[/quote]
,
be an isosceles triangle with 
.
be a point on 
.
be a point inside the triangle such that 
and![\[
CL \cdot BD = BL \cdot CD.
\]](http://latex.artofproblemsolving.com/1/6/7/16771838c95f86e92f79b8d049e46ab473e6287d.png)
Prove that the circumcenter of triangle 
lies on line 
.
satisfying 
and
for all 
.
Prove that
be a cyclic quadrilateral with 
being the midpoints of segments 
respectively. Suppose 
is the intersection of diagonals 
of quadrilateral 
Define 
to be the isogonal conjugate point of point 
Define 
similarly. Suppose 
intersects 
at a point 
Prove that: The Newton-Gauss line of quadrilateral 
Prove that
Prove that
be an odd prime number. How many 
of 
are there, the sum of whose elements is divisible by 
be a positive integer. Ana and Banana play a game. Banana thinks of a function 
and a prime number 
.
is nonconstant, 
,
for all integers 
.
.
in order. Ana wins if she can determine the value of 
be an acute triangle and 
its orthocenter. Let 
be a point on the parallel through 
to 
such that 
.
and 
as points on the parallels through 
to 
and through 
to 
similarly. If 
are positioned around the sides of 
,
it's circumcenter. Let 
and 
the intersection of the circumcircle of 
,
is the reflection of 
the midpoint of 
,
and the parallel through 
,
and the perpendicular through 
and 
a point on 
,
is the midpoint of 
.
lie on a line.
is inscribed in a circle 
.
— the circle inscribed in a segment 
circle of 
— the centers of circles 
and 
are parallel.
intersect at 
intersect at 
.
.
with the diagonals of 
and 
are tangent at a fixed point.
,
.
and 
,
is acute, and let 
and 
,
and 
and 
and 
Y by cubres
Kritesh manages traffic on a 
grid consisting of 2025 unit squares. Within each unit square is a car, facing either up, down, left, or right. If the square in front of a car in the direction it is facing is empty, it can choose to move forward. Each car wishes to exit the grid.
Kritesh realizes that it may not always be possible for all the cars to leave the grid. Therefore, before the process begins, he will remove
cars from the grid in such a way that it becomes possible for all the remaining cars to eventually exit the grid.
What is the minimum value of that guarantees that Kritesh's job is possible?
grid consisting of 2025 unit squares. Within each unit square is a car, facing either up, down, left, or right. If the square in front of a car in the direction it is facing is empty, it can choose to move forward. Each car wishes to exit the grid.Kritesh realizes that it may not always be possible for all the cars to leave the grid. Therefore, before the process begins, he will remove
cars from the grid in such a way that it becomes possible for all the remaining cars to eventually exit the grid.What is the minimum value of
that guarantees that Kritesh's job is possible?
This post has been edited 2 times. Last edited by Tony_stark0094, Apr 13, 2025, 3:12 AM
Reason: typo
Reason: typo
is the answer:
is the answer. To prove that it is minimal, consider the situation where cars in left 22 rows face right and right 22 rows face left, and top 22 cars in the middle row face down.
is the answer. To prove that it is minimal, consider the situation where cars in left 22 rows face right and right 23 rows face left.
.
.
rows face downwards.
.
.
.
cars in the middle row face left.
cars must be manually removed in this configuration, as desired.
cars are removed.
with cars pointing down. Fill the lower 
cells of the remaining row with cars pointing left and the remaining 
.
for any 
grid for odd 
cars must be removed. In each row, no cars facing opposite directions can be remaining as they bump into each other. Thus, in each row either all the right-facing or left-facing cars must be removed, mandating a removal of at least![\[\frac{n(n-1)}{2} + \frac{n-1}{2} = \frac{n^2-1}{2}\]](http://latex.artofproblemsolving.com/f/1/d/f1d343d222521b6fc5fd518f231f37e5cd7d0c61.png)
cars as desired.
, 
, 
and ![\[(U+L)+(U+R)+(D+R)+(D+L) = 2n^2\]](http://latex.artofproblemsolving.com/3/b/3/3b3e882bed42cba0df71636a9527bc79de076f47.png)
we have that,![\[\min\{U+L, U+R , D+R , D+L\} \le \left \lfloor \frac{2n^2}{4} \right \rfloor = \frac{n^2-1}{2}\]](http://latex.artofproblemsolving.com/5/6/5/5654607babab37133b3f5b5b06a71fe71ee7124d.png)
which finishes the proof of the bound and we are done.
