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.
by Thales. Hence, 
is cyclic implying the desired result.
by the Inscribed Angle Theorem, and 
since it's the intersection of the x and y axes. Therefore, is cyclic and it follows that 
. 
, cyclic quad BPOQ implies the result.
is cyclic. We go on to prove this claim.
is the diameter, 
. Also, since and 
are on the 
-axis, while 
is on the 
-axis, by definition of a right angle, 
.
, we have 
corresponds to and to , it's like a homothety with center A, and then somewhat flipped. Those, however, have the property that they create a cyclic quadrilateral, in this case 
, we have corresponds to and to , it's like a homothety with center A, and then somewhat flipped. Those, however, have the property that they create a cyclic quadrilateral, in this case 
is a cyclic quadrilateral, then it follows that 
We can show that this is is cyclic because 
Therefore, we get our result.
(By Inscribed Angle Theorem) and 
.
is cyclic. ( By Inscribed Angle Theorem)
such that 
,
are reals. Let 
be a circle with diameter 
meets the x-axis again at 
by AA similarity 
by vertical angles), 
.
,
is cyclic.
.
,
,
is a cyclic quadrilateral. Then by inscribed angles, we get 
,
.
is cyclic so 
follows.![[asy]
import geometry;
pair o, a, b, t, p, q; o = (0,0); a = (0,2); b = (0,4); t = (a+b)/2; real r = abs(a-b)/2; p = t + r * dir(40); q = extension(p, a, (0,0), (1,0));
dot(o); dot(p); dot(a); dot(b); dot(q); filldraw(circle(t,r),invisible, green);
dot("$A$", a, dir(90)); dot("$B$", b, dir(90)); dot("$T$", t); dot("$P$", p, dir(45)); dot("$Q$", q, dir(90)); dot("$O$", o, dir(45));
draw(q--o--b--p--cycle, blue); draw(q--b, dotted+blue); draw(o--p, dotted+blue);
markrightangle(q, o, b, 5); markrightangle(q, p, b, 5);
[/asy]](http://latex.artofproblemsolving.com/4/7/9/479fcac77f1aa9b052bd176c26b390dd858230eb.png)
and thus 
so 
are concyclic so 
is concylic. Thus we get our desired result.
is cyclic. 
,
as 
as desired.
denote directed angles. We have ![\[\angle BOA = \angle BPA = 90^{\circ}\]](http://latex.artofproblemsolving.com/5/d/b/5dbd5d9742f6deac22c00fa702111dbf50a23a23.png)
![\[\angle BOA = \angle QOA = 90^{\circ}\]](http://latex.artofproblemsolving.com/9/7/1/9718886147607bfdcee7854e6189e601e485eb1e.png)
![\[\angle BQA = \angle BQP = 90^{\circ}-\angle PBQ = \angle BAP = \angle BOP\]](http://latex.artofproblemsolving.com/2/2/3/22351b3179f055814c7c9f4861d9e315b819b5ca.png)
as desired.
,
because it is the intersections of the