and circumcenter 
.
be points on the line segments 
and 
respectively such that 
is a parallelogram. Prove that 
.
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?
that satisfy 
for all 
and 
satisfies the following recursion formulas: 
,
.
Y by Davi-8191, Amir Hossein, centslordm, Adventure10, mathematicsy, samrocksnature, jhu08, megarnie, HWenslawski, lneis1, Mango247, and 1 other user
Given a triangle 
satisfying 
. The incircle of triangle has center 
and touches the sides 
and 
at the points 
and 
, respectively. Let 
and 
be the reflections of the points and with respect to . Prove that the points 
, 
, , lie on one circle.
Proposed by Dimitris Kontogiannis, Greece
satisfying 
. The incircle of triangle has center 
and touches the sides 
and 
at the points 
and 
, respectively. Let 
and 
be the reflections of the points and with respect to . Prove that the points 
, 
, , lie on one circle.Proposed by Dimitris Kontogiannis, Greece
This post has been edited 2 times. Last edited by Amir Hossein, Jun 25, 2018, 1:59 AM
be a symmetrical point to 
with respect to 
lie on line 
and 
lie on line 
such that 
.
imply 
,
and 
.
and 
.
is congruent to 
and 
to 
.
and 
.
is congurent to 
and 
to 
.
and 
.
and 
are cyclic quadrilaterals thus 
is cyclic quadrilateral.
and 
we find that 
.
.
)
,
touches it at the points 
, 
, 
. Denote:
the centroid 
of the triangle 
, the B-exincircle 
, the C-exincircle 
;
; 
, 
, 
;
,
, 
;
, 
, 
of the points 
.
![$\frac 19\left\{\left[2\left(a^2+c^2\right)-b^2\right]-\left[2\left(a^2+b^2\right)-c^2\right]\right\}$](http://latex.artofproblemsolving.com/3/2/9/329f57c95e21b647aae442d05502ce94fa6aeac4.png)
.
and 
. Therefore, 
.
;
; 
and 
a.s.o.
in the triangle 
.
and obtain the relation 
. 
,
.
.
. Then
are cyclically 
and in this case 
belong to
, where the point 
is the second intersection between the line 
and the circumcircle of the triangle 
a.s.o.
,
,
.
.
meet the circumcircle at the point 
since 
.
,
.
similarly we have 
.
.
from point 
at 
so 
.
and 
( 
is a midpoint of the side 
.
.
all lie in a circle with centre 
See the my first reply where I offered a equivalent and complete enunciation of the proposed problem ! 
i think it is normal to consider most of the times one's solution as the most natural. it is as follows...
and 
.
,
,
is parallel to 
,
.
and 
.
.
.
.
.
is cyclic, and since 
,
.
we have ![\[\tan ALE=\frac{AE}{EL}=\frac{s-a}{2r}\]](http://latex.artofproblemsolving.com/e/d/8/ed86bc6419962568ccd8786ecd0be4054f1b91e6.png)
where 
is the semiperimeter of the triangle and 
is the inradius. Now we use the condition 
for finding r: ![\[r=\sqrt{\frac{(s-a)(s-b)(s-c)}{s}}=\sqrt{\frac{(s-a)(s-b)c}{2c}}=\sqrt{\frac{(s-a)(s-b)}{2}}\]](http://latex.artofproblemsolving.com/8/3/6/8363f3ea5310fdb009c4f8eb4df4bfa36567bf87.png)
whence ![\[\tan ALE=\frac{s-a}{2}\sqrt{\frac{2}{(s-a)(s-b)}}=\sqrt{\frac{2(s-a)^{2}}{4(s-a)(s-b)}}=\sqrt{\frac{s-a}{2(s-b)}}\]](http://latex.artofproblemsolving.com/7/7/0/77086cd9da3b1812ce47cfca95a951a059df10cf.png)
But we know that ![\[\tan \frac{B}{2}=\sqrt{\frac{(s-c)(s-a)}{s(s-b)}}=\sqrt{\frac{c(s-a)}{2c(s-b)}}=\sqrt{\frac{s-a}{2(s-b)}}\]](http://latex.artofproblemsolving.com/d/6/8/d68500838700781e2d0686e029d97d251d6422aa.png)
Thus 
,
.
