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.
every next number (from the fifth number) is the unit number of the sum of the four numbers preceding it. Is there any cases that we get the numbers 
and 
in this series?
, the diagonals 
and 
meet at the point 
. We know that 
and 
. If we have 
prove that 
.![$P \in \mathbb Q[x,y]$](http://latex.artofproblemsolving.com/a/3/8/a384cbf38fd3a7f812b22baaeb0860e7f6eb26c0.png)
with 
and ![$\forall (a,b) \in \mathbb Z^2 \cap [0,d]^2, P(a,b) \in \mathbb Z$](http://latex.artofproblemsolving.com/a/6/5/a65ecb4bb2d9b5e6096f322c2de8bd11b945ed6a.png)
.
?
. Find 
.
be nonnegative real numbers such that 
. Prove that![\[
\frac{a}{4 - b} + \frac{b}{4 - c} + \frac{c}{4 - a} + \frac{1}{16}(1 - a)^2(1 - b)^2(1 - c)^2 \leq 1,
\]](http://latex.artofproblemsolving.com/5/b/d/5bd3349071e075519bd986c845c500125b7d46f8.png)
and determine all such triples 
where the equality holds.
such that 
is unbounded and![\[2f(m)f(n)-f(n-m)-1\]](http://latex.artofproblemsolving.com/0/f/7/0f77b5addbf8f5e603eb1b6923cdac795cd132e9.png)
is a perfect square for all integer 
with orthocenter 
.Let 
and 
be the altitudes of triangle.Let 
and 
be reflections of points 
across the line 
, respectively.Let 
and 
be the midpoints of 
and 
, respectively.Let 
and 
.Prove that 
and 
are collinear.
let be the midpoint of 
. Let 
be a circle inside of and is tangent to 
at , respectively. The tangents from to meet at 
such that and 
lie on the same side of 
. Let 
and 
. If 
prove that 
is tangent to . and 
be the incircle and circumcircle of the acute triangle , respectively. Draw a square 
so that all of its sides are tangent to , and , are both on . Extend 
and 
, intersecting at and 
, respectively. Prove that 
and 
intersects on .
be real numbers such that 
Prove that if 
is a positive constant, then
is not always true
be a triangle with 
. We consider the closed disks of radius 
, centered in 
, , 
. Does there exist an equilateral triangle with one vertex in each of the 3 disks?
be a triangle with . We consider the closed disks of radius , centered in , , . Does there exist an equilateral triangle with one vertex in each of the 3 disks?
the centre and by 
the radius of the circle circumscribed to 
.
we get that 
, hence 
is equilateral.
): is a variable point on the disk centred at (with radius 
) and is a variable point on the disk centered at (with radius 
), then the locus of points 
such that 
is equilateral, is the disk centred at and with radius 
(and, of course, its reflection in , but that's redundant here).
, and I'm very intrested in this problem. 
in counterclockwise order.
be called 
respectively. be fixed and let vary on 
. For each , is simply the 
counterclockwise rotation of with respect to so, the locus of as varies is simply the circle that results from rotating 
degrees counterclockwise about . Let this circle have center , so 
is equilateral. Now if we let vary as well, the locus of is simply the clockwise rotation of with respect to , which is easily seen to be the circle centered at with radius . Then the locus of is just the union of all possible circles centered at , which is clearly a circle centered at with radius 
.
, . Then 
is an equilateral triangle. Suppose in 3 the circles centered around 
there are points 
forming an equilateral triangle. Consider the spiral similarity sending 
to 
, then that of to 
. Then it is easy to see that 
, 
, but 
, contradiction.
, and that 
collinear. Then using the same spiral similarities it is easy to determine that the triangle is determined by 
as reflection of across 
and similarly 
reflection of 
across 
, and by doing so we have a triangle that works.
