[/quote]
,
such that the equality 
holds for all pairs of real numbers 
.
is an obvious solution. Now, let 
.
or 
.
.
,
:
,
be an integer. Find all sequences 
satisfying the following conditions:![\[ \text{ (a) } a_i \in \left\{0,1\right\} \text{ for all } 1 \leq i \leq n^2 + n;
\]](http://latex.artofproblemsolving.com/3/c/5/3c509ec2e9013e8d3be492c8eb44a7c33841b74e.png)
![\[ \text{ (b) } a_{i + 1} + a_{i + 2} + \ldots + a_{i + n} < a_{i + n + 1} + a_{i + n + 2} + \ldots + a_{i + 2n} \text{ for all } 0 \leq i \leq n^2 - n.
\]](http://latex.artofproblemsolving.com/9/7/d/97d2a467d1c0dc8594ec024c3bb9b8c87ee85b19.png)
be an acute triangle with circumcircle 
.
be the midpoint of 
and let 
be the midpoint of 
.
be the foot of the altitude from 
and let 
be the centroid of the triangle 
be a circle through 
.
and 
are collinear.
with integer coefficients such that ![$P\left(\sqrt[3]5+\sqrt[3]{25}\right)=5+\sqrt[3]5$](http://latex.artofproblemsolving.com/5/8/e/58ec2a921cd988bd5ae017a8653544ddd758225f.png)
.
-
.
,
in 
such that the sets ![\[ A_j=\{x+t_j\mid x\in A\},\qquad j=1,2,\ldots,100 \]](http://latex.artofproblemsolving.com/d/5/c/d5c2a7b943e8c7200361698f3eb6f83700ad68d6.png)
are pairwise disjoint.
such that for every index 
,
,
is a multiple of 
.
and 
be the points in which the median and the angle bisector, respectively, at 
.
and 
meets 
and 
,
the point in which the perpendicular at 
produced.
is perpendicular to 
and 
such that for all 
if 
then 
Find the smallest of sum of all elements of 
,![\[
\sum_{\text{cyc}} \sqrt{1 + \left(x\sqrt{1 + y^2} + y\sqrt{1 + x^2}\right)^2} \geq \sum_{\text{cyc}} xy + 2\sum_{\text{cyc}} x
\]](http://latex.artofproblemsolving.com/9/5/5/955863a0cf8747ae45b736b0631f243d3908eb84.png)
such that there exist positive integers 
for which the following holds 
for all 
.
is a perfect cube
Y by nguyendangkhoa17112003, mathematicsy, Adventure10, Mango247
It is known that 
is the smallest angle in the triangle 
. The points 
and 
divide the circumcircle of the triangle into two arcs. Let 
be an interior point of the arc between and which does not contain 
. The perpendicular bisectors of 
and 
meet the line 
at 
and 
, respectively. The lines 
and 
meet at 
.
Show that
.
Alternative formulation:
Four different points
are chosen on a circle 
such that the triangle 
is not right-angled. Prove that:
(a) The perpendicular bisectors of and meet the line 
at certain points and 
respectively, and that the lines 
and 
meet at a certain point 
(b) The length of one of the line segments
and 
is the sum of the lengths of the other two.
is the smallest angle in the triangle 
. The points 
and 
divide the circumcircle of the triangle into two arcs. Let 
be an interior point of the arc between and which does not contain 
. The perpendicular bisectors of 
and 
meet the line 
at 
and 
, respectively. The lines 
and 
meet at 
.Show that
.Alternative formulation:
Four different points
are chosen on a circle 
such that the triangle 
is not right-angled. Prove that:(a) The perpendicular bisectors of
and meet the line 
at certain points and 
respectively, and that the lines 
and 
meet at a certain point 
(b) The length of one of the line segments
and 
is the sum of the lengths of the other two.
with the circle. Since 
are symmetric chords in the circle wrt a line which passes through the center (the perpendicular bisector of 
,
.
and 
are equal to 
,
,
,
.
![$TB+TC=\frac{R}{\cos A}\cdot [\sin (B-x)+\sin (C-y)]=$](http://latex.artofproblemsolving.com/3/3/8/338215a42c3ac53520634bcd874623b038c39b57.png)
,
to circumcircle of 
and by 
an image of 
in symetry with respect to an angle bisector of 
(obviosuly 
then by simple angle chasing we get that 
and 
.
is cyclic so by Ptolemy's theorem we get: 
.
