[/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.
Y by Davi-8191, A-Thought-Of-God, Adventure10, lian_the_noob12, Mango247, Rounak_iitr, bjump, cubres
Given trapezoid 
with parallel sides 
and 
, assume that there exist points 
on line 
outside segment , and 
inside segment 
such that 
. Denote by 
the point of intersection of and 
, and by 
the point of intersection of and . Let 
be the midpoint of segment , assume it does not lie on line . Prove that belongs to the circumcircle of 
if and only if belongs to the circumcircle of 
.
Proposed by Charles Leytem, Luxembourg
with parallel sides 
and 
, assume that there exist points 
on line 
outside segment , and 
inside segment 
such that 
. Denote by 
the point of intersection of and 
, and by 
the point of intersection of and . Let 
be the midpoint of segment , assume it does not lie on line . Prove that belongs to the circumcircle of 
if and only if belongs to the circumcircle of 
.Proposed by Charles Leytem, Luxembourg
This post has been edited 1 time. Last edited by April, Jul 10, 2009, 8:30 AM
and 
.
,
is cyclic, so 
.
since 
,
is cyclic. Thus, 
.
is cyclic if and only if 
and 
is cyclic if and only if 
.
,
,
,
.![\[ (IJ)(JK)=(FJ)(JE)\Leftrightarrow (a+b)(c)=(b)(c+b+c)\Leftrightarrow ac=b(b+c)\]](http://latex.artofproblemsolving.com/5/d/c/5dc862097522279e390ab106833905b7edca1e9e.png)
![\[ (IJ)(IK)=(IF)(IE)\Leftrightarrow (a+b)(a+b+c)=(a)(a+b+c+b+c)\Leftrightarrow b(b+c)=ac\]](http://latex.artofproblemsolving.com/b/6/7/b6763094598273e350f29788a1c969d7670ff41f.png)
if and only if 
.
,
is cyclic, so 
.
.
cyclic 
.
.
is cyclic. in deed 
![\[IJ.IK=JF.JE \Leftrightarrow IE.IF=IJ.IK\]](http://latex.artofproblemsolving.com/6/8/d/68d856c23c4f3d25197f25abcae8ce62152563a9.png)
so the equivalence is proved .
is cyclic.
the inversion centered at J and of ratio 
.
and 
and 
,
with 
)
must be a circle so 
is cyclic.
and 
,
is cyclic.![\[ABKI \ cyclic \Leftrightarrow K=K' \Leftrightarrow JDKC\ cyclic \]](http://latex.artofproblemsolving.com/0/9/6/096241c375fa14ba6a8671dacef81eaf1920e617.png)
and 
(As 
is concyclic
(
)
.
is cyclic or 
is cyclic,or 
is cyclic iff 
Also 
is cyclic iff 
From 
as desired.
is a cyclic quadrilateral. Consider the following
and lines 
and 
.
.
and 
Hence, both conditions are equivalent. 
implies the four points 
are concyclic. Also 
,
exists too. By PoP, we get 
.
and 
.
lies in the radical axis of 
and 
.
,
,
are concyclic, as desired.
are concyclic. Since 
are concylic, 
.
.
.
,
,
.
and 
.
.
,
.
and 
art 
,
.
be points on a line in that order. Let 
.
.
.
.
are inverses wrt 
,
is cyclic.
is cyclic 
,
,
Thus,
And so we are done. 
and 
.
as desired.
be a circle through 
are tangent to 
.
Hence, 
.
Let perpendicular at 
Hence, 
is cyclic. Now, here is a much more natural restatement of the problem:
lie on a circle. Let 
.
be the line through 
.
,
cyclic. We want to show 
.
,
gives us that we want to show 
.
.
,
hence 
is cyclic. Now we are ready to PoP:![\[I \in (ABK) \iff JI \cdot JK = JA \cdot JB = JF \cdot JE\]](http://latex.artofproblemsolving.com/e/a/f/eaf4731da2530cc1283036da23e1f14e9d054269.png)
![\[K \in (CDJ) \iff IF \cdot IE = ID \cdot IC = IK \cdot IJ\]](http://latex.artofproblemsolving.com/2/e/1/2e1125e70730d2ccfe7633e98e0e68e364b3b490.png)
and in fact both of these are the same condition as 
and we are done. 
to itself. Note that 
. By butterfly theorem, 
is the reflection of 
if and only if 
.
if and only if it is the inverse of 
are cyclic from the given angle conditions.
QED
so 
so 
is cyclic.
by Power of a Point, and that 
iff 
is cyclic by Power of a Point and it's converse. Thus 
is equivalent to 
by Power of a Point, and 
by Power of a Point and it's converse, so 
.
is equivalent to 
.
and 
respectively. Thus 
and 
,
.
is equivalent to 
However, 
which reduces to the same condition, so the claim is proved. 
.
and line 
)
(the PoP condition for 
to be cyclic needs this to be 
)
These are clearly equivalent.
is cyclic too!
,
Again if 
,
Since above conditions match,we are done 
By PoP, 
and by Reim 
cyclic.
