[/quote]
,
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?
interesting if 
of 
and 
is also a divisor of 
,
and 
.
is prime, 
with 
and 
be an acute triangle with orthocenter 
and circumcenter 
.
and 
be points on the line segments 
and 
respectively such that 
is a parallelogram. Prove that 
.
that satisfy 
for all 
and 
satisfies the following recursion formulas: 
,
.
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 
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Let 
be the orthocenter of an acute-angled triangle 
. The circle 
centered at the midpoint of 
and passing through intersects the sideline at points 
and 
. Similarly, define the points 
, 
, 
and 
.
Prove that the six points, , , , and are concyclic.
Author: Andrey Gavrilyuk, Russia
be the orthocenter of an acute-angled triangle 
. The circle 
centered at the midpoint of 
and passing through intersects the sideline at points 
and 
. Similarly, define the points 
, 
, 
and 
.Prove that the six points
, , , , and are concyclic.Author: Andrey Gavrilyuk, Russia
This post has been edited 4 times. Last edited by orl, Jul 20, 2008, 8:14 AM
the midpoints of 
and 
.
,
and 
are also perpendicular, and 
lies on the radical axis of 
and 
.
,
and 
all lie on a circle, whose center is clearly 
,
.
and 
also lie on the circle centered at 
and 
,
is perpendicular to the line connecting the centers of these circles, i.e. it is perpendicular to 
,
so 
lie on one circle. From the same reasoning 
and 
lie on a one circle as well. Suppose that this three circles doesn't coincide. Then we obtain a contradiction since radical axis of these circles - the sides of the triangle 
is cyclic consists in transforming the relation 
into 
(where 
and 
are the respective midpoints of ![$ [BC]$](http://latex.artofproblemsolving.com/3/5/5/3550468aa97af843ef34b8868728963dec043efe.png)
and ![$ [AC]$](http://latex.artofproblemsolving.com/5/b/0/5b08b2b92472414250205866b93405af4772e86c.png)
)
on both sides and applying the generalized version of Pythagoras' theorem (the one with cosines, I ignore the correct English name), we get a condition which is obviously true (simplifying cosines in right-angled triangles)...
is the midpoint of 
therefore 
are cyclic on circle 
lie on 
a
lie on circle 
.
,
be two isogonal points with respect to a given triangle 
,
,
be the orthogonal projections of 
and 
are concyclic. Let 
and 
.
.
circumcenters of the 
and 
respectively and 
midpoints of the segments 
.
.
.
.
.
and 
.
.
we have to prove that 
is symmetric in 
,
is the midpoint of 
is the midpoint 
.
.
is the radical axis of circle 
and 
.
cylcic, by similar argument on 
,
is independent from 
.
.
)
,
of the sides ![$ [CA]$](http://latex.artofproblemsolving.com/b/8/3/b8309224981a62a2c0df087562b92830bd7ad7b0.png)
, ![$ [AB]$](http://latex.artofproblemsolving.com/d/7/a/d7a8027c238eec9cf67de0f7ec6cb1df4df49a61.png)
respectively. Prove easily or is well-known that
. Thus, 
(remark that 
a.s.o.).
, 
, 
). In conclusion, 
.
obtain easily that 
.
be the nine-point circle of 
be the midpoint of 
be the midpoint of 
,
which is the square of the diameter of the nine-point circle, minus two times the power of 
is also equal to this value, and 
lie on a circle centered at 
is cyclic 
it is enough to show that 
and 
respectfully. Since 
,
,
.
,
.
,
and 
are cyclic, so 
are concyclic.
and 
is clearly 
cyclic. Similarly 
and 
,
,
don't concur, it follows that 
and 
respectively. Hence, from Midpoint Theorem, 
.
.
to be the circle centered at 
and 
.
is the radical axis of 
.
are collinear {i.e. 
lies on the radical axis of 
lie on a circle. Similarly, 
and 
lie on a circle.
)
are concyclic.
.
and 
are concyclic then rest of the cases follow similarly.
and 
to be 
and 
respectively.
and 
respectively.
and 
Also since line 
Hence, 
,
and 
,
.![\[
\begin{cases}
A_1HA_2 &: A_1C \cdot CA_2 = CI \cdot CH \\
B_1HB_2 &: B_1C \cdot CB_2 = CI \cdot CH.
\end{cases}
\]](http://latex.artofproblemsolving.com/f/f/d/ffd30b16d16de24106a803e6299aeedd89e50b38.png)
Equating yields 
which, by Power of Point, implies 
are both cyclic, thus 
,
,
is the same as the perpendicular bisector of 
and the same with the other points are equal. Let 
,
,
.
be the point on 
is tangent to 
be the point on 
is tangent to 
and 
,
.
because 
.
,
.
and the vertices be 
s.t. 
and the orthocenter is 
which is symmetric in 
and cyclic variants are all cyclic.
be the intersection of 
,
,
is just the 
and 
and 
at 
lies on 
is the radical axis of 
.
are concyclic. Similarly 
and 
are separately concyclic, which follows from symmetry. Hence 
be the circles centered at 
and 
and 
,
collinear.
,
are concyclic. Similarly, we can find that 
are concyclic, as are 
they are the same circle and we are done.
and 
by similar triangles. Thus, 
.
and 
;
.
is 
by power of a point on circle 
.
,
be the midpoint of 
and 
similarly. Note that we label the vertices as 
and 
are perpendicular to 
so it is also perpendicular to 
is cyclic. Similar conclusions follow for 
,
are distinct. This provides a contradiction as their radical axes will intersect at 
We begin by showing that 
are concyclic. Let 
but since 
so 
Therefore, 
and by Power of a Point our claim is proven.
are concyclic, and 
the circumcenter of 
so we are done. QED
.
are concyclic, which finishes.
