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.
a prime number with 
and 
when 
. then 
else 
.
be positive integers and 
be prime. We know that 
and 
. Prove that 
is a perfect square.
be a triangle with circumcentre 
. The points 
and 
are interior points of the sides 
and 
respectively. Let 
and 
be the midpoints of the segments 
and 
. respectively, and let 
be the circle passing through and . Suppose that the line is tangent to the circle . Prove that 
with 
such that ![\[ (a-1)(b-1)(c-1) \]](http://latex.artofproblemsolving.com/f/8/c/f8c33b59bdba1fdd12dec51f16a0b1b93e60754f.png)
is a divisor of 
is called good if there exists a permutation 
of the numbers 
, such that:
and 
have different parities for every 
is a quadratic residue modulo 
for every 
is considered a quadratic residue modulo if there exists an integer 
such that 
.
| 
with 
are positive integers and is prime (
), we get: 
be a real numbers such that 
Prove that
Where 
be non-negative real numbers. Prove that
Looking for an algebraic solution!
and 
have common external tangents 
and 
meeting at 
. Suppose touches at 
and touches at 
. A circle 
through and intersects again at 
and again at 
, such that quadrilateral 
is convex.
and 
meet at point 
, while lines 
and 
meet at point 
. Show that , , are collinear.
be an acute, scalene triangle with circumcenter 
, circumcircle 
, orthocenter 
. Line 
meets again at 
. Let 
be the midpoint of segments 
respectively. The line through and perpendicular to 
meets line at 
.
meets again at 
. Prove that the intersection of line 
and the perpendicular bisector of 
lies on the circumcircle of triangle 
. and perpendicular to 
meets line at 
. Line 
meets again at 
. Let 
be the intersection of lines 
; 
be the intersection of lines 
; 
be the intersection of lines 
. Let be the intersection of lines 
and 
be the line through and perpendicular to 
. Prove that bisects 
.
such that 
. Show that the equation 
has no integer solution.
, 
, 
, be positive numbers. Prove that
IMOC. Hope that there isn't any typo ... :blush:
IMOC. Hope that there isn't any typo ... 
Can anyone help me how to continue after showing injectivity ...
are 
satisfy the condition.
can be expressed as the sum of squares of positive divisors of .
, 
is valid expression, of course.
, must be even, so 
is valid expression.
, must be divisible by 
, so 
is valid expression.
, must be divisible by 
, so 
is valid expression..
, 
cannot be expressed as the sum of positive divisors of .
, 
cannot be expressed as the sum of positive divisors of .
, consider the following two cases (1),(2) is not divisible by 
such that 
and 
cannot be expressed as the sum of 
positive divisors of . is divisible by such that 
and 
cannot be expressed as the sum of positive divisors of .
sub-decks of cards rather than ? Since you pick cards in the end.?
then 
or 
, but I can't get more from it. Can you show the full solution?
sub-decks of cards rather than ? Since you pick cards in the end.? then or , but I can't get more from it. Can you show the full solution? sub-decks.
. Sorry about that, and I promise that I will blame FEcreater for that(X
sub-decks. is even, and each sub-deck consist of a single number? In this case the sum must be 
, which is not a multiple of .
sub-decks. is even, and each sub-deck consist of a single number? In this case the sum must be , which is not a multiple of .
be real numbers such that 
.
.
.
and 
,
and blah blah cauchy
,
,
and its root 
,
,
,
is non-constant,thus 
.
,
.
Now we will show that 
and will finish the problem thanks to Cauchy.
or 
.
or 
.
for any 
,
we have that 
only. Let 
be an arbitrarily small number, now it is easy to see that 
finishes the problem.
Now, by C-S we have that
)
.
and 
,
also, hence 
.
,
and 
are roots of 
.
is equivalent to 
,
(Try 
)
can be expressed as the sum of 
such that 
and 
.
be the smallest positive integer such that 
.
,
,
.
,
,
,
for sufficiently large 
for all 
.
such that 
implies 
.
there exists 
such that 
.![\[f^{(n+1)}(y) = f^{(n)}(f(y)) = f^{f(t_n)}(f(y)) = t_n+2f(y),\]](http://latex.artofproblemsolving.com/3/4/6/3461215efe906d8e0cb573c276e11ee979c7ae71.png)
![\[f^{(n+1)}(y) = f^{f(t_{n+1})}(y) = t_{n+1}+2y,\]](http://latex.artofproblemsolving.com/4/3/2/43233a7b2f6ec381cc35ae6c0f15c9858d96bdb2.png)
and so 
for all 
should be 
in order to have 
such that 
.
is periodic and thus bounded. Therefore for sufficiently large 
for all 
is such that 
implies 
.
for all ![\[f^{(n)}(y) = f^{f(t_n)}(y) = y,\]](http://latex.artofproblemsolving.com/f/1/2/f12be52b60560d6eac552db6d5c725845584c742.png)
![\[f^{(n+1)}(y) = f^{f(t_{n+1})}(y) = y,\]](http://latex.artofproblemsolving.com/7/a/9/7a9dd3043e18dc92af30c4ea7647a229b9dc70c7.png)
and so![\[f(y) = f(f^{(n)}(y)) = f^{(n+1)}(y) = y,\]](http://latex.artofproblemsolving.com/1/7/0/170c4300e7ab7b5cb5d127cbfbb0b8025bb7a7c0.png)
as desired.
.
and 
,
.
After that it is an easy infinite descent.
components (with ugly method), but I haven’t proved the opposite direction.
.
.
).It can be found that any two squares have Descartes distance at least 
over 
and let 
be the foot from 
to 
are collinear, and now we have that 
by Spiral Sim so if ![\[
\angle PIA = \angle PNM = \angle D_0NM = \angle D_1NM
\]](http://latex.artofproblemsolving.com/c/c/a/ccad96702781c56fff3970c14ed05b2ced3afad2.png)
where 
is the extouch point. Let 
meet 
at 
,![\[
\angle ND_1Q = \angle MND_1 = \angle PIA = \angle QIN,
\]](http://latex.artofproblemsolving.com/e/7/e/e7ef1d6eb0cc9be283936c6e99945928dde79e72.png)
so 
is cyclic. In particular, ![\[
\angle IQT = \angle IND_1 = \angle AIT = \angle AXT,
\]](http://latex.artofproblemsolving.com/5/b/2/5b2893c206e31ac21cffc85a84ad37eebf46c05d.png)
so 
.
by homothety centered at 
.
doesn’t require accurate construction. It can be generalised that for a 
chessboard, the final situation will be 
connected components. But I somehow found that the way that 8-mino tiling the plane is not optimal for special 
and I don't like to calculate the 
term :p
chessboard can be put 
.
,
is used). For Bob’s strategy, I am quite sure that the tiling with 
can be constructed by letting any 
that 
.
problem
is missing.