[/quote]
,
and 
such that
be a real number. Determine all natural numbers 
for which there exist distinct natural numbers 
,
,
,
denotes the greatest common divisor of natural numbers 
and 
)
,
lies on the circumcircle of 
.
intersects 
in 
.
lies on the circumcircle of triangle 
.
,
.
points and some chords between them is given.
In every step we can take two chords 
with a common point other than 
and erase exactly one of 
let 
be the minimum of chords after some steps. Find the maximum of 
In every step we can take two chords 
be a positive integer and let 
and 
be positive integers congruent to 1 modulo 4. Prove that there exists a positive integer 
such that at least one of the numbers 
and 
is divisible by 
be the lengths of the sides of a triangle. Prove that![\[\frac{1}{(a-4b)^2}+\frac{1}{(b-4c)^2}+\frac{1}{(c-4a)^2} \geqslant \frac{1}{ab+bc+ca}.\]](http://latex.artofproblemsolving.com/7/4/b/74b7aa5621bc52977d9acd448d74293dc27a8e00.png)
and 
meets for the second time the circumcircle of 
at point 
meets line 
at point 
meets line 
at point 
.
and 
pass through the same point.
.
and 
intersect the altitude from 
is the circumcenter of triangle 
and 
is the reflection of 
is the second intersection of circumcircles of triangles 
and 
.
are collinear.
and 
Find the value of 
and 
and 
is a prime number
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 
.
,
,
are collinear.
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 
,
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 
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)
