[/quote]
,
.
be reals such that 
and 
Find the value of 
be a triangle with circumcentre 
.
lie in the interiors of the sides 
respectively, such that 
is perpendicular to 
and 
is perpendicular to 
.
lies on the line 
and 
and 
on that line.)
be the circumcentre of triangle 
.
and 
,
to tangent 
disjoint from 
on 
and 
.
is tangent to two fixed circles.
as the set of all products of an element of 
and 
then 
We call a set 
of positive integers good if there do not exist sets 
of positive integers, each with at least two elements and such that the sets 
and 
is good.
,
,
and 
are integers.)
be the midpoint of side 
and 
be the feet of the altitudes from 
and 
intersect at a point 
is perpendicular to line 
and 
be prime numbers. Prove that if 
is a perfect square, then 
is also a perfect square.
and 
with conditions,
is a prime number.
is a perfect square.
is also perfect square.
be a sequence of positive real numbers such that for every 
,![\[
a_n = \max_{1 \leq i \leq 2025} a_{n-i} - \min_{1 \leq i \leq 2025} a_{n-i}
\]](http://latex.artofproblemsolving.com/e/3/0/e30e9e533eb9a41cf847484e676ac4522edda665.png)
Prove that there exists a natural number 
such that for all 
,![\[
a_n < \frac{1}{1404}
\]](http://latex.artofproblemsolving.com/0/1/f/01f1a81eed2230332d51a15501ef2a6ad3a7ec82.png)
such that:
for all 
for all 
is a natural number. A function 
is called \textit
-
with 
,
has a solution in natural numbers 
such that ![\( \frac{y_0}{x_0} \in \left[\frac{1}{100}, 100\right] \)](http://latex.artofproblemsolving.com/d/a/6/da67f464a934f890626d188b84940fc82c4f4ff9.png)
,
is a solution. Suppose 
,
is a polynomial with real coefficients, negative leading coefficients, and total degree greater than 2, and for every real number 
,
as ![\( \frac{y}{x} \in \left[\frac{1}{100}, 100\right] \)](http://latex.artofproblemsolving.com/6/5/d/65d4154dd5fb612f92f24f98406bc9bcfc713806.png)
.
is not 
such that for all 
Y by Rounak_iitr
Alice and Bob are playing game on an 
grid. Alice goes first, and they take turns drawing a black point from the coordinate set
![\[\{(i, j) \mid i, j \in \mathbb{N}, 1 \leq i, j \leq n\}\]](//latex.artofproblemsolving.com/4/3/f/43f7493d1dff43079eb5481ee17bc8f16d7b9553.png)
There is a constraint that the distance between any two black points cannot be an integer. The player who cannot draw a black point loses. Find all integers 
such that Alice has a winning strategy.
Proposed by chengbilly
grid. Alice goes first, and they take turns drawing a black point from the coordinate set![\[\{(i, j) \mid i, j \in \mathbb{N}, 1 \leq i, j \leq n\}\]](http://latex.artofproblemsolving.com/4/3/f/43f7493d1dff43079eb5481ee17bc8f16d7b9553.png)
There is a constraint that the distance between any two black points cannot be an integer. The player who cannot draw a black point loses. Find all integers 
such that Alice has a winning strategy.Proposed by chengbilly
This post has been edited 1 time. Last edited by chengbilly, Mar 5, 2025, 5:09 AM
.
.
then
is not a square. Alice can mark 
since points are symmetrical with respect to 
i
.
then Bob should mark 
.
is not a square.
.
then Bob wins.
Alice can play any point, but follow the rule; then she'll win. And if 
Bob can play any point, but follow the rule; then he'll win. Cuz the problem is equivalent to: pick any point and draw all point on its row and column by black. The game must end after 
both can't be marked.
then notice 
is a perfect square
.
,
must be either already colored black or an integer distance away from a black point. The former is impossible due to the symmetric nature of the moves until this point. For the latter to be possible the violating pair must be either 
, 
or 
where 
.
implying that Bob performed an illegal move in his previous move. Further if 
.
must be a violating pair where 
is the reflection of 
across the center square. But by the nature of the algorithm, 
, 
,
and 
.
for 
must be a violating pair. In the later case, the reflection of 
for odd positive integers 
which can never be a perfect square. Thus, this distance is not an integer and hence it cannot be a violating pair. Thus, as long as Alice has a legal move, so does Bob and since the number of cells on the board is even, it is Alice who runs out of legal moves first.
