[/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 
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.
Y by Davi-8191, mathematicsy, Adventure10, chessgocube, HWenslawski, ImSh95, Mango247, Amir Hossein
Let 
be the circumcenter and 
the orthocenter of an acute triangle 
. Show that there exist points 
, 
, and 
on sides 
, 
, and 
respectively such that ![\[ OD + DH = OE + EH = OF + FH\]](//latex.artofproblemsolving.com/a/8/7/a874af4cb802b9b341b055140a08e096aaf2112e.png)
and the lines 
, 
, and 
are concurrent.
be the circumcenter and 
the orthocenter of an acute triangle 
. Show that there exist points 
, 
, and 
on sides 
, 
, and 
respectively such that ![\[ OD + DH = OE + EH = OF + FH\]](http://latex.artofproblemsolving.com/a/8/7/a874af4cb802b9b341b055140a08e096aaf2112e.png)
and the lines 
, 
, and 
are concurrent.
.
equals 
.
are the intersection points between 
the reflections of 
across 
respectively. All we have to prove is that 
,
and 
are concurrent.
,
and 
,
is acute, so 
,
.
.
and 
,
and 
.
and 
,![\[ \dfrac {BD}{DC} = \dfrac { \sin B }{\sin C }.\]](http://latex.artofproblemsolving.com/4/1/b/41b575c3a62f879fe19923244df8edcf4426ed38.png)
and 
,
,
concur at the Gergonne point of 
.
be any line, and let 
be the intersection of 
and 
.
by the triangle inequality, since 
,
,
,
,
hit 
hit 
hit 
lie on the circumcircle of 
,
is the circumradius of 
(since 
are the reflections of 
I chose to scale because I felt it was the most elementary solution; anyone who knows anything about ellipses should be able to understand the solution I gave.
![\[ \frac { BD } { \sin B } = \frac { A'D } { \sin \angle{DBA'} } \quad \wedge \quad \frac { DC } { \sin C } = \frac { A'D } { \sin \angle{DCA'} } \]](http://latex.artofproblemsolving.com/d/c/a/dca38f07efd63ab71a099e4490506565de8f17eb.png)
![\[ \sin \angle{DBA'} = \sin \angle{DCA'} \]](http://latex.artofproblemsolving.com/6/0/5/6059a6e389664ba1602680ca40a1637f7d039913.png)
I unfortunately can't read the entire problems. Only half of the picture is displayed. Do you know why? 
be the reflections of 
respectively. Let 
and 
.
,
.
,
are the points we are looking for.
,
.
,
.
is an isosceles trapezoid, implying that 
and 
.
and 
.
is cyclic, so 
and 
.
and 
,![\[ \frac{BD}{DC}=\frac{\frac{DD'\sin\angle{DD'B}}{\sin\angle{DBD'}}}{\frac{DD'\sin\angle{DD'C}}{\sin\angle{DCD'}}}=\frac{\sin B\cos B}{\sin C\cos C}=\frac{\sin2B}{\sin2C}\]](http://latex.artofproblemsolving.com/c/b/c/cbc066a21c7fe1bb5180e2af1a97a2527bc45033.png)
and 
.![\[\frac{BD}{DC}\cdot\frac{CE}{EA}\cdot\frac{AF}{FB}=\frac{\sin2B}{\sin2C}\cdot\frac{\sin2C}{\sin2A}\cdot\frac{\sin2A}{\sin2B}=1,\]](http://latex.artofproblemsolving.com/9/f/c/9fc4f0c16807ee5e2f21e7ab5398ea5367a2667e.png)
are concurrent. //
.
is 
are points of tangency) or use a projective transformation taking the conic to a circle. Note that tangency is preserved for the latter since it is one-to-one). Now, since 
be the mirror images of 
and 
.
,
and 
.
.
is minimal, so the points 
,
,
at 
,
be the reflections of 
respectively. Let 
intersect 
.
,
.
,
.
and 
have the same circumradius, so 
Then, by Ratio Lemma, 
Taking the cyclic product of this over 
yields 1 as everything cancels, so by Ceva's, 
be the reflection of 
.
analogously. Now observe that we obviously have 
,
concur. For now, let 
be the ellipse with focii at 
.
