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 
with equal radii are given. Let 
,
be the intersection of the circles.
and 
are on and such that they are inside and respectively.
,
are on and respectively, such that 
and 
.Denote by 
as the other intersection of 
and 
. Prove that 
are concurrent.
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.
over to get 
and 
over to get 
it's sufficient to prove that 
.Let 
touch 
in 
.
touches and at and , respectively. Point 
is the excenter of . Points and are midpoints of 
and 
. Lines 
and 
cross at point . Prove that lies on the circumcircle of .
be the incenter and let 
be the midpoint of 
,it is well known that lies on the circumcircle of so it's sufficient to prove that 
is cyclic
are parallel to 
respectively so 
=
=
so 
=
and by 
and 
we get that 
=
so 
=
and so done.
.
. We can simply check that 
, 
and 
. Let 
and 
. Then 
. So we need check if:
.
We are done.
be complex numbers lying on unit circle centered at 
such that 
are vertices of the triangle and 
is the incenter. These numbers always exist. Then
Let 
be the intersection of line with circumcircle of . Then 
and
Since 
we have
Observe this is symmetric in 
so the same are coordinates for intersection of with circumcircle of other than . Thus 
qed
=
+ 
- 1/2.
and 2.
=
+
- 1/2. 
since JE=JD(easy to prove) it is enough to prove that 
-
+
- 
=0 we know that - 
= 
- 
-
+
=
- 
since 
= 
= 90 proof finishes
+ 
=180 B' and N and P' are colinear now it is enough to prove that B'N=CN=BM and that is true and proof is completed now
and be reflections of and across and . Note that 
and 
so we will prove 
is cyclic. Let incircle of meet at .
and 
, 
and 
are congruent.
, 
and 
so and are congruent. we prove the other one with same approach.
so is cyclic.
)
be the midpoint of minor arc 
is the midpoint of 
which means 
and 
. This gives us that that 
. Now 
which means that there is a spiral similarity at taking 
-excircle touch at . Construct rectangles 
and 
. 
intersects 
at . Note that: 
= 
, 
, 
so 
. Similarly, 
, thus 
, as desired.
such that 
.
.
.
.
It is easy to verify that 
with these coordinates.
and 
.
and thus 
are concyclic,as desird.
as the points such that quadrilaterals 
and 
are both parallelograms. It is now enough to show that lines 
and 
meet on 
.
is parallel to 
,
.
,
is cyclic. Similarily, 
is cyclic. Now ![\[\angle FBY + \angle FCZ = \angle FJY+\angle FJZ = \angle YJZ = 180^{\circ}-\angle A\]](http://latex.artofproblemsolving.com/7/c/8/7c8270223060f0e518f2e5d93e94ef826634ac99.png)
and hence lines 
meet on