[/quote]
,
such that 
.
be a sequence of positive integers satisfying the following property: for all positive integers 
,
and for all distinct integers 
,![\[
a_{m_1} + a_{m_2} + \cdots + a_{m_k} \leqslant a_{n_1} + a_{n_2} + \cdots + a_{n_\ell}.
\]](http://latex.artofproblemsolving.com/d/1/6/d1632ff7e5e55319cdbf56f45dca9ed84343aa47.png)
Prove that there exist two integers 
and 
such that 
for all 
.
on the board. Charlotte has two operations available: the GCD operation and the LCM operation.
and 
.
.
and, for each of them, determine the minimum number of GCD operations Charlotte must use.
be a triangle, 
its circumcircle and 
the center of 
and 
.
and the tangent to 
.
be an integer. We consider a square grid of size 
divided into 
unit squares. The grid is called balanced if:
,
or 
.
.
,
such that any balanced grid always contains an 
square whose absolute sum of the 
cells is less than or equal to 
.
of distinct positive real numbers is called radiant if it satisfies the following property: if 
is also an element of 
?
or 
.
.
and 
.
,
and 
.
be the circumcenter of triangle 
,
the orthocenter of triangle 
the point of intersection of the lines 
and 
.
to the system:![\[a_1^2 + a_1 - 1 = a_2\]](http://latex.artofproblemsolving.com/a/4/e/a4e26b83c929b82800576bac11e12de0c3aa1cf6.png)
![\[ a_2^2 + a_2 - 1 = a_3\]](http://latex.artofproblemsolving.com/9/9/5/995e8eec93828b82901b0639463f907f5ff95d08.png)
![\[\hspace*{3.3em} \vdots \]](http://latex.artofproblemsolving.com/e/e/1/ee15e1fefe2ad550d3c9b1c51207eaf66fbfa41f.png)
![\[a_{n}^2 + a_n - 1 = a_1\]](http://latex.artofproblemsolving.com/7/c/7/7c727d1f2044c1bf7c80981e1074bb3cde807b8f.png)
and 
and 
and 
are 
and 
respectively; the circumcenters of the triangles 
and 
are 
and 
respectively. Prove that 
and 
are parallel.
with 
prime and ![\[a^p+b^p=p!.\]](http://latex.artofproblemsolving.com/f/e/b/feb27894b1f6497559eeb094cfe92256bc26688e.png)
such that 
.
.
Y by Rounak_iitr, ItsBesi, farhad.fritl
Let 
be a triangle such that 
. Let the excircle opposite to A be tangent to the lines 
, and 
at points 
, and 
, respectively, and let 
be its centre. Let 
be a point on the side . The circumcircles of the triangles 
and 
intersect for the second time at 
. Let 
be the foot of the perpendicular from 
to the line 
. Prove that the points 
, and are collinear.
(The excircle of a triangle opposite to is the circle that is tangent to the line segment , to the ray 
beyond 
, and to the ray 
beyond 
.)
Proposed by Bozhidar Dimitrov, Bulgaria
be a triangle such that 
. Let the excircle opposite to A be tangent to the lines 
, and 
at points 
, and 
, respectively, and let 
be its centre. Let 
be a point on the side . The circumcircles of the triangles 
and 
intersect for the second time at 
. Let 
be the foot of the perpendicular from 
to the line 
. Prove that the points 
, and are collinear.(The excircle of a triangle
opposite to is the circle that is tangent to the line segment , to the ray 
beyond 
, and to the ray 
beyond 
.)Proposed by Bozhidar Dimitrov, Bulgaria
This post has been edited 2 times. Last edited by Lukaluce, Jun 28, 2024, 12:38 PM
lie on the circle 
.
,
we have 
.
,
,
are collinear.
so 
are concyclic.
so 
are concyclic.
are collinear
,
.
,
are concyclic in a circle with diameter 
are all concyclic. Now we have that:
.
and 
are the projections of 
so, 
is cyclic. Also
is cyclic.Also we know 
cyclic.So,by written some angles we deduce 
.
as the second intersection of 
and the cyclic 
.![\[\angle DQP=\angle RAD=\beta=\angle ABP\]](http://latex.artofproblemsolving.com/1/7/7/177b863c343f7ed2dd2a2d036a228e63cd807868.png)
hence 
is cyclic. Similarly, 
,
and 
.
Thus 
and 
are concyclic, so 
,
,
lie on a circle.
both 
,
is cyclic, and that ![\[\measuredangle DQE = \measuredangle DQP + \measuredangle PQE = \measuredangle ABC + \measuredangle BCA = \measuredangle BAC = \measuredangle DAE \]](http://latex.artofproblemsolving.com/0/5/e/05e96507e97d3285077eb9af3e4f299190f90306.png)
so 
.
quite clearly), we have![\[\measuredangle DQP' = \measuredangle DQR = \measuredangle DAR = \measuredangle CBA = \measuredangle P'BA\]](http://latex.artofproblemsolving.com/5/5/4/554ca59d2f667b3ce53e8d0d67e99558d1ecbdfa.png)
so 
lies on 
.
and it immediately follows that points 
and 
passes through a fixed point.
(the same argument works in general). Let 
be the point on 
such that 
,
then 
lies on 
.
,
as well.) Then
Now we find the coordinate of 
are concyclic, we have
Since 
,
Similarly, since 
are concyclic, we have
Setting these equal gives
Clearing denominators gives
Subtracting and factoring out 
gives
Thus by Vieta,
and so
Now we find the vectors
Then
We conjugate this, replacing each 
with its reciprocal, and then multiply across by 
,
Points 
and 
Points 
.
So 
Points 
and 
-
So 
Points 
and 
Points 
and 
Points 
and 
and 
we get that : 
Points 
lie on the circle with 
angle at 
and 
with 
.
.
and 
,
is clearly cyclic and moreover 
as they are both perpendicular to 
![\[\angle ADQ=\angle BDQ=\angle QPC=180^\circ- \angle CEQ=180^\circ- \angle AEQ \]](http://latex.artofproblemsolving.com/c/f/1/cf10249cdb40046e02a7a01af2fddeb50f3449a5.png)
cyclic 
cyclic. ![\[
\angle DQP = \angle ABP \overset{AR \parallel BC}{=} 180^\circ -\angle BAR = 180^\circ - \angle DAR = \angle DQR
\]](http://latex.artofproblemsolving.com/1/6/e/16ee4fa7c23a4a6e492c0cd3533c27995bb7d1db.png)
![\[\angle EQP = \angle ACB = \angle CAR = \angle EAR =\angle EQR\]](http://latex.artofproblemsolving.com/1/1/6/116d10b7a12040b81d12e22f547c3898ba9c6173.png)