,
.
in place of 
;
we used is that 
,
,
,
concur. By ellipse properties 
are the reflections of 
satisfy the length condition with a constant sum of 
.
So 
and analogously 
,
as desired. 
.
,![\[OD+DH=OD+DA'=OA'=R,\]](http://latex.artofproblemsolving.com/5/0/7/507540daa70c9da023b47267610d8e6f8cbf8116.png)
and similarly, we also find that ![\[\angle DHA'=\angle OAH=\angle BAC-\angle BAH-\angle OAC=a-(90-b)-(90-b)=180-(2b+a)=b-c,\]](http://latex.artofproblemsolving.com/c/c/8/cc8aa10ee4a5fafa8bf2c54ca9818c6d5964c06c.png)
by orthocenter properties. Additionally, we also get that 
,![\[\angle DHC=90-\angle DHA'-\angle DCH=c.\]](http://latex.artofproblemsolving.com/b/9/f/b9f77bee0cfc4facc018e89287231187be5c2aff.png)
Now using Law of Sines, we get that![\[BD=DH*\frac{\sin b}{\cos c},\]](http://latex.artofproblemsolving.com/c/0/2/c02fc3bd09ca6db8726f0e954cfc23d27013cc07.png)
and![\[CD=DH*\frac{\sin c}{\cos b},\]](http://latex.artofproblemsolving.com/8/c/a/8ca29ef1ef6f5bc8a1bad527ed2efb8b4bbaa561.png)
which gives that![\[\frac{BD}{CD}=\frac{\sin b\cos b}{\sin c\cos c}=\frac{2\sin b\cos b}{2\sin c\cos c}=\frac{\sin 2b}{\sin 2c}.\]](http://latex.artofproblemsolving.com/7/0/4/70486eb1aeeb0e297419c6c4a32a596fc9989ebd.png)
Similarly, we get that 
and 
.![\[\frac{BD}{CD}*\frac{CE}{EA}*\frac{AF}{FB}=1,\]](http://latex.artofproblemsolving.com/5/0/a/50ab301460b1a2a9751d4b2b441e0f6620938ef8.png)
which means that 
,![\[OD+DH = OD+DH_A = R,\]](http://latex.artofproblemsolving.com/d/c/5/dc5a0f4ba038e96d88062cd1c71ef9bce39085a3.png)
![\[\prod \frac{BD}{DC} = \prod \frac{BH_A \cdot \sin \angle BH_AD}{CH_A \cdot \sin \angle CH_AD} = \prod \frac{\sin
B \cos B}{\sin C \cos C} = 1. \quad \blacksquare\]](http://latex.artofproblemsolving.com/f/e/d/fed9f9800c1bc63b3ba77a34cf6e4f4ad03d73f0.png)
.
,
and 
.
is a reflection of H over BC, because of famous orthocenter property, similarly for 
.
,
and 
.
,
and 
DO + DH = EO + EH = FO + FH = R. So it is only left to show that AD, BE and CF are concurrent. Now using the ratio lemma 
,
and 
it follows that AD, BE and CF are concurrent, which was what we needed to show 
in 
, 
and 
the reflections of the orthocenter 
, 
and 
with sides 
.![\[OD+DH = OD + DH_a = R \]](http://latex.artofproblemsolving.com/1/e/f/1ef77f7c366de6a1405a75cf9c09447d36d366cc.png)
where ![\[OD+DH = OE + EH = OF + FH = R\]](http://latex.artofproblemsolving.com/c/8/f/c8f9318f90bc43991b4947ccba0339261902fddc.png)
which implies that ![\[\measuredangle ODC = \measuredangle H_a DB = \measuredangle BDH\]](http://latex.artofproblemsolving.com/4/b/5/4b52dc1040a3ddb8eb943efd89ddbf6324b66d86.png)
which implies that side 
and 
and 
.
.
which is constant, where 
is the circumradius.
and 
are isogonal lines wrt 
.
with the circumcircle we have,
which implies the isogonality. Now we can use Steiner's Ratio theorem and the law of sines to compute 
Applying a similar argument to the other points we obtain
and we're done.
and define 
and a similar expression holds for the other ratios 
But, 
so 
and we are done by Ceva's Theorem. QED
