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.
inscribed in 
with 
being the midpoint of 
. The tangents at 
of intersect at 
. Let 
be the projection of 
onto 
. On the perpendicular bisector of , take a point 
that is not on and different from M. Circle 
intersects 
at 
. Lines 
and 
intersect at 
. Prove that 
is an isosceles triangle.
be the set of all positive integers. We say that a function 
is Georgian if 
and, for every positive integer 
, there exists a positive integer 
such that![\[
f^{(k)}(n) = 1, \quad \text{where } f^{(k)} = f \circ f \cdots \circ f \quad \text{(applied } k \text{ times)}.
\]](http://latex.artofproblemsolving.com/5/3/c/53ca7c663237d939a7a049aa3569f8d595b259cf.png)
If 
is a Georgian function, we define, for each positive integer , 
as the smallest positive integer 
such that 
. Determine all positive real numbers 
for which there exists a Georgian function such that, for every positive integer 
, it holds that 
.
prove that
denotes the sum of the integers, 
denotes the sum of the square of the integers and 
denotes the sum of cubes of the integers(that is, if the three integers are 
, then 
, 
, 
). If 
and 
, find 
.
is inscribed in circle 
. The tangent lines to at 
and 
meet at 
. Point 
lies on ray 
such that 
. Points 
and 
lie on ray 
(with in between and ) such that 
. Prove that triangles and 
are similar to each other.
-gon. A move consists in choosing an edge of the -gon and swapping the two tokens placed at the endpoints of that edge. After a finite number of moves have been performed, it turns out that every two tokens have been swapped exactly once. Prove that some edge has never been chosen.
of the triangle touch its side at . Let the line intersect at 
and denote the excentre of opposite to by . Let and be the midpoints of and 
respectively.
and 
are concyclic.
be three mutually tangent circles such that 
are externally tangent at 
, 
are internally tangent at , and 
are internally tangent at . Let 
be the centers of , respectively. Given that 
is the foot of the perpendicular from to 
, prove that 
. be a scalene triangle with circumcircle 
, and suppose the incircle of touches at . The angle bisector of 
meets and at and . The circumcircle of 
intersects the -excircle at 
, 
, and at 
. Prove that line 
passes through either or . be an acute triangle with orthocentre 
, and let be the midpoint of 
. The point 
on is such that 
is an altitude of the triangle . Let 
be the reflection of in . The orthogonal projections of onto the lines 
, and are , 
and 
, respectively. Let 
be the point such that the circumcentre of triangle 
is the midpoint of the segment 
. lies on the segment 
.
be a quadrilateral inscribed in a circle 
. The lines and 
meet at , the lines and meet at , and the diagonals and 
meet at . Let be the midpoint of the segment 
, and let be the common point of the segment 
and the circle . Prove that the circumcircle of the triangle 
and are tangent to one another.
lies inside triangle so that 
. The point is such that 
is a parallelogram. Prove that 
.
the parallelogram internal angles, 
. Using the cosine theorem, for example, for the triangle 
together with 
, we get
, because their corresponding side lines are parallel: 
and 
. Hence,
into the last cosine theorem formula,
and substituting for 
,
together with 
, we get
be a parallelogram. Suppose a line passing through and lying outside the parallelogram meets and produced at and respectively. Show that ![\[ AC^2 + CE \cdot CF = AB \cdot AE + AD \cdot AF . \]](http://latex.artofproblemsolving.com/2/9/c/29ce31b24e798276b0e04b2b069a1e07501aba76.png)
.
.
.
, becomes 
. Division by AC transforms this into AC + CG = AU + AV.
.
.
.
, so that AU = VG, and we are done.
for the cevian-line 
and 
results: 
