[/quote]
,
such that for any two co-prime positive integers 
if 
then 
.
is a prime and 
,![\[
n! \sum_{pi+j=n} \frac{1}{p^i i! j!} \equiv 0 \pmod{p}.
\]](http://latex.artofproblemsolving.com/1/6/6/166f07aa9e69c0eb35833971acb9b04932fea087.png)
in 
.
be the symetric of 
the symetric of 
and 
![\[
f_1 = 1, f_{2n} = 3f_n, f_{2n+1} = f_{2n} + 1.
\]](http://latex.artofproblemsolving.com/d/1/0/d10022160679c8c49931c737d2bd04da8f186612.png)
Find the value of 
.
.
with 
is called historic if 
.
,
are written on a blackboard. The players take turns making moves with Ingrid starting. A move consists of one of the players crossing out a number on the board that has not yet been crossed out. If the product of all currently crossed out numbers is 
after the move, the player whose move it was receives one point, otherwise, zero points are awarded. The game ends after all numbers have been crossed out.
with 
elements. Prove that there exist three elements 
from 
.
be an acute triangle and 
its orthocenter. Let 
.
are positioned around the sides of 
be positive real numbers satisfying![\[ xy + yz + zx = 3xyz. \]](http://latex.artofproblemsolving.com/9/2/0/9200874fd6bf9a32ff0c4e82382a450be68cc444.png)
Prove that![\[
\sqrt{\frac{x}{3y^2z^2 + xyz}} + \sqrt{\frac{y}{3x^2z^2 + xyz}} + \sqrt{\frac{z}{3x^2y^2 + xyz}} \le \frac{3}{2}.
\]](http://latex.artofproblemsolving.com/f/5/f/f5fb85d81e93935601ceb5f30c25aea7aa3f23b2.png)
and 
be real numbers. Let 
be the matrix with entries ![\[a_{ij} = \begin{cases}1,&\text{if }x_i + y_j\geq 0;\\0,&\text{if }x_i + y_j < 0.\end{cases}\]](http://latex.artofproblemsolving.com/c/0/8/c0856fe7b2b113180659952c4d216a01d8849f64.png)
Suppose that 
matrix with entries 
,
such that the sum of the elements in each row and each column of 
.
.
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 
with diameter 
.
,
we have 
.
,
,
so 
are concyclic.
so 
are concyclic.
are collinear
passes through a fixed point as 
,
.
,
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.
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)
