[/quote]
,
has been partitioned into disjoint pairs 
)
,
equals 
or 
.![\[ |a_1-b_1|+|a_2-b_2|+\cdots +|a_{999}-b_{999}| \]](http://latex.artofproblemsolving.com/0/4/4/04422a7d2266b92a6b3f215b698f078cda85091f.png)
ends in the digit 
.
such that 
for all positive integers 
.
be a triangle with 
Line bisector of 
intersects 
at point 
.
.
,
,
by the recurrence![\[
a_{n+1} = a_n^2 + 1
\]](http://latex.artofproblemsolving.com/3/1/1/3114595ded8c8961d611afaab9ad19e254973001.png)
Prove that there is no natural number 
such that![\[
\prod_{k=1}^{n} \left( a_k^2 + a_k + 1 \right)
\]](http://latex.artofproblemsolving.com/a/4/f/a4f88ee7e3d4e53e369319f74cd9b12be26c9681.png)
is a perfect square.
be a triangle. 
is the midpoint of segment 
,
,
are selected on sides 
,
respectively such that 
and 
intersect at a point 
.
intersects line 
.
,
and the tangent to 
concur.
such that 
for all 
people playing a game at Cheesy's math casino, where 
be a positive integer. A subset of length 
from the set of integers from 
for 
.
be a positive integer and let 
,
,
be a sequence of non-negative real numbers. Find the maximum value of 
in terms of 
are randomly arranged in the cells of a 
square 
)
fractions. What is the highest possible value of the characteristic ?
be distinct real numbers such that![\[
a + b + c + \frac{1}{abc} = \frac{19}{2}
\]](http://latex.artofproblemsolving.com/5/1/e/51eef77358db2b6c98f0fb0fb4e30db837a03642.png)
Find the maximum possible value of 
.
.
at 
and 
.
Y by
Let 
be an acute triangle with orthocenter 
, and let 
denote the foot of the altitude from 
to 
. Let the circle 
with diameter 
intersect altitude 
at distinct points 
and 
. Suppose that the circle with center 
passing through point intersects segment 
at 
. Line 
meets at 
. If 
intersects 
at 
, and 
intersects at 
, prove that is tangent to the circumcircle of 
.
be an acute triangle with orthocenter 
, and let 
denote the foot of the altitude from 
to 
. Let the circle 
with diameter 
intersect altitude 
at distinct points 
and 
. Suppose that the circle with center 
passing through point intersects segment 
at 
. Line 
meets at 
. If 
intersects 
at 
, and 
intersects at 
, prove that is tangent to the circumcircle of 
.
.![$[CL]^2 = [CX]^2= [CY]^2 \Rightarrow L$](http://latex.artofproblemsolving.com/3/a/0/3a0d813ceb846745ae927892e0a6bbddb6eec0bd.png)
is the incenter of 
.![$[CL]^2 = [CX]^2 = [CE] \cdot [CB] = [CD] \cdot [CA] \Rightarrow \overline{AL} \perp \overline{CL}\Rightarrow R$](http://latex.artofproblemsolving.com/4/5/6/4567bb6a0790e2821b3a2b4ba1592e30649baf97.png)
and 
belong to the 
mixtilinear incircle of 
.
is the midpoint of arc 
and 
belongs to the 
be the feet of perpendicular from 
meet the line 
at points 
.
centered at 
.
and 
.
.
and similarly 
and we have proved the lemma. 
.
at 
.
,![\[CL'^2 = CD\cdot CA = CX^2 = CL^2.\]](http://latex.artofproblemsolving.com/d/c/9/dc9e1ea07a744020713f6dc1137a2caeb0135af7.png)
Also 
lie on same side of 
giving 
lies on 
or 
.
.
is the internal angle bisector of 
.
gives 
and 
m
is the 
and 
at 
and 
respectively. We know (well-known) that 
are collinear and 
.
and we are given 
are collinear. This gives 
and 
which further gives that 
is the 
by 
.
are tangent to 
is tangent to 
.
is parallel to 
,
.
and 
.![\[\measuredangle XRR' = \measuredangle XPB = \measuredangle R'YX,\]](http://latex.artofproblemsolving.com/e/d/0/ed0b04ceb8f66d55c4dd0f96740056f508bbb1a4.png)
so 
is cyclic. Similarly, 
is cyclic as well.
is acute) inversion centered at 
,
,
.
and 
.
is tangent to 
.
and 
are tangents to 
and 
are also tangents to 
form the 
.
,
by 
.
,
and 
are tangent to 
is harmonic. This means that the tangents to 
,
and 
.
,
.
,
lies on 
(as 
and 
)
are collinear (as 
)
,
(as 
and 
)
be the tangent to 
Thus, 
,
.
such that 
so 
is tangent to 
and also 
is tangent to 
and hence by PoP we have 
and so 
.
and see that 
and so 
and hence it is the 
is 
