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.
be real numbers such that 
Prove that
. Let 
and 
be the feet of the altitudes from 
to 
and from 
to 
respectively.
and 
as the reflections of across lines 
and , respectively. Let 
be the circumcircle of 
. Denote by 
the second intersection of line 
with , and by 
the intersection of ray 
with .
is the circumcenter of , prove that , , and are collinear if and only if quadrilateral 
can be inscribed within a circle.
of positive integers satisfying the following two conditions:
, if 
is in 
, then so are 
and 
; and contains an integer 
such that 
is not divisible by 
.
then
where 
Also, 
and 
is largest real root of the equation
such that the numbers 
si 
are simultaneously prime .
be a triangle, and let 
and 
be points on segment symmetric with respect to the midpoint of . Let 
denote the circle passing through and tangent to line at . Similarly, let 
denote the circle passing through and tangent to line at 
. Let the circles and intersect again at point 
(
). Prove that 
.
and 
, let us consider the equation
[list=a]
and 
, find the least positive integer 
satisfying this equation. and , there exist infinitely many positive integers satisfying this equation.
denotes the greatest common divisor of positive integers 
and 
.)
events 
that are each continuous and last on a certain time interval. Each event has a weight 
However, one can only choose to attend activities that do not overlap with each other. The goal is to maximize the sum of weights of all activities attended. Prove or disprove that the following algorithm allows for an optimal selection:
consider 
the sum of 
over all 
such that 
and are not compatible.
If there are multiple such events, delete the event with the minimal weight.
are 
spheres centered at , with radius of 
and , which is remarked as 
and 
, respectively. Given 
points 
outside of (not including the surface of ).
as the number of sets 
such that the midpoint of 
lies entirely inside of .
shown below has area . Find the area of the rectangle 
.
, to each of the vertices creating 
equilateral triangles. The area of 
is clearly equal to the area of 
. Therefore each of the triangles and 
is equal in area to one of the equilateral triangles. The remaining area, which is the rectangle, is just of the equilateral triangles or 
.
. Triangles and 
are both 
isosceles triangles, meaning that they have side lengths of 
with an altitude of 
, and the area of one of the triangles is 
, so the area of two is 
. And the area of the rectangle is 
. Putting these together we have 
, which means 
, and since we are looking for the value of , we just divde both sides by 
getting 
.
. Now connect 
. Note that ![$[FAB]=\frac12 x^2 \sin{120}=\frac{x^2\sqrt3}4$](http://latex.artofproblemsolving.com/a/7/8/a78c973decff9c59da89799fdb9e04bd02268264.png)
. Note that 
is a 
so ![$[FBC]=\frac{\sqrt3x}2$](http://latex.artofproblemsolving.com/a/4/1/a417561fa423eb04bb5b3f510cb2967e2d2f2778.png)
. Hence, ![$\frac{[FBC]}{[FAB]}=\frac{2}{1}$](http://latex.artofproblemsolving.com/4/7/6/4769345a29717e2c05759fcce0481e1b2ce6a3ba.png)
and the answer is 
.
such that 
.
?
then 
Equality holds iff 
and any cyclic permutations.
.