[/quote]
,
and circumcenter 
.
and 
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: 
,
.
interesting if 
of 
and 
is also a divisor of 
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 
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.
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 
such that 
Y by feridverdiyev, BEHZOD_UZ, tenplusten, Amir Hossein, mathleticguyyy, Adventure10, Mango247
Let 
be an isosceles triangle with 
, whose incentre is 
. Let 
be a point on the circumcircle of the triangle 
lying inside the triangle . The lines through parallel to 
and 
meet 
at 
and 
, respectively. The line through parallel to meets and at 
and 
, respectively. Prove that the lines 
and 
intersect on the circumcircle of the triangle .
Proposed by Hojoo Lee, Korea
be an isosceles triangle with 
, whose incentre is 
. Let 
be a point on the circumcircle of the triangle 
lying inside the triangle . The lines through parallel to 
and 
meet 
at 
and 
, respectively. The line through parallel to meets and at 
and 
, respectively. Prove that the lines 
and 
intersect on the circumcircle of the triangle .Proposed by Hojoo Lee, Korea
Attachments:
This post has been edited 1 time. Last edited by Amir Hossein, Aug 5, 2013, 6:55 AM
Reason: Note by Darij: This was also Problem 6 of the German pre-TST 2004, written in December 03.
Reason: Note by Darij: This was also Problem 6 of the German pre-TST 2004, written in December 03.
,
.
,
to be tangent to 
is also tangent to 
.
.
,
is to show that 
because 
.
turns the circle 
into the line 
in the invariant points 
.
.
,
,
.
,
is cyclic. similarly, 
is cyclic.
.
is cyclic. 
.
is also cyclic. 
.
.
.
.
.
where the point 
is the 
with the diameter ![$[II_{c}]$](http://latex.artofproblemsolving.com/4/d/3/4d360c2d7cf645c4c380fd8f59ddcee5827c26fd.png)
; the circumcircles 
, 
of the isosceles trapezoids 
, 
respectively ; the second intersection point 
between the line 
and the circle 
Prove easily that the lines 
From the relation 
obtain 
i.e. the quadrilaterals 
, 
are similarly as 
Thus,
and let 
, 
be two mobile points so that 
and exists a point 
for which 
Construct the points 
for which 
Ascertain the geometrical locus of the intersection 
where the point 
is the 
to meet the circumcircle of 
at 
.
intersect 
at 
,
meet 
.
,
is cyclic. Now extend 
,
to meet the circumcircle of 
at 
.
is cyclic; hence, 
.
at 
,
.
,
,
is parallel to 
,
is isosceles, as desired.
denote the intersection of 
and 
.
,
.
so 
is cyclic, and similarly 
is cyclic. The result then follows by direct angle chasing to show that 
,
and 
.
and 
.
and 
are similar. This follows from 
.
.
.
and 
are cyclic. Hence, 
and 
are cyclic too. Thus we have to show that circles 
,
,
have a point in common. Therefore, consider 
.
and 
.
is cyclic too. The conclusion follows.
meet 
,
is harmonic. Now 
bisects 
(because 
is a paralellogram). So 
harmonic, which yields 
.
,
,
.
.
and it's easy to verify that 
.
.
,
.
.
![[asy]
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draw((1.660000000000002,3.580000000000004)--(4.220000000000004,1.340000000000002));
draw((4.220000000000004,1.340000000000002)--(1.347581935409076,0.1927304575931286));
draw((1.347581935409076,0.1927304575931286)--(1.660000000000002,3.580000000000004));
draw(circle((2.368300313660502,1.806628929897715), 1.909590083396774));
draw((1.438901086131840,1.182822302271518)--(2.291169545721799,-0.1014028128268362));
draw((2.291169545721799,-0.1014028128268362)--(3.471717322567828,1.994747342753156));
draw((1.438901086131840,1.182822302271518)--(3.471717322567828,1.994747342753156));
draw((2.030481395477170,1.419104784674090)--(2.778764072909347,0.7643574419209349));
draw(circle((2.002030691392678,0.6316259215254196), 0.7879926411260609));
draw(circle((3.105051331405495,0.8207367229440963), 1.229936943302180));
draw((2.030481395477170,1.419104784674090)--(1.939162244754406,0.4290129399956997));
/* dots and labels */
dot((1.660000000000002,3.580000000000004),dotstyle);
label("$C$", (1.718017351869851,3.665804292213880), NE * labelscalefactor);
dot((4.220000000000004,1.340000000000002),dotstyle);
label("$B$", (4.273985312412097,1.416019993611593), NE * labelscalefactor);
dot((1.347581935409076,0.1927304575931286),dotstyle);
label("$A$", (1.398521356802070,0.2711593446187132), NE * labelscalefactor);
dot((2.030481395477170,1.419104784674090),dotstyle);
label("$P$", (2.077450346321104,1.495893992378538), NE * labelscalefactor);
dot((1.939162244754406,0.4290129399956997),dotstyle);
label("$D$", (1.997576347554159,0.5107813409195485), NE * labelscalefactor);
dot((2.778764072909347,0.7643574419209349),dotstyle);
label("$E$", (2.836253334607084,0.8435896691151531), NE * labelscalefactor);
dot((1.438901086131840,1.182822302271518),dotstyle);
label("$F$", (1.491707688696840,1.256271996077703), NE * labelscalefactor);
dot((3.471717322567828,1.994747342753156),dotstyle);
label("$G$", (3.528494657253942,2.068324316874978), NE * labelscalefactor);
dot((2.291169545721799,-0.1014028128268362),dotstyle);
label("$X$", (2.343697008877588,-0.02171198419341888), NE * labelscalefactor);
clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle);
/* end of picture */ [/asy]](http://latex.artofproblemsolving.com/f/4/1/f411f0672b5772972e351f5d6cda57dce9971a02.png)
so 
and 
.
