[/quote]
,
and some point 
on 
intersects the circumcircle of 
at 
and the perpendicular from 
intersects the circumcircle of 
at 
.
does not depend on the choice of 
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 
cars from the 
coins, all of which show 
on the opposite side. The coins are laid out in a row on the table. Peer and Solveig alternate taking turns. On his turn, Peer may turn over one or more coins, so long as he does not turn over two adjacent coins. On her turn, Solveig picks precisely two adjacent coins and turns them over. When the game begins, all the coins are showing 
for which Solveig can keep Peer from winning.
be an angle in the interval 
.
is irrational, and that 
and ![$ \cos[(k + 1)\theta ]$](http://latex.artofproblemsolving.com/b/3/a/b3a14c57a1a8b5a6585195c394181c5124f8c471.png)
are both rational for some positive integer 
,
.
be three positive reals such that 
.![\[ \frac { x^5-x^2 }{x^5+y^2+z^2} + \frac {y^5-y^2}{x^2+y^5+z^2} + \frac {z^5-z^2}{x^2+y^2+z^5} \geq 0 . \]](http://latex.artofproblemsolving.com/a/1/4/a14adf0f1e35df57c734c7df701d5a67da22dc3c.png)
.
at 
and 
.
and 
and 
such that quadrilateral 
is a rhombus.
be an integer. Two players, Alice and Bob, play the following game on the complete graph 
:
,![\[
\left( \frac{1}{4} - \varepsilon \right) n^2 + n.
\]](http://latex.artofproblemsolving.com/a/9/c/a9cca00e87c643af32eb75a2501b59060f3504fb.png)
such that the following inequality holds for all 
:![\[
ab + bc + ca \leq \frac{(a + b + c)^2}{3} + k \cdot \max \{ (a - b)^2, (b - c)^2, (c - a)^2 \} \leq a^2 + b^2 + c^2
\]](http://latex.artofproblemsolving.com/1/c/e/1ce66163068b499383adb4a4ecec86c5f4edaec3.png)
be positive real numbers such that : 
. Prove that : 
where 
are lengths of parallel sides and 
are other two sides.
,
,
,
,
in some order.
board that is entirely covered by dominoes, with no overlap. He jumps along the vertical lines of the board, always going from the midpoint of the vertical segment of a 
square to another midpoint of the vertical segment, according to the rules:
When the domino is horizontal, the cricket jumps to the opposite vertical segment (such as from 
to 
)
When the domino is vertical downwards in relation to its position, the cricket jumps diagonally downwards (such as from 
to 
When the domino is vertically upwards relative to its position, the cricket jumps diagonally upwards (such as from 
)
board.
Prove that
candies. A starts first. On a move, if in the box there are 
so that 
and 
,
Y by nguyendangkhoa17112003, AFSA, Adventure10
Let 
be a triangle with 
and let 
be its incenter. Let 
be the circumcircle of . Lines 
and 
intersect in two new points, 
and 
respectively. Let 
be another point on lying on arc 
not containing 
, and let 
be the intersections of 
with and , respectively. Let 
be the intersections of 
with and of 
with respectively.
(i) Prove that
lie on the same circle 
(ii) Prove that lines
and 
intersect on
be a triangle with 
and let 
be its incenter. Let 
be the circumcircle of . Lines 
and 
intersect in two new points, 
and 
respectively. Let 
be another point on lying on arc 
not containing 
, and let 
be the intersections of 
with and , respectively. Let 
be the intersections of 
with and of 
with respectively.(i) Prove that
lie on the same circle 
(ii) Prove that lines
and 
intersect on 
.
.
,
is cyclic. Analogeously, 
is cyclic. So 
So 
,
and 
are cyclic. Now 
.
,
.
are concyclic. So 
are concyclic. Done.
are collinear.
are not concyclic
.
.
.
.
c
.
.
is cyxlic. Similarly, quadrilateral 
is also cyclic.
,
hence, quadrilateral 
is cyclic.From quadrilaterals 
are cyclic implies 
so, quadrilateral 
cyclic.If 
,
.
,
.
,
.
and quadrilateral 
,
so
.
.
implies quadrilateral 
is cyclic. Hence points 
are concyclic. Proved.
