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.
be a point in the plane of triangle 
such that the segments 
, 
, and 
are the sides of an obtuse triangle. Assume that in this triangle the obtuse angle opposes the side congruent to . Prove that 
is acute.
+1 w
and 
be two distinct rays not lying on the same line, and let 
be a circle with center 
that is tangent to ray at 
and ray at 
. Let 
be a point on segment 
. The line through parallel to intersects line at . Let 
be the intersection of lines 
and , and let 
be the intersection of line and the line through parallel to . Prove that line 
is tangent to . be an acute triangle with point 
lie on the plane such that 
is a parallelogram. 
is the orthocenter of 
intersects 
at 
and 
intersects 
at 
The circle with diameter 
intersects the circumcircle of 
again at 
Prove that: The circumcircle of 
passes through the reflection point of over 
with 
, let 
denote the incenter and the midpoint of side 
. The line through 
perpendicular to 
intersects the tangent from to the incircle (different from line ) at a point > Show that is tangent to the circumcircle of triangle 
.
and 
. At each step, it is allowed to add to one of the polynomials and to multiply another one by the polynomial 
. Can the polynomials become equal at some point?
such that for any positive integer number 
and any arrangement of the numbers from 1 to 
on a circle, there exists a pair of adjacent numbers whose ratio (when dividing the larger number by the smaller one) is less than .
from the nonnegative integers to the integers satisfying 
and ![\[ f(mn) = f(m) + f(n) + f(m)f(n) \]](http://latex.artofproblemsolving.com/e/f/f/eff10a32cffaec742403ebfc46ff09bc8ef7b7c5.png)
for all nonnegative integers 
and 
.
make the equation ![\[\sum_{i=1}^n \sum_{j=1}^n \left\lfloor \frac{ij}{n+1} \right\rfloor=\frac{n^2(n-1)}{4}\]](http://latex.artofproblemsolving.com/e/7/f/e7fb8c20a43535b5aaed5ab254f1ef043263c62b.png)
true?
be the binary representation of 
. Prove that for any positive integer , at least one of the digits 
, 
, 
, 
equals 
.
, an arbitrary point 
is chosen on the side 
. In triangles 
and 
, the angle bisectors 
and 
are drawn, respectively. The point 
is the circumcenter of 
. Prove that the second intersection point of the circumcircles of triangles 
and 
lies on the line 
.
such that![\[
4k^3 + 4k + 1
\]](http://latex.artofproblemsolving.com/f/3/5/f355e456cd47ecfdb13597d52fec3d848b67c594.png)
numbers, none of which appears more than 
times. Prove that they can be divided into 
groups of numbers each in such a way that the sums of the numbers in these groups are pairwise distinct.
be a cyclic quadrilateral. and intersect at . Let and 
be the projections of onto and and and be the midpoints of and 
, respectively. Prove that 
is a kite.
. At each move, Bob erases arbitrary two numbers 
and 
, where 
, and writes the single number 
. After 
such moves the blackboard will contain a single number. What is its maximum possible value? (Reminder that 
is the maximum integer not exceeding 
.)
with real coefficients is given. Prove that there is a positive integer such that the equation 
has no rational roots.
as a counterexample. By Lagrange Interpolating, has rational coefficients. Thus let 
for 
integers. Take a prime 
. Let 
. It is easy to show 
divides the denominator of 
. But remark that then 
for all so we are done.
for each positive integer . (
is rational)
, so 
is rational for 
.
is rational, so 
is rational, so is rational. Then we easy get that and 
are also rational.
that 
, 
, 
are integers and is not a perfect square. Then for each positive integer 
there exists such rational that 
is a perfect square for each positive integer . Let 
. Then 
, which is not a perfect square, because 
. Contradiction.
(non constant) exists. has rational coefficients. Thus, we may write 
for some integer 
so that has integer coefficients. Now, we see that there exists a sequence of rational numbers 
such that 
for all . Let 
where 
is an integer. Now, write 
for coprime integers 
and non zero denominator. Choose 
to be a prime number. We have 
and thus, 
. Therefore, 
and so 
, a contradiction. The conclusion holds. and the same argument may be applied.
and we assume the contrary that for all 
has a rational root 
.
and 
,
.
(for the contrary would result in 
)
.![\[ a(u_k+u_l) + b \]](http://latex.artofproblemsolving.com/4/2/c/42c96c338aa38d48a338b4e1095925db0f18c214.png)
.
and 
are rationals, so their difference is also a rational number. Therefore, 
is a rational number. Since 
,
has a rational solution, 
with 
be a prime that does not divide either numerator nor denominator of 
,
.
is a solution, where 
when written in lowest form. Then we have 
so we have 
.
.
is divisible by 
,
,
.
is the denominator of