and 
Prove that
![$$ \left( 2a^2+\frac{2}{a}+\frac{1}{b}-\frac{1}{c}\right)\left( 2a^2+\frac{2}{a}-\frac{1}{b}+\frac{1}{c}\right) \ge 9\sqrt[3]{4}$$](http://latex.artofproblemsolving.com/f/a/a/faa1119662e27f6862ae52372f68ffb409e20964.png)
be the set of non-negative integers and 
be the set of positive real numbers. Let 
be a function such that 
and 
for all integers 
,
for all integers 
.
.
be an algebraic extension of 
,
be an algebraic closure of 
is an embedding such that 
for all 
,
extends to an automorphism of 
is an embedding. Then, there exists an embedding 
that extends 
,
is an isomorphism.
and 
that extends 
is also an algebraically closed field.
,
be the field of algebraic numbers, 
.
defined by
Then, according to the exercise above, 
How should we interpret 
?
,
.
such that![\[\mathcal{E}(B \cup C) = B \cup C.\]](http://latex.artofproblemsolving.com/e/c/3/ec34115c9e6e169c19c739220065f647934a1a5f.png)
.
and ![$f:[a,b]\rightarrow\mathbb{R}$](http://latex.artofproblemsolving.com/6/7/c/67c320754a36aaec178e61f04bfcbac41c0bf675.png)
be a strictly monotonous function such that 
.
.
for which there exists a scrictly monotonous function 
such that 
,
, ∀x∈[0,1].
Y by FadingMoonlight, noob_mathematician, TFIRSTMGMEDALIST, tiendung2006, Adventure10, Mango247
In an acute angled triangle 
with 
, let 
denote the incenter and 
the midpoint of side 
. The line through 
perpendicular to 
intersects the tangent from to the incircle (different from line ) at a point 
> Show that is tangent to the circumcircle of triangle 
.
Proposed by Tejaswi Navilarekallu
with 
, let 
denote the incenter and 
the midpoint of side 
. The line through 
perpendicular to 
intersects the tangent from to the incircle (different from line ) at a point 
> Show that is tangent to the circumcircle of triangle 
.Proposed by Tejaswi Navilarekallu
and 
be the points where the incircle touches the lines 
respectively. Let 
denote the antipode of 
e![[asy]
import cse5;import olympiad;size(7cm);
pair A=(3,6),B=(0,0),C=(10,0),I=incenter(A,B,C),M=(B+C)/2,P,D=foot(I,B,C),X,Dp,Ee;
P=extension(A,rotate(90,A)*I,M,reflect(I,M)*B);X=foot(I,P,M);Dp=2*I-D;Ee=2*M-D;
fill(A--B--C--cycle,cyan+opacity(0.25));
filldraw(circle(I,abs(I-D)),cyan+opacity(0.25),green);
draw(A--Dp--X--MP("E",Ee,S)^^X--D,dotted);
draw(MP("I",I,W)--A--MP("P",P,E)--MP("X",X,SE)--M--MP("D",D,S)--I--X,brown);
draw(M--I--P,darkcyan);
draw(MP("A",A,NW)--MP("B",B,S)--MP("M",M,S)--MP("C",C,S)--cycle,blue);
draw(I--MP("D'",Dp,SE),dashed);
dot(A^^B^^C^^M^^I^^M^^P^^D^^X^^Dp^^Ee);
[/asy]](http://latex.artofproblemsolving.com/e/8/b/e8b15a90f69b0bbb8b3ca6e875efe70f5d224cb8.png)
are collinear. Note that 
,
,
.
because 
are collinear. Now, the positive homothety centered at 
are collinear, proving the claim.
and 
are both cyclic quadrilaterals, so
Here the last equality follows from the fact that 
and 
are congruent right triangles. Thus 
,
.
and 
point of tangency to incircle from 
cyclic so 
.
so it suffices to prove 
or 
.
where 
.
collinear and since 
thus 
and 
so we are done
.
be tangent to incircle 
is cyclic. Since, 
is cyclic. Let 
be the 
.
collinear. Hence, 
.
Hence, 
- midline in 
, so 
.
,
meet it at 
where 
.
since, evidently, 
.
.
,
incircle, as we need to prove 
(as 
is cyclic) 
.
,
(as 
is cyclic and 
,
,
circumcenter of 
is midpoint of 
.
.
,
,
,
is cyclic, and let 
.
Then just angle chase:
be the intersection of 
be the tangency point of 
collinear. and 
.
are collinear. Also, since 
,
is cyclic.
is also cyclic, we have 
,
,
be the tangency point mentioned.
is the 
which we call 
is cyclic .
.
,
.
such that 
,
.
and 
passes through the antipode of 
is the intersection of mixtilinear incircle and the circumcircle and furthermore 
are isogonal. Hence,
where we used the fact that 
is cyclic due to proper right angles. Done.
i
and since 
,
also 
,
,
or 
.
incircl
,
diameter.So 
,
.
at 
,
,
(since 
is cyclic)
is parallel to 
are collinear.
.
.
and 
.
Are the touch points of the incircle with the sides
And notice that 
because of the 90 degree angles which gives us that the condition is equivalent to prove 
tangent to 
of 
of 
,
is cyclic.
,
,
is a diameter of the incircle and ray 
and 
to 
be the circumcenter of 
,
passes through the intersection points of 
and incircle , Thus 
, say 
and since 
are collinear we can say 
but 
, Inverting back we get 
. 
excircle touch point on 
be the reflection of 
be the circle with centre 
.
.
is cyclic
.
is what we want.
lies on 
is the diameter of 
.
is the diameter of the incircle 
are collinear.
are collinear, thus 
all are collinear.
w.r.t. base 
and we are done 
.
and 
is cyclic so 
because 
and 
.
is tangent to the incircle and that 
.
,
being cyclic with diameter 
.
be the point other than 
is tangent to the incircle. (So 
cyclic.
.
intersect the incircle at 
.
be the excenter. Note that 
,
,
and 
are also 
,
(tangency).
.
,
.
and 
,
.
is tangent to 
and let 
.
are collinear.
meet the incircle at points 
where 
is the one further from 
are collinear. Thus, 
which means that 
which means that the circle with diameter 
meets the incircle at point 
(since 
is tangent to the incircle, and thus 
completing the proof.
(where the last equality follows from the fact that 
is cyclic).
and 
.
are concyclic.
.
K is the second tangency point of 
.
and we are done
be the antipode of 
such that 
.
.
.
thus, 
which gives 
as desired
.![\[\measuredangle MPI = \measuredangle EPI = \measuredangle EAI = \measuredangle XAI = \measuredangle MIA\]](http://latex.artofproblemsolving.com/9/8/6/9863ba079aae4edcb20b63dd540ff6eb91eb89cf.png)
which implies the desired tangency.
.
.
,
is cyclic. The rest is trivial, as now, 
,
is tangent to 
Let 
.
and it is well known that 
is a line and is parallel to 
.
is cyclic.
And done. 
(by 
inversion) and so 
is tangent to 
.
