AoPS Academy
[/quote]
, do not answer with "
is a solution" only. Either you post any kind of proof or at least something unexpected (like "
is the smallest solution). Someone that does not see that is a solution of the above without your post is completely wrong here, this is an IMO-level forum.
and some point 
on 
. The perpendicular from to 
intersects the circumcircle of 
at 
and the perpendicular from to 
intersects the circumcircle of 
at 
. Show that the line 
does not depend on the choice of .(Shining Sun, USA)
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.
cars from the grid in such a way that it becomes possible for all the remaining cars to eventually exit the grid. that guarantees that Kritesh's job is possible?
coins, all of which show 
on one side and 
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 . Peer plays first, and he wins if all the coins show simultaneously at any time. Find all 
for which Solveig can keep Peer from winning.
be an angle in the interval 
. Given that 
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 
, show that 
.
be three positive reals such that 
. Prove 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)
be a triangle with altitude 
. The 
-excircle touches 
at 
, and intersects the circumcircle at two points 
and 
. Prove that one can select points 
and 
on lines 
and 
such that quadrilateral 
is a rhombus.
be an integer. Two players, Alice and Bob, play the following game on the complete graph 
: They take turns to perform operations, where each operation consists of coloring one or two edges that have not been colored yet. The game terminates if at any point there exists a triangle whose three edges are all colored.
, Alice has a strategy such that, no matter how Bob colors the edges, the game terminates with the number of colored edges not exceeding![\[
\left( \frac{1}{4} - \varepsilon \right) n^2 + n.
\]](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 to 
).
board.
candies. A starts first. On a move, if in the box there are candies, the player chooses positive integer 
so that 
and 
, and eats candies from the box. The player who eats the last candy wins. Who has winning strategy, in terms of .
with 
is the foot of the altitude of 
. Let 
and be the reflections of with respect to and . Let 
and 
be the circumcenters of 
and 
. Show that: 
with is the foot of the altitude of . Let and be the reflections of with respect to and . Let and be the circumcenters of and . Show that: be the midpoint of .
. Now, ![\[\frac{AC}{\sin \alpha} = \frac{CD}{\sin (B-\alpha)} = \frac{BC\sin B}{\sin(B-\alpha)}\implies \cos B = \frac{\sin(B-\alpha)}{\sin \alpha} = \sin B\cot \alpha - \cos B.\]](http://latex.artofproblemsolving.com/e/5/7/e57349b3fbac04e0b0454c52d5a4bb35d1c6763d.png)
This gives 
. Therefore, ![\[\frac{AC}{2MO_2}=\frac{AC}{2BC} \implies MO_2 = BC,\]](http://latex.artofproblemsolving.com/6/9/f/69f9ecd5de0b53c79b5a68f3f22c2da87be5c7b2.png)
which gives 
and 
and we are done.
linear.
be midpoints of segments 
respectively.
, 
.
perpendicular trapezoid and 
its midsegment so 
. Similarly 
.
. Done.
with is the foot of the altitude of . Let and be the reflections of with respect to and . Let and be the circumcenters of and . Show that: 
be the medial triangle of and let 
be the midpoints of 
respectively.
forms a parallelogram.
are collinear. Also, 
and 
are collinear. Also, 
and 
. So, 
. Also, is the midpoint of 
. Thus, 
and 
. So, the claim is true.
.
be the point such that 
is a rectangle. Denote by 
and 
the intersections of 
with 
and 
with 
, respectively. Because 
is the perpendicular bisector of 
and 
is the perpendicular bisector of 
, we have that 
is a rhombus, thus 
, but 
is precisely the -midline of triangle 
, because 
and 
are cyclic.
Prove that
,
;
,
be midpoint of 
,
= 
+ 
= 
+ 
= 
= 
then: 
,
= 
,
= 
= 
)
and 
is 
so 
are cyclic.
are cyclic.![\[\angle BAO_1=\angle BEO_1=90-\alpha\]](http://latex.artofproblemsolving.com/b/5/2/b5232a7a1a18a3fc7e6a77a0fb60bb4a935a1dd0.png)
![\[\angle O_2BA=\angle O_2FA=\alpha\]](http://latex.artofproblemsolving.com/4/9/8/49899eb78aafaf6f24d1204c0d4c950d470e3de1.png)
So we have 
is a rectangle.
and 
and 
and 
which means 
and 
and this gives us the desired conclusion