,
hence 
is similar to 
.
with cicrle around 
is isosceles so is 
. Now we have 
and 
.
).
.
.
.
.
.
is isosceles and so 
.
.
is isosceles.
.
is isosceles. Therefore 
is not the smallest then 
may lie outside 
.
meet 
.
.
.
Hence, triangle 
is isosceles. Thus, 
as wanted.
meet 
.
like reflections of the same chord 
.
.
be the midpoints of segments 
,
of 
.
are the external angle bisectors of 
,
is the foot of perpendicular from 
,
.
This completes the proof of the problem. 
![[asy]
size(300);
pair C = dir(-30), B = dir(-140), A = dir(110), U = dir(-110) ;
dot("$A$",A,dir(90));
dot("$B$",B,dir(B));
dot("$C$",C);
draw(unitcircle);
pair O = (0,0), M = 0.5*A + 0.5*B, N = 0.5*A + 0.5*C ;
dot("$O$",O);
dot("$M$",M,dir(180));
dot("$N$",N);
draw(B--A--C);
dot("$U$",U,dir(-90));
draw(A--U);
pair V = extension(A,U,O,M), W = extension(A,U,O,N), T = extension(B,V,C,W);
dot("$V$",V,dir(30));
dot("$W$",W,dir(-140));
dot("$T$",T,dir(100));
draw(B--V);
draw(C--T);
draw(O--M);
draw(W--N);
pair L = 0.5*A + 0.5*U ;
dot("$L$",L,dir(-40));
draw(O--L);
draw(B--C);
[/asy]](http://latex.artofproblemsolving.com/5/2/5/525cca61352cb9c0a6af046ff157c5151178efb0.png)
and 
.
,
,
.
is a isosceles trapezoid, hence 
We are done.
is an isosceles trapezoid because 
Now 
which follows trivially from angle chasing. This implies that 
Now 
Since 
is a cyclic isosceles trapezoid. Thus, 
and 
Then, 
.
and 
so 
which yields 
.
so it suffices to show 
is an isosceles trapezoid. Indeed, we have 
so 
,
,
to be an exterior angle in order to form some isosceles triangle(s).
and 
.
and 
are isosceles trapezoids. Furthermore, by a simple arc chase, 
is an isosceles trapezoid too. Ergo 
and 
.
and 
Note that 
We have 
and 
which means that 
so 
is cyclic.
is 
.
etc. Thus, by law of sines, 
and 
Since 
, it suffices to show that 
By sum to product, we have 
Therefore, the desired equation simplifies to just 
However, we have 
hence done.
intersect 
So 
is an Isosceles trapezoid, similarly, 
is an Isosceles trapezoid. Since 
,
is an isosceles trapezoid.
So it suffices to show 
,
,
to 
and 
denote the perpendicular bisectors of 
over 
and over 
.
and 
are isosceles trapeziums.
and 
are collinear from a litlil bit of angel chessing.
is also an isosceles trapezium 
.
,
is an isoceles trapezoid :
and the fact that 
and thus 
,
is isoceles at 
![[asy]
size(300);
pair C = dir(-30), B = dir(-140), A = dir(110), U = dir(-110) ;
dot("$A$",A,dir(90));
dot("$B$",B,dir(B));
dot("$C$",C);
draw(unitcircle);
pair O = (0,0), M = 0.5*A + 0.5*B, N = 0.5*A + 0.5*C ;
dot("$O$",O);
dot("$M$",M,dir(180));
dot("$N$",N);
draw(B--A--C);
dot("$U$",U,dir(-90));
draw(A--U);
pair V = extension(A,U,O,M), W = extension(A,U,O,N), T = extension(B,V,C,W);
dot("$V$",V,dir(30));
dot("$W$",W,dir(-140));
dot("$T$",T,dir(100));
draw(B--V);
draw(C--T);
draw(O--M);
draw(W--N);
pair L = 0.5*A + 0.5*U ;
dot("$L$",L,dir(-40));
draw(O--L);
draw(B--C);
[/asy]](http://latex.artofproblemsolving.com/1/4/a/14a1012a0a3936166aa909ae20daa27e028540c2.png)
as desired.
is between 
Then due to symmetry from the perpendicular bisectors we have that arcs 
and arcs 
Adding these yields arcs 
therefore 
But note that 
so we are done. QED
