AoPS Academy
eggs, each with a hardness value 
, where 
is a permutation of the set 
. He is throwing the eggs from an 
-floor building.
-th egg is dropped from the 
-th floor, its new hardness becomes![\[
\left\lfloor \frac{k_i}{j+1} \right\rfloor.
\]](http://latex.artofproblemsolving.com/7/5/8/7583af9655a43eb3f6167f955148984a468d4e70.png)
If 
, then the egg breaks and cannot be used again. and can Arthur always determine the correct ordering of the eggs according to their initial hardness values?
. The interior bisector of 
intersects 
at 
and the minor arc of 
in circumcircle of at 
. Suppose that 
is a point on the minor arc of such that 
. 
is a point on the line segment of 
such that 
. 
is the circumcenter of . Prove that 
.
+2 w![$f \in C^1([0,1]), g \in C^2(f([0;1]))$](http://latex.artofproblemsolving.com/f/5/2/f52369171ef8571734950120c43aacac10154fcd.png)
.
?
![$\{x_n\} \subseteq [0,1]$](http://latex.artofproblemsolving.com/8/e/4/8e4206325387485d73765c9d7070ecb1ce18ff60.png)
, there exists some constant 
, for any 
, among the sequence 
and 
could be chosen to satisfy 
and 
.
, where 
are positive integers and is odd.
and 
.
with multiplication of matrixes operation is making an isomorphic-group structure with 
.
.
plane centered on the 
axis which are tangent to both the parabola 
and the line 
. Determine the lengths of the circles' diameters.
by 
and 
for all 
, where 
Show that 
is an integer for all 
.
, is defined in plane polar coordinates 
by the inequality
, where 
and 
.
from 
to 
and 
from 
to 
but in vain.
there exist nonnegative integers 
and integers 
such that 
in 
. Then let ![\[ u = |WA| = |AU|, \qquad v = |UB| = |BV|, \qquad w = |VC| = |CW|. \]](http://latex.artofproblemsolving.com/d/5/d/d5d5674f274fd362b304a3972aac9e0754d7e05a.png)
To prove 
, proceed as Tintarn did. For the rest we'll just resort to some length bashing. Let 
be the intersection of the altitude onto 
with .
.
in two ways. (Heron and 1/2 times base times altitude) 
as the midpoint of . By Pythagoras and 
it is easy to compute 
respectively. We can conclude 
. So there is a homothety with center that takes 
to 
. That implies 
, so we're done. (Watch out for different cases while length bashing.) 
and 
giving us 
. Compute 
by Pythagoras. Now we can use the law of cosines to prove 
, so ![\[ x_S = \frac{x_A+x_B+x_C}{3} = \frac{0+(u+v)+(2u-v)}{3}=u \]](http://latex.artofproblemsolving.com/7/b/3/7b36bf81fad098dbe6b7eb709f19f4c6a16deb2d.png)
with which the result follows. 
.
and 
,
and centroid 
.
,
and line 
is perpendicular to one of the sides of the triangle.
is equilateral. But clearly, this is impossible.
be the point where the incircle touches 
.
.
and 
and 
and 
and we are done.