so 
is cyclic. Similarly, 
is also cyclic. Let 
and 
so ![\[\angle AEX + \angle BDX = 180^{\circ} - \theta \implies \angle DXE = \theta = \angle FXE,\]](http://latex.artofproblemsolving.com/6/9/4/694c56154046571699259a205eb3be9ddfea102c.png)
are collinear. Similarly, 
are also collinear. It hence suffices to show that ![\[\angle AXB = \angle AXP + \angle BXP = \theta + \theta = 180^{\circ} - \angle ACB,\]](http://latex.artofproblemsolving.com/4/4/3/443ac421d8c0f13462a5d31e24a6f939c7f5592a.png)
.
so, by p.o.p. 
,
and the two parallelograms 
are similar, 
,
is cyclic; since 
is an isosceles trapezoid, 
is cyclic as well and 
.
is cyclic and 
.
and 
side by side we get 
,
are collinear.
and 
are isosceles trapezoids, hence they are cyclic. Moreover, since 
,
,
hence 
so that 
are collinear. By symmetry, 
are collinear, so the result is obtaine
where 
.
.
we have 
.
.
and 
then 
in 
So 
.
.
.
which is true since 
.
,
.
is a parallelogram, 
passes through the midpoint 
and 
;
.
and 
,
is cyclic, so 
.
,
is cyclic, so 
.
true since 
is cyclic. Thus 
intersect on 
intersect on 
we find 
are concurrent.
and 
are collinear.
where we use parallel lines.
is cyclic
is cyclic
and 
is isosceles cyclic trapezoid
is parallelogram 
and 
is isosceles cyclic triangle
we find 
is cyclic and from 
we have 
is cyclic.
is cyclic.As desired.
,
lying inside the triangle ![[asy]
unitsize(100);
pair A, B, C, M, P, D, E, F, G, K;
A = dir(90); M = dir(270); B = dir(230); C = reflect(A, M) * B;
P = M + abs(M - B) * dir(100);
K = 2 * foot(origin, A, P) - A;
D = extension(P, P + C - A, B, C);
E = extension(P, P + B - A, B, C);
F = extension(P, P + C - B, A, C);
G = extension(P, P + B - C, A, B);
draw(F--G, gray(0.5));
draw(P--D^^P--E, gray(0.5));
draw(A--K, gray(0.5));
draw(F--K^^G--K, gray(0.5) + dashed);
draw(B--P^^C--P, gray(0.5));
draw(A--B--C--cycle);
draw(arc(M, C, B));
draw(circumcircle(B, P, K)^^circumcircle(C, P, K));
draw(circumcircle(B, E, K)^^circumcircle(C, D, K), dotted);
draw(unitcircle);
dot(A^^B^^C^^P^^D^^E^^F^^G^^K);
label("$A$", A, dir(A));
label("$B$", B, dir(200));
label("$C$", C, dir(340));
label("$P$", P, dir(90));
label("$D$", D, dir(290));
label("$E$", E, dir(230));
label("$F$", F, dir(40));
label("$G$", G, dir(140));
label("$K$", K, dir(K));
[/asy]](http://latex.artofproblemsolving.com/a/c/f/acf45e5b89b545cde48d385cc25227a6c2860297.png)
again at 
and 
are tangent to 
.
,
,
,
is tangent to 
.
is tangent to 
.
,
.
bisects 
is a parallelogram, 
also bisects 
followed by reflection in the angle bisector of 
.
):-
be the 
be its circumcircle. Suppose the tangent to 
and 
at 
and 
meet at a point.
.
Also 
lies on the perpendicular bisector of 
,
are collinear. And, 
,
.
.
and 
,
is a parallelogram. Thus, 
is cyclic. Similarly, 
is cyclic. Let 
is cyclic. 
tangent to 
,
and similarly, 
are tangents to 
,
,
and 
are cyclic. So we see that pentagon 
is cyclic too. Thus we have 
. Now note that 
.
.
concur at a point, call it 
,
.
,
,
where 
,
since
is cyclic. Therefore, it suffices to show given 
,
is cyclic. I got stuck here, so if anyone has any ideas, that would be awesome.
,
.
is cyclic.
.
lie on 
and it is well-known that centre of 
Hence, we have 
since 
and 
.
.
to 
.
and 
all concur and since 
and 
,
intersect on the 
