[/quote]
,
be an acute triangle with 
and let 
be its orthocenter. Let 
be a point on the perpendicular bisector of 
such that 
and 
are on different sides of 
a point on the perpendicular bisector of 
such that 
and 
a point on the perpendicular bisector of 
such that 
and 
lie on the opposite side of 
w.r.t 
.
and 
square is filled with numbers 
.
squares is summed,and arranged in an increasing order. Is it possible to obtain the following sequences as a result of this operation?
be a triangle with points 
lie on the perpendicular bisector of 
lie on a circle. Suppose 
are perpendicular to sides 
at points 
The tangent lines from points 
to the circumcircle of 
intersects at point 
Prove that: 
are parallel.
,
be four points in the plane, with 
and 
.![\[\frac{AB \cdot CD}{AC \cdot BD}, \]](http://latex.artofproblemsolving.com/f/3/c/f3cf9729b32b210faeea6e2d23da3582be2ae061.png)
and 
are orthogonal. (Intersecting circles are said to be orthogonal if at either common point their tangents are perpendicuar. Thus, proving that the circumcircles of the triangles 
such that for any positive real numbers 
and 
,
Proposed by Athanasios Kontogeorgis, Grecce, and Dorlir Ahmeti, Kosovo
of positive real numbers satisfies ![\[a_{k+1}\geq\frac{ka_k}{a_k^2+(k-1)}\]](http://latex.artofproblemsolving.com/0/3/9/03909a5f60d944063182ec97cf73f283178e8d4d.png)
for every positive integer 
for every 
.
,
and 
be the respective intersections with BC of the internal angle bisector, external angle bisector, and the median from A. The circumcircle of 
intersects 
a second time at point 
different from A. Define 
and 
analogously. Prove that the circumcenter of 
lies on the Euler line of ABC.
such that 
for all 
of the acute-angled triangle 
is tangent to its side 
at a point 
.
be an altitude of triangle 
be the midpoint of the segment 
is the common point of the circle 
(distinct from 
are tangent to each other at the point 
is called lucky if it has at least two distinct prime divisors and can be written in the form:![\[
n = p_1^{\alpha_1} + \cdots + p_k^{\alpha_k}
\]](http://latex.artofproblemsolving.com/7/4/4/744a5ccaeb9476ebd7d999c395762cb6e99a7a71.png)
where 
are distinct prime numbers that divide 
,
.
,
,
be a non-isosceles triangle with incenter 
Let 
be the smaller circle through 
tangent to 
and 
(the addition of indices being mod 3). Let 
be the second point of intersection of 
and 
Prove that the circumcentres of the triangles 
are collinear.
such that 
coprime, 
and 
.
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 
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 
,
and 
.
is the bisector of 
,
,
.
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 
be a triangle and 
and 
are cevians concurrent with this altitude at 
is cyclic if and only if 
and 
.
intersect 
.
are collinear if and only if 
from which we find 
which means that 
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 
.
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 
(possibly at infinity). By Pascal's on 
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 
expoint in 
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 
,
