[/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 
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 
(distinct from 
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.
be a set of 2025 positive real numbers. For a subset 
,
as the median of 
when all elements of 
.![\[
P(A) = \sum_{\substack{T \subseteq A \\ |T| \text{ odd}}} M_T, \quad Q(A) = \sum_{\substack{T \subseteq A \\ |T| \text{ even}}} M_T.
\]](http://latex.artofproblemsolving.com/1/d/f/1df2ea8487289111939a53dcc49576fbe6fdef37.png)
Find the smallest real number 
such that for any set ![\[
P(A) - Q(A) \leq C \cdot \max(A),
\]](http://latex.artofproblemsolving.com/c/d/e/cdea343d11c722a423e857aeb01202863405773e.png)
where 
denotes the largest element in 
in 
colors, there exists an arithmetic progression of size 
.
from this?
.
be foots of perpendicular lines from 
.
is cyclic and 
.
bisects 
Y by buratinogigle, Mualpha7, Amir Hossein, iarnab_kundu, Davi-8191, math1181, AlastorMoody, A-Thought-Of-God, jhu08, megarnie, HWenslawski, ImSh95, Adventure10, Tastymooncake2, ItsBesi, and 1 other user
Given triangle 
the point 
is the centre of the excircle opposite the vertex 
This excircle is tangent to the side 
at 
, and to the lines 
and 
at 
and 
, respectively. The lines 
and 
meet at 
, and the lines 
and 
meet at 
Let 
be the point of intersection of the lines 
and , and let 
be the point of intersection of the lines 
and 
Prove that is the midpoint of 
(The excircle of opposite the vertex 
is the circle that is tangent to the line segment , to the ray beyond 
, and to the ray beyond 
.)
Proposed by Evangelos Psychas, Greece
the point 
is the centre of the excircle opposite the vertex 
This excircle is tangent to the side 
at 
, and to the lines 
and 
at 
and 
, respectively. The lines 
and 
meet at 
, and the lines 
and 
meet at 
Let 
be the point of intersection of the lines 
and , and let 
be the point of intersection of the lines 
and 
Prove that is the midpoint of 
(The excircle of
opposite the vertex 
is the circle that is tangent to the line segment , to the ray beyond 
, and to the ray beyond 
.)Proposed by Evangelos Psychas, Greece
and so 
is cyclic. But 
and so 
.
and hence 
.
,
,
are concyclic 
are concyclic 
.
be the midpoints of 
.
and 
so 
is cyclic. Thus 
is cyclic. Angle-chasing shows that this angle is actually 
,
and so 
is clearly the diameter of 
,
are concyclic.
and 
,
.
(since 
and 
.
,
.
,
.
,
.
.
.
which inplies 
we get 
[ASA] implying 
which means 
is cyclic. (1)
is cyclic. (2)
,
from (1)
.
is cyclic (3)
is cyclic (4)
to be cyclic.
.
.
.
,
(from 
and 
)
is the perpendicular bisector of 
,
,
is cyclic and similarly for 
.
,
bisects 
and 
bisects 
,
.
.
trivially, we have 
are cyclic, which concludes that 
.
and 
.
and easily get 
,
is cyclic because 
.
,
,
.
.
,
and similarly 
is the midpoint of 
,
is the medial triangle of 
and 
and 
are parallel to the internal bisectors of 
,
,
.
from tangency, and 
from the orthocenter, 
.
and 
with any of the methods above, or by considering the excenters 
,
is antiparallel to 
is the medial triangle of 
and we are done.
is cyclic. Firstly, note that since 
and 
are the perpendicular bisectors of 
and 
,
and 
are isosceles. Angle chasing gives us 
,
is cyclic. Furthermore, 
is cyclic with diameter 
.
and 
are angle bisectors, it follows that 
and 
are isosceles, so![\[ SB + BM = AB + BM = \text{semiperimeter} = AC + CM = TC + CM.\]](http://latex.artofproblemsolving.com/6/6/d/66d3a87090ce6266e3e7b9dac3c0154c5f682d6b.png)
:
,
,
is isosceles and 
is the angle bisector of 
,
.
