[/quote]
,
be an acute triangle with orthocenter 
and circumcenter 
.
and 
be points on the line segments 
and 
respectively such that 
is a parallelogram. Prove that 
.
be an integer. Show that, if 
is an integer, then it is a perfect square.[/quote]
.
is at least rational, so that 
must be a perfect square then. Using Conway's topograph method, I have found out that the smallest non-negative pairs 
for which this happens are 
and 
,
,
The eigenvalues of that matrix are irrational, however, so that any calculation which uses powers of that matrix is a little cumbersome. There must be an easier way, but I cannot find it. Can you?
that satisfy 
for all 
and 
satisfies the following recursion formulas: 
,
.
interesting if 
of 
and 
is also a divisor of 
and 
be points on segment 
denote the circle passing through 
.
denote the circle passing through 
at 
.
)
.
the feet of the altitudes lying on 
respectively. One of the intersection points of the line 
and the circumcircle is 
The lines 
and 
meet at point 
Prove that 
airline companies in a country. Between any two cities, there is exactly one 
and circumscribed about the incircle 
,
.
,
,
,
,
,
,
,
(distinct from 
intersects 
(distinct from 
be the midpoint of the arc 
of 
at 
and 
,
at 
.
.
but doesn't tell it to anyone. Bernaldo thens pick an arbitrary positive integer 
(not necessarily in 
.
chosen by Arnaldo.
be a set of 2025 positive real numbers. For a subset 
,
as the median of 
when all elements of 
.![\[
P(A) = \sum_{\substack{T \subseteq A \\ |T| \text{ odd}}} M_T, \quad Q(A) = \sum_{\substack{T \subseteq A \\ |T| \text{ even}}} M_T.
\]](http://latex.artofproblemsolving.com/1/d/f/1df2ea8487289111939a53dcc49576fbe6fdef37.png)
Find the smallest real number 
such that for any set ![\[
P(A) - Q(A) \leq C \cdot \max(A),
\]](http://latex.artofproblemsolving.com/c/d/e/cdea343d11c722a423e857aeb01202863405773e.png)
where 
denotes the largest element in 
in 
colors, there exists an arithmetic progression of size 
.
from this?
.
be foots of perpendicular lines from 
.
is cyclic and 
.
bisects 
such that 
Y by nguyendangkhoa17112003, Adventure10, Mango247
Let 
be a triangle, and 
its orthocenter. Let 
be a point on the circumcircle of triangle (distinct from the vertices 
, 
, 
), and let 
be the foot of the altitude of triangle from the vertex . Let the parallel to the line 
through the point meet the parallel to the line 
through the point at a point 
. Let the parallel to the line 
through the point meet the parallel to the line through the point at a point 
. The lines 
and 
intersect at some point 
. Prove that the lines 
and are parallel.
be a triangle, and 
its orthocenter. Let 
be a point on the circumcircle of triangle (distinct from the vertices 
, 
, 
), and let 
be the foot of the altitude of triangle from the vertex . Let the parallel to the line 
through the point meet the parallel to the line 
through the point at a point 
. Let the parallel to the line 
through the point meet the parallel to the line through the point at a point 
. The lines 
and 
intersect at some point 
. Prove that the lines 
and are parallel.
,
and 
,
.
.
and 
are the reflections of the point 
.
and 
.
,
,
.
be the foot of altitudes from 
respectively and wlog let 
is a parallelogram so 
.
are concyclic 
.
are concyclic.So 
.
so from 
we get that 
.
are concyclic 
.
across 
be 
and the reflection of 
is cyclic 
(
,
(simple transformation geometry tricks)
and 
are cyclic(simple angle chasing)
f
and some simple angle chasing)
and 
solves the problem.
,
cyclic.
,
and 
.
.
is cyclic.
.
.
.
,
& similarly 
are cyclic. 
and then note that,
Hence, 
is cyclic and By Reim's Theorem 
.
,
is a projective map. Similarly 
is also projective. Also, taking perspectivity from 
is a projective map. Let 
As 
are fixed, so we get that the points 
are all projective related to 
for three positions of 
and when 
(It requires some work, but I am really not in the mood of writing all that 
). Hence, done. 
![[asy]
size(10cm);
defaultpen(fontsize(10pt));
pair A = dir(125), B = dir(210), C = dir(330), P = dir(280), Q = A+B-P, R = A+C-P, H = orthocenter(A,B,C), E = foot(B,C,A), X = extension(A,Q,R,H);
dot("$A$", A, dir(90));
dot("$B$", B, dir(260));
dot("$C$", C, dir(330));
dot("$P$", P, dir(280));
dot("$Q$", Q, dir(90));
dot("$R$", R, dir(90));
dot("$X$", X, dir(0));
dot("$H$", H, dir(270));
dot("$E$", E, dir(45));
draw(unitcircle);
draw(A--B--C--A, linewidth(1.2));
draw(R--H--B--P--C--H--A--X--E--H^^C--R--A--Q--B^^A--P);
draw(circumcircle(A,B,H)^^circumcircle(A,C,H)^^circumcircle(A,E,H), dotted);
[/asy]](http://latex.artofproblemsolving.com/9/3/9/939a4fb451694e35f03503c067717c1e1a97608f.png)
denote a directed angle modulo 
.
and 
so 
.
,
as desired. 
is cyclic. Therefore 
and we are done. 
Now, since 
and 
,
But remark that 
so it follows that 
.
is cyclic, so 
which means 
,
and 
are congruent and corresponding sides are parallel. Let the orthocenter of 
;
so that a translation bringing 
and 
are concyclic. So 
as desired.
are concyclic. This implies that 
so 
which implies the result.
