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 cyclic quadrilateral. Let 
be the midpoint of the arc 
of its circumcircle which does not contain 
or 
. Let the lines 
and 
meet at 
and the lines 
and 
meet at 
. Prove that the lines 
and are parallel.
be an acute triangle with circumcenter 
and orthocenter 
. Let 
denote the circumcircle of triangle , and 
the midpoint of 
. The tangents to at 
and , and the line through perpendicular to line 
, determine a triangle whose circumcircle we denote by 
. Define 
and 
similarly., and are concurrent on line .
such that for all positive integers 
and 
,![\[f(m+nf(m))=f(n)^m+2024! \cdot m.\]](http://latex.artofproblemsolving.com/8/b/9/8b90a9326c0fa448a5a574809c184f43513f768f.png)
Jaedon Whyte
, and an odd prime 
not dividing any of them, and an integer 
such that if one considers the infinite sequence 
and looks at the highest power of that divides each of them, these powers are not all zero, and are all at most . Prove that there exists some 
(which may depend on 
) such that whenever divides an element of this sequence, the maximum power of that divides that element is exactly 
.
, let 
be the least prime that does not divide 
, and define 
to be the product of all primes less than . In particular, 
For having 
, define 
. Consider the sequence 
defined by 
and ![\[ x_{n+1} = \frac{x_n p(x_n)}{q(x_n)} \]](http://latex.artofproblemsolving.com/8/2/5/825be3bbef6a29de69aead03392ca7215093f320.png)
for 
. Find all such that 
.
, 
, 
be an infinite sequence of positive integers. Suppose that there is an integer 
such that, for each 
, the number
is an integer. Prove that there is a positive integer such that 
for all 
. be a right triangle with 
. Point lies on the line 
such that is between and . Let 
be the midpoint of 
and let be the seconf intersection point of the circumcircle of 
and the circumcircle of 
. Prove that as varies, the line 
passes through a fixed point.
and 
intersect each other at 
and 
. The common tangent to two circles nearer to touch and at 
and 
respectively. Let 
and 
be the reflection of and respectively with respect to . The circumcircle of the triangle 
intersect circles and respectively at points 
and 
(both distinct from ). Show that the line 
is the second tangent to and .
to 
face-down on a table. On each move, the Detective can point to two cards and inquire if the numbers on them are consecutive. The Magician replies truthfully. After a finite number of moves, the Detective points to two cards. She wins if the numbers on these two cards are consecutive, and loses otherwise.
questions.
= 2∠AMP.
is the midpoint of the arc 
, containing 
, in the circumcircle of 
, and let 
be the -excircle's center. Assume that the external angle bisector of 
intersects at 
. Prove that 
is perpendicular to 
, where 
is the incenter of .
with 
and the circle 
of radius centered at 
Let be the midpoint of 
The line 
intersects a second time at 
Let be a point on such that 
Let be the intersection of 
and Prove that 
cities. For every two cities there exists exactly one direct one-way road between them. We say that a path from 
to 
is a sequence of roads such that one can move from to along this sequence without returning to an already visited city. A collection of paths is called diverse if no road belongs to two or more paths in the collection.
and be two distinct cities in Anisotropy. Let 
denote the maximal number of paths in a diverse collection of paths from to . Similarly, let 
denote the maximal number of paths in a diverse collection of paths from to . Prove that the equality 
holds if and only if the number of roads going out from is the same as the number of roads going out from .
years ago! for solving it, extend 
to intersect the circle with center 
...
is the closer tangent to , which I think the original problem states (it doesn't matter though).
, 
, and 
be the reflection of over . Clearly 
. But 
, so it suffices to show 
(equivalent since 
are the midpoints of 
respectively), which is just a simple angle chase.
to get a geometric relation. perpendicular to intersect at point 
.
) yields 
are concyclic, which leads to 
the other intersection point of two circles,let be the intersection point of and and let be the circumcenter of 
.Now,let 
be the intersection of the perpendicular from to 
and the perpendicular from to 
.Now,we have that 
is similar to 
so we get 
is similar to 
.Now,from an easy angle chase and using the previos similarity we get 
is similar to 
and from this we get 
so 
and that is it.
and let 
cut 
for a second time at 
, respectively. Let be the projection of onto 
Note that 
is the external center of homothety that maps 
It is clear that this homothety takes 
and 
, so 
Hence, 
is a diameter of 
are collinear. Furthermore, note that 
, where the last step follows from Power of a Point. Therefore, 
are concyclic, so 
, where the angles are directed. Hence, 
are concyclic, so 
, as desired. 
be the points of intersection between circles with circumcenters 
with closer to line than . Let 
, and 
denote the reflections of 
over , respectively. Also let denote the midpoint of line . We then make the following claim:
.
, that ![\[\measuredangle A_1OC = \measuredangle BO_2C = \measuredangle AMP, \]](http://latex.artofproblemsolving.com/d/c/e/dce3a79a273352aae2d04c26e2c22d320f8328d8.png)
as 
is also cyclic. 
are collinear: are collinear.
. It then suffices to equivalently show . Indeed, we have that ![\[\measuredangle AA_1C = \measuredangle MAP, \]](http://latex.artofproblemsolving.com/a/2/f/a2f53ca1321b99a2e7c1bc9e2d025cc90c002ab3.png)
and by the above claim/lemma, we obtain the desired similarlity. are collinear, in particular 
, as desired. 
the symmetric of wrt 
; be the ex-similicenter ; the intersection of and 
;
the antipodes of 
; the intersection of 
and ; the intersection of 
and we will show that :
i.e. 
is parallelogram.
but 
hence 
are collinear.
then 
.
be . A common tangent touches the circles at 
respectively. Let the perpendicular from to the line meet at . Prove that 
.
, the problem becomes equivalent to a statement regarding the which is quite neat: In triangle with orthocentre and point on arc 
such that 
(hence 
bisects 
), the line through parallel to , the line , and the perpendicular bisector of line concur. Let 
be points on 
such that 
is a parallelogram. Then line 
passes through the midpoint of 
, i.e, the centre of the circumcircle of 
. Also, 
by projecting all on line , hence 
or 
, hence is the perpendicular bisector of , and proving lies on it was all we needed.
where 
.
,
is the 
-
after inversion.
be a triangle and 
a
to 
.
such that, 
.
be a point such that 
is rectangle. Prove, 
.
