[/quote]
,
are coprime. Consider the sequence 
given by 
.
such that 
is even and 
is an integer. Prove that the numerator in its simplest form of 
is an odd integer greater than 
.
be an integer and let 
be 
distinct subsets of 
.
such that the n subsets 
are also disjoint.
s
.
satisfies 
.
,
.
such that for all irrational numbers 
and 
,
where 
and 
.
satisfy that 
.![$$\sum_{cyc} \sqrt[4]{\frac{a^3}{b+c}}$$](http://latex.artofproblemsolving.com/e/f/0/ef089c4f2915bbf75b0f60b044bbe9273b301a81.png)
compute the smallest possible value of ![\[\sum_{k=1}^{n}\left\lfloor\frac{a_k}{k}\right\rfloor\]](http://latex.artofproblemsolving.com/5/6/2/562c27c7779285b6439c4087f52933a41f29ee2e.png)
over all permutations 
of 
with incenter 
,
touches sides 
and 
and 
,
and 
touches 
.
meets 
.
is parallel to 
.
-
.
and 
Prove that
Let 
Prove that
Let 
Prove that
of positive integers such that ![\[ k!=(2^n-1)(2^n-2)(2^n-4)\cdots(2^n-2^{n-1}). \]](http://latex.artofproblemsolving.com/3/5/3/353bbb15bac205053c26208af8cac7c32a296f55.png)
Proposed by Gabriel Chicas Reyes, El Salvador
Y by Adventure10, Mango247, and 1 other user
Let 
be a triangle with semiperimeter 
and inradius 
. The semicircles with diameters 
, 
, 
are drawn on the outside of the triangle . The circle tangent to all of these three semicircles has radius 
. Prove that
![\[\frac{s}{2}<t\le\frac{s}{2}+\left(1-\frac{\sqrt{3}}{2}\right)r. \]](//latex.artofproblemsolving.com/1/1/7/117063afcc60a443e6af8bd5f789af0827af58db.png)
Alternative formulation. In a triangle, construct circles with diameters , , and , respectively. Construct a circle 
externally tangent to these three circles. Let the radius of this circle be .
Prove:
, where is the inradius and is the semiperimeter of triangle .
Proposed by Dirk Laurie, South Africa
be a triangle with semiperimeter 
and inradius 
. The semicircles with diameters 
, 
, 
are drawn on the outside of the triangle . The circle tangent to all of these three semicircles has radius 
. Prove that![\[\frac{s}{2}<t\le\frac{s}{2}+\left(1-\frac{\sqrt{3}}{2}\right)r. \]](http://latex.artofproblemsolving.com/1/1/7/117063afcc60a443e6af8bd5f789af0827af58db.png)
Alternative formulation. In a triangle
, construct circles with diameters , , and , respectively. Construct a circle 
externally tangent to these three circles. Let the radius of this circle be .Prove:
, where is the inradius and is the semiperimeter of triangle .Proposed by Dirk Laurie, South Africa
Attachments:
instead of the < sign in the left member, but yet nobody gave a complete solution):
.
. Besides the Soddy circles (which I never used at any problem 
) I didn't learn anything out of it.
the circles with diameters 
respectively.Let 
be the circle externally tangent to these three circles.Let 
be the point of tangency of 
with 
analogously.Let 
be the midpoints of 
the center of 
are collinear.Thus 
.
.
gives
the midpoints of segments 
respectively. Let the incircle of 
touch segments 
at 
respectively. Consider the circle internally tangent to the circles with centers 
respectively. This is called the inner Soddy circle of 
.
be the circumradius of 
,
.
.
respectively.
.
.
which is trivial by some applications of Muirhead's inequality.
can be solved somewhat nicely.
be 
,
.
.
,
.
,
are of the form 
(
)
.
,
,
)
.
is half of that of 
.
which is 
denote the mid points of 
.
and 
.
,
.
.
which readily proves the left part of the inequality because 
.
.
.
.
, so the bound holds in this case.
.
such that, under the inversion 
, the circles 
and 
become con-centric and 
and let the points where 
with 
.
(for a point 
.
and 
.
.
makes the circles 
,
and orthogonal to 
as noted in the previous case.
,
,
intersect the circle at 
,
,
,
at 
at 
.
,
,
.![\[t=OC'=OF+FC'=OF+AF=OF+ED=OE+FD=OD+EF\]](http://latex.artofproblemsolving.com/f/2/1/f2161fa0f3aa26c4d6e1085889c967a0e52b155a.png)
so 
.
,
and 
.
centered at 
and 
analogously. Then clearly there exists a circle centered at 
.
so it is the perimeter of 
.
is the semiperimeter of 
so the radius of the small circle is 
as desired.
,
where 
is the inradius of ![\[\frac{1}{u}=\frac{1}{a}+\frac{1}{b}+\frac{1}{c}-2\sqrt {\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca}}\]](http://latex.artofproblemsolving.com/2/1/4/214893f86c2d16aa2398f8ba5a9b9bd59ecf260c.png)
and we have![\[\frac{1}{v}=\frac{s}{A}=\frac{a+b+c}{\sqrt{abc(a+b+c)}}=\sqrt{\frac{a+b+c}{abc}}\]](http://latex.artofproblemsolving.com/f/c/2/fc20f00bc890065cbcf072d17b1091497f6502f4.png)
We want to show that![\[\frac{1}{a}+\frac{1}{b}+\frac{1}{c}-2\sqrt {\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca}}\ge \sqrt{\frac{a+b+c}{abc}}\]](http://latex.artofproblemsolving.com/9/e/f/9ef631ba70971885f6c802dab2e85bb81aaa4b16.png)
which is equivalent to![\[\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\ge 3\sqrt {\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca}}\]](http://latex.artofproblemsolving.com/d/8/5/d854981ae3f5d5336f8a8321677bee668ea34ea0.png)
and this is a trivial inequality.
