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 an acute triangle with orthocenter 
and circumcenter 
. Let 
and 
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]
. 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?
that satisfy 
for all 
and 
satisfies the following recursion formulas: 
, 
and 
. Find 
and 
interesting if is composite and for every divisor 
of at least one of 
and 
is also a divisor of with 
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 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 
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
and 
be non-collinear points. Prove that there is a unique point 
in the plane of 
such that ![\[ XA^2 + XB^2 + AB^2 = XB^2 + XC^2 + BC^2 = XC^2 + XA^2 + CA^2.\]](http://latex.artofproblemsolving.com/3/e/e/3eec6741ec0d9d940a42ccd391db7c9d828af912.png)
and be non-collinear points. Prove that there is a unique point in the plane of such that![\[ XA^2 + XB^2 + AB^2 = XB^2 + XC^2 + BC^2 = XC^2 + XA^2 + CA^2.
\]](http://latex.artofproblemsolving.com/0/9/9/099ee5d31517d10cd428a8215dba073d564b20b5.png)
, and let 
be the corresponding complex numbers to 
respectively. (Then 
)
be a 
rotation of , so 
.
be the real product, then we have:
(Using that )
, 
, so there is equality iff is the orthocenter. is the orthocenter rotated around the circumcenter 
is equivalent to 
Which clearly means that belows to the reflection of the B-altitude with respect to the perpendicular bisector of 
By similar reasoning, we conclude that is the reflection of the orthocenter with respect to the circuncenter 
In other words, is the De Longchamps point 
of 
, radius segment AB
, radius segment AC
, radius segment BC.
. Since 
not collinear we have 
. We have
with:
. From (II), we have 
. Then (I) and (III) are of the form:
. Multiply (I) by 
and (III) by 
, and subtract:
. But 
,
,
, 
, so this is obviously the case.
such that 
