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.
has circumcircle 
and incenter 
. The incircle of triangle touches side 
at 
respectively. Circumcircle of triangle 
intersects again at 
distinct from 
, circumcircle of triangle 
intersects again at 
distinct from 
. Prove that the 
points 
are concyclic.
is defined by 
and for all positive integers 
, 
. Show that for every positive integer 
, there exists a suitable positive integer 
such that 
is a multiple of .
lies inside triangle so that 
. The point is such that 
is a parallelogram. Prove that 
.
, point 
on 
satisfies 
and 
.
be the point where the perpendicular from 
to line 
meets the angle bisector of 
.
holds. is an isosceles triangle.
be a positive integer, and let its positive divisors be![\[
d_1 < d_2 < \cdots < d_k.
\]](http://latex.artofproblemsolving.com/6/3/e/63ec82aba0d6c4eece89af6a1247a89498616281.png)
Define 
to be the number of ordered pairs 
with 
such that 
.
. when![\[
n = p_1^{e_1} p_2^{e_2} \cdots p_k^{e_k},
\]](http://latex.artofproblemsolving.com/2/7/2/272dc2cc33ebbf695311cb61567fa0297e1299b6.png)
where the 
are distinct primes and the 
are positive integers.
such that 
.
that verify the following equation for all 
:
be a positive integer. Determine all sequences 
of positive integers such that
for all positive integers 
.
and 
be two circles with respective centres 
and 
intersecting in two distinct points 
and such that 
is an obtuse angle. Let the circumcircle of 
intersect and respectively in points 
and 
. Let the line 
intersect in 
; let the line 
intersect in 
. Prove that, the points 
are concyclic.
be an acute scalene triangle with circumcenter 
. Let 
be the feet of the altitudes from 
to 
respectively. Let 
be the midpoint of 
. Let 
be the circle with diameter 
. Let 
be the intersection of 
and 
. Let 
be the orthocenter of . Let 
be the intersection of 
and . Let 
be the lines through tangent to 
respectively. Let be the intersection of 
and 
. Let be the intersection of 
and 
. Let 
be the line through parallel to 
and let 
be the reflection of across . Prove that 
is tangent to 
.
be a positive integer. The ordered pairs 
where 
are integers in 
are each labeled with a positive integer less than or equal to 
such that every label is used exactly times and there exist integers 
and 
such that the following property holds: For any two lattice points 
and 
that are both labeled 
, there exists an integer such that 
and 
are both divisible by . How many such labelings exist?
be an odd prime. Define 
over the positive integers as follows:
be chosen such that there exists an ordered pair of positive integers 
where 
such that 
. Prove that there exists at least three integers 
such that 
and 
is a perfect square.
denote the set of all positive integers less than 
that are relatively prime to . Let 
. Is the polynomial 
reducible over the rational numbers, given that it has integer coefficients?
, 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 
. 
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 
and everything follows
and solving with AB+BD=AC+CD, we get AB=CD