AoPS Academy
[/quote]
, do not answer with "
is a solution" only. Either you post any kind of proof or at least something unexpected (like "
is the smallest solution). Someone that does not see that is a solution of the above without your post is completely wrong here, this is an IMO-level forum.
be a triangle, and let 
and 
be points on segment 
symmetric with respect to the midpoint of . Let 
denote the circle passing through and tangent to line 
at 
. Similarly, let 
denote the circle passing through and tangent to line 
at 
. Let the circles and intersect again at point 
(
). Prove that 
.
be an integer. Show that, if 
is an integer, then it is a perfect square.[/quote]
. If is an integer, then 
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 
, and that, for every such pair 
, the "next" such pair can be calculated as
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?
be an acute triangle with 
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 
be an acute triangle with orthocenter 
and circumcenter 
. Let 
and 
be points on the line segments and respectively such that 
is a parallelogram. Prove that 
.
cities, 
airline companies in a country. Between any two cities, there is exactly one -way flight connecting them which is operated by one of the two companies. A female mathematician plans a travel route, so that it starts and ends at the same city, passes through at least two other cities, and each city in the route is visited once. She finds out that wherever she starts and whatever route she chooses, she must take flights of both companies. Find the maximum value of .
be inscribed in the circumcircle 
and circumscribed about the incircle 
, with 
. The incircle touches the sides , 
, and at 
, , and , respectively. A line through 
, perpendicular to 
, intersects , , and at 
, 
, and 
, respectively. The line meets at 
(distinct from ). The circumcircle of triangle 
intersects at 
(distinct from ). Let 
be the midpoint of the arc 
of . The line cuts segments and 
at 
and 
, respectively, and the tangents to the circle 
at and intersect at 
. Prove that 
.
known by both players, Arnaldo picks a number 
but doesn't tell it to anyone. Bernaldo thens pick an arbitrary positive integer 
(not necessarily in ). Then Arnaldo tells the number of divisors of 
. Show that Bernaldo can choose in a way that he can find out the number 
chosen by Arnaldo.
be a set of 2025 positive real numbers. For a subset 
, define 
as the median of 
when all elements of are arranged in increasing order, with the convention that 
. Define![\[
P(A) = \sum_{\substack{T \subseteq A \\ |T| \text{ odd}}} M_T, \quad Q(A) = \sum_{\substack{T \subseteq A \\ |T| \text{ even}}} M_T.
\]](http://latex.artofproblemsolving.com/1/d/f/1df2ea8487289111939a53dcc49576fbe6fdef37.png)
Find the smallest real number 
such that for any set of 2025 positive real numbers, the following inequality holds:![\[
P(A) - Q(A) \leq C \cdot \max(A),
\]](http://latex.artofproblemsolving.com/c/d/e/cdea343d11c722a423e857aeb01202863405773e.png)
where 
denotes the largest element in .
in 
colors, there exists an arithmetic progression of size 
. How can we derive Van der Waerden's theorem for 
from this?
be circumcenter of a non-isosceles triangle and be a point in the interior of 
. Let 
be foots of perpendicular lines from to 
. Suppose that 
is cyclic and is the circumcenter of , 
. Prove that 
bisects
interesting if is composite and for every divisor 
of at least one of 
and 
is also a divisor of with 
has two parallel tangents 
and
. A circle 
touches at 
and at 
. A circle 
touches at 
, at 
and at 
. The lines 
and 
meet at 
. Show that is the circumcenter of 
(very nice !);
, where 
is the inradius of the triangle 
(easily);
(easily).
be tangent to 
at . A homothety centered at maps to 
. Since and 
are parallel, is mapped to 
, so , , and are collinear. Similarly, , , and are collinear. and 
intersect at . is the external homothety center of and . and are the internal homothety centers of and and and , respectively. Hence, by d'Alembert's theorem, , , and are collinear.
intersect at 
. Since the homothety centered at maps to , 
, so 
. By power of a point on , 
, so 
. Substituting this into , so 
is cyclic. Since 
, 
, so lies on the radical axis of and . By symmetry, must lie on the radical axis of and , so 
must be the radical center of , , and . Since 
, 
, and 
are tangents and has equal power with respect to each of these circles, 
, as desired.
touch and at and repectively. be 
, the centre of be 
and centre of 
.
and 
as 
are collinear.
are collinear and 
are collinear.
(alternate segment, is tangent to )
as is tangent to and at diametrically opposite points.
is cyclic, 
lies on the radical axis of the circles and , but as and touch each other 
is the direct common tangent to and .
is the direct common tangent to and .
and 
are the points where the incircle of 
touches the sides.
the incentre of 
passes through the incentre of 
passes through the incentre of 
such that 
