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.
days later (but perhaps earlier, as well). Hedvig received the trophy today. Determine all integers 
for which we can be absolutely certain that she cannot receive the trophy again in 
days, given the above information.
be real numbers such that : 
. Prove that : 
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.
where 
and 
.
![[asy]size(10cm);
import markers;
pair X1=(0,2),X2=(0,-2),Y1=(sqrt(7),1),Y2=(-sqrt(7),1),Z1=(sqrt(7),-1),Z2=(-sqrt(7),-1);
fill(X1--Y1--Z1--cycle^^X1--Y1--Z2--cycle,cyan);draw(X1--Y1--Z1--cycle^^X1--Y1--Z2--cycle,blue);
draw(anglemark(Y1,Z2,X1,10)^^anglemark(Z1,X1,Y1,10),blue);
draw(anglemark(X1,Y1,Z2,10,12)^^anglemark(Y1,Z1,X1,10,12),blue);
D(MP("X_1\;(0,2)",X1,N)--MP("Y_1\;(\sqrt{7},1)",Y1,NE)--MP("Z_1\;(\sqrt{7},-1)",Z1,SE)--MP("X_2\;(0,-2)",X2,S)--MP("Z_2\;(-\sqrt{7},-1)",Z2,SW)--MP("Y_2\;(-\sqrt{7},1)",Y2,NW)--X1);
label("$\triangle X_1Y_1Z_1\sim\triangle Z_2X_1Y_1$",(0,-3));
dot(X1^^X2^^Y1^^Y2^^Z1^^Z2);
[/asy]](http://latex.artofproblemsolving.com/7/a/0/7a0905fae32e34c8667c745d10a5f5e9d6735020.png)
where and .
(though slightly annoying, I'll follow the notation of the diagram from the official solution). Now let 
be the point such that 
is a parallelogram. From this we already know that 
.
such that 
( on the same side of 
with respect to 
). In other words, we consider the spiral similarity with center 
that maps to , and define as the image of 
via this transformation. Now, by definition we get that 
is similar to our original triangle . We shall now prove that 
is similar to the original triangle as well. More precisely, we'll show that 
.
, where in the last step we have used that 
, due to the similarity. So it is enough to prove that 
. be the lengths of the sides opposite to 
in our original triangle. Now 
, hence the ratio of the similarity is 
. Therefore, 
, and we have 
, as wanted.
are similar, and this holds without any extra requirements on the initial triangle .
is constructed such that 
is a parallelogram. Notice this implies that 
is a parallelogram as well, because it has a pair of opposite sides that are parallel and of equal length.
(respective sides are equal), and also 
(two halves of a parallelogram). So we have 6 similar triangles now.
and 
are congruent (two halves of a parallelogram). So if happens to be similar to the initial triangle , then all 8 triangles 
are similar and we are done.
if and only if 
in the starting triangle. Provided that this is the case, we further need 
, that is 
. Now, 
, and because of the similarity we have that 
, hence 
. So we need 
, and if we set 
we have 
, which is the same to saying that 
and 
for some 
. 
AND 
??
such that 
in the plane such that all of the triangles 
?
