[/quote]
,
be a prime and 
a nonempty set. Suppose that for each element 
there exist elements 
,
such that 
is divisible by 
,
.
different point-like cities were founded one by one; from the second onwards, each newly founded city and the nearest existing city (the older one, if there were two) were declared sister cities. The surviving map of the empire shows the cities and the distances between them, but not the order in which they were founded. Historians have tried to deduce from the map that each city had at most 41 sister cities.
,
,
is divided into cyclic quadrilaterals with pairwise disjoint interiors. None of the vertices of the cyclic quadrilaterals in the decomposition is an interior point of a side of any cyclic quadrilateral in the decomposition or of a side of the quadrilateral 
be an acute triangle with incenter 
and circumcenter 
.
and 
at 
and 
respectively, and 
is the reflection of 
over 
and 
meet at 
,
and 
,
is the midpoint of 
.
and 
,
.
has an inscribed circle with center 
.
and 
be 
and 
,
and 
be 
.
be a point on the Neuberg cubic. Show that as 
be the circumcircle of a non-isosceles triangle 
be a tangent line to 
)
be midpoint of 
and 
.
and 
intersect at points 
.
and 
intersects 
,
intersects 
.
,
.
and 
are chosen on lines 
and 
respectively such that 
is a parallelogram. Points 
and 
and 
are the angle bisectors of angles 
and 
respectively. Let 
past 
past 
.
at 
.
on 
and 
on 
is cyclic, then 
is tangent to the circumcircle of 
.
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 
denote the number of cars facing each direction. Since,![\[(U+L)+(U+R)+(D+R)+(D+L) = 2n^2\]](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.
