[/quote]
,
points on the plane. No three of them lie on the same straight line and no four lie on the same circle. Prove that it is possible to split these points into 
pairs and cover each pair of points with a circle containing no other points.
is called pretty if it has finite expression in bas
is a positive integer in ![$[2;2022]$](http://latex.artofproblemsolving.com/f/0/5/f05e359b3b158e7a4a742cbe662e8d1417fd7926.png)
.
that with every 
in 
then there is at least one pretty number between 
and 
if there exists a positive integer 
and the decimal representation of 
consists only of digits among 
.
?
such that 
divides![$\sum_{0 \leq i < j \leq 2024} \left[\binom{2024}{i}\binom{2034}{j} - \binom{2024}{j}\binom{2034}{i}\right]^2$](http://latex.artofproblemsolving.com/4/c/4/4c42b9f80506262b06ff41a89c911784d28e9de2.png)
,
.
with 
,
to the sides 
are 
,
is the orthocenter. 
is the midpoint of segment 
.
and 
intersect at 
.
from 
intersect at 
.
are colinear
,
as the greatest exponent 
such that 
divides 
.
and infinitely many 
.
prime and 
satisfying the equation: 
be positive real numbers with 
.
and 
satisfy 
,
is an even number.
Y by samrocksnature, Adventure10, Mango247, Rounak_iitr
Let 
and 
be two circles with respective centres 
and 
intersecting in two distinct points 
and 
such that 
is an obtuse angle. Let the circumcircle of 
intersect and respectively in points 
and 
. Let the line 
intersect in 
; let the line 
intersect in 
. Prove that, the points 
are concyclic.
and 
be two circles with respective centres 
and 
intersecting in two distinct points 
and 
such that 
is an obtuse angle. Let the circumcircle of 
intersect and respectively in points 
and 
. Let the line 
intersect in 
; let the line 
intersect in 
. Prove that, the points 
are concyclic.This post has been edited 1 time. Last edited by integrated_JRC, Jan 22, 2018, 2:10 AM
is a diameter of 
.
is a diameter of 
.
is a straight line parrallel to 
.
meet 
,
.
is cyclic implying 
giving 
are collinear. Similarly, 
are collinear.
.
.
and 
.
is not isosceles, 
and 
gives 
is cyclic. So, 
.
.
and 
are co-linear.
we get,
intersects 
at 
are co-linear. (Proved)
are co-linear.
are co-linear.
are concyclic [Q.E.D.]
,
and 
,
and 
.
is obtuse, 
,
and 
,
,
.
is the reflection of 
is the reflection of 
because there is only one point other than 
whose reflections in 
(why?) 
and the orthocenter clearly satisfies this condition. Hence, 
is the orthocenter of 
and 
are the feet of altitudes through 
is cyclic.
meet the circle 
again at 
.
.
)
and 
are collinar in that order .
and since 
and 
are the radiuses of their respective circles , we get that 
wich completes the proof.
which can never be equal to 
so your claim is wrong
I think my solution works fine.
are collinear. Similarly 
are collinear.
(Angle in a semicircle). By inscribed angles theorem, 
and 
passes through 
which implies the conclusion
are coliner. Similarly 
and 
.Note that 
since it's radial axis of 
.
is isosceles .And hence 
and by similar argument we have 
and hence 
.
.
.
are concyclic . 
.
are Coliner and 
parallel to 
which gives 
.
parallel to 
parallel to 
are concyclic so 
.
hence 
.
we get 
so we are done .
is cyclic..
are collinear, so is 
and 
and we are done.
are collinear, hece 
,
are concyclic
are collinear.
so 
so 
is cyclic.
and 
are the reflection of 
and 
respectively, then note that since both 
,
is the orthocenter of 
and therefore, 
are in fact collinear, the conclusion now follows immediately.
Let 
be the perpendicular bisector of segment 
be the reflection of 
.
.
such that 
.
be the perpendicular bisector of segment 
.
be the mid-point of 
and let 
be the mid-point of 
.
This gives
Now, if we let 
be the mid-point of 
,
.
.
be the perpendicular bisector of 
.
are parallel, it follows by Equal Intercept Theorem that 
Now, 
Also, since 
Combining, we get 
This means that 
are concyclic.
and 
,
(exterior angle of cyclic quadrilateral equals interior opposite angle). This proves that 
is the perpendicular bisector of 
,
.
which has been left as an exercise to the reader (you can send your proof as a reply to this message).
Please note that this proof may seem overly extended, however I have made use of many properties so that the reader recalls them and gains benefit from reading this proof.
,
,
.
since 
,
.
,
.
,
.
as desired.
.
.
?
).
a
and 
and so you get 
which finishes.
and 
become two lines that intersect at 
becomes the line 
,
and 
,
are collinear as 
;
is cyclic by a quick angle chase.
.
are collinear, 
are collinear, 
are collinear, 
are concyclic, and 
are concyclic.
.![\[\angle AF^*C^* = \angle C^*F^*O_1^* = \angle C^*D^*B^*\]](http://latex.artofproblemsolving.com/9/1/b/91b6741d9e16784dd56e24881e4c35ee658cc428.png)
![\[\measuredangle AO_1D = \measuredangle AO_2D = 2\measuredangle ADO_2 = 2\measuredangle AO_1O_2 = \measuredangle AO_1B\]](http://latex.artofproblemsolving.com/a/0/9/a09b3d926c5041c685b4efc06635f8b14a6b233e.png)
from which it is clear that ![\[\measuredangle FCE = \measuredangle FCB = 90^\circ = \measuredangle BDE = \measuredangle FDE\]](http://latex.artofproblemsolving.com/4/7/e/47ee4a61c9c7358236097e5c38bf28ba788daacc.png)
from which it is clear that the points 
As 
,
,
,
,
are collinear,
and 
,
is a cyclic quadrilateral.
is collinear.
Then, we have that 
goes to a line 
goes to the reflection of 
goes to the reflection of 
going to the intersections of 
with 
We also have that 
go to the intersections of 
and 
since 
Thus, 
so 
is cyclic. However, since 
we have that 
is cyclic. Inverting back, we get our desired claim.
![[asy]
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By Angle Chasing we get, 
Also, 
Since, 
and 
are collinear and 
and 
are collinear and parallel to 
Now By Angle Chasing we get, 
And 
Therefore 
are concyclic points.
is the perpendicular bisector of 
.
.
which gives 
is a circle so 
collinear, finishes by PoP.
