
?
where 
,
be a scalene triangle with circumcircle 
and circumcircle 
.
and 
as the midpoints of 
and 
,
to 
.
meets 
,
intersects 
.
at 
and 
at 
.
intersect 
,
intersect 
.
is the reflection of 
and 
lies on the circumcircle of 
.
be real numbers such that 
and 
and determine the equality cases.
be positive real integer such that 
Prove that
such that 
and 
satisfying :![\[n^{2p}=m^2+n^2+p+1\]](http://latex.artofproblemsolving.com/7/4/e/74ee592cb8b3f3a3996ea34f40b78fde9d3672fc.png)
and 
are integers and 
is a prime number.
Approach: I'm trying to factor 
to get 
.
such that 
divides 
.
Y by Adventure10, Mango247
Let 
be triangle with circumcircle 
of fixed 
, 
and not a diameter. Let 
be the incenter of the triangle and 
. The circle passing through 
and tangent to 
cuts for second time at 
(
). 
cut also at 
respectively.
a) Let
. Prove that 
goes through a fixed point.
b) Suppose
cut also at 
respectively and 
. On 
take 
for 
. Let 
be a point of the circimcircle of triangle 
so that 
. Prove that the midpoint of 
belongs to a fixed circle.
be triangle with circumcircle 
of fixed 
, 
and not a diameter. Let 
be the incenter of the triangle and 
. The circle passing through 
and tangent to 
cuts for second time at 
(
). 
cut also at 
respectively.a) Let
. Prove that 
goes through a fixed point.b) Suppose
cut also at 
respectively and 
. On 
take 
for 
. Let 
be a point of the circimcircle of triangle 
so that 
. Prove that the midpoint of 
belongs to a fixed circle.Attachments:
This post has been edited 1 time. Last edited by parmenides51, Aug 27, 2018, 11:37 AM
Reason: added figure, made by Tran Quang Hung
Reason: added figure, made by Tran Quang Hung
is supposed to be fixed as well?
(WLOG that 
is bigger). This implies that 
passes through the midpoint of arc 
.
in other words that 
is harmonic. Now, by inversion at 
with power 
,
is also harmonic. Let 
.
onto line 
we get that 
As 
also, we have that 
and 
and 
.
,
be the reflections of 
respectively. Then, let 
.
by the condition. Also, it's well-known that 
is the perpendicular bisector of 
.
and 
,
by projecting 
from 
onto line 
.
,
.
.
,
,
is 
. 
. 
= 
. 
. 
. 
= 
= 
= 
= 1 or 
,
,
concurrent at 
which not containing 
,
,
,
,
then: 
be the center of circle through 
be the midpoint of arc 
is the 
s
s
.
.
concur , say at 
is the midpoint of arc 
.
be 
and 
which is a fixed line.
suffices which can be done using harmonic bundles and Butterfly theorem (Same as Pathological)
inversion 
are concurrent at 
and point 
are pairs of lines that involution swaps. By projecting involution from 
.
.![\[ -1=Z(I,P;Q,\infty_{IQ}) \overset{\text{rotate} 90^{\circ}}{=} O(T,\infty_{AT};I_a,P') \implies T \ \text{is the midpoint of} \ I_aP' \implies P'=I\]](http://latex.artofproblemsolving.com/8/c/d/8cd50d2c7cb7afd55671535b4b7366b9ed7d2cc4.png)
)
and the midpoint 
