[/quote]
,
be a trapezium inscribed in a circle 
with diameter 
. The circle with center 
and radius 
and 
(where 
)
at 
.
is perpendicular to 
.
![\[ x + y^2 = p^m \]](http://latex.artofproblemsolving.com/e/c/f/ecf553036f22327dd457dc06611b6bc1b0ef072c.png)
![\[ x^2 + y = p^n \]](http://latex.artofproblemsolving.com/5/b/d/5bd672934a395aabd6865ca4b50d8da75c0fe07e.png)
For 
positive integers and 
being a prime.
and 
.
Equality holds when 
or 
,
such that 
.
intersect again at 
.
Equality holds when 
intersect its circumcircle 
.
determine a triangle 
,
is relative to 
are concurrent at a point 
and that 
and 
.
lie in the interiors of segments 
respectively, such that 
Let 
intersect 
at 
,
intersect 
at 
.
and 
are either parallel or coincide.
and 
.
Where 
such that :
.
be reals such that 
Prove that 
Let 
Prove that 
be a fixed integer. There are 
beads on a circular necklace. You wish to paint the beads using 
colors, such that among any 
consecutive beads every color appears at least once. Find the largest value of 
for which this task is 
possible.
Y by Adventure10, Mango247, ddami
On sides 
and 
of 
given are 
and 
, respectively. Let 
(
) be a point of intersection of circumcircle of 
and line that is parallel to 
and passing through C. Let 
be a point of intersection of line 
and side , and let 
be on line such that 
and 
. If 
, prove that point of intersection of 
and 
lie on angle bisector of 
.
Proposed by Milos Milosavljevic
and 
of 
given are 
and 
, respectively. Let 
(
) be a point of intersection of circumcircle of 
and line that is parallel to 
and passing through C. Let 
be a point of intersection of line 
and side , and let 
be on line such that 
and 
. If 
, prove that point of intersection of 
and 
lie on angle bisector of 
.Proposed by Milos Milosavljevic
This post has been edited 1 time. Last edited by mihajlon, May 22, 2015, 10:28 PM
and 
.
are concyclic ,
,
are concyclic 
are collinear (Reim theorem (C-D-B and F-E-H)) .
,
,
is the bisector of 
. i.e. the intersection of 
and 
lie on the bisector of 
are cyclic, 
.
.
,
are cyclic.
to 
(the midarc 
say 
concur by radical at say 
and let 
.
and since 
is exterior bisector 
must be the interior bisector.
.
.
and 
are cyclic, since
and
Therefore, we have 
and since 
,
be the incentre and 
.
,
and 
multiplying those together we obtain 
where the first is true, since 
and we used the fact that 
and we conclude by Ceva's theorem, that 
are concurrent, i.e. 
and 
is cyclic.
,
,
.
.
. So this claim is also done. 
and 
,
and 
.
,
lies on the angle bisector of 
b
denote the major arc midpoint of 
in 
.![[asy]
import geometry;
size(12cm);
pair foot(pair P, pair A, pair B) {
return foot(triangle(A,B,P).VC);
}
path rect = (2,2)--(2,-1.1)--(-3.7,-1.1)--(-3.7,2)--cycle;
pair C = dir(140);
pair A = dir(220);
pair B = dir(320);
pair X = intersectionpoint(bisector(line(C,A),line(C,B)),line(A,B));
pair P = (2C+7X)/9;
pair D = intersectionpoint(line(A,P),line(B,C));
pair E = intersectionpoint(line(B,P),line(A,C));
pair F = intersectionpoints(parallel(C,line(A,B)),circle(C,D,E))[1];
pair Y = intersectionpoint(line(D,E),line(B,A));
pair G = intersectionpoint(line(F,D),line(A,B));
pair H = intersectionpoint(line(F,E),line(A,B));
pair N = intersectionpoints(line(Y,C),circle(C,D,E))[1];
draw(A--B--C--cycle , red+1.3);
draw(C--X,dotted+blue);
draw(D--Y,red);
draw(A--Y,red);
draw(N--Y,red);
draw(circumcircle(A,E,D),green);
draw(circumcircle(B,E,D),green);
draw(circle(F,G,H),purple+dashed);
draw(circumcircle(A, B, C) , blue);
draw(circumcircle(C, E, D) , orange);
dot("$A$", A, dir(A));
dot("$B$", B, dir(B));
dot("$C$", C, dir(C));
dot("$X$", X, dir(X));
dot("$D$", D, dir(20));
dot("$E$", E, dir(E));
dot("$F$", F, dir(F));
dot("$G$", G, dir(G));
dot("$H$", H, dir(H));
dot("$Y$", Y, dir(Y));
dot("$N$", N, dir(N));
clip(currentpicture,rect);
[/asy]](http://latex.artofproblemsolving.com/7/c/1/7c1d1bb9d2261749774af55fd598bb94924944c5.png)
![\[\measuredangle AGF = \measuredangle CFG = \measuredangle CFD = \measuredangle CED\]](http://latex.artofproblemsolving.com/0/1/1/0113421f818db7c2c5ef8cdaece2817c51aad2a1.png)
which implies that ![\[\measuredangle HDE = \measuredangle ADE + \measuredangle HDA = \measuredangle AGE + \measuredangle GEB = \measuredangle GBE = \measuredangle HBE\]](http://latex.artofproblemsolving.com/7/4/2/7427fd92ba380dab9f1470235edc77888020ffcd.png)
from which it follows that ![\[\measuredangle FED = \measuredangle FDC = \measuredangle FCB = \measuredangle HBD\]](http://latex.artofproblemsolving.com/9/7/f/97fba7d2088746378209be407038191558cf7b0c.png)
which since 
and 
,
and 
.![\[\measuredangle HEN = \measuredangle FEN = \measuredangle FDN = \measuredangle GDN\]](http://latex.artofproblemsolving.com/d/6/7/d674ec5f6541b52a64ccb8fcd569dcc73e0843ed.png)
which implies that 
.![\[\measuredangle NHF = \measuredangle NHE = \measuredangle NGD = \measuredangle NGF\]](http://latex.artofproblemsolving.com/1/1/e/11ed70dbbd42c1f4b89ac92bb17b80841c5ab6e9.png)
from which it follows that 
is a cyclic quadrilateral. Further,![\[\measuredangle AGN = \measuredangle HGN = \measuredangle HFN = \measuredangle EFN = \measuredangle ECN = \measuredangle ACN\]](http://latex.artofproblemsolving.com/2/3/e/23e4bf301dd77c29e829451d3085dd0469ed2a45.png)
and thus 
must also be cyclic. Now, by the Radical Center Theorem on circles 
, 
and 
,
, 
and 
intersect at a point 
,![\[-1=(YZ;ED)\overset{C}{=}(YX;AB)\]](http://latex.artofproblemsolving.com/d/9/f/d9f5d181f0a46e75ea455a2c8cc3a411899084a7.png)
Now, say 
.
,![\[-1=(YX';BC)\]](http://latex.artofproblemsolving.com/4/f/b/4fb351ec875c8904f91fb07d23e11838cc1eb2aa.png)
which implies that 
,
b
are collinear
intersect extended 
at 
.
.
.
Hence, 
Hence 
Hence 
intersect 
.
,
From Ceva's Theorem
