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.
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 
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?
interesting if is composite and for every divisor 
of at least one of 
and 
is also a divisor of with 
such that 
holds for all real numbers 
and 
.
such that the numbers ![\[ (a_1)^{2018}+a_2, (a_2)^{2018}+a_3, \dots, (a_{2018})^{2018}+a_1 \]](http://latex.artofproblemsolving.com/6/1/a/61ad92b49bd9b4b8839e6f6c4e81c758a11835a3.png)
are all powers of 
be a triangle with 
. Let 
be the midpoint of 
, and let 
be a point on segment such that 
. Let 
be the point of intersection, different from 
, of the circumcircle of triangle 
and line 
. Define 
and 
as the points of intersection of line 
with 
and 
, respectively. Prove that is the midpoint of 
.
such that 
be a point inside a triangle 
such that![\[ \angle APB - \angle ACB = \angle APC - \angle ABC.
\]](http://latex.artofproblemsolving.com/a/6/e/a6e3dc27d0457682fa22af4f918f58d9cd8403bc.png)
,
be the incenters of triangles 
,
,
,
,
meet at a point.
.
and 
the intersection between 
and the circumcircle 
.
,![$\frac{[ABT]}{[ACT]}=\frac{AB\cdot BT}{AC\cdot CT}\ (*)$](http://latex.artofproblemsolving.com/a/8/1/a81cb199852c88a551e31356dda2e0b9165964b7.png)
![$[XYZ]$](http://latex.artofproblemsolving.com/c/7/6/c7608a3e97e44fc5bcf0cd62a1f7dfc0a0a2e7d4.png)
)![$\frac{[ABT]}{[ACT]}=\frac{BS}{CS}=\frac{[PBT]}{[PCT]}=\frac{PB\cdot BT}{PC\cdot CT}$](http://latex.artofproblemsolving.com/c/d/5/cd523e6f5c3943ab5a1a15e10ae961e9096f3d48.png)
.
,
,
in the triangles 
respectively cut 
.
.
.
.
.
be the circumcirlce of 
.
.
and 
,
,
.
and 
,
and 
are the same straight line.
)
and 
,
and 
.
.
.
.
and 
,
,
(in short multiply the sides of traingles 
by 
)
are the sides
The given condition is equivalent to assuming that 
After doing inversion from 
,
Then do some angle chasing using the modified to prove that condition and deduce that
Use the fact that if 
are any two points then
Deduce the to prove that condition 
Notice 
so 
If 
and 
we see 
and 
and 
in the display equation with 
and 
,
for some fixed points 
(such a structure effectively defines a circle), while a line is defined by 
.
hence 
is an isosceles triangle. The desired condition is equivalent to 
using inversion distance formula, but this is evident.
point 
(
and 
,
is issosilate 
concur.
.
and denote images with 
;
,
.
which is now clearly true. 
and 
meet 
respectively. Then 
and 
.
.
is isosceles.![\[\angle P^*C^*B^* = \angle P^*C^*A - \angle AC^*B^* = \angle AP^*C^* - \angle AB^*C^*,\]](http://latex.artofproblemsolving.com/7/a/9/7a9cf8d8c7291d2ebf34172c68a0b36fdd735005.png)
and ![\[\angle P^*B^*C^* = \angle P^*B^*A - \angle AB^*C^* = \angle AP^*B^* - \angle AC^*B^*. \square\]](http://latex.artofproblemsolving.com/a/f/e/afecf21cb9ad7b6b53128b45ac59cdd87d7937a9.png)
,![\[\frac{BP}{AB \cdot AP} = \frac{CP}{AC \cdot AP}.\]](http://latex.artofproblemsolving.com/d/e/5/de5a415c39d0b019dbfc71146c2dc75e14a45312.png)
We thus obtain the desired result. 
.
.
.![\[\angle ABP - \angle ABC = \angle ACP - \angle ACB \iff \angle CBP = \angle BCP \iff BP = CP.\]](http://latex.artofproblemsolving.com/0/3/f/03fd3c6425e665e08d2daada30dcb5ef8bbccc0c.png)
,![\[\frac{AC}{BP \cdot \frac{AB \cdot AC}{AB \cdot AP}} = \frac{AB}{CP \cdot \frac{AB \cdot AC}{AC \cdot AP}} \iff AB \cdot AC = AB \cdot AC. \quad \blacksquare\]](http://latex.artofproblemsolving.com/d/e/2/de22d64f22a311e916f3767f289dd89e44d6c9c8.png)
.
,
,
.
is the midpoint of arc 
,
,
,
concur at the incenter of 
.
,
must lie on the line connecting 
and 
,
and 
.
and 
be the internal and external angle bisector of 
,
and 
are isogonal lines.
and 
,
are part of the same involution, which implies 
and the result follows by Apollonius circle.
,