[/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 
and circumcenter 
.
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 
.
)
.
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 
Y by Davi-8191, nguyendangkhoa17112003, Adventure10, jhu08, Mango247, Rounak_iitr, and 1 other user
Let 
be a triangle such that 
and 
. The tangent at 
to the circumcircle 
of triangle meets the line 
at 
. Let 
be the reflection of in the line , let 
be the foot of the perpendicular from to 
, and let 
be the midpoint of the segment 
. Let the line 
intersect the circle again at 
.
Prove that the line
is tangent to the circumcircle of triangle 
.
be a triangle such that 
and 
. The tangent at 
to the circumcircle 
of triangle meets the line 
at 
. Let 
be the reflection of in the line , let 
be the foot of the perpendicular from to 
, and let 
be the midpoint of the segment 
. Let the line 
intersect the circle again at 
.Prove that the line
is tangent to the circumcircle of triangle 
.Attachments:
This post has been edited 1 time. Last edited by ThAzN1, Oct 17, 2004, 7:33 PM
.
s.t. 
.
for a harmonic quadruple, so their intersections with the circle (other than 
)
is harmonic. This means that the tangents from 
to the circle meet on the diagonal 
,
are collinear. We will keep this in mind in what follows.
and 
,
to be similar. At the same time, the triangles 
are similar, so 
are similar. What we have to prove is now reduced to a simple angle chase: 
,
we have 
and also 
hence AZ bisect DH and cause 
hence 
be the midpoint of ![$ [BC]$](http://latex.artofproblemsolving.com/3/5/5/3550468aa97af843ef34b8868728963dec043efe.png)
and let 
be the orthogonal projection of 
in 
.
,
and ![$ [ABC]$](http://latex.artofproblemsolving.com/8/0/4/804b9de1e7a0a77c9e7b272da41c8e1cdff3c514.png)
.![$ [ADZ]$](http://latex.artofproblemsolving.com/2/2/7/227c9ee11f0a57c19ec4ba2b12a15f408fc3d361.png)
is sent to the circumcircle of ![$ [AFZ]$](http://latex.artofproblemsolving.com/d/4/d/d4d74200a3fe100f10b9488eafe1ca6ae6373b48.png)
.
.
be the complex numbers corresponding to 
,
and 
.
are on the unit circle and 
is on the real axis. Therefore we have:
belongs to line 
:
,
and 
.
:
,
this problem was used for one friendly competition of geometry (in witch each participant propose 2 problems) and when I find out that it is a G8... xD
,
,
and 
(we will find out that this point doesn't exist, by now assume that it exists and we goal one contradiction) where 
is the circuncircle of ![$[AXB]\sim [DQA]$](http://latex.artofproblemsolving.com/a/8/2/a82cbf743afa88b57b6c3110dc96f496031650b2.png)
and ![$[AYB]\sim [DZA]$](http://latex.artofproblemsolving.com/6/5/8/658c098e7f061269ebeac85d2c560d65244e31a1.png)
so, since 
we can conclude 
and so ![$[DQ]$](http://latex.artofproblemsolving.com/6/6/c/66ccd722466ee2037ba25bd8ecedb458ea09ec5e.png)
.
is one harmonic conjugate, just take the pencil from 
are in harmonic division, but since 
we find that the point 
must be in the infinity, and 
,
.
so AMXB is cyclic with circumcircle 
.
to 
,
,
,
is a diameter of 
intersects 
,
is a right triangle. Thus 
is another circle 
.
,
.
,
with nice tangents: 
tangent to 
and it is easy to show that 
is the midpoint of 
,
.
.
be the intersection of 
.
is the inverse of the circumcircle of 
.
.
.
,
,
and finally of 
)
as 
It suffices to show that 
is cyclic.
which implies 
where the last equality came from cyclic quadrilateral.
and 
intersect at 
,
,
.
,
and on the Apollonian circle of 
.
is tangent to 
.![\[\mathcal{YUH}\]](http://latex.artofproblemsolving.com/b/2/5/b257916d2f4721901d29fbeb9a78164bfb2b1166.png)
intersect 
.
it means that 
Hence, 
, so 
concyclic.
which implies that 
concyclic
and that ![\[\angle ADC=\angle ACB-\angle CAD=\angle ACB-\angle BAC=90^\circ-2\angle ABC\]](http://latex.artofproblemsolving.com/a/a/d/aadeb7edc85a239c372146b71567ef2c999a92a0.png)
so 
be intersections of 
respectively. 
so 
Since 
so 
is midpoint of 
Since 
It remains to show that 
so 
and 
as desired.
Constructing 
and 
Finally 
yields the conclusion.
lie on 
,
.
we know 
is a diameter of 
implies 
,
so 
.
It follows that 
,
which finishes. 
is well-known and commonly seen. At this point, the definition of 
the earlier G problems were weird length chasing stuff which I hated.
,
.![\[WBE=90=AXE\Rightarrow AX\parallel BW\implies(DMVP_{\infty})\stackrel{A}{=}(AZEW)\stackrel{B}{=}(AYXP_{\infty})=-1,\]](http://latex.artofproblemsolving.com/3/5/e/35ef5aeb7501cf783223b333f3b53cc742194cf9.png)
so M is the midpoint of DV. It follows that 
,
is tangent to (ADZ), as needed.
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denote the point at infinity along 
,
,
,
and let 
.
.
,
.
,
.
,
is the median in 
.
and since 
is the midpoint of 
we obtain that 
and we are done.
as 
and 
.![\[ 90^{\circ}=\measuredangle EAT=\measuredangle EBT \implies TB\perp XE \implies XA \parallel BT \]](http://latex.artofproblemsolving.com/a/e/9/ae9c045bd581dba4a9b7d11004c7e23689673511.png)
where the last implication follows due to the fact that 
.
is the 
.
which gives us that 
are collinear.
and we are done.
inversion.
and 
.
.
with 
.
are concyclic.
be the antipode of 
are collinear.![\[\measuredangle TEA'=2\measuredangle CAE=2\measuredangle ECA=2\measuredangle DAC=\measuredangle DEC=\measuredangle DET\]](http://latex.artofproblemsolving.com/2/0/7/207fdac1ba1ccd7d226cc0bc1077b5bc7ee1be22.png)
And 
hence 
is tangent to 
and 
,
thus, 
are collinear.![\[180-\measuredangle ACT=\measuredangle ECA=\measuredangle CAE=\measuredangle CAA'=\measuredangle CA'T=\measuredangle XA'Y=\measuredangle A'YX=\measuredangle TYA\]](http://latex.artofproblemsolving.com/c/7/e/c7ec23a6bb5dc62323a11e735b041af034caadc2.png)
Which completes the proof as desired
,
collinear.
,
.
is an isosceles trapezium with 
.
.
,
.
.
(by considering 
)
,
is a parallelogram.
passes through the midpoint 
.
,
as desired. 
and let line through 
meet 
Now it is enough to show 
is cyclic
as then we also have
giving 
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