[/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 
.
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 
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, tenplusten, anantmudgal09, OlympusHero, Adventure10, HWenslawski, Jc426, math_comb01, Mango247, Rounak_iitr, and 3 other users
In a convex quadrilateral 
, the diagonal 
bisects neither the angle 
nor the angle 
. The point 
lies inside and satisfies ![\[\angle PBC=\angle DBA\quad\text{and}\quad \angle PDC=\angle BDA.\]](//latex.artofproblemsolving.com/c/2/0/c20761f3eadd054958f40259f3d1c05f26279783.png)
Prove that is a cyclic quadrilateral if and only if 
.
, the diagonal 
bisects neither the angle 
nor the angle 
. The point 
lies inside and satisfies ![\[\angle PBC=\angle DBA\quad\text{and}\quad \angle PDC=\angle BDA.\]](http://latex.artofproblemsolving.com/c/2/0/c20761f3eadd054958f40259f3d1c05f26279783.png)
Prove that is a cyclic quadrilateral if and only if 
.Attachments:
This post has been edited 2 times. Last edited by djmathman, Aug 1, 2015, 2:53 AM
Reason: formatting
Reason: formatting
,
,
,
be the orthogonal projections of the point P on the lines AB, BC, CD, DA, and let 
,
,
,
be the reflections of P in these lines, or, equivalently, the reflections of P in the points 
is the line BD. Hence, the points 
, we have 
.
and 
,
.
.
.
and 
,
has the segment AP as diameter. Therefore, by the Extended Law of Sines, we have
,
.
.
is the perpendicular bisector of 
are the symmetrics of 
respectively wrt 
is an isosceles trapezoid which has 
to 
and 
and 
since they subtend equal arcs. The circle also gives us 
by equal arcs 
.
since they are vertical, and 
by subtraction. The equal angles and two adjacent equal sides force 
and 
at 
and 
.
and triangles 
and 
are similar in ratio 
.
by subtraction, and triangles 
and 
are also similar in ratio 
and 
are similar in ratio 
,
.
is 
by hypothesis, so 
.
is cyclic, its reflection 
is also cyclic and the original quadrilateral 
. ]
to 
,
.
,
and by an analogous argument that 
.
so that by definition, 
.
,
be the associated intersection point for quadrilateral 
.
is cyclic, so you agree that 
.
,
,
while 
.
.
Let 
and 
Notice 
so 
and 
Similarly, 
Since 
it lies on the perpendicular bisector of 
which is equivalent to the perpendicular bisector of 
Construct 
such that 
Similarly construct 
on 
such that 
Notice 
as 
are isogonal conjugates wrt 
so 
and 
Hence, 
and 
is cyclic isosceles trapezoid. Also, 
and 
so 
Similarly, 
is cyclic isosceles trapezoid and so 
is cyclic. 
intersect the circumcircle of 
Let 
Note that 
so 
This implies 
Similarly, 
is a cyclic trapezoid so 
Thus it must also lie on the perpendicular bisector of 
be reflections of 
Since 
is a reflection of 
over 
so 
is cyclic. Also, 
Thus, 
is cyclic. This implies 
so 
is isosceles trapezoid. Thus, 
so the reflection of 
and 
and 
are both isosceles trapezoids and therefore 
(ii) Assume 
,
which implies 
,
bisects 
.
is cyclic. Since 
,
and 
or 
is cyclic. However, the latter is absurd because 
and 
,
on 
to the corresponding point 
.
is cyclic, then define 
.
such that 
and 
,
so one direction is completed
.
.
both belong to the 
,
.
and thus 
is also cylcic. Hence 
is cyclic.
denote the reflection of 
and 
be 
.
,
,
are cyclic with diameters 
,
respectively.
and 
Now, because 
.
,
.
In addition, we have 
Thus, 
is cyclic if and only if 
.
It follows that 
is cyclic, as this is the harder portion.
and 
and 
.
and 
to be isogonal in this context.
and 
,
and 
excenters, respectively. Now comes the key claim:
and 
are both cyclic.
is cyclic, as the others are analogous. Observe that 
is the angle bisector of 
,
,
!
on 
,
.
bisects 
.
is the external angle bisector of 
,
.
,
.
and 
.
,
is an isoceles trapezoid, so 
.
.