Note that ![\[JF\cdot IK-IF^2=FJ\cdot FJ-IF\cdot IJ=FJ(JE-IK)-IJ^2=JF\cdot JE+IK\cdot IF-JI\cdot JK\]](http://latex.artofproblemsolving.com/0/5/4/05466e2b186f63230b9150402d5c3d88d5ac4ab7.png)
which when 
is equal to 
which means that 
![\[IF^2-IK\cdot FJ=IF\cdot FK-IK\cdot IJ\]](http://latex.artofproblemsolving.com/d/5/6/d569c20bed7f4679114045351d6c8f745a9b7c2f.png)
which when 
is equal to 
which means 
Thus, we are done.
Which is equivalent to DECF to be cyclic .
must also be cyclic. From PoP,
if and only if![\[ JI \cdot JK = JA \cdot JB = JE \cdot JF \]](http://latex.artofproblemsolving.com/1/5/0/150fef320c0cd8f26a2ee5a7277d11170ed45122.png)
which is equivalent to 
(from a lemma regarding harmonic bundles).![\[ IE \cdot IF = IC \cdot ID = IK \cdot IJ \]](http://latex.artofproblemsolving.com/f/8/8/f88bc5a8b66b74e1dfd2acd924317dee232c6154.png)
But 
from PoP, so the condition is true if and only if
,
cyclic.
.
so 
which by 
is equivalent to 
,
are cyclic. 
cyclic is equivalent to 
which due to 
.
.
and 
.
.
,
respectively on the complex plane.
.
The third condition is equivalent to
.
thus both cyclicities imply each other.
,
,
,
,
which implies its cyclicity.
we have ABEF cyclic. Since 
we have CDEF cyclic by Reim's. Then, ![\[ I \in (ABK) \iff JB \cdot JA = JI \cdot JK \]](http://latex.artofproblemsolving.com/8/5/a/85a9475be494a5d76aab4e12225aaf7b86080b41.png)
But since we also have 
by PoP, this is equivalent to 
.
intersects 
,
.
.
,
.![\[\angle DAE=\angle CBF,\]](http://latex.artofproblemsolving.com/f/7/b/f7b162c8b3a21bbc58d7da818a1da158e44c3ac4.png)
we must have that 
,
is cyclic. Now, by Power of a Point, we have that![\[K\in (CDJ) \iff IC*ID=IK*IJ,\]](http://latex.artofproblemsolving.com/8/d/7/8d71552b8156ad6c01425a1a3e481555aa5c7142.png)
however, since ![\[IC*ID=IE*IF,\]](http://latex.artofproblemsolving.com/5/5/a/55adc0a75f14f7be31aed4783a8eb7360a153b9c.png)
so our problem becomes proving that 
.
(by 
and 
,![[asy]
draw((0,0)--(12,0));
dot((0,0)); dot((6,0)); dot((8,0)); dot((4,0)); dot((12,0));
label("$E$", (0,0), S); label("$I$", (6,0), S); label("$F$",(8,0),S); label("$J$",(12,0),S); label("$K$", (4,0),S);
label("$\frac{a+1}{2}$", (2,0), N, red); label("$\frac{a-1}{2}$", (5,0), N, red); label("$1$", (7,0), N, red); label("$\frac{a+1}{a-1}$", (10,0), N, red);
[/asy]](http://latex.artofproblemsolving.com/9/9/6/996f681d3e97298ee8861e3b6fb4a66324d28408.png)
![\[IE*JF=a*1=\frac{a-1}{2}*\frac{2a}{a-1}=IJ*IK,\]](http://latex.artofproblemsolving.com/3/6/f/36f1517f2353baff44b8db5b1889bc5c298d8a89.png)
as desired. This means that 
.
.
,
also.![\[I''C*I''D=I''K*I''J.\]](http://latex.artofproblemsolving.com/b/e/e/bee67f50518ddcbf4a0fdce7a1e4064b8cd37e22.png)
Again, by Reim's, note that ![\[I''C*I''D=I''K*I''J=I''E*I''F \implies \frac{I''E}{I''J}=\frac{I''K}{I''F}.\]](http://latex.artofproblemsolving.com/4/a/9/4a9833b944f7b50acd40561c395574dcb024bb8b.png)
Now, knowing this, and that 
and 
and set up a "number line" with directed lengths and label the (directed) lengths as follows:![[asy]
draw((0,0)--(12,0));
dot((0,0)); dot((6,0)); dot((8,0)); dot((4,0)); dot((12,0));
label("$E$", (0,0), S); label("$I$", (6,0), S); label("$F$",(8,0),S); label("$J$",(12,0),S); label("$K$", (4,0),S);
label("$a'+1$", (2,0), N, red); label("$a'$", (5,0), N, red); label("$1$", (7,0), N, red); label("$\frac{a'+1}{a'}$", (10,0), N, red);
[/asy]](http://latex.artofproblemsolving.com/2/6/6/2662c416200890717aad11a8757e178505d5583b.png)
![\[I''E*JF=\frac{a'+1}{a'}*(2a'+1)=1*\frac{(a'+1)(2a'+1)}{a'}=I''F*JE,\]](http://latex.artofproblemsolving.com/3/f/d/3fd29c4e5d05fb96ac446c2c10a7d1f818c9184e.png)
which means that 
,
.
,
.
as desired.