,
,
is also an altitude of 
:
is an altitude, meaning that 
,
.
and 
,
Now, applying Claims 
so we are done. 
and 
,
.
and 
implies that 
is cyclic. Similarly, 
is cyclic. Thus with 
.
and 
.
,
.
and 
This implies that 
too.
is a right triangle and therefore 
and 
and 
are congruent too.
and 
where 
is the semiperimeter of ABC.
and we are done.
.
is cyclic.
and
is cyclic. Since 
.
are concylic.
is a parallelogram.
.
bisects 
.
excircle.
![\[\angle LFJ = \angle XBJ - \angle CXL = 90 - \frac B2 - \frac C2 = \frac A2 = \angle LAJ,\]](http://latex.artofproblemsolving.com/c/e/2/ce2b5041cd68f9f6c62d9fd65bbadc8fb3bbff3f.png)
implies 
is also cyclic. Instead exploiting symmetry like a normal person, we can instead notice that the converse of Pascal's on hexagon 
implies these six points lie on a conic, which must be a circle as we know five of the points define a circle.
,
and 
,![\[XS = AK = AL = XT. \quad \blacksquare\]](http://latex.artofproblemsolving.com/c/d/6/cd617e38f059ff16229999f531460be70a3331d9.png)
![[asy]
size(330);
pair A, B, C, J, K, L, F, G, S, T, X;
A = dir(120);
B = dir(210);
C = dir(330);
J = extension(A, incenter(A, B, C), B, rotate(90, B)*incenter(A, B, C));
K = foot(J, A, B);
L = foot(J, A, C);
X = foot(J, B, C);
F = extension(B, J, X, L);
G = extension(C, J, X, K);
S = extension(A, F, B, C);
T = extension(A, G, B, C);
draw(A--K--J--L--A--J--X^^K--G--J--F--L^^A--S--T--cycle);
draw(circumcircle(A, K, L), dashed);
draw(circumcircle(X, K, L), dotted);
draw(A--K^^S--X, red+linewidth(1.5));
draw(A--L^^T--X, blue+linewidth(1.5));
dot("$A$", A, N);
dot("$B$", B, SW);
dot("$C$", C, SE);
dot("$K$", K, W);
dot("$L$", L, E);
dot("$J$", J, SE);
dot("$X$", X, N);
dot("$F$", F, NW);
dot("$G$", G, NE);
dot("$S$", S, W);
dot("$T$", T, E);
[/asy]](http://latex.artofproblemsolving.com/f/a/4/fa4ee7384a25b233f79911623a08bb2aedc050b4.png)
and 
and 
.
and 
.
,
,
and 
,
so 
,
.
is a paralelogram. We want to prove that 
but 
so we´re done. 
so 
and 
and 
so 
and 
,
and 
,
and 
.
so 
,
.
and 
,
,
,
,
.
,![\[\angle JF'L=\angle JML=90^{\circ}-\angle K,\]](http://latex.artofproblemsolving.com/c/6/9/c69aee0c018419aa54650797ae20dc15d69b51ba.png)
so since 
lies on 
,
Therefore 
Thus 
is cyclic. Similarly 
is cyclic and it follows that pentagon 
is cyclic.
Also, 
so 
But 
so by homothety 
Therefore 
where 
too, and we are done. QED
.
for some 
where![\[
\begin{vmatrix}
-a & t & c\\
0 & s-b & s-c\\
b-s & 0 & s
\end{vmatrix}=0.
\]](http://latex.artofproblemsolving.com/8/c/5/8c5952e88f4a12e4fd563abfed377801983d675f.png)
It follows that 
so 
.![\[
S=(0:-a-c:c)=\left(0,\,\frac{a+c}{a},\,-\frac{c}{a}\right)
\]](http://latex.artofproblemsolving.com/a/6/a/a6a5c581322fb97e76df6b7ff1191b4a6faa6e9b.png)
and similarly![\[
T=\left(0,\,-\frac{b}{a},\,\frac{a+b}{a}\right).
\]](http://latex.artofproblemsolving.com/1/7/3/173aa9fd78970606124602ff9b0f362e8a965d5c.png)
Thus 
,
,
