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.
identical cars on a circular track. Among all of them, they have just enough gas for one car to complete a lap. Show that there is a car that can complete a lap by collecting gas from the other cars on its way around
, which doesn't have right angle, let 
be its circumcenter. Let 
and 
be orthocenters of triangle 
and 
, respectively. Let 
be circumcenter of triangle 
. If circle 
and circle 
intersect at 
for the second time, prove that lies on 
.
is a polynomial with real coefficients such that![\[ P(x)^2 - 1 = 4 P(x^2 - 4x + 1). \]](http://latex.artofproblemsolving.com/b/4/2/b425a938a1dbbac25e0e0793b30c9e58845d45ab.png)
Find .
and 
be altitudes in an acute triangle 
which meet at 
. Suppose that 
meets the circumcircle of at 
and 
such that lies on the shorter arc of 
and lies on the shorter arc of 
. Let 
and 
meet at 
. Show that the circumcircles of 
and 
and the line 
concur.
be a diameter of a circle 
, and let 
be the tangent at 
. Furthermore, let 
be a fixed, positive real, and consider all pairs of points 
and 
lying on , on opposite sides of , such that 
. The lines 
and 
intersect at points 
and 
, respectively. Show that all the lines 
pass through a common point.
such that
for all 
satisfying 
, your right-neighbor’s is 
, and the next person over is 
, then for every trio in the circle they see
with circumcenter . 
and 
intersect the altitude from 
to at points and 
respectively. 
is the circumcenter of triangle 
and 
is the reflection of over . 
is the second intersection of circumcircles of triangles 
and 
. Show that 
are collinear.
be the circumcenter of an acute triangle . Line 
intersects the altitudes of through 
and 
at and , respectively. The altitudes meet at 
. Prove that the circumcenter of triangle 
lies on a median of triangle .
and 
.
with multiplication of matrixes operation is making an isomorphic-group structure with 
.
.![$f \in C([0,1];[0,1])$](http://latex.artofproblemsolving.com/4/7/5/475518b7db0b787dd90b93f602c7b912c663fba3.png)
bijective, 
and ![$(a_k) \in [0,1]^\mathbb N$](http://latex.artofproblemsolving.com/a/d/8/ad8a80d847c60ee8e916a1f9b08b7b03ec17c814.png)
with 
converge.
converge?
on (1): 
on (2): 
, so we have 
on (2): 
so we have 
.
, so 
is commutative.
and the commutative binary operation such that 
(
exists) for all 
and 
for all 
, 
(the other cases are analogous) and 
(the other cases are analogous). Notice that 
, so isn't associative.
be a set with a binary operation 
such that
for all 
.
for all 
