[/quote]
,
board, 
pieces are placed (vertically or horizontally) so that they occupy entirely 
boxes on the board and do not have a common point.
,
be the sum of the digits of 
such that there are infinite positive integers 
.
,
be fixed positive integers. There are 
ducks sitting in a
be the circumcenter of triangle 
.
intersects the altitude from 
at point 
.
be the midpoints of 
,
respectively. If 
intersects 
at 
,
meets 
at 
,
is cyclic.
and circumcentre 
be the touchpoints of the incircle with 
,
is the Euler line of 
.
with its incircle 
.
and 
be the points of tangency of 
and 
,
and 
such that 
is tangent to 
.
is the intersection point of the lines 
and 
.
and 
are parallel or coincide.
and let 
be the midpoint of 
Suppose that 
different from 
such that 
Prove that the orthocenter of 
lies on the circumcircle of 
be a scalene oblique triangle with circumcenter 
,
)
,
be the circumcevian triangles of 
be the pedal triangle of 
be the reflection in 
of 
.
,
cyclically.
be the reflection in 
.
,
cyclically.
,
be the circumcenters of 
,
,
)
,
is 
.
with an obtuse angle 
, let 
be feet of altitudes from 
on sides 
respectively. The tangents from 
at points 
respectively and we know that 
.
lies on segment 
in such a way that 
and let 
be a point on line 
and 
.
is tangent to circumcircle of triangle 
.
is tangent to sides 
and 
,
bisects segment 
.
.
such that 
and 
.
be the feet of the perpendiculars from 
be the intersection point of lines 
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.
, ![\[(U+L)+(U+R)+(D+R)+(D+L) = 2n^2\]](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.
