[/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 
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 
,
.
,
,
,
,
,
,
intersects 
be the midpoint of the arc 
of 
at 
,
at 
.
.
but doesn't tell it to anyone. Bernaldo thens pick an arbitrary positive integer 
(not necessarily in 
.
chosen by Arnaldo.
Y by Adventure10, Mango247
Let two circles 
and 
meet at the points 
and 
. A line through meets again at 
and again at 
. Let 
, 
, 
be three points on the line segments 
, 
, 
respectively, with 
parallel to and 
parallel to . Let 
and 
be points on those arcs of and of respectively that do not contain . Given that 
is perpendicular to and 
is perpendicular to prove that 
.
and 
meet at the points 
and 
. A line through meets again at 
and again at 
. Let 
, 
, 
be three points on the line segments 
, 
, 
respectively, with 
parallel to and 
parallel to . Let 
and 
be points on those arcs of and of respectively that do not contain . Given that 
is perpendicular to and 
is perpendicular to prove that 
.Attachments:
This post has been edited 1 time. Last edited by orl, Oct 25, 2004, 12:16 AM
.
is ccylic because 
,
.
are similar, then we would have 
,
,
,
,
.
are similar, no matter which line 
cantered at 
.
,
,
and 
,
is the symmetric of 
,
because 
and 
,
.
so that 
.
be 
and the parallel from 
meet 
at 
.![\[ \angle NDF = \angle BAF = 90 - \angle BAE' = 90 - \angle E'CK' = \angle K'E'C\]](http://latex.artofproblemsolving.com/1/4/a/14a69d755331e7fd6c3c6cf31497c8d8a0c45367.png)
and 
.
,
.
,
,
.
,
,
,
and 
.
.
,
from parallelogram 
.
and ![\[ \angle EKM = 90 + \angle MKB = 90 + \angle MNB = \angle MNF\]](http://latex.artofproblemsolving.com/5/d/2/5d2fba2eafcc8e907550b9774b5e5eda6a86386d.png)
implying that 
.![\[ \angle EMF = \angle KMN - (\angle KME + \angle NMF) = \angle KBN - (\angle EMK + \angle KEM)\]](http://latex.artofproblemsolving.com/3/d/5/3d521cf736e6251f37bdec84a0dc0c70c0eec36c.png)
Noting that![\[ \angle EMK + \angle KEM = 180 - \angle EKM = 180 - (90 + 180 - \angle CBD) = \angle KBN - 90\]](http://latex.artofproblemsolving.com/9/3/a/93acdfd6cb7fa20382d1c8b4fe7240925950869e.png)
we have that 
,
in my figure.
has to be on the major arc 
.
,
.
,
By symmetry,
Thus 
gives us![\[1=\frac{CN}{CB}+\frac{BN}{BC}=\frac{NM}{BD}+\frac{BN}{BC}=\frac{BK}{BD}+\frac{BN}{BC},\]](http://latex.artofproblemsolving.com/4/8/2/482696ba7baae3fe91633ca8836f20aa4235d6f8.png)
whence plugging in and simplifying yields![\[\cos(\angle{FBK}+\angle{EBN})\sin(\angle{BAD}-\angle{FBK}+\angle{EBN})=0.\]](http://latex.artofproblemsolving.com/d/c/c/dcc410ce3373dfb3e4ade8b2ff85595712a480b8.png)
But![\[0=\sin(\angle{BAD}-\angle{FBK}+\angle{EBN})=\sin(180^\circ-\angle{BCE}-\angle{KBF})\]](http://latex.artofproblemsolving.com/9/4/1/94101dbf16bcc73c6f8b68d6fddcb5e0cb9f9ed0.png)
is impossible since 
.
,
.
at 
.
,
,
,
.
,
,
.
and 
.
,
,
,
(by lemma 1), and 
.
,
is cyclic.
denote the image of 
,
are collinear. 
and 
,![\[ \frac{C'E'}{E'B'} \cdot \frac{B'F'}{F'D'} \cdot \frac{D'M'}{M'C'} = -1. \]](http://latex.artofproblemsolving.com/2/5/b/25bd16de3745fcce1d997d70034c924f007f97f2.png)
.
.![\[ \frac{MC}{MD} = (M', A; C', D') = \frac{\frac{M'C'}{M'D'}}{\frac{AC'}{AD'}}, \]](http://latex.artofproblemsolving.com/2/a/b/2abdc6f73abf5c38299caa2e7cac7d6c019dbc4b.png)
.
.
;
.![\[\angle EMF=\angle EMN+\angle NMK+\angle KMF=\angle EMN+\angle MNC+\angle KMF=90^{\circ}-\angle MEN+\angle KMF\]](http://latex.artofproblemsolving.com/a/1/2/a1229dd981624b091380023c4a0d0f9f5b1a1c8b.png)
so 
.
.
.
,
.
.
.![\[\angle ECB=\angle CEN+\angle BEN=\angle FDK+\angle FBK=180^{\circ}-\angle BFD\]](http://latex.artofproblemsolving.com/f/d/6/fd66eb7da985e96373c0bfd0a1915a2758488f2d.png)
as desired.
.
.
is cyclic. Let 
such that 
and let 
be the projections of 
and 
are parallel, it follows that![\[\frac{CN}{NB}=\frac{BK}{KD}=\frac{BY}{YC}=\frac{CY'}{BY'}\]](http://latex.artofproblemsolving.com/1/a/4/1a4ef100b5848e4d29c0182cdf0dcf4105c97795.png)
is an isosceles trapezoid and 
,
and 
are similar. The above equality implies that 
and therefore that 
and that 
is the reflection of 
around the bisector of 
.
