AoPS Academy
[/quote]
, do not answer with "
is a solution" only. Either you post any kind of proof or at least something unexpected (like "
is the smallest solution). Someone that does not see that is a solution of the above without your post is completely wrong here, this is an IMO-level forum.
. Let 
be the midpoint of the arc 
not containing 
and define 
similarly. Show that the orthocenter of 
is the incenter 
of 
.
be a positive integer and consider the set 
.
of numbers not previously chosen such that for every distinct 
, neither 
divides 
nor divides .
, all multiples of every element in , including those in itself, are removed from 
. until no moves are possible., which player has a winning strategy if any
table filled with natural numbers, we say it is a divisor table if:
-th row are exactly all the divisors of some natural number 
,
-th column are exactly all the divisors of some natural number 
,
for every 
.
is given. Determine the smallest natural number 
, divisible by , such that there exists an divisor table, or prove that such does not exist.
such that 
,![\[
f(x^2+f(y))=f(xf(x))+y
\]](http://latex.artofproblemsolving.com/d/e/d/dedd2338779fcc6723077ce586713fb4a68b9931.png)
be a triangle with incenter and let 
be an arbitrary point on the side 
. Let the line through perpendicular to 
intersect 
at 
. Let the line through perpendicular to intersect at 
. Prove that the reflection of across the line 
lies on the line .
be an acute triangle. Let 
be the reflection of point over the line . Let 
and 
be the circumcenter and the orthocenter of triangle , respectively, and let be the midpoint of segment 
. Let and 
be the points where the reflection of line 
with respect to line 
intersects the circumcircle of triangle , where point lies on the arc not containing . If 
is a point on the line 
such that 
, prove that 
.
![$p(x)\in\mathbb{R}[x]$](http://latex.artofproblemsolving.com/9/a/e/9aecb18edc17f7391a44db6e4f9d1502ab1d92f7.png)
such that for all 
:
such that:
be a triangle. Circle 
passes through , meets segments 
and 
again at points and respectively, and intersects segment at and 
such that lies between 
and . The tangent to circle 
at and the tangent to circle 
at meet at point 
. Suppose that points and are distinct. Prove that line 
is parallel to . with circumcircle and incenter , point 
bisects arc 
and line 
meets at 
. The excircle opposite to touches 
at point . Point 
on the circumcircle of 
is such that 
. Prove that the lines 
and 
meet on .
be an inscribed circle of 
and touching , 
, at , , respectively. Let 
and 
be diameters of . Let and 
be the pole of 
and 
with respect to , respectively. 
cuts again at 
. 
cuts again at . The tangent at of cuts 
at . The tangent at of cuts 
at 
. Let 
and 
be midpoint of 
and 
, respectively.
, 
and perpendicular bisector of 
are concurrent.
grid. The rabbit's starting square is 
(unknown to the hunter), and after 
rounds, the rabbit is at square 
. In the -th round of the game, two things occur in order:
which shares a point with (There are up to eight of these).
squares of the hunter's choosing. If the rabbit is in one of these squares, the rabbit is captured and the game ends. can the rabbit avoid capture indefinitely?
be an acute triangle with orthocenter . Let be the point such that the quadrilateral 
is a parallelogram. Let be the point on the line 
such that bisects 
. Suppose that the line intersects the circumcircle of the triangle 
at 
and 
. Prove that 
.
be a non-cyclic quadrilateral. Let 
and 
be the circumcentre and the circumradius of the triangle 
. Define 
and 
in a similar way. Prove that![\[\frac{1}{O_1A_1^2-r_1^2}+\frac{1}{O_2A_2^2-r_2^2}+\frac{1}{O_3A_3^2-r_3^2}+\frac{1}{O_4A_4^2-r_4^2}=0.\]](http://latex.artofproblemsolving.com/f/f/5/ff575c3012e0b86c5aeeec32aae5268a4dad74d2.png)
for all integers 
. But now 
, so we must have 
.
is attainable; consider 
. 
on both sides, we see that the RHS is 
. Since perfect cubes are only 
, we see that 
.
, we have the following construction: 
. Our answer is 
. 
. This is achievable when 
Now, to show that this is optimal, note that taking 
gives that 
, but cubes are 
. Therefore, it takes at least numbers to get to , so we are done.
for all integers . Note that 
.
( As if 
, then 
, then 
. But we want 
which is not possible)
is mentioned by many posts above
gives 
but since each 
clearly 
Now we construct a solution for 
showing that this is the answer. Note that we can just do 
so this finishes off the problem.
is the minimum. The construction is:
Now I prove minimality.
By taking our original equation modulo 9, we get that 
Thus, 
Prove that
such that there exist integers 
with ![\[x^3_1+x^3_2+\,\ldots\,+x^3_t=2002^{2002}\,?\]](http://latex.artofproblemsolving.com/e/f/a/efaa98e77ac0273284ee3446e4335d8cc4f681cb.png)
is congruent to 4 (mod9)., we need at least 4 terms.
so 
.
.
are coprime, 
,
by Euler's Theorem. Therefore 
or 
.
,
.
.
,
is not satisfies. We have ![\[ 2002^{2002}\equiv 4^{2002}\equiv 4\pmod 9.\]](http://latex.artofproblemsolving.com/5/e/6/5e672ff0642825fc556062006af9ebb4f5642969.png)
Since 
,
for 
and 
,
from Euler's Theorem. Since the set of cubic residues mod 
,
.
so 
.
works.
where 
.
works.
.
.
,
.
.
is indeed a really big number, so it would be quite nice if we could somehow reduce it to something feasible, at least something to the order of a thousand... wait this is a no-brainer, let the 
of our numbers be 
,
to get:
.
must satisfy this equation- aha so we have found that 
(aka 
and 
)
will be quite useless here, so let's start with 
.
:
-
,
,
we get that![\[x_1^3 + x_2^3 + \cdots + x_t^3 = 2002^{2002} \equiv 4 \pmod{9}\]](http://latex.artofproblemsolving.com/0/3/0/030592aae3cd928965b8458eb339b21232320616.png)
and since the only cubic residues modulo 
,
.
:![\[(2002^{667})^3 + (2002^{667})^3 + (10\cdot 2002^{667})^3 + (10\cdot 2002^{667})^3 = 2002^{2002}\]](http://latex.artofproblemsolving.com/4/c/3/4c3ba5157d0469fa46b02bd79f8aed1a3cde2646.png)
Using 
we get 
so we need at least 
to end the problem.
and cubes are 
in 
.
.
,
cubes.
Taking the equation 
and 
.
and 
:
.
.