.
and 
are similar. Indeed, we have 
and ![\[\frac{AE}{LE}=\frac{EI}{EX}\Leftrightarrow \frac{s-a}{2r}=\frac{r}{s-b}\Leftrightarrow r=\sqrt{\frac{(s-a)(s-b)}{2}}\]](http://latex.artofproblemsolving.com/f/6/1/f61783e60f694cc16cea29821e69dfde105b2314.png)
Hence 
.
is isosceles, 
;
and therefore 
,
is bisected by 
(a short sine-rule bash). But the midpoint of 
,
,
touches the incircle at 
and 
.
.
since 
.
.
.
in terms of the sides. Use: 
,
and 
.
.
]
,
is cyclic by the converse of the power of a point theorem.
.
.
we have:
Define 
to be a reflection of 
.
,
and 
are congruent isosceles triangles.
and 
combined with 
.
,
.
all lie on the same circle centred at the point 
.
We can unhomogenize 
We can unhomogenize 
The equation of a circle is 
and since 
Therefore, if we can show that both 
,
so all that's left is verifying that 
of 
resp. and let 
be the 
be defined as usual. It is easy to see that 
.
is tangent to the incircle by Pitot, in fact at 
and 
.
through direct usage of lengths
where the last line follows from 
,
.
are collinear. However, 
so 
.
.
.
as well, so 
is cyclic. Similarly, 
is cyclic. Hence, 
is cyclic
;
are collinear.
On the other hand, 
,
symmetrically (the length condition still applies.) This is enough to prove the concyclicity.
and let 
We further claim that 
is cyclic.
of 
and the point 
of 
Thus it suffices to prove that 
are collinear. For now, we will only focus on the point 
A rephrasing is to prove that if 
then 
Projecting through the point 
we get 
Thus it suffices to prove that 
is harmonic. To do so, since the tangents at 
intersect at 
we must prove that 
are collinear. It turns out that 
and this can easily be proven using barycentric coordinates and the given condition on the triangle.
KIB=90+B/2. On the other hand, the fact that 3AB=AE+BD+2CD 
and 
are reflections about 
and 
cyclic implies 
cyclic.
as our reference triangle.
.
and 
it can now be computed that 
and similarly 
.
and let 
.
.
and 
over the perpendicular bisector of 
gives us segments 
and 
respectively, so![\[ PI = IM = EM = DM = PK = PL,\]](http://latex.artofproblemsolving.com/9/a/8/9a8184c17af522ebc03bed82261e99f6aa4a6919.png)
hence the result follows.
as the unit circle.
If 
which is clearly wrong.
By 
Which is true as desired. 
to meet 
using the length condition.
and 
give us enough information to angle chase.
Thus we get that 
.
so 
,
,![\[AM = MI = MB,\]](http://latex.artofproblemsolving.com/3/a/d/3ad3288719d2b65589a70cdf3bdfc992d4c9836c.png)
and by Ptolemy's Theorem on 
we have 
,
,
so![\[CN=NI=IM=AM=MB.\]](http://latex.artofproblemsolving.com/b/9/b/b9b35ae577944afa532a4af0976cc1d2a5bd8894.png)
Now, we will show that 
lie on a circle centered at 
.
lie on this circle, so we must show 
.
through the point 
respectively. Now, we have 
from 
being the median from 
,
we have 
,
be the incircle and let 
be the circle inversion through it. Notice it suffices to prove that 
and 
concur because
We prove the former argument and the latter follows with a symmetric argument.
.
.
and 
.
.
.
Let 
So,
.![$$[ABC]=r\frac{(1+x)+(1+1-x)+(x+1-x)}{2}=2r,\ [ABC]=\sqrt{2(2-1-x)(2-2+x)(2-1)}\implies 2r^2=x(1-x)$$](http://latex.artofproblemsolving.com/7/d/9/7d9a9590dfdaf4db0ea75bdfd2bff57849765cc6.png)
Thus we're done.
