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 
be two triangles inscribed in the same circle, centered at 
, and sharing the same orthocenter 
. The Simson lines of the points 
with respect to triangle form a non-degenerate triangle 
. lies on the circle with diameter 
.
lie in this order on the circle and form a convex, non-degenerate hexagon.
and 
be positive integers posessing the following property: the equation
holds for all positive integers 
. Prove that 
for some integer 
.
be a triangle with circumcircle 
and 
a line without common points with . Denote by 
the foot of the perpendicular from the center of to . The side-lines 
intersect at the points 
different from . Prove that the circumcircles of the triangles 
, 
and 
have a common point different from or are mutually tangent at .
have the property that both![\[
\left\lfloor \sqrt{2n - 1} \right\rfloor \quad \text{and} \quad \left\lfloor \sqrt{3n + 2} \right\rfloor
\]](http://latex.artofproblemsolving.com/d/e/c/dec527c4673470e7023d6aed00fac90268bd2fcc.png)
are consecutive perfect squares.
such that
is an integer.
be a triangle with 
, and let 
be the foot of the altitude from 
. Let 
be a point in the interior of the segment 
. Let 
be the point on the segment 
such that 
. Similarly, let 
be the point on the segment 
such that 
. Let 
be the point of intersection of 
and 
.
.
be a point on segment 
. Let 
be a fixed circle passing through , and let 
be a variable point on . Let 
be the intersection of the tangent to the circumcircle of 
at 
and the tangent to the circumcircle of 
at 
. Show that as varies, lies on a fixed line.
, 
, with 
, 
. Prove that there exists 
such that for every 
![\[
f_n(x) = \frac{x}{1 + nx}, \quad \text{if and only if } f(x) = \frac{x}{1 + x}, \quad \forall x \geq 0.
\]](http://latex.artofproblemsolving.com/0/f/c/0fc36d9264eb7e103128c489aeae521a859c1fd4.png)
(For and , the notation 
represents 
. ), let 
be the sum of the remainders when is divided by 
. For example, 
. Find all positive integers such that 
.
be the set of all positive integers. Find all functions 
such that 
and 
have the same number of 
's in their binary representations, for any 
.
is called beautiful if, for every integer 
, the base-
representation of contains the consecutive digits 
, 
, , 
(in this order, from left to right). Determine whether the set of all beautiful integers is finite.
that are relatively prime be called intertwined if among any two divisors of greater than , there exists a divisor of and among any two divisors of greater than , there exists a divisor of . For example, pair 
is intertwined. for which there exists an intertwined pair such that the product 
is equal to the product of the first prime numbers. such that the product is the product of 
distinct prime numbers. such that the number of divisors of is greater than .
without further conclusions.
is possible for a function - this does not mean that 
.
or the "cancellation" of is called injectivity. You have to prove the function is injective first before cancellation.
from the equation 
in any way - so 
is wrong for that function as well. (thus is not injective)
without further conclusions. is possible for a function - this does not mean that . or the "cancellation" of is called injectivity. You have to prove the function is injective first before cancellation.
,we can divide by 
on both sides to get 
(or simply avoid pointwise trap).
.Let 
for 
.
gives 
.Take some 
such that 
.Then we have 
or 
.But 
gives that is even which means 
or 
, contradiction. Thus we are done.
be the assertion of the problem.
value we get that 
or 
(Just 
works).
such that the equality 
holds for all pairs of real numbers 
.
is an obvious solution. Now, let 
.
or 
.
Adding these two together we get:
