[/quote]
,
and 
.![\[P=\frac{1}{5-2a}+\frac{1}{5-2b}+\frac{1}{5-2c}.\]](http://latex.artofproblemsolving.com/c/3/7/c3708768a84c2bbb1e4fae502a9bb1e92cb13d02.png)
be the incentre of acute-angled triangle 
.
,
,
respectively. Let line 
intersect the circumcircle of the triangle at 
and 
,
lies between 
.
so that for every 
in 
.
.
and 
intersect at a point 
.
be the circumcircle of acute triangle 
and 
intersect at 
and 
.
and 
.
are concyclic.
the centre of the circle passing through 
,
are difined similarly. Prove that 
,
,
are concurrent.
> 
that satisfy 
,
€ 
be a fixed positive integer. Given a set 
of 
points in the plane such that no three are collinear and no four concyclic, let 
be the number of circles 
that contain 
in their interior, and let ![\[m(S)=a_1+a_2+\cdots + a_n.\]](http://latex.artofproblemsolving.com/4/9/b/49b928b1f20e2d1799a1234c7f25a2bc3d62d0dd.png)
Prove that there exists a positive integer 
depending only on 
such that the points of 
are the vertices of a convex polygon if and only if 
such that 
.
and 
.
,![\[
\sum_{\text{cyc}} \sqrt{1 + \left(x\sqrt{1 + y^2} + y\sqrt{1 + x^2}\right)^2} \geq \sum_{\text{cyc}} xy + 2\sum_{\text{cyc}} x
\]](http://latex.artofproblemsolving.com/9/5/5/955863a0cf8747ae45b736b0631f243d3908eb84.png)
?
and 
is a subset of 
such that if 
then 
Find the smallest of 
that satisfy the following property: if 
,
,
,
are all the positive divisors of 
,
divides 
for every 
.
,
:
,
Y by estoyanovvd, Amir Hossein, Mathuzb, Pluto04, HWenslawski, Adventure10, Mango247, KhaiMathAddict, ItsBesi
A quadrilateral 
is inscribed in a circle 
where 
,and is not paralel to .Point 
is the intersection of diagonals 
and 
, and the perpendicular from to intersects the segment at a point 
.If 
bisects the angle 
prove that is diameter of .
Proposed by Emil Stoyanov,Bulgaria
is inscribed in a circle 
where 
,and is not paralel to .Point 
is the intersection of diagonals 
and 
, and the perpendicular from to intersects the segment at a point 
.If 
bisects the angle 
prove that is diameter of .Proposed by Emil Stoyanov,Bulgaria
This post has been edited 2 times. Last edited by microsoft_office_word, May 9, 2018, 7:02 PM
,
.
and 
,
,
.
,
,
,
collinear.So 
Letting 
and 
,
,
,
,
,
coincide, so 
are altitudes, so 
and obtain 
is cyclic 
so as 
it follows 
lie on the internal angle bisector. Let:
From this we get:
And as 
.
.
on the circle we get 
.
using 
gives:
And using 
we see:
But as 
which would mean 
contradiciting the problem statement we must have 
.
:
But this gives 
and hence as 
And hence 
,
,
and 
,
,
.
be the intersection of 
and 
,
.
,
are collinear. Let 
be the center of 
,
.
,
.
Then dropping the perpendicular from 
The angle bisector condition is equivalent to:
Now focussing on 
we see as 
:
:
Combining these we get:
And similarly for 
by interchanging 
:
Which we can expand out cancel a lot of terms then factorise to give:
Clearly 
.
Similarly:
.
.
is isogonal to the 
a
,
.
.
so 
is cyclic 
we get that 
and 
and 
are cevians concurrent with this altitude at 
.
is cyclic if and only if 
intersect 
and 
.
intersect 
.
are collinear if and only if 
from which we find 
which means that 
as needed 
and 
be points on the diagonals 
.
is cyclic we get that 
which further implies 
.
.
be the point of intersection of 
and 
are cyclic by angle chasing we get that 
.
triangles 
and 
must be equal, or in other words 
.
is isosceles and since line 
What's the term for this in English? I think I know what it means but I want to be sure.
by Lemma 9.18, we have 
Now let 
From Lemma 9.11, 
projecting through 
we have 
This is enough to show that 
By Brocard, we have 
but since 
are collinear.
is diameter of 
.
,
and 
(we use that 
and 
are cyclic, where 
is the midpoint of 
.
and 
are equal and since their circumcircles are congruent, we have that 
.