,
,![\[\angle{DBF}=\angle{CBX}=90^\circ - \angle{CBE}=\angle{NEB}\]](http://latex.artofproblemsolving.com/a/f/4/af4c59c132f3fa26c34256f2f3e466ab09d83bd8.png)
which implies that![\[\frac{EN}{KB} = \frac{BK}{KF} \quad \Rightarrow \quad \frac{EN}{NM}=\frac{MK}{KF}\]](http://latex.artofproblemsolving.com/1/6/b/16b0f7762c359ae75f7fd13e2cb2ae4a87c8645b.png)
is a parallelogram. This also implies that 
.
which implies that![\[\angle{EMF}=\angle{NMK}+\angle{EMN}+\angle{MFK}=\angle{CBD}+\angle{EMN}+\angle{MEN}=90^\circ\]](http://latex.artofproblemsolving.com/5/6/c/56c646fea5df62ccb8c489bb9283f459a039f7ed.png)
as desired. 
.
about 
implies arcs 
and 
have the same measure.
.![\[
\frac{K'C}{K'B} = \frac{DK}{KB} = \frac{AD}{AC} = \frac{NB}{NC},
\]](http://latex.artofproblemsolving.com/4/1/8/418516e681d20f2d294a00ffbc8438abd92cce6c.png)
.
.
.
(because 
.
as well. We have
.
are distinct line segments in that order such that 
and if we let points 
and 
,
and 
.
on the perpendicular from 
.
,
as 
from definition. Then 
,
.
from the similar triangles equal angles, thus 
as there is a unique 
is a parallelogram, we have 
,
with some ratio manipulating. Also, 
,
,
.
,
and as 
,
and![\[\angle EMF=180-\angle BNM-\angle EMN-\angle FMK=180-\angle BNM-\angle EMN-\angle MEN=180-\angle BNM-(180-90-\angle BNM)=90\]](http://latex.artofproblemsolving.com/1/f/1/1f1ca434d8a4b673cda5239930ef4f384adb40e2.png)
,
.
,
on 
,
.
and 
,
is given, then we have 
.
.
.
are concyclic.
.
,
are concyclic.
,![\[\frac{EB}{BF} = \frac{BN}{FK} = \frac{MK}{KF}\]](http://latex.artofproblemsolving.com/6/d/1/6d10f55b5fb3bff876b8a8b38421fc1b15358010.png)
Combining with 
,
.
,
concyclic, and we're done.
,
be the antipodes of 
with respect to their corresponding circles. It's not hard to see that the spiral similarity centered about 
to 
also sends 
are collinear and 
.![\[
\frac{BN}{CN} = \frac{\cot\angle EBN}{\cot \angle ECN} = \frac{\sin \angle ECN}{\sin \angle EBN}\cdot \frac{\sin \angle CBE_1}{\sin \angle BCE_1} = (E,E_1;C,B)=(AE,AF;CD,AB)
\]](http://latex.artofproblemsolving.com/c/7/6/c76697c08a0e392ca511c3f1cdf03b96c38a96ec.png)
Now 
.
on lines 
and ![\[
(C',D';A,M')=\frac{DM}{CM}=(AE,AF;CD,AB)=(AE',AF';C'D',AB')=(E',F';AB' \cap E'F', C'D' \cap E'F') = (C',D';A,C'D' \cap E'F')
\]](http://latex.artofproblemsolving.com/a/3/f/a3f334122258f3d890870d03670a41c0686ae57a.png)
so 
meets 
at 
again at 
be a diameter of the circumcircle of 
and 
.
.
,
and 
are similar. Now, note that 
,
we have that 
.
,
and 
are similar. But since 
and 
,
as desired.
,
.
,
determines the location of 
.
,
.
.
to intersect 
to intersect 
,
Hence, 
.
a
.
![[asy]
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[/asy]](http://latex.artofproblemsolving.com/1/6/1/161fd8ac0d658c4879a422d71fc0f897f98ca40b.png)
so 
Extend 
Note that 
Let the transformation 
be from 
to 
respectively consisting of one reflection, preserving all angles and, by extension, length ratios. 
Since 
Therefore, 
This implies that 
so 
Additionally, by angle chasing 
so 
Now, 
so we are done.
is reflection of 
and 
so by spiral similarity 
sending 
.![\[\frac{BN}{NC} = \frac{DK}{KB} = \frac{DM}{MC}\]](http://latex.artofproblemsolving.com/0/b/9/0b9a58630e957236dc7de383c26ce34697646f8c.png)
it follows that 
collinear.
and note that since ![\[\angle BFD + \angle BE'D = \angle BAC + \angle BEC = 180^{\circ}\]](http://latex.artofproblemsolving.com/1/9/5/195d1ed08cc3eae9c27c20c43a7713f5d5d88692.png)
implying that 
form an orthocentric system. Thus we have ![\[\angle EBN = \angle E'DK = \angle BFK\]](http://latex.artofproblemsolving.com/c/7/f/c7f271e5ac004f6c44eeb67d469251ee028095bd.png)
This implies that triangles 
and 
are similar. We therefore obtain ![\[\frac{EN}{BN} = \frac{BK}{KF}\]](http://latex.artofproblemsolving.com/9/d/8/9d807eec4dcb059befb002ce033e8ff2c22a3c96.png)
Since 
along with 
.![\[\frac{EN}{NM} = \frac{EN}{BK} = \frac{BN}{KF} = \frac{MK}{KF}\]](http://latex.artofproblemsolving.com/7/c/c/7cc814bb190a8b62b5e52ce1dfeeee054bbeb6ad.png)
Since ![\[\angle ENM = 90^{\circ} + \angle CBD = \angle MKF\]](http://latex.artofproblemsolving.com/3/f/0/3f0bffb849653f587f6fd5745de96d59a377b350.png)
we have 
.
with each other.
,
,
concylic. 