,
is an isoceles trapezoid as well. This means that 
and 
share a perpendicular bisector.
and 
.
to meet the circumcircle of 
respectively.
so it follows that 
and 
which implies that 
is an isosceles trapezoid with legs 
we have 
so 
It follows that 
is an isosceles trapezoid with legs 
so 
is parallel with 
we have 
so 
Since 
so 
as desired. 
such that the fourth point is not outside the circumcircle of those three points.
then 
and let 
be the second intersections of the lines 
with 
(a property we proved in the necessity that it must be isosceles), but since 
intersects that circle at only two points, either 
is an isosceles triangle. We proceed when 
because the other case is analogous.
Since P lies on the perpendicular bisector of 
,
,
which implies that 
It follows that P lies on the perpendicular bisector of 
,
is an isosceles trapezoid (perpendicular bisectors are parallel). Therefore, since 
by SSS congruence, we have 
Hence, 
,
.
share a perpendicular bisector, and the result follows.
.
and 
and 
are congruent as 
,
.
,
is cyclic, and similarly 
is cyclic. It follows that all the points 
are concyclic, as needed.
and 
,
and 
.
and 
are isosceles trapezoids (with 
)
are collinear, as are 
,
which lies on the perpendicular bisector of 
by symmetry.
be points and 
such that 
are isogonal in 
,
etc. The isogonality condition is equivalent to
Upon expansion and simplification this means that any 
)
,
,
and 
.
,
.
is placed at 
,
so 
but 
.
are valid positions for 
, then construct 
is cyclic. Let 
intersect again a
.
and 
are collinear.
.
which finishes.
.
are isogonaly conjugate with respect to three of the four angle, so 
means orange, 
means green and 
and 
are like the diagram that is (hopefully) attached.
,![\[ \measuredangle PAC = \measuredangle ACP \iff \measuredangle BAC + \measuredangle CBD + \measuredangle DCA + \measuredangle ADB = 0. \qquad (\spadesuit). \]](http://latex.artofproblemsolving.com/7/e/5/7e57e142c8f1fa0c11c8588dfc18cc3f5af78fa0.png)
A careless reader would be forgiven for thinking that 
implies the problem or at least one direction, but it turns out the situation is more subtle. The second part analyzes the angle conditions more carefully and provides a complete proof.
.
.
,![[asy]
import geometry;
size(9cm); pair A = (-0.50099,1.69422); pair B = (-1.8,-0.6); pair C = (0.09182,-1.79600); pair D = (1.55802,-0.60384); pair P = (-1.04359,-0.19363);
draw(circumcircle(A,B,C), gray+dashed); filldraw(A--B--C--D--cycle, invisible, black); draw(B--D); draw(A--C);
draw(C--P--A, brown); draw(B--P--D, gray);
markangle(radius=15, n=2, P, A, C, red, StickIntervalMarker(1,1, red)); markangle(radius=15, n=2, A, C, P, red, StickIntervalMarker(1,1, red)); markangle(radius=10, n=3, A, P, B, deepgreen, StickIntervalMarker(1,2, deepgreen)); markangle(radius=10, n=3, C, P, D, deepgreen, StickIntervalMarker(1,1, deepgreen)); markangle(radius=12, n=1, B, A, C, blue); markangle(radius=12, n=1, C, B, D, blue); markangle(radius=12, n=1, D, C, A, blue); markangle(radius=12, n=1, A, D, B, blue);
dot("$A$", A, dir(A)); dot("$B$", B, dir(B)); dot("$C$", C, dir(C)); dot("$D$", D, dir(D)); dot("$P$", P, dir(45));
[/asy]](http://latex.artofproblemsolving.com/0/e/c/0eca105f1f7fddffcbb253a96d5434767c875c18.png)
and 
in the claim (colored green with three rings) so that their only dependence on 
and 
(colored red with two rings), as follows: 
Since the claim says 
,
in terms of only 
These four latter angles are colored blue with one ring in the figure. This proves 
.![\[ \measuredangle BAC + \measuredangle CBD + \measuredangle DCA + \measuredangle ADB = 0 \]](http://latex.artofproblemsolving.com/b/8/6/b86d4b46e9417c6e66aa186a00663048a68a1057.png)
is equivalent to 
.
meets line 
.
meets line 
,
.
(that is, 
)
if and only if 
is not degenerate) and the second lemma turns our angle condition into 
and hence 
be the point for which 
is cyclic and harmonic; then 
.
(the corresponding Apollonian circle only meets line 