,
and 
,
we get our desired conclusion 
then 
is cyclic.
so 
is cyclic and 
so 
and 
so 
so 
so 
![[asy]
size(12cm);
pair B = dir(0); pair C = dir(60); pair D = dir(110); pair A = dir(180); pair M = extension(A, C, B, D); pair E = foot(M, A, B); pair X = extension(A, C, D, E); pair Y = IP(Line(D, E, 10), unitcircle, 1);
draw(unitcircle, blue); draw(A--B--C--D--cycle, blue); markscalefactor=0.03; draw(anglemark(C,E,M), cyan); draw(anglemark(M,E,D), cyan); draw(C--E--M, cyan+blue); draw(D--Y, cyan+blue); draw(A--C, cyan+blue); draw(A-(0, 0.5)--A+(0, 1.25), cyan+blue+dashed); draw(B-(0, 0.5)--B+(0, 1.25), cyan+blue+dashed); draw(Y--C+(0, 0.4), cyan+blue+dashed); draw(E--M+(0, 0.6), cyan+blue+dashed);
dot("$A$", A, W); dot("$B$", B, E); dot("$C$", C, NE); dot("$D$", D, N); dot("$E$", E, SW); dot("$M$", M, NE); dot("$X$", X, NE); dot("$Y$", Y, SE);
[/asy]](http://latex.artofproblemsolving.com/6/1/2/612a54a191cdefc5e5cbfca6b880b0189a996233.png)
and 
.
Thus, 
is harmonic so 
,
,
concur at some point 
we find that 
,
by symmetry, contradiction. Therefore 
so 
and 
.
is an altitude of 
,
collinear.
and 
.![\[(CD; NY) = (CD; N'Y) = -1,\]](http://latex.artofproblemsolving.com/3/e/c/3ec4b170ef2a82ecd496d73101fdbd078b5b1f9a.png)
.
.
.![\[\angle MBA = \angle A'XM = \angle MXA = 90 - \angle XAB,\]](http://latex.artofproblemsolving.com/a/6/6/a66dbd0191b4b374a33ab6bfa7355603cb43914f.png)
,![[asy]
size(300); defaultpen(fontsize(10));
pair A, B, C, D, E, M, N, X, Y;
A = dir(0);
B = dir(180);
C = dir(100);
D = dir(50);
M = extension(A, C, B, D);
E = foot(M, A, B);
N = extension(E, X, C, D);
X = extension(A, D, B, C);
Y = extension(A, B, C, D);
draw(circumcircle(A, B, C));
draw(B--X--A--C--Y--B--D--E--C--X--E);
dot("$A$", A, dir(315));
dot("$B$", B, dir(225));
dot("$C$", C, dir(120));
dot("$D$", D, dir(45));
dot("$E$", E, dir(270));
dot("$M$", M, dir(170));
dot(extension(E, X, C, D));
label("$N$", extension(E, X, C, D), dir(225));
dot("$X$", X, dir(90));
dot("$Y$", Y, dir(270));
[/asy]](http://latex.artofproblemsolving.com/a/4/b/a4b010fa98e669170c5eba228890806e6ba3ee24.png)
and 
.
and 
which implies that 
so 
passes through 
is a diameter as desired.
and 
.
and 
be the intersections of 
with 
and 
respectively. Then, due to the Right Angles/Bisectors picture, we know that ![\[(DC;R_1X)=-1\]](http://latex.artofproblemsolving.com/e/5/8/e58ff49b92916c3b6eae84a9524c61e549764232.png)
Further, from the Ceva/Menelaus picture we also have that![\[(DC;R_2X)=-1\]](http://latex.artofproblemsolving.com/7/9/a/79a15ec513bcdec05d53ede5cfcfbcdb2daeaabf.png)
Thus, 
and indeed points 
.
.
.
hinted on pascal 
,
.
.
,
be the tangent lines to 
must concur at a point call it 
is harmonic. Pascal on 
gives us 
contradiction.
, 
and 
.
due to the famous right angle-angle bisector-harmonic lemma.
lies on the polar of P with respect to 
hence 
due to brokard's.
which implies 
which coupled with the given conditions implies that 
= 
and 
= 
from the fact that 
bisects 
and 
we obtain 
then project from 
the line 
onto 
to get 
then do the same for the line 
to get 
concurrent the rest just finishes by brocard on 
.
and 
intersects at 
and 
and 
intersects at 
collinear, since 
we know that 
and 
,
and 
concurrent, let that concurrency point be 
,
and 
and 
,
.
,
.
is perpendicular to 
,
