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.
are non-negative integers.
doesn't have any solution on integers.
is strictly increasing : 
such that:
be a polynomial with real coefficients such that 
for all 
. Prove that 
for some real constant 
and 
.
be a continuous function. A chord is defined as a segment of integer length, parallel to the x-axis, whose endpoints lie on the graph of 
. It is known that the graph of contains exactly 
chords, one of which has length 2025. Find the minimum possible value of .
be a hyperboloid of one sheet and let 
be another quadric that intersects the hyperboloid at the curve 
. Let 
be a point on , and let 
be a line through in one specific ruling of the hyperboloid. Let this line intersect again at 
, now define 
to be the line through in the opposite ruling. Similarily define 
. Prove that if 
then this is true for all initial choices of . be positive integers such that 
is a square. Prove that 
is not prime.
be a real polynomial of degree 
that has only simple roots. It is possible to determine a positive quantity 
so that for every pair of positive real numbers 
, 
with 
, every transformed polynomial of the form 
has exactly 
or 
sign variations.
, let 
be the point on the segment 
such that 
. Suppose that the points 
, 
and the centroids of triangles 
and 
lie on a circle. Prove that 
.
at origin and on the 
axis. be 
and be 
, be 
, D be 
are 
means
-
hence the 
- coordinate of must be 
- coordinate of the circumcentre of 
, I got two relations,
be the centroids of 
, 
respectively. Note that as 
, we have 
. Since 
is a cyclic quadrilateral, it must be isosceles, so 
. As 
and 
concur at the midpoint 
of 
, it must hold true that 
.
It is given that , so through some manipulation 
. Thus the roots of 
and 
concur, whence 
. However, if 
and 
, then 
, contradiction (hmm maybe this is what TUAM meant?). Therefore as desired. 
, 
be respectively .
and 
cut at 
and resp.
. hence , 
is parallel to is a cyclic trapezoid . hence , it is isosceles . hence , 
[and ] cut at 
[it's mid pt.]
. Now , using Appolnius theorem gives , 
[as 
] 
+
= 
+
, apply stewart's theorem and then law of cosines to get 
= 
. This proves the result.
. Let the centroids be 
and be the midpoint of .
), we get 
and so, by appolonius theorem, we get 
.
then we are already done. So assume both of these lengths are distinct. For the sake of contradiction, assume that 
. Call the sum condition in the problem to be 
.
be the midpoint of 
and let 
and 
be the centroid of and 
respectively. Since medians divide each other in the ratio of 
we know that![\[ \frac{ZX_1}{ZB} = \frac{ZX_2}{ZC} = \frac{1}{3}\]](http://latex.artofproblemsolving.com/0/2/7/027794e1b8964bf20c7acec46b09350cc0861257.png)
which gives 
. But since 
is cyclic, we get 
. Invoking ``Appolonius Theorem'' in and , we get that
Divide 
from to get 
which violates the triangle inequality which means the assumption is false and thus we are done. 
)
, therefore 
. Hence 
. is given by 
and the centroid of is given by 
has equation of the form 
. For both centroids, 
and 
, therefore 
must be the same.
should be equal if we interchange 
. This reduces immediately to 
, which implies 
so 
.
as 
. It is trivial that 
As 
are concylic by power of point we get,
,
,
and 
where 
denotes semi-perimeter of the triangle . Applying Apollonius Theorem in triangle and and equating the median lengths we get:
Clearly the later violates triangle inequality, hence 
is an isosceles triangle as desired. (QED) 
. Thus, is a isosceles trapezoid which implies . Therefore the medians passes through and are equal. By Median Theorem
and the result follows easily.
.
. and , we have 
. But note that 
, contradicting the triangle inequality. So we have 
. 
such that: 
.
.
for which 
is a perfect square.
be the centroid of 
and 
meet at 
is cyclic, 
.
is one-third 
and 
is one-third 
.
.
touches to the side 
.
.
we get that 
.
let Midpoint of 
midpoints of 
and 
perpendicular 
lies on 
and everything follows
and solving with AB+BD=AC+CD, we get AB=CD